Why is negative times negative = positive?












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Someone recently asked me why a negative $times$ a negative is positive, and why a negative $times$ a positive is negative, etc.



I went ahead and gave them a proof by contradiction like so:



Assume $(-x) cdot (-y) = -xy$



Then divide both sides by $(-x)$ and you get $(-y) = y$



Since we have a contradiction, then our first assumption must be incorrect.



I'm guessing I did something wrong here. Since the conclusion of $(-x) cdot (-y) = (xy)$ is hard to derive from what I wrote.



Is there a better way to explain this? Is my proof incorrect? Also, what would be an intuitive way to explain the negation concept, if there is one?










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  • 30




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    You haven't proven that -xy = (-x)y.
    $endgroup$
    – Qiaochu Yuan
    Nov 12 '10 at 2:24






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    It is not an uncommon question, and it's never not easy to show.
    $endgroup$
    – J. M. is a poor mathematician
    Nov 12 '10 at 4:06






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    J.M. is pointing out that logical negation works the same way as multiplying negative numbers (two negatives make a positive), not belittling your question. You understood his double negative statements, and so you have another intuitive path to see that negative*negative=positive.
    $endgroup$
    – Jonas Kibelbek
    Nov 12 '10 at 4:48






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    Does he understand why a negative times a positive is negative? And if so, how?
    $endgroup$
    – Paul
    Nov 8 '15 at 21:00






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    Have you shown him a sequence, to see a pattern? So for example $2×-3=-6$, $1×-3=-3$, $0×-3=0$, $-1×-3=3$...?
    $endgroup$
    – Paul
    Nov 8 '15 at 21:03
















111












$begingroup$


Someone recently asked me why a negative $times$ a negative is positive, and why a negative $times$ a positive is negative, etc.



I went ahead and gave them a proof by contradiction like so:



Assume $(-x) cdot (-y) = -xy$



Then divide both sides by $(-x)$ and you get $(-y) = y$



Since we have a contradiction, then our first assumption must be incorrect.



I'm guessing I did something wrong here. Since the conclusion of $(-x) cdot (-y) = (xy)$ is hard to derive from what I wrote.



Is there a better way to explain this? Is my proof incorrect? Also, what would be an intuitive way to explain the negation concept, if there is one?










share|cite|improve this question











$endgroup$








  • 30




    $begingroup$
    You haven't proven that -xy = (-x)y.
    $endgroup$
    – Qiaochu Yuan
    Nov 12 '10 at 2:24






  • 60




    $begingroup$
    It is not an uncommon question, and it's never not easy to show.
    $endgroup$
    – J. M. is a poor mathematician
    Nov 12 '10 at 4:06






  • 13




    $begingroup$
    J.M. is pointing out that logical negation works the same way as multiplying negative numbers (two negatives make a positive), not belittling your question. You understood his double negative statements, and so you have another intuitive path to see that negative*negative=positive.
    $endgroup$
    – Jonas Kibelbek
    Nov 12 '10 at 4:48






  • 12




    $begingroup$
    Does he understand why a negative times a positive is negative? And if so, how?
    $endgroup$
    – Paul
    Nov 8 '15 at 21:00






  • 9




    $begingroup$
    Have you shown him a sequence, to see a pattern? So for example $2×-3=-6$, $1×-3=-3$, $0×-3=0$, $-1×-3=3$...?
    $endgroup$
    – Paul
    Nov 8 '15 at 21:03














111












111








111


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$begingroup$


Someone recently asked me why a negative $times$ a negative is positive, and why a negative $times$ a positive is negative, etc.



I went ahead and gave them a proof by contradiction like so:



Assume $(-x) cdot (-y) = -xy$



Then divide both sides by $(-x)$ and you get $(-y) = y$



Since we have a contradiction, then our first assumption must be incorrect.



I'm guessing I did something wrong here. Since the conclusion of $(-x) cdot (-y) = (xy)$ is hard to derive from what I wrote.



Is there a better way to explain this? Is my proof incorrect? Also, what would be an intuitive way to explain the negation concept, if there is one?










share|cite|improve this question











$endgroup$




Someone recently asked me why a negative $times$ a negative is positive, and why a negative $times$ a positive is negative, etc.



I went ahead and gave them a proof by contradiction like so:



Assume $(-x) cdot (-y) = -xy$



Then divide both sides by $(-x)$ and you get $(-y) = y$



Since we have a contradiction, then our first assumption must be incorrect.



I'm guessing I did something wrong here. Since the conclusion of $(-x) cdot (-y) = (xy)$ is hard to derive from what I wrote.



Is there a better way to explain this? Is my proof incorrect? Also, what would be an intuitive way to explain the negation concept, if there is one?







algebra-precalculus arithmetic education faq






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edited Dec 20 '16 at 2:24









BLAZE

6,164112857




6,164112857










asked Nov 12 '10 at 2:11









SevSev

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  • 30




    $begingroup$
    You haven't proven that -xy = (-x)y.
    $endgroup$
    – Qiaochu Yuan
    Nov 12 '10 at 2:24






  • 60




    $begingroup$
    It is not an uncommon question, and it's never not easy to show.
    $endgroup$
    – J. M. is a poor mathematician
    Nov 12 '10 at 4:06






  • 13




    $begingroup$
    J.M. is pointing out that logical negation works the same way as multiplying negative numbers (two negatives make a positive), not belittling your question. You understood his double negative statements, and so you have another intuitive path to see that negative*negative=positive.
    $endgroup$
    – Jonas Kibelbek
    Nov 12 '10 at 4:48






  • 12




    $begingroup$
    Does he understand why a negative times a positive is negative? And if so, how?
    $endgroup$
    – Paul
    Nov 8 '15 at 21:00






  • 9




    $begingroup$
    Have you shown him a sequence, to see a pattern? So for example $2×-3=-6$, $1×-3=-3$, $0×-3=0$, $-1×-3=3$...?
    $endgroup$
    – Paul
    Nov 8 '15 at 21:03














  • 30




    $begingroup$
    You haven't proven that -xy = (-x)y.
    $endgroup$
    – Qiaochu Yuan
    Nov 12 '10 at 2:24






  • 60




    $begingroup$
    It is not an uncommon question, and it's never not easy to show.
    $endgroup$
    – J. M. is a poor mathematician
    Nov 12 '10 at 4:06






  • 13




    $begingroup$
    J.M. is pointing out that logical negation works the same way as multiplying negative numbers (two negatives make a positive), not belittling your question. You understood his double negative statements, and so you have another intuitive path to see that negative*negative=positive.
    $endgroup$
    – Jonas Kibelbek
    Nov 12 '10 at 4:48






  • 12




    $begingroup$
    Does he understand why a negative times a positive is negative? And if so, how?
    $endgroup$
    – Paul
    Nov 8 '15 at 21:00






  • 9




    $begingroup$
    Have you shown him a sequence, to see a pattern? So for example $2×-3=-6$, $1×-3=-3$, $0×-3=0$, $-1×-3=3$...?
    $endgroup$
    – Paul
    Nov 8 '15 at 21:03








30




30




$begingroup$
You haven't proven that -xy = (-x)y.
$endgroup$
– Qiaochu Yuan
Nov 12 '10 at 2:24




$begingroup$
You haven't proven that -xy = (-x)y.
$endgroup$
– Qiaochu Yuan
Nov 12 '10 at 2:24




60




60




$begingroup$
It is not an uncommon question, and it's never not easy to show.
$endgroup$
– J. M. is a poor mathematician
Nov 12 '10 at 4:06




$begingroup$
It is not an uncommon question, and it's never not easy to show.
$endgroup$
– J. M. is a poor mathematician
Nov 12 '10 at 4:06




13




13




$begingroup$
J.M. is pointing out that logical negation works the same way as multiplying negative numbers (two negatives make a positive), not belittling your question. You understood his double negative statements, and so you have another intuitive path to see that negative*negative=positive.
$endgroup$
– Jonas Kibelbek
Nov 12 '10 at 4:48




$begingroup$
J.M. is pointing out that logical negation works the same way as multiplying negative numbers (two negatives make a positive), not belittling your question. You understood his double negative statements, and so you have another intuitive path to see that negative*negative=positive.
$endgroup$
– Jonas Kibelbek
Nov 12 '10 at 4:48




12




12




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Does he understand why a negative times a positive is negative? And if so, how?
$endgroup$
– Paul
Nov 8 '15 at 21:00




$begingroup$
Does he understand why a negative times a positive is negative? And if so, how?
$endgroup$
– Paul
Nov 8 '15 at 21:00




9




9




$begingroup$
Have you shown him a sequence, to see a pattern? So for example $2×-3=-6$, $1×-3=-3$, $0×-3=0$, $-1×-3=3$...?
$endgroup$
– Paul
Nov 8 '15 at 21:03




$begingroup$
Have you shown him a sequence, to see a pattern? So for example $2×-3=-6$, $1×-3=-3$, $0×-3=0$, $-1×-3=3$...?
$endgroup$
– Paul
Nov 8 '15 at 21:03










40 Answers
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Well if I were to explain this in an intuitive way to someone (or at least try), I would like to think of an analogy with walking over the real line, by agreeing that walking left will be walking in the negative direction and walking right in the positive direction.



Then I will try to convey the idea that if you are multiplying to numbers (let's suppose they are integers to make things easier to picture) then a product as $2*3$ would just mean that you have to walk right (in the positive direction) a distance of $2$ (say miles for instance) three times, that is, first you walk $2$ miles, then another $2$ miles and finally another $2$ miles to the right.



Now you picture where you're at? Well, you're at the right of the origin so you are in the positive section. But in the same way you can play this idea with a negative times a positive.



With the same example in mind, what would $-2*3$ mean? First, suppose that the $-2$ just specifies that you will have to walk left a distance of $2$ miles. Then how many times you will walk that distance? Just as before $3$ times and in the end you'll be $6$ miles to the left of the origin so you'll be in the negative section.



Finally you'll have to try to picture what could $(-2)*(-3)$ mean. Maybe you could think of the negative sign in the second factor just making you change direction, that is, it makes you turn around and start walking the specified distance. So in this case the $-2$ tells you to walk left a distance of $2$ miles but the $-3$ tells you to fist turn around and then walk $3$ times the $2$ miles in the other direction, so you'll end up walking right and end in the point that is $6$ miles to the right of the origin, so you'll be in the positive section, and $(-2)*(-3) = 6$.



I don't know if this will help but it's the only way I can think of this in some intuitive sense.






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    This is pretty soft, but I saw an analogy online to explain this once.



    If you film a man running forwards ($+$) and then play the film forward ($+$) he is still running forward ($+$). If you play the film backward ($-$) he appears to be running backwards ($-$) so the result of multiplying a positive and a negative is negative. Same goes for if you film a man running backwards ($-$) and play it normally ($+$) he appears to be still running backwards ($-$). Now, if you film a man running backwards ($-$) and play it backwards ($-$) he appears to be running forward ($+$). The level to which you speed up the rewind doesn't matter ($-3x$ or $-4x$) these results hold true.
    $$text{backward} times text{backward} = text{forward}$$
    $$ text{negative} times text{negative} = text{positive}$$
    It's not perfect, but it introduces the notion of the number line having directions at least.






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      This is a good answer, because the rate at which the individual is running in the video multiplied by the rate of the playback equals the apparent rate at which they run in the playback. That little piece of nuance is probably a bit tricky to explain to an 8-year-old though.
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      – Dustan Levenstein
      Nov 9 '15 at 3:43










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      @MatthewGraves The video is gone - do you have another link?
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      – max_zorn
      Feb 4 '18 at 1:21










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      @max_zorn youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse"
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      – Matthew Graves
      Feb 12 '18 at 22:19










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      youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse"
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      – Matthew Graves
      Feb 12 '18 at 22:19



















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    Informal justification of $text{positive} times text{negative} = text{negative}$



    Continue the pattern:



    $$
    begin{array}{r}
    2 & times & 3 & = & 6\
    2 & times & 2 & = & 4\
    2 & times & 1 & = & 2\
    2 & times & 0 & = & 0\
    2 & times & -1 & = & ? & (text{Answer} = -2 )\
    2 & times & -2 & = & ? & (text{Answer} = -4 )\
    2 & times & -3 & = & ? & (text{Answer} = -6 )\
    end{array}
    $$



    The number on the right-hand side keeps decreasing by 2.





    Informal justification of $text{negative} times text{negative} = text{positive}$



    Continue the pattern:



    $$
    begin{array}{r}
    2 & times & -3 & = & -6\
    1 & times & -3 & = & -3\
    0 & times & -3 & = & 0\
    -1 & times & -3 & = & ? & (text{Answer} = 3 )\
    -2 & times & -3 & = & ? & (text{Answer} = 6 )\
    -3 & times & -3 & = & ? & (text{Answer} = 9 )\
    end{array}
    $$



    The number on the right-hand side keeps increasing by 3.






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    • 5




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      Simple and just what I needed. Thank you.
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      – John H
      Feb 18 '12 at 21:07










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      And very informal :) But I like intuitive approaches. Separates us from the nerds who live their lives without intuition.
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      – Alfe
      Nov 12 '13 at 11:28






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      @Alfe I would say the opposite. "Nerds" might not talk "intuitive" to you, but I think their intuition is much better than "ours" and that might be what makes them nerds.:-)
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      – quapka
      Nov 20 '14 at 0:25










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      I find it nice to extend this to all the reals by drawing a graph.
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      – Michael Anderson
      Jun 12 '15 at 7:41










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      This is by far the best answer, +1.
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      – Zubin Mukerjee
      Jun 13 '15 at 3:25



















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    Someone sent this to me recently:



    I give you three $20 notes: +3 * +20 = +60 for you
    I give you three $20 debts: +3 * -20 = -60 for you
    I take three $20 notes from you: -3 * +20 = -60 for you
    I take three $20 debts from you: -3 * -20 = +60 for you





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    • 12




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      That's from the reddit ELI5 from a few days ago reddit.com/r/explainlikeimfive/comments/3r90cw/…
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      – StuperUser
      Nov 9 '15 at 15:45



















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    I think a lot of answers are either too simple or stray away from mathematics too much. Just remember that multiplication is repeated addition. When dealing with negative numbers, it becomes repeated subtraction.



    I'd simply put it in this context:




    1. The equation:
      $$begin{equation*}begin{array}{c}
      phantom{times9}2\
      underline{timesphantom{9}2}\
      phantom{times9}4\
      end{array}end{equation*}$$

      is just adding postive $2$, two times.



    2. The equation:
      $$begin{equation*}begin{array}{c}
      phantom{times999}2\
      underline{timesphantom{1}-2}\
      phantom{times9}-4\
      end{array}end{equation*}$$
      is just adding positive $2$, negative two times, which means instead of adding in the positive direction, you add in the negative direction (subtraction).


      You could also just say that you're adding $-2$ (or subtracting $2$), two times.



    3. The equation:
      $$begin{equation*}begin{array}{c}
      phantom{times9}-2\
      underline{timesphantom{1}-2}\
      phantom{times999}4\
      end{array}end{equation*}$$
      is adding $-2$, negative two times. Adding $-2$, two times, yields the diagram in (2). Since you have to add negative two times, you reverse the direction you are adding in.



      You could also say that you are subtracting $-2$, two times.








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    • $begingroup$
      My memory of how I learned this is somewhat uncertain, because I was only 8 or 9 years old at the time, but I think this is very similar to how Isaac Asimov explained multiplication of negative numbers in either Realm of Numbers or Realm of Algebra. His explanation made sense to me at the time and I don't recall ever being confused as to why the product of negatives is positive. But I think it was a few years before this fact came up in the school curriculum.
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      – David K
      Nov 9 '15 at 23:30






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      @homersimpson - I was thinking of this kind of logic but what knocks me out is how can we assume minus minus 2 is positive 2 - aren't we supposed to prove this?
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      – KGhatak
      Jan 11 '16 at 8:00



















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    $overbrace{bf Law of Signs}^{rmLarge {(-x)(-y)} = xy} $ proof: $rm, (-x)(-y) = (-x)(-y) + color{#c00}x(overbrace{color{#c00}{-y} + y}^{Large =,0}) = (overbrace{-x+color{#c00}x}^{Large =,0})(color{#c00}{-y}) + xy = xy$



    Equivalently, $ $ evaluate $rm, overline{(-x)(-y) + } overline{ underline {color{#c00}{x(-y)}}}underline{phantom{(}! +,xy}, $ in $,2,$ ways, noting over/underlined terms $ = 0$



    Said more conceptually $rm (-x)(-y) $ and $rm:xy:$ are both inverses of $rm color{#c00}{x(-y)} $ so they are equal by uniqueness of inverses: $ $ if $,color{#c00}a,$ has two additive inverses $,{-a},$ and $,color{#0a0}{-a},,$ then



    $$rm {-a}, =, {-a}+overbrace{(color{#c00}a+color{#0a0}{-a})}^{large =,0}, =, overbrace{({-a}+color{#c00}a)}^{large =,0}+color{#0a0}{-a}, =, color{#0a0}{-a} $$



    Equivalently, $ $ evaluate $rm, overline{-a +!} overline{phantom{+} underline {color{#c00}{a}}}underline{ + color{#0a0}{-a}}, $ in $,2,$ ways, noting over/underlined terms $ = 0$



    This proof of the Law of Signs uses well-known laws of positive integers (esp. the distributive law), so if we require that these laws persist in the larger system of positive and negative integers, then the Law of Signs is a logical consequence of these basic laws of positive integers.



    These fundamental laws of "numbers" are axiomatized by the algebraic structure known as a ring, and various specializations thereof. Since the above proof uses only ring laws (most notably the distributive law), the Law of Signs holds true in every ring, e.g. rings of polynomials, power series, matrices, differential operators, etc. In fact every nontrivial ring theorem (i.e. one that does not degenerate to a theorem about the underlying additive group or multiplicative monoid), must employ the distributive law, since that is the only law that connects the additive and multiplicative structures that combine to form the ring structure. Without the distributive law a ring degenerates to a set with two completely unrelated additive and multiplicative structures. So, in a sense, the distributive law is a keystone of the ring structure.



    Remark $ $ More generally the Law of Signs holds for any odd functions
    under composition, e.g. polynomials with all terms having odd power. Indeed we have



    $qquadrm f(g) = (-f) (-g) = -(f(-g)) iff f(-g) = -(f(g)) $



    $qquadrm phantom{ (-f) (-g) = -(f(-g)) = f(g) }!!overset{ large g(x)=x}iff f(-x) = -f(x), $ ie. $rm:f:$ is odd



    Generally such functions enjoy only a weaker near-ring structure.
    In the above case of rings, distributivity implies that multiplication
    is linear hence odd (viewing the ring in Cayley-style
    as the ring of endormorphisms of its abelian additive group,
    i.e. representing each ring element $rm r $ by the linear map $rm x to r x, $
    i.e. as a $1$-dim matrix).






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    • 4




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      +1. The simplest way to prove why is by using the group axioms and thier consequences.
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      – Mathemagician1234
      Jun 13 '15 at 3:18










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      What's the logical thinking skills or tactic used to jump in one step from xy or (−x)(−y) to xy+x (−y)+(−x)(−y)?
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      – Pacerier
      Feb 2 '18 at 7:15












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      @Pacerier It's a common way to prove uniqueness of inverses. I edited the answer to elaborate on that.
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      – Bill Dubuque
      Oct 16 '18 at 17:48



















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    Simple Answer:
    $$
    (-a)b + ab = (-a)b + ab
    $$
    $$(-a)b + ab = b(a-a)
    $$
    $$(-a)b + ab = b(0)
    $$
    $$(-a)b + ab = 0 $$
    $$(-a)b = -ab $$
    $$(-a)(-b) + (-ab) = (-a)(-b) + (-a)b $$
    $$(-a)(-b) + (-ab) = (-a)(b-b) $$
    $$(-a)(-b) + (-ab) = (-a)(0) $$
    $$(-a)(-b) + (-ab) = 0 $$
    $$*(-a)(-b) = ab $$
    Hope this helps (Credit to Michael Spivak's Calculus)



    ~ Alan






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    • $begingroup$
      Well,I wouldn't exactly call that a SIMPLE answer. A DIRECT and STRAIGHTFORWARD one,yes,absolutely. But not simple.
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      – Mathemagician1234
      Nov 20 '14 at 0:30










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      Easily the best and most satisfying answer here.
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      – Derek 朕會功夫
      Aug 17 '15 at 20:42



















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    This is really one of those important questions that leads many people to say "Math sucks!". In fact, for many students, mathematics stopped making sense somewhere along the way. Either slowly or dramatically, they gave up on the field as hopelessly baffling and difficult, and they grew up to be adults who — confident that others share their experience — nonchalantly announce, “Math was just not for me
    or “I was never good at it.
    Usually the process is gradual, but for Ruth
    McNeill, the turning point was clearly defined. In an
    article in the Journal of Mathematical Behavior, she
    described how it happened:




    What did me in was the idea that a negative number
    times a negative number comes out to a positive
    number. This seemed (and still seems) inherently
    unlikely — counter intuitive, as mathematicians say.
    I wrestled with the idea for what I imagine to be
    several weeks, trying to get a sensible explanation
    from my teacher, my classmates, my parents, any-
    body. Whatever explanations they offered could not
    overcome my strong sense that multiplying intensifies something, and thus two negative numbers
    multiplied together should properly produce a very
    negative result. I have since been offered a moderately convincing explanation that features a film of a
    swimming pool being drained that gets run back-
    wards through the projector. At the time, however,
    nothing convinced me. The most commonsense of
    all school subjects had abandoned common sense;
    I was indignant and baffled.


    Meanwhile, the curriculum kept rolling on, and I
    could see that I couldn’t stay behind, stuck on nega-
    tive times negative. I would have to pay attention to
    the next topic, and the only practical course open to
    me was to pretend to agree that negative times nega-
    tive equals positive. The book and the teacher and
    the general consensus of the algebra survivors of so-
    ciety were clearly more powerful than I was. I capitu-
    lated. I did the rest of algebra, and geometry, and
    trigonometry; I did them in the advanced sections,
    and I often had that nice sense of “aha!” when I
    could suddenly see how a proof was going to come
    out. Underneath, however, a kind of resentment and
    betrayal lurked, and I was not surprised or dismayed
    by any further foolishness my math teachers had up
    their sleeves.... Intellectually, I was disengaged, and
    when math was no longer required, I took German
    instead.






    I will show in this answer that: negative $times$ negative $=$ positive, is in fact not counter-intuitive at all! There are many ways that we can use to show that result, but I'd like to show my personal way of thinking about the latter.



    Let's imagine we're sitting near a road, and there is a car that is moving with a constant speed. We also have a clock, and so we can measure time.



    Before going any further, we should first specify some assumptions like if the car is moving in the right, then its velocity will be positive, and if it's moving in the left direction, then its velocity will be negative.



    enter image description here



    Imagine now that you have a video of the above scene, and time is positive if you play the video normally but is negative if you play it backwards. We also know the following: $$rm Velocity=dfrac{rm Distance}{rm Time}.$$
    Solving for distance we get: $$rm Velocitytimes{rm Time}={rm Distance}.$$
    Here the important part comes, if the car is moving in the $+$ direction and the time the video is played is positive, i.e. the video is played normally, then you'll see that the car moves along the $+$ direction and you'll calculate that it moves "a positive distance". Therefore the following holds: $$rm positivetimes positive=positive.$$



    If the car moves along the $-$ direction and the time the video is played is positive, i.e. the video is played normally, then you'll see the car going along the $-$ direction, you'll then calculate that it moves "a negative distance". Thus: $$rm negativetimes positive=negative.$$



    If however, the car moves along the $+$ direction but the time the video is played is negative, i.e. the video is played backwards, then you'll see that the car is moving in the $-$ direction, and you'll calculate that it moves "a negative distance". Thus: $$rm positivetimes negative=negative.$$



    If the car moves along the $-$ direction but the time the video is played is negative, i.e. the video is played backwards, you'll see that the car moves along the $+$ direction! Thus it moves "a positive distance". And therefore: $$color{grey}{boxed{color{white}{underline{overline{color{black}{displaystylerm, negativetimes negative=positive.,}}}}}}$$ As you have seen, it takes only a little bit of imagination for it to make sense.



    I hope this helps.
    Best wishes, $mathcal H$akim.






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    • 6




      $begingroup$
      The material you quoted contains the interesting statement "Whatever explanations they offered could not overcome my strong sense that multiplying intensifies something". I wonder whether anyone tried the very short answer "yes, but negative intensification is attenuation"?
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      – Andreas Blass
      Jun 29 '14 at 21:43










    • $begingroup$
      @AndreasBlass Or in another form: "negative multiplication is reverse intensification".
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      – Hakim
      Feb 22 '15 at 19:56










    • $begingroup$
      @Hakim, Doesn't Velocity×Time equal displacement instead of distance?
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      – Pacerier
      Feb 2 '18 at 7:39



















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    The elementary intuition behind the product of two negatives can be thought of as follows. You have a bank account. You pay 3 bills for 40 dollars each, $3 cdot (-40) = -120$ is added to your account.



    The opposite of being billed would be billing someone else.



    So, if you bill 3 people for 40 dollars each, $(-3) cdot (-40)$ is added to your account. This value should be positive since it results in you receiving money.






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      $begingroup$

      Here's a proof. First, for all $x$, $xcdot 0=xcdot(0+0)=xcdot 0 +xcdot 0$. Subtracting $xcdot0$ from each side, $xcdot0=0$. Now, for all $x$ and $y$, $0=xcdot0=xcdot(-y+y)=xcdot(-y)+xcdot y$. Subtracting $xcdot y$ from both sides, $xcdot(-y)=-(xcdot y)$. Applying this twice along with the identity $-(-a)=a$, $(-x)cdot(-y)=-(-x)cdot y=-(-(xcdot y))=xcdot y$.



      Your proof implicitly uses the fact that $-xy=(-x)y$, and assumes that there are only two possibilities, $xy$ or $-xy$, then shows that the latter is impossible. These seem like plausible assumptions, but I tried to be very careful in my proof above (thus using $-(xcdot y)$ rather than simply $-xy$ to not be confused with $(-x)cdot y$).



      I only have a vague intuitive notion that I probably can't explain well, but I sometimes think of a negative number like $-5$ as being "$5$ in the other direction", and so multiplying by $-5$ means "multiply by $5$ and switch direction", i.e., sign. This means if you multiply $-5$ by a negative number, you should switch its direction back to positive.






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        I have always viewed negative numbers as a "flip" on the number line.



        For example, -2 is the same as 2, but flip-mirrored to the other side of the zero.



        Multiplication then works as follows:



        2 x 3 has no flips, so it's just 2x3 = 6.



        -2 x 3 has one flip, so you start with 2x3 = 6 but with one flip so it becomes -6 instead.



        2 x -3 is the same, only one flip, so it's 2x3 = 6 but flipped to -6.



        -2 x -3 has two flips, so you start with 2x3 = 6. When you apply the two flips it gets you back to where you started because you flip to negative and then flip back. So -3 x -2 = 6.






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          I think the x and y get in the way a bit; you can see the crucial steps using just 1 and -1. What you've really shown is that (-1)(-1)=-1 leads to a contradiction. If we divide by -1, we get -1=1, which is not true!



          Getting the right answer, (-1)(-1)=1, uses a couple more steps: First, you must agree that (1)+(-1)=0, (1)(-1)=-1, and (0)(-1)=0.



          Now, we multiply the first equation by (-1) and use the distributive property to get (-1)(-1)+(-1)(1)=(-1)(0). Now, we simplify the parts we know to get (-1)(-1)+(-1)=0. Solve for (-1)(-1), and you get (-1)(-1)=1.



          So, we must have (-1)(-1)=1 if we accept basic rules of arithmetic: 0 is the additive identity, 1 is the multiplicative identity, -1 is the additive inverse of 1, and multiplication distributes over addition.



          One physical explanation people often like for negative*negative=positive is multiplying rates. You can film someone walking forwards (positive rate) or walking backwards (negative rate). Now, play the film back, but in reverse (another negative rate). What do you see if you play a film backwards of someone walking backwards? You see the person walking forwards, because negative*negative=positive!






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            Quite a good explanation is that one wants the distributive law to work in general with positive quantities when you add (smaller) negative ones:
            If $x>age0$ and $y>bge0$ then
            $$
            (x-a)(y-b)=(x+(-a))(y+(-b))=xy+(-a)y+x(-b)+(-a)(-b)
            $$
            For this to always work, one needs $(-a)y=-(ay)$ in case $b=0$, $x(-b)=-xb$ in case $a=0$, and $(-a)(-b)=ab$.






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              $begingroup$

              One thing that must be understood is that this law cannot be proven in the same way that the laws of positive rational and integral arithmetic can be. The reason for this is that negatives lack any "external" (external to mathematics, ie. pre-axiomatic, intuitive, conceptuel, empirical, physical, etc.) definition.



              For example. Without even getting into the Peano axioms, I can prove that, where $a$ and $b$ are positive integers, $ab=ba$. Indeed, $ab$ is just the process of taking $a$ sets of $b$. Take one element from each of these sets, thus forming a set of $a$ elements. Repeat this $b$ times: you will clearly use up exactly all of the elements and obtain $b$ sets of $a$ elements, in other words, $ba$. Similar informal (but entirely convincing, reasonable, and I would say irrefutable) reasoning can be used to demonstrate the rules for manipulating positive fractions, say.



              Notice that in the above paragraph I used the fact that both positive integers and positive integer multiplication have pre-axiomatic, "physical" definitions.



              Ask someone why the product of two negatives is positive, and the best they can do is explain, not prove. "Well, negative kind of means 'opposite', so doing the opposite twice means doing the usual, ie positive" does not constitute a proof, but merely an explanation serving to make the accepted mathematical axiom less surprising. Another common one begins with "we would like the usual properties of arithmetic to hold, so assume they do...", but then it remains to be explained why it's so important that the usual laws of arithmetic hold. Euler himself, in an early chapter of his textbook on algebra, gave the following supremely questionable justification. After justifying $(-a)b=-(ab)$ by analogies with debts, he writes:




              It remains to resolve the case in which - is multiplied by -; or, for example, -a by -b. It is evident, at first sight, with regard to the letters, that the product will be ab; but it is doubtful whether the sign + or the sign - is to be placed before it, all we know is, that it must be one or the other of these signs. Now, I say that it cannot be the sign -: for -a by +b gives -ab, and -a by -b cannot produce the same result as -a by +b...




              With no disrespect to Euler (especially consdiering this was intended as an introductory textbook), I think we can agree that this is a pretty philosophically dubious argument.



              The reason it is impossible is because there is no pre-axiomatic definition for what a negative number or negative multiplication really is. Oh, you could probably come up with one involving opposite "directions", and notions of symmetry, but it would be quite artificial and not at all obviously "the best" definition. In my opinion, negatives are ultimately best understood as purely abstract objects. It so happens - and this is quite myseterious - that these utterly abstract laws of calculation lead to physically meaningful results. This was nicely expressed in 1778 by the mathematician John Playfair, when addressing the then controversial issues of negative and complex numbers:




              Here then is a paradox which remains to be explained. If the operations of this imaginary arithmetic are unintelligible, why are they not altogether useless? Is investigation an art so mechanical, that it may be conducted by certain manual operations? Or is truth so easily discovered, that intelligence is not necessary to give success to our researches?




              Quoted in Negative Math: How Mathematical Rules Can be Positively Bent by Alberto A. Martínez.



              One way of approching the problem is with the idea that negative numbers are a different name for subtraction. The differences between subtraction and addition force us, if we reject negatives, to create many different rules covering all the different possibilities ($a - b$, $b - a$, and $a + b$, and if the particular theorem or problem involves more than two variables, the difficulty is compounded further...). The idea of negatives could be described as the insight that rather than having two operations and one type of number, we can have one operation and two types of number. Indeed, if you start with some perfectly physically meaningful axioms about subtraction, you will find that the $(-1)(-1)=1$ law seems to be implicit within them. Hint: starting from the very reasonable axioms $a(b-c)=ab-ac , a - (b - c) = a - b + c$, consider the product $(a-b)(c-d)$.



              But even that explanation doesn't altogether satisfy me. I've become convinced that my education cheated me on how deep an idea negative numbers are, and I expect to remain puzzled by them for many years. Anyway, I hope some of the above is useful to someone.






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              • 1




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                This $(-a)(-b)=ab$ stuff started when we gave up on distinguishing multiplier and multiplied in a multiplication. I can have three unities of apples, but I can't have apple unities of threes. In this sense, it's possible to have 3 unities of an 1D vector backwards $3 times -1$ but not -1 unities of a vector forwards $-1 times 3$, unless we define that a negative multiplier change the sense of the vector. I think a negative-free algebra can be created by distinguish three kind of numbers: scalar (no sign), up and down.
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                – Leonardo Castro
                Sep 9 '16 at 12:22












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                If you add an up and a down, you cancel out the amount they have in common keep the remaining up or down part. This would the subtraction of this algebra. Multiplying by $-1$ should be replaced by a simple inversion operator. Actually, $+1$ and $-1$ as multipliers are usually taken as the identity and inversion operators. In that way, they form a simple identity-inversion group and so $-1 times -1 = 1$ since two inversions are equivalent to identity. I don't think, however, that algebra should necessarily be that way.
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                – Leonardo Castro
                Sep 9 '16 at 12:33










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                I used up and down just to avoid all the assumptions about positive and negative, but they could be right and left, yin and yang, whatever... I wonder if yin times yin is yang or the other way around.
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                – Leonardo Castro
                Sep 9 '16 at 12:35





















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              Extend reals to the complex plane. Multiplication by $-1$ is a rotation by $pi$ radians. When you multiply by two negatives, you rotate by $2pi$. :-)






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                Why don't we tell a story! The dastardly Dalton gang is on the loose but Al Catchem is hot on their trail and nearly catches them at their latest heist. When pulling out of the parking lot, there are a few possibilities:




                • Al is quick, up to the task and catches the Daltons,

                • Al's car stalls and the Daltons get away,

                • The Dalton's are tipped off early and escape, or

                • The Dalton's car stalls, they are captured and the city is saved.


                In summary:




                • positive $times$ positive: If a good thing happens to a good person, that's good! :)

                • negative $times$ positive: If a bad thing happends to a good person, that's bad. :(

                • positive $times$ negative: If a good thing happens to a bad person, that's also bad. :(

                • negative $times$ negative: But, if a bad thing happens to a bad person, that's good!! :)






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                • 22




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                  While I can understand the sentiment, I'm not sure this is the kind of morality we should be teaching 8-year-old kids.
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                  – MichaelS
                  Nov 9 '15 at 3:32






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                  Downvoted because this isn't really a mathematical explanation.
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                  – Daniel R. Collins
                  Nov 9 '15 at 3:39






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                  @DanielR.Collins That is funny - the question is how to explain the idea to an 8 year old. :)
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                  – Mark McClure
                  Nov 9 '15 at 3:40






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                  @MichaelS You are, of course, correct. When I've actually done this, which I have many times, I've tried to tell a story with it that (hopefully) makes the morality reasonably clear. Turns out, you can have great fun with it!
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                  – Mark McClure
                  Nov 9 '15 at 4:10






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                  Why is this immoral? From a naturalistic philosophical viewpoint, this is arguably the most moral, since babies have been shown to prefer "good" characters over "evil" ones.
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                  – March Ho
                  Nov 9 '15 at 13:48



















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                I prefer the explanation by my favorite mathematician , V. I. Arnold (physicist really, since in his own words, "mathematics is a part of physics" and "an experimental science"). I believe it's the most natural (yet totally mathematical) explanation of a basic notion like multiplication of negative numbers.



                This is an excerpt from Arnold's wonderful memoir "Yesterday and Long Ago" (3d ed., available in English from Springer), full of world history, drama and ingenious storytelling. The 2007 translation into English, I believe, is not of best quality, but it's the only one so far.



                from the short story The Arnold Family




                I faced a real difficulty with school mathematics several years after
                the multiplication table: it was necessary to leam that “minus
                multiplied by minus is plus” I wanted to know the proof of this rule;
                I have never been able to leam by heart what is not properly
                understood. I asked my father to explain the reason why (—1) • (—2) =
                (+2). He, being a student of great algebraists, S. O. Shatunovsky and
                E. Noether, gave the following “scientific explanation”: “The point
                is,” he said: “that numbers form a field such that the distributive
                law (x+y)z=xz+yz holds. And if the product of minus by minus had not
                been plus, this law would be broken”.



                However, for me this “deductive” (actually juridical) explanation did
                not prove anything - what of it! One can study any axioms! Since
                that day I have preserved the healthy aversion of a naturalist to the
                axiomatic method with its non-motivated axioms.



                The axiomophile Rene Descartes stated that “neither experimental tests
                that axioms reflect a reality, nor comparison of theoretical results
                with reality should be a part of science” (why should results
                correspond to reality if the initial principles do not correspond to
                it?).



                Another thesis of Descartes’ theory and methods of education is even
                more peculiar and contemporary: “It is necessary to forbid all other
                methods of teaching except mine because only this method is
                politically correct
                : with my purely deductive method any dull
                student can be taught as successfully as the most gifted one
                , while
                with other methods imagination and even drawings are used
                unavoidably, and for this reason geniuses advance faster
                than dunces
                ”.



                Contrary to the deductive theories of my father and Descartes, as a
                ten year old, I started thinking about a naturally-scientific sense
                of the rule of signs
                , and I have come to the following conclusion. A
                real (positive or negative) number is a vector on the axis of
                coordinates (if a number is positive the corresponding vector is
                positively directed along this axis).



                A product of two numbers is an area of a rectangle whose sides
                correspond to these numbers (one vector is along one axis and the
                other is along a perpendicular axis in the plane). A rectangle, given
                by an ordered pair of vectors, possesses, as a part of the plane, a
                definite orientation (rotation from one vector to another can be
                clockwise or anti-clockwise). The anti-clockwise rotation is
                customarily considered positive and the clockwise rotation is then
                negative. And lastly, the area of a parallelogram (for example, a
                rectangle) generated by the two vectors x and у (taken in a
                definite order) emanating from one point in the plane is considered
                to be positive
                if the pair of vectors (taken in this order) defines
                positive orientation of the plane (and negative if the rotation from the direction of the vector x to the direction of the vector у is
                negative).



                Thus, the rule of signs is not an axiom taken out of the blue, but
                becomes a natural property of orientation which is easily verified
                experimentally.




                from the short story Axiomatic Method




                My first trouble in school was caused by the rule for multiplication
                of negative numbers, and I asked my father to explain this peculiar
                rule.



                He, as a faithful student of Emmy Noether (and consequently of Hilbert
                and Dedekind) started explaining to his eleven-year-old son the
                principles of axiomatic science: a definition is chosen such that the
                distributive identity a(b+c)=ab+ac holds.



                The axiomatic method requires that one should accept any axiom with a
                hope that its corollaries would be fruitful (probably this could be
                understood by the age of thirty when it would be possible to read and
                appreciate “Anna Karenina”). My father did not say a word either about
                the oriented area of a rectangular or about any non-mathematicai
                interpretation of signs and products.



                This “algebraic” explanation was not able to shake either my hearty
                love for my father or a deep respect of his science. But since that
                time I have disliked the axiomatic method with its non-motivated
                definitions. Probably it was for this reason that by this time I got
                used to talking with non-algebraists (like L. I. Mandel’shtam, I. E.
                Tamm, P. S. Novikov, E. L. Feinberg, M. A. Leontovich, and A. G.
                Gurvich) who treated an ignorant interlocutor with full respect and
                tried to explain non-trivial ideas and facts of various sciences such
                as physics and biology, astronomy and radiolocation.



                Negative numbers I came to understand a year later while deriving an
                “equation of time”, which takes into account a correction for the
                length of a day corresponding to the time of year. It is not possible
                to explain to algebraists that their axiomatic method is mostly
                useless for students.



                One should ask children: at what time will high tide be tomorrow if
                today it is at 3 pm? This is a feasible problem, and it helps children
                to understand negative numbers better than algebraic rules do. Once I
                read from an ancient author (probably from Herodotus) that the tides
                "always occur three and nine o'clock". To understand that the monthly
                rotation of the Moon about the Earth affects the tide timetable, there
                is no need to live near an ocean. Here, not in axioms, is laid true
                mathematics.







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                • $begingroup$
                  2 very good and insightful anecdotes-although I wouldn't really call either of them PROOFS, would anyone here?They're more intuitive explanations why it's so.Of course, the natural next question is:Can either of these arguments be precisely reformulated as proofs?
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                  – Mathemagician1234
                  Jun 13 '15 at 3:21








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                  @Mathemagician1234 you know, I agree with Arnold on the idea that explanations or intuition like this are much much more powerful than any proofs or definitions, because they are real and without artificial restrictions of formal systems. Using them, one can come up with a correct system of definitions in no time, if they have to. However, having the definitions in the first place does not add anything to one's true understanding of the subject.
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                  – level1807
                  Jun 13 '15 at 7:36










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                  With the interpretation in terms of oriented areas we would have $xy=-yx$ which is kind of problematic.
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                  – Sergio
                  Feb 2 '18 at 22:55



















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                Perhaps some intuition can be gained by plotting each number's position on the number line. Taking the inverse of any number is visualized by taking the mirror-image of the original plot. So the inverse of a positive number (a point to the right of zero) is a negative number (a point to the left of zero, at the same distance from zero). Likewise, the inverse of a negative number is a positive number. If we agree that multiplying a number by -1 is the same as finding the inverse, then we can see that the product of two negatives must be a positive, because the mirror-image of a mirror-image is the original image.






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                  As for the product of two negatives being a positive, simply consider the multiplicative inverse:



                  $$-acdot -b$$
                  $$(-1)acdot (-1)b$$
                  $$(-1)(-1)acdot b$$



                  Note that $(-1)^{-1} = -1$. Any number times its multiplicative inverse is $1$.



                  $$(-1)(-1)^{-1}acdot b$$
                  $$(1)acdot b$$
                  $$=acdot b$$






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                  • 1




                    $begingroup$
                    Great answer! Can you explain how you came up with this? I mean it's pretty self explanatory that a number multiplied by it's multiplicative inverse is 1 but using it here is pretty clever. How did you think to use it here?
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                    – user500668
                    Jul 28 '18 at 13:27










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                    @user500668 Good question. I used to be clever once upon a time. I believe at the time I was just making use of multiplicative inverses while solving "contest style" math problems.
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                    – daOnlyBG
                    Oct 22 '18 at 14:53










                  • $begingroup$
                    For what it's worth, I wish whoever gave me a "-1" score would have the decency to justify their action.
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                    – daOnlyBG
                    Oct 22 '18 at 14:54



















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                  A bit late, but I have written a formal version of Jonas' proof here. The "applying this twice" was the only tricky part.



                  See Lemma 7 starting at line 224 in http://www.dcproof.com/FieldLemmas.htm






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                    I would explain it by number patterns.



                    First, to establish that a positive times a negative is negative: $3 times 2 = 6, 3 times 1 = 3, 3 times 0 = 0$. Notice in each case, as we reduce the second factor by 1, the product is being reduced by 3. So for consistency the next product in the pattern must be $0 - 3 = -3$. Therefore we have $3 times (-1) = -3, 3 times (-2) = -6$, and likewise a negative for any other other positive times a negative.



                    Second, to establish that a negative times a negative is positive: we now know that $3 times (-2) = -6, 2 times (-2) = -4, 1 times (-2) = -2, 0 times (-2) = 0$. Notice in each case, as we reduce the first multiplier by 1, the product is being increased by 2. So for consistency the next product in the pattern must be $0 + 2 = 2$. Therefore we have $(-1) times (-2) = 2, (-2) times (-2) = 4$, and likewise a positive for any other other negative times a negative.






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                    • $begingroup$
                      I tried explaining this to him and he looked even more confused than when I narrated the other examples, I just think this is a bit too complicated for someone like him to understand yet.
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                      – yuritsuki
                      Nov 9 '15 at 7:20










                    • $begingroup$
                      Funny - this is the second argument that I typically present to elementary teachers and their students after my cops and robbers story! Note that it has been presented on this site before as well.
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                      – Mark McClure
                      Nov 9 '15 at 15:16



















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                    I would go for the flipping explanation of the negative numbers: multiplying with a negative number flips from positive to negative and from negative to positive.

                    Imagine he understands that multiplying with 1 makes no difference, then it's very simple:

                    -1 * 1 = -1 can mean two things (for children, the fact that multiplication is commutative is obvious):

                    - either it keeps -1 as -1

                    - either it flips +1 (positive number) to the negative side



                    -1 * (-1) then simply flips it back from the negative to the positive side.



                    Good luck






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                      It might be easiest to explain using whole numbers. Suppose $P$ is some positive number. Then $-P$ is negative. Now $-2P$ is $P$ subtracted from $-P$, so is still negative. Subtract another $P$ and you get $-3P$, which is still negative. Similarly for $-4P, -5P$, and so on. So negative times positive is positive. Same idea for positive times negative.



                      When it comes to negative times negative, it's a little harder... But how about... $-P$ is negative, so $-(-P)$ is now positive, flipping around $0$. So $-2(-P)$, which is $-(-P)$ added to itself, is still positive. In general adding $-(-P)$ to itself $Q$ times gives $(-Q)(-P)$, which is therefore positive as well.






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                        If your son is clear on the concept of money and knows what a credit card is, this might be a good explanation:



                        Imagine you tell your son that you will buy him $color{green}{mathrm{seven}}$ gift vouchers worth £$color{green}{5}$ each and pay for them using your credit card. Explain to your son that you now owe money, and say that it is $7 times -5=-mathrm{£}35$. Not part of your question; but this covers the $ text{positive} times text{negative} = text{negative}$ case.



                        You are having dinner with your best friend when the bill arrives from the credit card company, when your friend sees the bill he generously insists on paying it for you. You now have £$color{blue}{35}$ worth of gift vouchers without having paid anything.



                        So you tell your son that your best friend $color{red}{fbox{took-away}}$ seven $color{red}{fbox{debts}}$ of £$5$ ($color{red}{-7}timescolor{red}{-5}$) and this equals a gain of £$color{blue}{35}$.






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                        • 2




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                          (+1) I've seen this taught particularly effectively by drawing the money in a "bag" (I imagine physical cards with + or - numbers on work well for the same reason). The basic premise is that if a bag contains 7, -3, and 5 (total = 9) we see that "taking away" or "crossing out" the -3 makes the new total 12, so "taking away -3 is the same as adding 3". If the bag contains two 6s, among other things, we can easily taking that "minus two lots of six from the bag is the same as minus 12 from the total". Or if the bag contains two -5s, among others, we see "minus two lots of -5 is the same as +10"
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                          – Silverfish
                          Nov 9 '15 at 21:17



















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                        Your child may have been introduced to the minus sign by means of the word opposite. This is a great term to use in your conversations. Indeed, $a$ and $-a$ are opposites in that they are additive inverses. On the number line, opposite numbers are mirrored in their distances from zero, which provides a nice visual aid as well. We can use the term to describe arithmetic operations:



                        The opposite of three times five is the opposite of 15.
                        $$-3 times 5 = -15$$



                        The opposite of three times the opposite of five is the opposite of the opposite of fifteen... which is just fifteen.
                        $$-3 times -5 = 15$$



                        Okay, so we can use language to better cultivate our understanding of negative numbers. But what about a physical example? Well, here's a cute one that a friend once told me. It's a bit contrived, but I think it gets the point across. You can turn this into a demonstration too.




                        Suppose an ice cube lowers the temperature of a drink by $1$ degree.



                        Placing three ice cubes in a glass will lower the temperature by $3$ degrees, or
                        $$ 3 times -1 = -3$$



                        Removing two of the ice cubes will raise the temperature by $2$ degrees, or
                        $$ -2 times -1 = 2$$







                        share|cite|improve this answer











                        $endgroup$









                        • 1




                          $begingroup$
                          Can you really warm up a drink by removing ice cubes from it? In reality, an ice cube in a drink absorbs some heat from the drink, and some of the ice may melt and become part of the liquid in the drink, then mix with the other liquid, lowering its temperature. In either case, when you remove the ice, you do not undo either of those effects; the drink remains cooler than before you put the ice in it. So you either have to make some complicated "averaging" argument or you have to rely on a false physical intuition to be "unlearned" later.
                          $endgroup$
                          – David K
                          Nov 9 '15 at 13:49










                        • $begingroup$
                          @DavidK Just use the word "pretend." I doubt the child is ready for a talk about specific heat.
                          $endgroup$
                          – zahbaz
                          Nov 9 '15 at 15:55












                        • $begingroup$
                          What should we pretend? Should we pretend that when you remove some of the ice, the liquid in the glass warms up? That's a "false intuition" of the sort I meant. There are so many analogies that actually work in real life, why not just use one of them instead of making up plausible (to a naive person) but incorrect physics?
                          $endgroup$
                          – David K
                          Nov 9 '15 at 19:23










                        • $begingroup$
                          @DavidK No, we should suppose that the scenario is grossly simplified. Children can understand that sort of disclaimer. The point isn't to teach physics, but to paint a mental image of what opposites mean in terms of multiplication. We can use a metaphorical example as a heuristic, no? I get your point, but I find this example works well enough for pedagogical purposes.. If you want a physical example, then take the ratio of the mass of your ice cube and water to be approx 1:100. Then, yes, one ice cube will (eventually) lower the temperature of the drink by about 1 degree C.
                          $endgroup$
                          – zahbaz
                          Nov 9 '15 at 20:42












                        • $begingroup$
                          Or if you will, replace the word ice cube by weight, glass by buoy, and temperature by height above sea level. It's just an architecture to express opposites.
                          $endgroup$
                          – zahbaz
                          Nov 9 '15 at 21:05



















                        5












                        $begingroup$

                        One way to picture this is to imagine a number line. Then rotate it $180^{circ}$. Each number will now be superimposed over its negative: $-1$ will be where $+1$ was; $+2$ will be where $-2$ was. Rotation of the number line by $180^{circ}$ is the equivalent of multiplying by $-1$.



                        Now do the rotation twice. The number line is unchanged. So, multiplying by $-1$ twice is the same as multiplying by $+1$.



                        This approach has applications with Complex numbers. In these scenarios, the number line is rotated $90^{circ}$ counter clockwise to multiply by $i$.



                        But that's another story.






                        share|cite|improve this answer









                        $endgroup$





















                          5












                          $begingroup$


                          1. Explain the definition of negative numbers.

                          2. Point out that the definition of $-x$ implies that $-(-x) = x$.

                          3. Explain that $-x = (-1)times x$.

                          4. Point out that (2) and (3) imply that $(-1)times(-1) = 1$.






                          share|cite|improve this answer









                          $endgroup$









                          • 4




                            $begingroup$
                            "Huh??? x is a letter! you can't multiply numbers and letters!"
                            $endgroup$
                            – Paul Sinclair
                            Nov 9 '15 at 4:28






                          • 4




                            $begingroup$
                            @PaulSinclair The children of today are used to using computers, they can easily be taught to think of $x$ as storing some arbitrary number.
                            $endgroup$
                            – Count Iblis
                            Nov 9 '15 at 4:45






                          • 7




                            $begingroup$
                            @CountIbis I wouldnt take that bet. I teach engineers and for many years this has been the assumption. Net result is that i get Bachelors level students that can not do anything with a computer, not even frigging graph a function. Some dont even know what a file is.
                            $endgroup$
                            – joojaa
                            Nov 9 '15 at 5:08












                          • $begingroup$
                            @PaulSinclair Indeed! With kids of that age, a common response is that $x$ means 24. Because $a$ is one, and $b$ is two, and so on, and in that way the more inventive kids discover they can add numbers and letters. It's just that they've made the "wrong" extension into symbols. And explaining that $x$, or indeed $a$ might represent an arbitrary number is surprisingly hard if the kid hasn't developed that kind of abstraction yet. You probably would have more luck with "an empty box that a number can go in", rather than using a letter.
                            $endgroup$
                            – Silverfish
                            Nov 9 '15 at 21:21



















                          4












                          $begingroup$

                          Symbology: $$begin{align*}
                          -a times -b &= (-1 times a)times (-1 times b) \
                          &= -1 times a times -1 times b \
                          &= -1 times -1 times a times b \
                          &= (-1 times -1) times (a times b) \
                          &= a times b.
                          end{align*}$$



                          What's going on?:

                          Consider the number line. When I multiply something by $2$, I double its distance from $0$. This happens whether the "something" is positive or negative. When I multiply something by $-2$, I double its distance from $0$ and flip to the other side of the number line. For instance $-2 times 3 : 3 rightarrow 6 rightarrow -6$. If I start with a negative number, it's already on the negative half of the number line and will be flipped to the positive half: $-2 times -3 : -3 rightarrow -6 rightarrow 6$.



                          That is, multiplication by a negative is the same as two steps: multiplying by the thing as if it had no negative, then applying the negative sign. That's what the symbology above says: multiplying by $-a$ is the same as multiplying by $a$ then by $-1$ and similarly for $-b$. But then the two "flip to the other half of the number line"s, the two "$-1$"s, cause two flips. But two flips takes anything away and then right back to itself, so two flips really does nothing. In this way $-1 times -1$ is the same as doing nothing, or is the same as multiplying by $1$. That's why in the symbols above, we can drop the "$-1 times -1$" -- they're the same as multiplying by $1$.






                          share|cite|improve this answer









                          $endgroup$









                          • 1




                            $begingroup$
                            "When I multiply something by −2, I double its distance from 0 and flip to the other side of the number line": I like this answer a lot, but I think to serve as a good explanation for a kid at this level it would help if you could explain why multiplying works on the number line like this (it's safe to assume you can't use the symbology directly with the kid, although you're clearly trying to get across that train of logic). Until you can explain why negative multiplication reflects the number line, we have replaced one "why's does it work like that?" question by another.
                            $endgroup$
                            – Silverfish
                            Nov 9 '15 at 21:14



















                          4












                          $begingroup$

                          Well.... this always made sense to me (but I've found it doesn't for others)



                          A) Multiplication is adding a number a bunch of times.



                          1) Negative numbers are numbers that are less than zero. The cancel out positive numbers. They are anti-numbers



                          So)i) positive x positive: add a bunch of positive numbers a positive number of times. Result: A big positive gain.



                          ii) positive x negative: add a bunch of negative anti-numbers. The result is a big amount of potential cancelling. Result: negative.



                          iii) negative x positive: take a bunch of positive numbers and take them away. Result: a loss; negative.



                          iv) negative x negative: take a bunch of anti-numbers and take them away. By taking away the taking away, what's left is putting things back. If you annihilite the annihitation the result is a net gain: Result: positive.






                          share|cite|improve this answer









                          $endgroup$





















                            3












                            $begingroup$

                            Well, the way I think about it is this. We have the non-negative integers (0,1,2,3,4, etc.).



                            We introduce the negative numbers and need to define multiplication with negative numbers so that we have internal consistency.



                            We wish to keep the property that 0*anything = 0, negative or positive.



                            We also want to keep the distributive property.



                            In order to keep the above two properties, we're forced to define the product of two negatives as a positive.



                            0*(-3) = 0



                            (5 + (-5))*(-3) = 0 (I'm plugging in 5+ (-5) for zero)



                            5*(-3) + (-5)*(-3) = 0 (distributive property)



                            add 5*3 to both sides. 5*3 cancels with 5*(-3)



                            (-5)*(-3) = 5*3






                            share|cite|improve this answer









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                              67












                              $begingroup$

                              Well if I were to explain this in an intuitive way to someone (or at least try), I would like to think of an analogy with walking over the real line, by agreeing that walking left will be walking in the negative direction and walking right in the positive direction.



                              Then I will try to convey the idea that if you are multiplying to numbers (let's suppose they are integers to make things easier to picture) then a product as $2*3$ would just mean that you have to walk right (in the positive direction) a distance of $2$ (say miles for instance) three times, that is, first you walk $2$ miles, then another $2$ miles and finally another $2$ miles to the right.



                              Now you picture where you're at? Well, you're at the right of the origin so you are in the positive section. But in the same way you can play this idea with a negative times a positive.



                              With the same example in mind, what would $-2*3$ mean? First, suppose that the $-2$ just specifies that you will have to walk left a distance of $2$ miles. Then how many times you will walk that distance? Just as before $3$ times and in the end you'll be $6$ miles to the left of the origin so you'll be in the negative section.



                              Finally you'll have to try to picture what could $(-2)*(-3)$ mean. Maybe you could think of the negative sign in the second factor just making you change direction, that is, it makes you turn around and start walking the specified distance. So in this case the $-2$ tells you to walk left a distance of $2$ miles but the $-3$ tells you to fist turn around and then walk $3$ times the $2$ miles in the other direction, so you'll end up walking right and end in the point that is $6$ miles to the right of the origin, so you'll be in the positive section, and $(-2)*(-3) = 6$.



                              I don't know if this will help but it's the only way I can think of this in some intuitive sense.






                              share|cite|improve this answer











                              $endgroup$


















                                67












                                $begingroup$

                                Well if I were to explain this in an intuitive way to someone (or at least try), I would like to think of an analogy with walking over the real line, by agreeing that walking left will be walking in the negative direction and walking right in the positive direction.



                                Then I will try to convey the idea that if you are multiplying to numbers (let's suppose they are integers to make things easier to picture) then a product as $2*3$ would just mean that you have to walk right (in the positive direction) a distance of $2$ (say miles for instance) three times, that is, first you walk $2$ miles, then another $2$ miles and finally another $2$ miles to the right.



                                Now you picture where you're at? Well, you're at the right of the origin so you are in the positive section. But in the same way you can play this idea with a negative times a positive.



                                With the same example in mind, what would $-2*3$ mean? First, suppose that the $-2$ just specifies that you will have to walk left a distance of $2$ miles. Then how many times you will walk that distance? Just as before $3$ times and in the end you'll be $6$ miles to the left of the origin so you'll be in the negative section.



                                Finally you'll have to try to picture what could $(-2)*(-3)$ mean. Maybe you could think of the negative sign in the second factor just making you change direction, that is, it makes you turn around and start walking the specified distance. So in this case the $-2$ tells you to walk left a distance of $2$ miles but the $-3$ tells you to fist turn around and then walk $3$ times the $2$ miles in the other direction, so you'll end up walking right and end in the point that is $6$ miles to the right of the origin, so you'll be in the positive section, and $(-2)*(-3) = 6$.



                                I don't know if this will help but it's the only way I can think of this in some intuitive sense.






                                share|cite|improve this answer











                                $endgroup$
















                                  67












                                  67








                                  67





                                  $begingroup$

                                  Well if I were to explain this in an intuitive way to someone (or at least try), I would like to think of an analogy with walking over the real line, by agreeing that walking left will be walking in the negative direction and walking right in the positive direction.



                                  Then I will try to convey the idea that if you are multiplying to numbers (let's suppose they are integers to make things easier to picture) then a product as $2*3$ would just mean that you have to walk right (in the positive direction) a distance of $2$ (say miles for instance) three times, that is, first you walk $2$ miles, then another $2$ miles and finally another $2$ miles to the right.



                                  Now you picture where you're at? Well, you're at the right of the origin so you are in the positive section. But in the same way you can play this idea with a negative times a positive.



                                  With the same example in mind, what would $-2*3$ mean? First, suppose that the $-2$ just specifies that you will have to walk left a distance of $2$ miles. Then how many times you will walk that distance? Just as before $3$ times and in the end you'll be $6$ miles to the left of the origin so you'll be in the negative section.



                                  Finally you'll have to try to picture what could $(-2)*(-3)$ mean. Maybe you could think of the negative sign in the second factor just making you change direction, that is, it makes you turn around and start walking the specified distance. So in this case the $-2$ tells you to walk left a distance of $2$ miles but the $-3$ tells you to fist turn around and then walk $3$ times the $2$ miles in the other direction, so you'll end up walking right and end in the point that is $6$ miles to the right of the origin, so you'll be in the positive section, and $(-2)*(-3) = 6$.



                                  I don't know if this will help but it's the only way I can think of this in some intuitive sense.






                                  share|cite|improve this answer











                                  $endgroup$



                                  Well if I were to explain this in an intuitive way to someone (or at least try), I would like to think of an analogy with walking over the real line, by agreeing that walking left will be walking in the negative direction and walking right in the positive direction.



                                  Then I will try to convey the idea that if you are multiplying to numbers (let's suppose they are integers to make things easier to picture) then a product as $2*3$ would just mean that you have to walk right (in the positive direction) a distance of $2$ (say miles for instance) three times, that is, first you walk $2$ miles, then another $2$ miles and finally another $2$ miles to the right.



                                  Now you picture where you're at? Well, you're at the right of the origin so you are in the positive section. But in the same way you can play this idea with a negative times a positive.



                                  With the same example in mind, what would $-2*3$ mean? First, suppose that the $-2$ just specifies that you will have to walk left a distance of $2$ miles. Then how many times you will walk that distance? Just as before $3$ times and in the end you'll be $6$ miles to the left of the origin so you'll be in the negative section.



                                  Finally you'll have to try to picture what could $(-2)*(-3)$ mean. Maybe you could think of the negative sign in the second factor just making you change direction, that is, it makes you turn around and start walking the specified distance. So in this case the $-2$ tells you to walk left a distance of $2$ miles but the $-3$ tells you to fist turn around and then walk $3$ times the $2$ miles in the other direction, so you'll end up walking right and end in the point that is $6$ miles to the right of the origin, so you'll be in the positive section, and $(-2)*(-3) = 6$.



                                  I don't know if this will help but it's the only way I can think of this in some intuitive sense.







                                  share|cite|improve this answer














                                  share|cite|improve this answer



                                  share|cite|improve this answer








                                  edited Sep 30 '17 at 4:39









                                  tatan

                                  5,79962761




                                  5,79962761










                                  answered Nov 12 '10 at 2:42









                                  Adrián BarqueroAdrián Barquero

                                  10.8k23983




                                  10.8k23983























                                      133












                                      $begingroup$

                                      This is pretty soft, but I saw an analogy online to explain this once.



                                      If you film a man running forwards ($+$) and then play the film forward ($+$) he is still running forward ($+$). If you play the film backward ($-$) he appears to be running backwards ($-$) so the result of multiplying a positive and a negative is negative. Same goes for if you film a man running backwards ($-$) and play it normally ($+$) he appears to be still running backwards ($-$). Now, if you film a man running backwards ($-$) and play it backwards ($-$) he appears to be running forward ($+$). The level to which you speed up the rewind doesn't matter ($-3x$ or $-4x$) these results hold true.
                                      $$text{backward} times text{backward} = text{forward}$$
                                      $$ text{negative} times text{negative} = text{positive}$$
                                      It's not perfect, but it introduces the notion of the number line having directions at least.






                                      share|cite|improve this answer











                                      $endgroup$









                                      • 26




                                        $begingroup$
                                        This is a good answer, because the rate at which the individual is running in the video multiplied by the rate of the playback equals the apparent rate at which they run in the playback. That little piece of nuance is probably a bit tricky to explain to an 8-year-old though.
                                        $endgroup$
                                        – Dustan Levenstein
                                        Nov 9 '15 at 3:43










                                      • $begingroup$
                                        @MatthewGraves The video is gone - do you have another link?
                                        $endgroup$
                                        – max_zorn
                                        Feb 4 '18 at 1:21










                                      • $begingroup$
                                        @max_zorn youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse"
                                        $endgroup$
                                        – Matthew Graves
                                        Feb 12 '18 at 22:19










                                      • $begingroup$
                                        youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse"
                                        $endgroup$
                                        – Matthew Graves
                                        Feb 12 '18 at 22:19
















                                      133












                                      $begingroup$

                                      This is pretty soft, but I saw an analogy online to explain this once.



                                      If you film a man running forwards ($+$) and then play the film forward ($+$) he is still running forward ($+$). If you play the film backward ($-$) he appears to be running backwards ($-$) so the result of multiplying a positive and a negative is negative. Same goes for if you film a man running backwards ($-$) and play it normally ($+$) he appears to be still running backwards ($-$). Now, if you film a man running backwards ($-$) and play it backwards ($-$) he appears to be running forward ($+$). The level to which you speed up the rewind doesn't matter ($-3x$ or $-4x$) these results hold true.
                                      $$text{backward} times text{backward} = text{forward}$$
                                      $$ text{negative} times text{negative} = text{positive}$$
                                      It's not perfect, but it introduces the notion of the number line having directions at least.






                                      share|cite|improve this answer











                                      $endgroup$









                                      • 26




                                        $begingroup$
                                        This is a good answer, because the rate at which the individual is running in the video multiplied by the rate of the playback equals the apparent rate at which they run in the playback. That little piece of nuance is probably a bit tricky to explain to an 8-year-old though.
                                        $endgroup$
                                        – Dustan Levenstein
                                        Nov 9 '15 at 3:43










                                      • $begingroup$
                                        @MatthewGraves The video is gone - do you have another link?
                                        $endgroup$
                                        – max_zorn
                                        Feb 4 '18 at 1:21










                                      • $begingroup$
                                        @max_zorn youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse"
                                        $endgroup$
                                        – Matthew Graves
                                        Feb 12 '18 at 22:19










                                      • $begingroup$
                                        youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse"
                                        $endgroup$
                                        – Matthew Graves
                                        Feb 12 '18 at 22:19














                                      133












                                      133








                                      133





                                      $begingroup$

                                      This is pretty soft, but I saw an analogy online to explain this once.



                                      If you film a man running forwards ($+$) and then play the film forward ($+$) he is still running forward ($+$). If you play the film backward ($-$) he appears to be running backwards ($-$) so the result of multiplying a positive and a negative is negative. Same goes for if you film a man running backwards ($-$) and play it normally ($+$) he appears to be still running backwards ($-$). Now, if you film a man running backwards ($-$) and play it backwards ($-$) he appears to be running forward ($+$). The level to which you speed up the rewind doesn't matter ($-3x$ or $-4x$) these results hold true.
                                      $$text{backward} times text{backward} = text{forward}$$
                                      $$ text{negative} times text{negative} = text{positive}$$
                                      It's not perfect, but it introduces the notion of the number line having directions at least.






                                      share|cite|improve this answer











                                      $endgroup$



                                      This is pretty soft, but I saw an analogy online to explain this once.



                                      If you film a man running forwards ($+$) and then play the film forward ($+$) he is still running forward ($+$). If you play the film backward ($-$) he appears to be running backwards ($-$) so the result of multiplying a positive and a negative is negative. Same goes for if you film a man running backwards ($-$) and play it normally ($+$) he appears to be still running backwards ($-$). Now, if you film a man running backwards ($-$) and play it backwards ($-$) he appears to be running forward ($+$). The level to which you speed up the rewind doesn't matter ($-3x$ or $-4x$) these results hold true.
                                      $$text{backward} times text{backward} = text{forward}$$
                                      $$ text{negative} times text{negative} = text{positive}$$
                                      It's not perfect, but it introduces the notion of the number line having directions at least.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Sep 15 '16 at 18:25

























                                      answered Nov 8 '15 at 21:08









                                      miradulomiradulo

                                      3,15521425




                                      3,15521425








                                      • 26




                                        $begingroup$
                                        This is a good answer, because the rate at which the individual is running in the video multiplied by the rate of the playback equals the apparent rate at which they run in the playback. That little piece of nuance is probably a bit tricky to explain to an 8-year-old though.
                                        $endgroup$
                                        – Dustan Levenstein
                                        Nov 9 '15 at 3:43










                                      • $begingroup$
                                        @MatthewGraves The video is gone - do you have another link?
                                        $endgroup$
                                        – max_zorn
                                        Feb 4 '18 at 1:21










                                      • $begingroup$
                                        @max_zorn youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse"
                                        $endgroup$
                                        – Matthew Graves
                                        Feb 12 '18 at 22:19










                                      • $begingroup$
                                        youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse"
                                        $endgroup$
                                        – Matthew Graves
                                        Feb 12 '18 at 22:19














                                      • 26




                                        $begingroup$
                                        This is a good answer, because the rate at which the individual is running in the video multiplied by the rate of the playback equals the apparent rate at which they run in the playback. That little piece of nuance is probably a bit tricky to explain to an 8-year-old though.
                                        $endgroup$
                                        – Dustan Levenstein
                                        Nov 9 '15 at 3:43










                                      • $begingroup$
                                        @MatthewGraves The video is gone - do you have another link?
                                        $endgroup$
                                        – max_zorn
                                        Feb 4 '18 at 1:21










                                      • $begingroup$
                                        @max_zorn youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse"
                                        $endgroup$
                                        – Matthew Graves
                                        Feb 12 '18 at 22:19










                                      • $begingroup$
                                        youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse"
                                        $endgroup$
                                        – Matthew Graves
                                        Feb 12 '18 at 22:19








                                      26




                                      26




                                      $begingroup$
                                      This is a good answer, because the rate at which the individual is running in the video multiplied by the rate of the playback equals the apparent rate at which they run in the playback. That little piece of nuance is probably a bit tricky to explain to an 8-year-old though.
                                      $endgroup$
                                      – Dustan Levenstein
                                      Nov 9 '15 at 3:43




                                      $begingroup$
                                      This is a good answer, because the rate at which the individual is running in the video multiplied by the rate of the playback equals the apparent rate at which they run in the playback. That little piece of nuance is probably a bit tricky to explain to an 8-year-old though.
                                      $endgroup$
                                      – Dustan Levenstein
                                      Nov 9 '15 at 3:43












                                      $begingroup$
                                      @MatthewGraves The video is gone - do you have another link?
                                      $endgroup$
                                      – max_zorn
                                      Feb 4 '18 at 1:21




                                      $begingroup$
                                      @MatthewGraves The video is gone - do you have another link?
                                      $endgroup$
                                      – max_zorn
                                      Feb 4 '18 at 1:21












                                      $begingroup$
                                      @max_zorn youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse"
                                      $endgroup$
                                      – Matthew Graves
                                      Feb 12 '18 at 22:19




                                      $begingroup$
                                      @max_zorn youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse"
                                      $endgroup$
                                      – Matthew Graves
                                      Feb 12 '18 at 22:19












                                      $begingroup$
                                      youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse"
                                      $endgroup$
                                      – Matthew Graves
                                      Feb 12 '18 at 22:19




                                      $begingroup$
                                      youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse"
                                      $endgroup$
                                      – Matthew Graves
                                      Feb 12 '18 at 22:19











                                      89












                                      $begingroup$

                                      Informal justification of $text{positive} times text{negative} = text{negative}$



                                      Continue the pattern:



                                      $$
                                      begin{array}{r}
                                      2 & times & 3 & = & 6\
                                      2 & times & 2 & = & 4\
                                      2 & times & 1 & = & 2\
                                      2 & times & 0 & = & 0\
                                      2 & times & -1 & = & ? & (text{Answer} = -2 )\
                                      2 & times & -2 & = & ? & (text{Answer} = -4 )\
                                      2 & times & -3 & = & ? & (text{Answer} = -6 )\
                                      end{array}
                                      $$



                                      The number on the right-hand side keeps decreasing by 2.





                                      Informal justification of $text{negative} times text{negative} = text{positive}$



                                      Continue the pattern:



                                      $$
                                      begin{array}{r}
                                      2 & times & -3 & = & -6\
                                      1 & times & -3 & = & -3\
                                      0 & times & -3 & = & 0\
                                      -1 & times & -3 & = & ? & (text{Answer} = 3 )\
                                      -2 & times & -3 & = & ? & (text{Answer} = 6 )\
                                      -3 & times & -3 & = & ? & (text{Answer} = 9 )\
                                      end{array}
                                      $$



                                      The number on the right-hand side keeps increasing by 3.






                                      share|cite|improve this answer











                                      $endgroup$









                                      • 5




                                        $begingroup$
                                        Simple and just what I needed. Thank you.
                                        $endgroup$
                                        – John H
                                        Feb 18 '12 at 21:07










                                      • $begingroup$
                                        And very informal :) But I like intuitive approaches. Separates us from the nerds who live their lives without intuition.
                                        $endgroup$
                                        – Alfe
                                        Nov 12 '13 at 11:28






                                      • 14




                                        $begingroup$
                                        @Alfe I would say the opposite. "Nerds" might not talk "intuitive" to you, but I think their intuition is much better than "ours" and that might be what makes them nerds.:-)
                                        $endgroup$
                                        – quapka
                                        Nov 20 '14 at 0:25










                                      • $begingroup$
                                        I find it nice to extend this to all the reals by drawing a graph.
                                        $endgroup$
                                        – Michael Anderson
                                        Jun 12 '15 at 7:41










                                      • $begingroup$
                                        This is by far the best answer, +1.
                                        $endgroup$
                                        – Zubin Mukerjee
                                        Jun 13 '15 at 3:25
















                                      89












                                      $begingroup$

                                      Informal justification of $text{positive} times text{negative} = text{negative}$



                                      Continue the pattern:



                                      $$
                                      begin{array}{r}
                                      2 & times & 3 & = & 6\
                                      2 & times & 2 & = & 4\
                                      2 & times & 1 & = & 2\
                                      2 & times & 0 & = & 0\
                                      2 & times & -1 & = & ? & (text{Answer} = -2 )\
                                      2 & times & -2 & = & ? & (text{Answer} = -4 )\
                                      2 & times & -3 & = & ? & (text{Answer} = -6 )\
                                      end{array}
                                      $$



                                      The number on the right-hand side keeps decreasing by 2.





                                      Informal justification of $text{negative} times text{negative} = text{positive}$



                                      Continue the pattern:



                                      $$
                                      begin{array}{r}
                                      2 & times & -3 & = & -6\
                                      1 & times & -3 & = & -3\
                                      0 & times & -3 & = & 0\
                                      -1 & times & -3 & = & ? & (text{Answer} = 3 )\
                                      -2 & times & -3 & = & ? & (text{Answer} = 6 )\
                                      -3 & times & -3 & = & ? & (text{Answer} = 9 )\
                                      end{array}
                                      $$



                                      The number on the right-hand side keeps increasing by 3.






                                      share|cite|improve this answer











                                      $endgroup$









                                      • 5




                                        $begingroup$
                                        Simple and just what I needed. Thank you.
                                        $endgroup$
                                        – John H
                                        Feb 18 '12 at 21:07










                                      • $begingroup$
                                        And very informal :) But I like intuitive approaches. Separates us from the nerds who live their lives without intuition.
                                        $endgroup$
                                        – Alfe
                                        Nov 12 '13 at 11:28






                                      • 14




                                        $begingroup$
                                        @Alfe I would say the opposite. "Nerds" might not talk "intuitive" to you, but I think their intuition is much better than "ours" and that might be what makes them nerds.:-)
                                        $endgroup$
                                        – quapka
                                        Nov 20 '14 at 0:25










                                      • $begingroup$
                                        I find it nice to extend this to all the reals by drawing a graph.
                                        $endgroup$
                                        – Michael Anderson
                                        Jun 12 '15 at 7:41










                                      • $begingroup$
                                        This is by far the best answer, +1.
                                        $endgroup$
                                        – Zubin Mukerjee
                                        Jun 13 '15 at 3:25














                                      89












                                      89








                                      89





                                      $begingroup$

                                      Informal justification of $text{positive} times text{negative} = text{negative}$



                                      Continue the pattern:



                                      $$
                                      begin{array}{r}
                                      2 & times & 3 & = & 6\
                                      2 & times & 2 & = & 4\
                                      2 & times & 1 & = & 2\
                                      2 & times & 0 & = & 0\
                                      2 & times & -1 & = & ? & (text{Answer} = -2 )\
                                      2 & times & -2 & = & ? & (text{Answer} = -4 )\
                                      2 & times & -3 & = & ? & (text{Answer} = -6 )\
                                      end{array}
                                      $$



                                      The number on the right-hand side keeps decreasing by 2.





                                      Informal justification of $text{negative} times text{negative} = text{positive}$



                                      Continue the pattern:



                                      $$
                                      begin{array}{r}
                                      2 & times & -3 & = & -6\
                                      1 & times & -3 & = & -3\
                                      0 & times & -3 & = & 0\
                                      -1 & times & -3 & = & ? & (text{Answer} = 3 )\
                                      -2 & times & -3 & = & ? & (text{Answer} = 6 )\
                                      -3 & times & -3 & = & ? & (text{Answer} = 9 )\
                                      end{array}
                                      $$



                                      The number on the right-hand side keeps increasing by 3.






                                      share|cite|improve this answer











                                      $endgroup$



                                      Informal justification of $text{positive} times text{negative} = text{negative}$



                                      Continue the pattern:



                                      $$
                                      begin{array}{r}
                                      2 & times & 3 & = & 6\
                                      2 & times & 2 & = & 4\
                                      2 & times & 1 & = & 2\
                                      2 & times & 0 & = & 0\
                                      2 & times & -1 & = & ? & (text{Answer} = -2 )\
                                      2 & times & -2 & = & ? & (text{Answer} = -4 )\
                                      2 & times & -3 & = & ? & (text{Answer} = -6 )\
                                      end{array}
                                      $$



                                      The number on the right-hand side keeps decreasing by 2.





                                      Informal justification of $text{negative} times text{negative} = text{positive}$



                                      Continue the pattern:



                                      $$
                                      begin{array}{r}
                                      2 & times & -3 & = & -6\
                                      1 & times & -3 & = & -3\
                                      0 & times & -3 & = & 0\
                                      -1 & times & -3 & = & ? & (text{Answer} = 3 )\
                                      -2 & times & -3 & = & ? & (text{Answer} = 6 )\
                                      -3 & times & -3 & = & ? & (text{Answer} = 9 )\
                                      end{array}
                                      $$



                                      The number on the right-hand side keeps increasing by 3.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited May 2 '13 at 13:16









                                      Jordan Gray

                                      10316




                                      10316










                                      answered Nov 15 '10 at 6:41









                                      Dan ChristensenDan Christensen

                                      8,74321835




                                      8,74321835








                                      • 5




                                        $begingroup$
                                        Simple and just what I needed. Thank you.
                                        $endgroup$
                                        – John H
                                        Feb 18 '12 at 21:07










                                      • $begingroup$
                                        And very informal :) But I like intuitive approaches. Separates us from the nerds who live their lives without intuition.
                                        $endgroup$
                                        – Alfe
                                        Nov 12 '13 at 11:28






                                      • 14




                                        $begingroup$
                                        @Alfe I would say the opposite. "Nerds" might not talk "intuitive" to you, but I think their intuition is much better than "ours" and that might be what makes them nerds.:-)
                                        $endgroup$
                                        – quapka
                                        Nov 20 '14 at 0:25










                                      • $begingroup$
                                        I find it nice to extend this to all the reals by drawing a graph.
                                        $endgroup$
                                        – Michael Anderson
                                        Jun 12 '15 at 7:41










                                      • $begingroup$
                                        This is by far the best answer, +1.
                                        $endgroup$
                                        – Zubin Mukerjee
                                        Jun 13 '15 at 3:25














                                      • 5




                                        $begingroup$
                                        Simple and just what I needed. Thank you.
                                        $endgroup$
                                        – John H
                                        Feb 18 '12 at 21:07










                                      • $begingroup$
                                        And very informal :) But I like intuitive approaches. Separates us from the nerds who live their lives without intuition.
                                        $endgroup$
                                        – Alfe
                                        Nov 12 '13 at 11:28






                                      • 14




                                        $begingroup$
                                        @Alfe I would say the opposite. "Nerds" might not talk "intuitive" to you, but I think their intuition is much better than "ours" and that might be what makes them nerds.:-)
                                        $endgroup$
                                        – quapka
                                        Nov 20 '14 at 0:25










                                      • $begingroup$
                                        I find it nice to extend this to all the reals by drawing a graph.
                                        $endgroup$
                                        – Michael Anderson
                                        Jun 12 '15 at 7:41










                                      • $begingroup$
                                        This is by far the best answer, +1.
                                        $endgroup$
                                        – Zubin Mukerjee
                                        Jun 13 '15 at 3:25








                                      5




                                      5




                                      $begingroup$
                                      Simple and just what I needed. Thank you.
                                      $endgroup$
                                      – John H
                                      Feb 18 '12 at 21:07




                                      $begingroup$
                                      Simple and just what I needed. Thank you.
                                      $endgroup$
                                      – John H
                                      Feb 18 '12 at 21:07












                                      $begingroup$
                                      And very informal :) But I like intuitive approaches. Separates us from the nerds who live their lives without intuition.
                                      $endgroup$
                                      – Alfe
                                      Nov 12 '13 at 11:28




                                      $begingroup$
                                      And very informal :) But I like intuitive approaches. Separates us from the nerds who live their lives without intuition.
                                      $endgroup$
                                      – Alfe
                                      Nov 12 '13 at 11:28




                                      14




                                      14




                                      $begingroup$
                                      @Alfe I would say the opposite. "Nerds" might not talk "intuitive" to you, but I think their intuition is much better than "ours" and that might be what makes them nerds.:-)
                                      $endgroup$
                                      – quapka
                                      Nov 20 '14 at 0:25




                                      $begingroup$
                                      @Alfe I would say the opposite. "Nerds" might not talk "intuitive" to you, but I think their intuition is much better than "ours" and that might be what makes them nerds.:-)
                                      $endgroup$
                                      – quapka
                                      Nov 20 '14 at 0:25












                                      $begingroup$
                                      I find it nice to extend this to all the reals by drawing a graph.
                                      $endgroup$
                                      – Michael Anderson
                                      Jun 12 '15 at 7:41




                                      $begingroup$
                                      I find it nice to extend this to all the reals by drawing a graph.
                                      $endgroup$
                                      – Michael Anderson
                                      Jun 12 '15 at 7:41












                                      $begingroup$
                                      This is by far the best answer, +1.
                                      $endgroup$
                                      – Zubin Mukerjee
                                      Jun 13 '15 at 3:25




                                      $begingroup$
                                      This is by far the best answer, +1.
                                      $endgroup$
                                      – Zubin Mukerjee
                                      Jun 13 '15 at 3:25











                                      60












                                      $begingroup$

                                      Someone sent this to me recently:



                                      I give you three $20 notes: +3 * +20 = +60 for you
                                      I give you three $20 debts: +3 * -20 = -60 for you
                                      I take three $20 notes from you: -3 * +20 = -60 for you
                                      I take three $20 debts from you: -3 * -20 = +60 for you





                                      share|cite|improve this answer









                                      $endgroup$









                                      • 12




                                        $begingroup$
                                        That's from the reddit ELI5 from a few days ago reddit.com/r/explainlikeimfive/comments/3r90cw/…
                                        $endgroup$
                                        – StuperUser
                                        Nov 9 '15 at 15:45
















                                      60












                                      $begingroup$

                                      Someone sent this to me recently:



                                      I give you three $20 notes: +3 * +20 = +60 for you
                                      I give you three $20 debts: +3 * -20 = -60 for you
                                      I take three $20 notes from you: -3 * +20 = -60 for you
                                      I take three $20 debts from you: -3 * -20 = +60 for you





                                      share|cite|improve this answer









                                      $endgroup$









                                      • 12




                                        $begingroup$
                                        That's from the reddit ELI5 from a few days ago reddit.com/r/explainlikeimfive/comments/3r90cw/…
                                        $endgroup$
                                        – StuperUser
                                        Nov 9 '15 at 15:45














                                      60












                                      60








                                      60





                                      $begingroup$

                                      Someone sent this to me recently:



                                      I give you three $20 notes: +3 * +20 = +60 for you
                                      I give you three $20 debts: +3 * -20 = -60 for you
                                      I take three $20 notes from you: -3 * +20 = -60 for you
                                      I take three $20 debts from you: -3 * -20 = +60 for you





                                      share|cite|improve this answer









                                      $endgroup$



                                      Someone sent this to me recently:



                                      I give you three $20 notes: +3 * +20 = +60 for you
                                      I give you three $20 debts: +3 * -20 = -60 for you
                                      I take three $20 notes from you: -3 * +20 = -60 for you
                                      I take three $20 debts from you: -3 * -20 = +60 for you






                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 9 '15 at 7:47









                                      EraesrEraesr

                                      46743




                                      46743








                                      • 12




                                        $begingroup$
                                        That's from the reddit ELI5 from a few days ago reddit.com/r/explainlikeimfive/comments/3r90cw/…
                                        $endgroup$
                                        – StuperUser
                                        Nov 9 '15 at 15:45














                                      • 12




                                        $begingroup$
                                        That's from the reddit ELI5 from a few days ago reddit.com/r/explainlikeimfive/comments/3r90cw/…
                                        $endgroup$
                                        – StuperUser
                                        Nov 9 '15 at 15:45








                                      12




                                      12




                                      $begingroup$
                                      That's from the reddit ELI5 from a few days ago reddit.com/r/explainlikeimfive/comments/3r90cw/…
                                      $endgroup$
                                      – StuperUser
                                      Nov 9 '15 at 15:45




                                      $begingroup$
                                      That's from the reddit ELI5 from a few days ago reddit.com/r/explainlikeimfive/comments/3r90cw/…
                                      $endgroup$
                                      – StuperUser
                                      Nov 9 '15 at 15:45











                                      47












                                      $begingroup$

                                      I think a lot of answers are either too simple or stray away from mathematics too much. Just remember that multiplication is repeated addition. When dealing with negative numbers, it becomes repeated subtraction.



                                      I'd simply put it in this context:




                                      1. The equation:
                                        $$begin{equation*}begin{array}{c}
                                        phantom{times9}2\
                                        underline{timesphantom{9}2}\
                                        phantom{times9}4\
                                        end{array}end{equation*}$$

                                        is just adding postive $2$, two times.



                                      2. The equation:
                                        $$begin{equation*}begin{array}{c}
                                        phantom{times999}2\
                                        underline{timesphantom{1}-2}\
                                        phantom{times9}-4\
                                        end{array}end{equation*}$$
                                        is just adding positive $2$, negative two times, which means instead of adding in the positive direction, you add in the negative direction (subtraction).


                                        You could also just say that you're adding $-2$ (or subtracting $2$), two times.



                                      3. The equation:
                                        $$begin{equation*}begin{array}{c}
                                        phantom{times9}-2\
                                        underline{timesphantom{1}-2}\
                                        phantom{times999}4\
                                        end{array}end{equation*}$$
                                        is adding $-2$, negative two times. Adding $-2$, two times, yields the diagram in (2). Since you have to add negative two times, you reverse the direction you are adding in.



                                        You could also say that you are subtracting $-2$, two times.








                                      share|cite|improve this answer











                                      $endgroup$













                                      • $begingroup$
                                        My memory of how I learned this is somewhat uncertain, because I was only 8 or 9 years old at the time, but I think this is very similar to how Isaac Asimov explained multiplication of negative numbers in either Realm of Numbers or Realm of Algebra. His explanation made sense to me at the time and I don't recall ever being confused as to why the product of negatives is positive. But I think it was a few years before this fact came up in the school curriculum.
                                        $endgroup$
                                        – David K
                                        Nov 9 '15 at 23:30






                                      • 5




                                        $begingroup$
                                        @homersimpson - I was thinking of this kind of logic but what knocks me out is how can we assume minus minus 2 is positive 2 - aren't we supposed to prove this?
                                        $endgroup$
                                        – KGhatak
                                        Jan 11 '16 at 8:00
















                                      47












                                      $begingroup$

                                      I think a lot of answers are either too simple or stray away from mathematics too much. Just remember that multiplication is repeated addition. When dealing with negative numbers, it becomes repeated subtraction.



                                      I'd simply put it in this context:




                                      1. The equation:
                                        $$begin{equation*}begin{array}{c}
                                        phantom{times9}2\
                                        underline{timesphantom{9}2}\
                                        phantom{times9}4\
                                        end{array}end{equation*}$$

                                        is just adding postive $2$, two times.



                                      2. The equation:
                                        $$begin{equation*}begin{array}{c}
                                        phantom{times999}2\
                                        underline{timesphantom{1}-2}\
                                        phantom{times9}-4\
                                        end{array}end{equation*}$$
                                        is just adding positive $2$, negative two times, which means instead of adding in the positive direction, you add in the negative direction (subtraction).


                                        You could also just say that you're adding $-2$ (or subtracting $2$), two times.



                                      3. The equation:
                                        $$begin{equation*}begin{array}{c}
                                        phantom{times9}-2\
                                        underline{timesphantom{1}-2}\
                                        phantom{times999}4\
                                        end{array}end{equation*}$$
                                        is adding $-2$, negative two times. Adding $-2$, two times, yields the diagram in (2). Since you have to add negative two times, you reverse the direction you are adding in.



                                        You could also say that you are subtracting $-2$, two times.








                                      share|cite|improve this answer











                                      $endgroup$













                                      • $begingroup$
                                        My memory of how I learned this is somewhat uncertain, because I was only 8 or 9 years old at the time, but I think this is very similar to how Isaac Asimov explained multiplication of negative numbers in either Realm of Numbers or Realm of Algebra. His explanation made sense to me at the time and I don't recall ever being confused as to why the product of negatives is positive. But I think it was a few years before this fact came up in the school curriculum.
                                        $endgroup$
                                        – David K
                                        Nov 9 '15 at 23:30






                                      • 5




                                        $begingroup$
                                        @homersimpson - I was thinking of this kind of logic but what knocks me out is how can we assume minus minus 2 is positive 2 - aren't we supposed to prove this?
                                        $endgroup$
                                        – KGhatak
                                        Jan 11 '16 at 8:00














                                      47












                                      47








                                      47





                                      $begingroup$

                                      I think a lot of answers are either too simple or stray away from mathematics too much. Just remember that multiplication is repeated addition. When dealing with negative numbers, it becomes repeated subtraction.



                                      I'd simply put it in this context:




                                      1. The equation:
                                        $$begin{equation*}begin{array}{c}
                                        phantom{times9}2\
                                        underline{timesphantom{9}2}\
                                        phantom{times9}4\
                                        end{array}end{equation*}$$

                                        is just adding postive $2$, two times.



                                      2. The equation:
                                        $$begin{equation*}begin{array}{c}
                                        phantom{times999}2\
                                        underline{timesphantom{1}-2}\
                                        phantom{times9}-4\
                                        end{array}end{equation*}$$
                                        is just adding positive $2$, negative two times, which means instead of adding in the positive direction, you add in the negative direction (subtraction).


                                        You could also just say that you're adding $-2$ (or subtracting $2$), two times.



                                      3. The equation:
                                        $$begin{equation*}begin{array}{c}
                                        phantom{times9}-2\
                                        underline{timesphantom{1}-2}\
                                        phantom{times999}4\
                                        end{array}end{equation*}$$
                                        is adding $-2$, negative two times. Adding $-2$, two times, yields the diagram in (2). Since you have to add negative two times, you reverse the direction you are adding in.



                                        You could also say that you are subtracting $-2$, two times.








                                      share|cite|improve this answer











                                      $endgroup$



                                      I think a lot of answers are either too simple or stray away from mathematics too much. Just remember that multiplication is repeated addition. When dealing with negative numbers, it becomes repeated subtraction.



                                      I'd simply put it in this context:




                                      1. The equation:
                                        $$begin{equation*}begin{array}{c}
                                        phantom{times9}2\
                                        underline{timesphantom{9}2}\
                                        phantom{times9}4\
                                        end{array}end{equation*}$$

                                        is just adding postive $2$, two times.



                                      2. The equation:
                                        $$begin{equation*}begin{array}{c}
                                        phantom{times999}2\
                                        underline{timesphantom{1}-2}\
                                        phantom{times9}-4\
                                        end{array}end{equation*}$$
                                        is just adding positive $2$, negative two times, which means instead of adding in the positive direction, you add in the negative direction (subtraction).


                                        You could also just say that you're adding $-2$ (or subtracting $2$), two times.



                                      3. The equation:
                                        $$begin{equation*}begin{array}{c}
                                        phantom{times9}-2\
                                        underline{timesphantom{1}-2}\
                                        phantom{times999}4\
                                        end{array}end{equation*}$$
                                        is adding $-2$, negative two times. Adding $-2$, two times, yields the diagram in (2). Since you have to add negative two times, you reverse the direction you are adding in.



                                        You could also say that you are subtracting $-2$, two times.









                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Nov 9 '15 at 8:20

























                                      answered Nov 9 '15 at 8:05









                                      homersimpsonhomersimpson

                                      46745




                                      46745












                                      • $begingroup$
                                        My memory of how I learned this is somewhat uncertain, because I was only 8 or 9 years old at the time, but I think this is very similar to how Isaac Asimov explained multiplication of negative numbers in either Realm of Numbers or Realm of Algebra. His explanation made sense to me at the time and I don't recall ever being confused as to why the product of negatives is positive. But I think it was a few years before this fact came up in the school curriculum.
                                        $endgroup$
                                        – David K
                                        Nov 9 '15 at 23:30






                                      • 5




                                        $begingroup$
                                        @homersimpson - I was thinking of this kind of logic but what knocks me out is how can we assume minus minus 2 is positive 2 - aren't we supposed to prove this?
                                        $endgroup$
                                        – KGhatak
                                        Jan 11 '16 at 8:00


















                                      • $begingroup$
                                        My memory of how I learned this is somewhat uncertain, because I was only 8 or 9 years old at the time, but I think this is very similar to how Isaac Asimov explained multiplication of negative numbers in either Realm of Numbers or Realm of Algebra. His explanation made sense to me at the time and I don't recall ever being confused as to why the product of negatives is positive. But I think it was a few years before this fact came up in the school curriculum.
                                        $endgroup$
                                        – David K
                                        Nov 9 '15 at 23:30






                                      • 5




                                        $begingroup$
                                        @homersimpson - I was thinking of this kind of logic but what knocks me out is how can we assume minus minus 2 is positive 2 - aren't we supposed to prove this?
                                        $endgroup$
                                        – KGhatak
                                        Jan 11 '16 at 8:00
















                                      $begingroup$
                                      My memory of how I learned this is somewhat uncertain, because I was only 8 or 9 years old at the time, but I think this is very similar to how Isaac Asimov explained multiplication of negative numbers in either Realm of Numbers or Realm of Algebra. His explanation made sense to me at the time and I don't recall ever being confused as to why the product of negatives is positive. But I think it was a few years before this fact came up in the school curriculum.
                                      $endgroup$
                                      – David K
                                      Nov 9 '15 at 23:30




                                      $begingroup$
                                      My memory of how I learned this is somewhat uncertain, because I was only 8 or 9 years old at the time, but I think this is very similar to how Isaac Asimov explained multiplication of negative numbers in either Realm of Numbers or Realm of Algebra. His explanation made sense to me at the time and I don't recall ever being confused as to why the product of negatives is positive. But I think it was a few years before this fact came up in the school curriculum.
                                      $endgroup$
                                      – David K
                                      Nov 9 '15 at 23:30




                                      5




                                      5




                                      $begingroup$
                                      @homersimpson - I was thinking of this kind of logic but what knocks me out is how can we assume minus minus 2 is positive 2 - aren't we supposed to prove this?
                                      $endgroup$
                                      – KGhatak
                                      Jan 11 '16 at 8:00




                                      $begingroup$
                                      @homersimpson - I was thinking of this kind of logic but what knocks me out is how can we assume minus minus 2 is positive 2 - aren't we supposed to prove this?
                                      $endgroup$
                                      – KGhatak
                                      Jan 11 '16 at 8:00











                                      42












                                      $begingroup$

                                      $overbrace{bf Law of Signs}^{rmLarge {(-x)(-y)} = xy} $ proof: $rm, (-x)(-y) = (-x)(-y) + color{#c00}x(overbrace{color{#c00}{-y} + y}^{Large =,0}) = (overbrace{-x+color{#c00}x}^{Large =,0})(color{#c00}{-y}) + xy = xy$



                                      Equivalently, $ $ evaluate $rm, overline{(-x)(-y) + } overline{ underline {color{#c00}{x(-y)}}}underline{phantom{(}! +,xy}, $ in $,2,$ ways, noting over/underlined terms $ = 0$



                                      Said more conceptually $rm (-x)(-y) $ and $rm:xy:$ are both inverses of $rm color{#c00}{x(-y)} $ so they are equal by uniqueness of inverses: $ $ if $,color{#c00}a,$ has two additive inverses $,{-a},$ and $,color{#0a0}{-a},,$ then



                                      $$rm {-a}, =, {-a}+overbrace{(color{#c00}a+color{#0a0}{-a})}^{large =,0}, =, overbrace{({-a}+color{#c00}a)}^{large =,0}+color{#0a0}{-a}, =, color{#0a0}{-a} $$



                                      Equivalently, $ $ evaluate $rm, overline{-a +!} overline{phantom{+} underline {color{#c00}{a}}}underline{ + color{#0a0}{-a}}, $ in $,2,$ ways, noting over/underlined terms $ = 0$



                                      This proof of the Law of Signs uses well-known laws of positive integers (esp. the distributive law), so if we require that these laws persist in the larger system of positive and negative integers, then the Law of Signs is a logical consequence of these basic laws of positive integers.



                                      These fundamental laws of "numbers" are axiomatized by the algebraic structure known as a ring, and various specializations thereof. Since the above proof uses only ring laws (most notably the distributive law), the Law of Signs holds true in every ring, e.g. rings of polynomials, power series, matrices, differential operators, etc. In fact every nontrivial ring theorem (i.e. one that does not degenerate to a theorem about the underlying additive group or multiplicative monoid), must employ the distributive law, since that is the only law that connects the additive and multiplicative structures that combine to form the ring structure. Without the distributive law a ring degenerates to a set with two completely unrelated additive and multiplicative structures. So, in a sense, the distributive law is a keystone of the ring structure.



                                      Remark $ $ More generally the Law of Signs holds for any odd functions
                                      under composition, e.g. polynomials with all terms having odd power. Indeed we have



                                      $qquadrm f(g) = (-f) (-g) = -(f(-g)) iff f(-g) = -(f(g)) $



                                      $qquadrm phantom{ (-f) (-g) = -(f(-g)) = f(g) }!!overset{ large g(x)=x}iff f(-x) = -f(x), $ ie. $rm:f:$ is odd



                                      Generally such functions enjoy only a weaker near-ring structure.
                                      In the above case of rings, distributivity implies that multiplication
                                      is linear hence odd (viewing the ring in Cayley-style
                                      as the ring of endormorphisms of its abelian additive group,
                                      i.e. representing each ring element $rm r $ by the linear map $rm x to r x, $
                                      i.e. as a $1$-dim matrix).






                                      share|cite|improve this answer











                                      $endgroup$









                                      • 4




                                        $begingroup$
                                        +1. The simplest way to prove why is by using the group axioms and thier consequences.
                                        $endgroup$
                                        – Mathemagician1234
                                        Jun 13 '15 at 3:18










                                      • $begingroup$
                                        What's the logical thinking skills or tactic used to jump in one step from xy or (−x)(−y) to xy+x (−y)+(−x)(−y)?
                                        $endgroup$
                                        – Pacerier
                                        Feb 2 '18 at 7:15












                                      • $begingroup$
                                        @Pacerier It's a common way to prove uniqueness of inverses. I edited the answer to elaborate on that.
                                        $endgroup$
                                        – Bill Dubuque
                                        Oct 16 '18 at 17:48
















                                      42












                                      $begingroup$

                                      $overbrace{bf Law of Signs}^{rmLarge {(-x)(-y)} = xy} $ proof: $rm, (-x)(-y) = (-x)(-y) + color{#c00}x(overbrace{color{#c00}{-y} + y}^{Large =,0}) = (overbrace{-x+color{#c00}x}^{Large =,0})(color{#c00}{-y}) + xy = xy$



                                      Equivalently, $ $ evaluate $rm, overline{(-x)(-y) + } overline{ underline {color{#c00}{x(-y)}}}underline{phantom{(}! +,xy}, $ in $,2,$ ways, noting over/underlined terms $ = 0$



                                      Said more conceptually $rm (-x)(-y) $ and $rm:xy:$ are both inverses of $rm color{#c00}{x(-y)} $ so they are equal by uniqueness of inverses: $ $ if $,color{#c00}a,$ has two additive inverses $,{-a},$ and $,color{#0a0}{-a},,$ then



                                      $$rm {-a}, =, {-a}+overbrace{(color{#c00}a+color{#0a0}{-a})}^{large =,0}, =, overbrace{({-a}+color{#c00}a)}^{large =,0}+color{#0a0}{-a}, =, color{#0a0}{-a} $$



                                      Equivalently, $ $ evaluate $rm, overline{-a +!} overline{phantom{+} underline {color{#c00}{a}}}underline{ + color{#0a0}{-a}}, $ in $,2,$ ways, noting over/underlined terms $ = 0$



                                      This proof of the Law of Signs uses well-known laws of positive integers (esp. the distributive law), so if we require that these laws persist in the larger system of positive and negative integers, then the Law of Signs is a logical consequence of these basic laws of positive integers.



                                      These fundamental laws of "numbers" are axiomatized by the algebraic structure known as a ring, and various specializations thereof. Since the above proof uses only ring laws (most notably the distributive law), the Law of Signs holds true in every ring, e.g. rings of polynomials, power series, matrices, differential operators, etc. In fact every nontrivial ring theorem (i.e. one that does not degenerate to a theorem about the underlying additive group or multiplicative monoid), must employ the distributive law, since that is the only law that connects the additive and multiplicative structures that combine to form the ring structure. Without the distributive law a ring degenerates to a set with two completely unrelated additive and multiplicative structures. So, in a sense, the distributive law is a keystone of the ring structure.



                                      Remark $ $ More generally the Law of Signs holds for any odd functions
                                      under composition, e.g. polynomials with all terms having odd power. Indeed we have



                                      $qquadrm f(g) = (-f) (-g) = -(f(-g)) iff f(-g) = -(f(g)) $



                                      $qquadrm phantom{ (-f) (-g) = -(f(-g)) = f(g) }!!overset{ large g(x)=x}iff f(-x) = -f(x), $ ie. $rm:f:$ is odd



                                      Generally such functions enjoy only a weaker near-ring structure.
                                      In the above case of rings, distributivity implies that multiplication
                                      is linear hence odd (viewing the ring in Cayley-style
                                      as the ring of endormorphisms of its abelian additive group,
                                      i.e. representing each ring element $rm r $ by the linear map $rm x to r x, $
                                      i.e. as a $1$-dim matrix).






                                      share|cite|improve this answer











                                      $endgroup$









                                      • 4




                                        $begingroup$
                                        +1. The simplest way to prove why is by using the group axioms and thier consequences.
                                        $endgroup$
                                        – Mathemagician1234
                                        Jun 13 '15 at 3:18










                                      • $begingroup$
                                        What's the logical thinking skills or tactic used to jump in one step from xy or (−x)(−y) to xy+x (−y)+(−x)(−y)?
                                        $endgroup$
                                        – Pacerier
                                        Feb 2 '18 at 7:15












                                      • $begingroup$
                                        @Pacerier It's a common way to prove uniqueness of inverses. I edited the answer to elaborate on that.
                                        $endgroup$
                                        – Bill Dubuque
                                        Oct 16 '18 at 17:48














                                      42












                                      42








                                      42





                                      $begingroup$

                                      $overbrace{bf Law of Signs}^{rmLarge {(-x)(-y)} = xy} $ proof: $rm, (-x)(-y) = (-x)(-y) + color{#c00}x(overbrace{color{#c00}{-y} + y}^{Large =,0}) = (overbrace{-x+color{#c00}x}^{Large =,0})(color{#c00}{-y}) + xy = xy$



                                      Equivalently, $ $ evaluate $rm, overline{(-x)(-y) + } overline{ underline {color{#c00}{x(-y)}}}underline{phantom{(}! +,xy}, $ in $,2,$ ways, noting over/underlined terms $ = 0$



                                      Said more conceptually $rm (-x)(-y) $ and $rm:xy:$ are both inverses of $rm color{#c00}{x(-y)} $ so they are equal by uniqueness of inverses: $ $ if $,color{#c00}a,$ has two additive inverses $,{-a},$ and $,color{#0a0}{-a},,$ then



                                      $$rm {-a}, =, {-a}+overbrace{(color{#c00}a+color{#0a0}{-a})}^{large =,0}, =, overbrace{({-a}+color{#c00}a)}^{large =,0}+color{#0a0}{-a}, =, color{#0a0}{-a} $$



                                      Equivalently, $ $ evaluate $rm, overline{-a +!} overline{phantom{+} underline {color{#c00}{a}}}underline{ + color{#0a0}{-a}}, $ in $,2,$ ways, noting over/underlined terms $ = 0$



                                      This proof of the Law of Signs uses well-known laws of positive integers (esp. the distributive law), so if we require that these laws persist in the larger system of positive and negative integers, then the Law of Signs is a logical consequence of these basic laws of positive integers.



                                      These fundamental laws of "numbers" are axiomatized by the algebraic structure known as a ring, and various specializations thereof. Since the above proof uses only ring laws (most notably the distributive law), the Law of Signs holds true in every ring, e.g. rings of polynomials, power series, matrices, differential operators, etc. In fact every nontrivial ring theorem (i.e. one that does not degenerate to a theorem about the underlying additive group or multiplicative monoid), must employ the distributive law, since that is the only law that connects the additive and multiplicative structures that combine to form the ring structure. Without the distributive law a ring degenerates to a set with two completely unrelated additive and multiplicative structures. So, in a sense, the distributive law is a keystone of the ring structure.



                                      Remark $ $ More generally the Law of Signs holds for any odd functions
                                      under composition, e.g. polynomials with all terms having odd power. Indeed we have



                                      $qquadrm f(g) = (-f) (-g) = -(f(-g)) iff f(-g) = -(f(g)) $



                                      $qquadrm phantom{ (-f) (-g) = -(f(-g)) = f(g) }!!overset{ large g(x)=x}iff f(-x) = -f(x), $ ie. $rm:f:$ is odd



                                      Generally such functions enjoy only a weaker near-ring structure.
                                      In the above case of rings, distributivity implies that multiplication
                                      is linear hence odd (viewing the ring in Cayley-style
                                      as the ring of endormorphisms of its abelian additive group,
                                      i.e. representing each ring element $rm r $ by the linear map $rm x to r x, $
                                      i.e. as a $1$-dim matrix).






                                      share|cite|improve this answer











                                      $endgroup$



                                      $overbrace{bf Law of Signs}^{rmLarge {(-x)(-y)} = xy} $ proof: $rm, (-x)(-y) = (-x)(-y) + color{#c00}x(overbrace{color{#c00}{-y} + y}^{Large =,0}) = (overbrace{-x+color{#c00}x}^{Large =,0})(color{#c00}{-y}) + xy = xy$



                                      Equivalently, $ $ evaluate $rm, overline{(-x)(-y) + } overline{ underline {color{#c00}{x(-y)}}}underline{phantom{(}! +,xy}, $ in $,2,$ ways, noting over/underlined terms $ = 0$



                                      Said more conceptually $rm (-x)(-y) $ and $rm:xy:$ are both inverses of $rm color{#c00}{x(-y)} $ so they are equal by uniqueness of inverses: $ $ if $,color{#c00}a,$ has two additive inverses $,{-a},$ and $,color{#0a0}{-a},,$ then



                                      $$rm {-a}, =, {-a}+overbrace{(color{#c00}a+color{#0a0}{-a})}^{large =,0}, =, overbrace{({-a}+color{#c00}a)}^{large =,0}+color{#0a0}{-a}, =, color{#0a0}{-a} $$



                                      Equivalently, $ $ evaluate $rm, overline{-a +!} overline{phantom{+} underline {color{#c00}{a}}}underline{ + color{#0a0}{-a}}, $ in $,2,$ ways, noting over/underlined terms $ = 0$



                                      This proof of the Law of Signs uses well-known laws of positive integers (esp. the distributive law), so if we require that these laws persist in the larger system of positive and negative integers, then the Law of Signs is a logical consequence of these basic laws of positive integers.



                                      These fundamental laws of "numbers" are axiomatized by the algebraic structure known as a ring, and various specializations thereof. Since the above proof uses only ring laws (most notably the distributive law), the Law of Signs holds true in every ring, e.g. rings of polynomials, power series, matrices, differential operators, etc. In fact every nontrivial ring theorem (i.e. one that does not degenerate to a theorem about the underlying additive group or multiplicative monoid), must employ the distributive law, since that is the only law that connects the additive and multiplicative structures that combine to form the ring structure. Without the distributive law a ring degenerates to a set with two completely unrelated additive and multiplicative structures. So, in a sense, the distributive law is a keystone of the ring structure.



                                      Remark $ $ More generally the Law of Signs holds for any odd functions
                                      under composition, e.g. polynomials with all terms having odd power. Indeed we have



                                      $qquadrm f(g) = (-f) (-g) = -(f(-g)) iff f(-g) = -(f(g)) $



                                      $qquadrm phantom{ (-f) (-g) = -(f(-g)) = f(g) }!!overset{ large g(x)=x}iff f(-x) = -f(x), $ ie. $rm:f:$ is odd



                                      Generally such functions enjoy only a weaker near-ring structure.
                                      In the above case of rings, distributivity implies that multiplication
                                      is linear hence odd (viewing the ring in Cayley-style
                                      as the ring of endormorphisms of its abelian additive group,
                                      i.e. representing each ring element $rm r $ by the linear map $rm x to r x, $
                                      i.e. as a $1$-dim matrix).







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Oct 16 '18 at 17:38

























                                      answered Nov 12 '10 at 2:40









                                      Bill DubuqueBill Dubuque

                                      214k29197659




                                      214k29197659








                                      • 4




                                        $begingroup$
                                        +1. The simplest way to prove why is by using the group axioms and thier consequences.
                                        $endgroup$
                                        – Mathemagician1234
                                        Jun 13 '15 at 3:18










                                      • $begingroup$
                                        What's the logical thinking skills or tactic used to jump in one step from xy or (−x)(−y) to xy+x (−y)+(−x)(−y)?
                                        $endgroup$
                                        – Pacerier
                                        Feb 2 '18 at 7:15












                                      • $begingroup$
                                        @Pacerier It's a common way to prove uniqueness of inverses. I edited the answer to elaborate on that.
                                        $endgroup$
                                        – Bill Dubuque
                                        Oct 16 '18 at 17:48














                                      • 4




                                        $begingroup$
                                        +1. The simplest way to prove why is by using the group axioms and thier consequences.
                                        $endgroup$
                                        – Mathemagician1234
                                        Jun 13 '15 at 3:18










                                      • $begingroup$
                                        What's the logical thinking skills or tactic used to jump in one step from xy or (−x)(−y) to xy+x (−y)+(−x)(−y)?
                                        $endgroup$
                                        – Pacerier
                                        Feb 2 '18 at 7:15












                                      • $begingroup$
                                        @Pacerier It's a common way to prove uniqueness of inverses. I edited the answer to elaborate on that.
                                        $endgroup$
                                        – Bill Dubuque
                                        Oct 16 '18 at 17:48








                                      4




                                      4




                                      $begingroup$
                                      +1. The simplest way to prove why is by using the group axioms and thier consequences.
                                      $endgroup$
                                      – Mathemagician1234
                                      Jun 13 '15 at 3:18




                                      $begingroup$
                                      +1. The simplest way to prove why is by using the group axioms and thier consequences.
                                      $endgroup$
                                      – Mathemagician1234
                                      Jun 13 '15 at 3:18












                                      $begingroup$
                                      What's the logical thinking skills or tactic used to jump in one step from xy or (−x)(−y) to xy+x (−y)+(−x)(−y)?
                                      $endgroup$
                                      – Pacerier
                                      Feb 2 '18 at 7:15






                                      $begingroup$
                                      What's the logical thinking skills or tactic used to jump in one step from xy or (−x)(−y) to xy+x (−y)+(−x)(−y)?
                                      $endgroup$
                                      – Pacerier
                                      Feb 2 '18 at 7:15














                                      $begingroup$
                                      @Pacerier It's a common way to prove uniqueness of inverses. I edited the answer to elaborate on that.
                                      $endgroup$
                                      – Bill Dubuque
                                      Oct 16 '18 at 17:48




                                      $begingroup$
                                      @Pacerier It's a common way to prove uniqueness of inverses. I edited the answer to elaborate on that.
                                      $endgroup$
                                      – Bill Dubuque
                                      Oct 16 '18 at 17:48











                                      36












                                      $begingroup$

                                      Simple Answer:
                                      $$
                                      (-a)b + ab = (-a)b + ab
                                      $$
                                      $$(-a)b + ab = b(a-a)
                                      $$
                                      $$(-a)b + ab = b(0)
                                      $$
                                      $$(-a)b + ab = 0 $$
                                      $$(-a)b = -ab $$
                                      $$(-a)(-b) + (-ab) = (-a)(-b) + (-a)b $$
                                      $$(-a)(-b) + (-ab) = (-a)(b-b) $$
                                      $$(-a)(-b) + (-ab) = (-a)(0) $$
                                      $$(-a)(-b) + (-ab) = 0 $$
                                      $$*(-a)(-b) = ab $$
                                      Hope this helps (Credit to Michael Spivak's Calculus)



                                      ~ Alan






                                      share|cite|improve this answer











                                      $endgroup$













                                      • $begingroup$
                                        Well,I wouldn't exactly call that a SIMPLE answer. A DIRECT and STRAIGHTFORWARD one,yes,absolutely. But not simple.
                                        $endgroup$
                                        – Mathemagician1234
                                        Nov 20 '14 at 0:30










                                      • $begingroup$
                                        Easily the best and most satisfying answer here.
                                        $endgroup$
                                        – Derek 朕會功夫
                                        Aug 17 '15 at 20:42
















                                      36












                                      $begingroup$

                                      Simple Answer:
                                      $$
                                      (-a)b + ab = (-a)b + ab
                                      $$
                                      $$(-a)b + ab = b(a-a)
                                      $$
                                      $$(-a)b + ab = b(0)
                                      $$
                                      $$(-a)b + ab = 0 $$
                                      $$(-a)b = -ab $$
                                      $$(-a)(-b) + (-ab) = (-a)(-b) + (-a)b $$
                                      $$(-a)(-b) + (-ab) = (-a)(b-b) $$
                                      $$(-a)(-b) + (-ab) = (-a)(0) $$
                                      $$(-a)(-b) + (-ab) = 0 $$
                                      $$*(-a)(-b) = ab $$
                                      Hope this helps (Credit to Michael Spivak's Calculus)



                                      ~ Alan






                                      share|cite|improve this answer











                                      $endgroup$













                                      • $begingroup$
                                        Well,I wouldn't exactly call that a SIMPLE answer. A DIRECT and STRAIGHTFORWARD one,yes,absolutely. But not simple.
                                        $endgroup$
                                        – Mathemagician1234
                                        Nov 20 '14 at 0:30










                                      • $begingroup$
                                        Easily the best and most satisfying answer here.
                                        $endgroup$
                                        – Derek 朕會功夫
                                        Aug 17 '15 at 20:42














                                      36












                                      36








                                      36





                                      $begingroup$

                                      Simple Answer:
                                      $$
                                      (-a)b + ab = (-a)b + ab
                                      $$
                                      $$(-a)b + ab = b(a-a)
                                      $$
                                      $$(-a)b + ab = b(0)
                                      $$
                                      $$(-a)b + ab = 0 $$
                                      $$(-a)b = -ab $$
                                      $$(-a)(-b) + (-ab) = (-a)(-b) + (-a)b $$
                                      $$(-a)(-b) + (-ab) = (-a)(b-b) $$
                                      $$(-a)(-b) + (-ab) = (-a)(0) $$
                                      $$(-a)(-b) + (-ab) = 0 $$
                                      $$*(-a)(-b) = ab $$
                                      Hope this helps (Credit to Michael Spivak's Calculus)



                                      ~ Alan






                                      share|cite|improve this answer











                                      $endgroup$



                                      Simple Answer:
                                      $$
                                      (-a)b + ab = (-a)b + ab
                                      $$
                                      $$(-a)b + ab = b(a-a)
                                      $$
                                      $$(-a)b + ab = b(0)
                                      $$
                                      $$(-a)b + ab = 0 $$
                                      $$(-a)b = -ab $$
                                      $$(-a)(-b) + (-ab) = (-a)(-b) + (-a)b $$
                                      $$(-a)(-b) + (-ab) = (-a)(b-b) $$
                                      $$(-a)(-b) + (-ab) = (-a)(0) $$
                                      $$(-a)(-b) + (-ab) = 0 $$
                                      $$*(-a)(-b) = ab $$
                                      Hope this helps (Credit to Michael Spivak's Calculus)



                                      ~ Alan







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited May 13 '13 at 2:35









                                      Vigneshwaren

                                      308111




                                      308111










                                      answered May 7 '13 at 20:39









                                      Alan GarciaAlan Garcia

                                      46942




                                      46942












                                      • $begingroup$
                                        Well,I wouldn't exactly call that a SIMPLE answer. A DIRECT and STRAIGHTFORWARD one,yes,absolutely. But not simple.
                                        $endgroup$
                                        – Mathemagician1234
                                        Nov 20 '14 at 0:30










                                      • $begingroup$
                                        Easily the best and most satisfying answer here.
                                        $endgroup$
                                        – Derek 朕會功夫
                                        Aug 17 '15 at 20:42


















                                      • $begingroup$
                                        Well,I wouldn't exactly call that a SIMPLE answer. A DIRECT and STRAIGHTFORWARD one,yes,absolutely. But not simple.
                                        $endgroup$
                                        – Mathemagician1234
                                        Nov 20 '14 at 0:30










                                      • $begingroup$
                                        Easily the best and most satisfying answer here.
                                        $endgroup$
                                        – Derek 朕會功夫
                                        Aug 17 '15 at 20:42
















                                      $begingroup$
                                      Well,I wouldn't exactly call that a SIMPLE answer. A DIRECT and STRAIGHTFORWARD one,yes,absolutely. But not simple.
                                      $endgroup$
                                      – Mathemagician1234
                                      Nov 20 '14 at 0:30




                                      $begingroup$
                                      Well,I wouldn't exactly call that a SIMPLE answer. A DIRECT and STRAIGHTFORWARD one,yes,absolutely. But not simple.
                                      $endgroup$
                                      – Mathemagician1234
                                      Nov 20 '14 at 0:30












                                      $begingroup$
                                      Easily the best and most satisfying answer here.
                                      $endgroup$
                                      – Derek 朕會功夫
                                      Aug 17 '15 at 20:42




                                      $begingroup$
                                      Easily the best and most satisfying answer here.
                                      $endgroup$
                                      – Derek 朕會功夫
                                      Aug 17 '15 at 20:42











                                      28












                                      $begingroup$

                                      This is really one of those important questions that leads many people to say "Math sucks!". In fact, for many students, mathematics stopped making sense somewhere along the way. Either slowly or dramatically, they gave up on the field as hopelessly baffling and difficult, and they grew up to be adults who — confident that others share their experience — nonchalantly announce, “Math was just not for me
                                      or “I was never good at it.
                                      Usually the process is gradual, but for Ruth
                                      McNeill, the turning point was clearly defined. In an
                                      article in the Journal of Mathematical Behavior, she
                                      described how it happened:




                                      What did me in was the idea that a negative number
                                      times a negative number comes out to a positive
                                      number. This seemed (and still seems) inherently
                                      unlikely — counter intuitive, as mathematicians say.
                                      I wrestled with the idea for what I imagine to be
                                      several weeks, trying to get a sensible explanation
                                      from my teacher, my classmates, my parents, any-
                                      body. Whatever explanations they offered could not
                                      overcome my strong sense that multiplying intensifies something, and thus two negative numbers
                                      multiplied together should properly produce a very
                                      negative result. I have since been offered a moderately convincing explanation that features a film of a
                                      swimming pool being drained that gets run back-
                                      wards through the projector. At the time, however,
                                      nothing convinced me. The most commonsense of
                                      all school subjects had abandoned common sense;
                                      I was indignant and baffled.


                                      Meanwhile, the curriculum kept rolling on, and I
                                      could see that I couldn’t stay behind, stuck on nega-
                                      tive times negative. I would have to pay attention to
                                      the next topic, and the only practical course open to
                                      me was to pretend to agree that negative times nega-
                                      tive equals positive. The book and the teacher and
                                      the general consensus of the algebra survivors of so-
                                      ciety were clearly more powerful than I was. I capitu-
                                      lated. I did the rest of algebra, and geometry, and
                                      trigonometry; I did them in the advanced sections,
                                      and I often had that nice sense of “aha!” when I
                                      could suddenly see how a proof was going to come
                                      out. Underneath, however, a kind of resentment and
                                      betrayal lurked, and I was not surprised or dismayed
                                      by any further foolishness my math teachers had up
                                      their sleeves.... Intellectually, I was disengaged, and
                                      when math was no longer required, I took German
                                      instead.






                                      I will show in this answer that: negative $times$ negative $=$ positive, is in fact not counter-intuitive at all! There are many ways that we can use to show that result, but I'd like to show my personal way of thinking about the latter.



                                      Let's imagine we're sitting near a road, and there is a car that is moving with a constant speed. We also have a clock, and so we can measure time.



                                      Before going any further, we should first specify some assumptions like if the car is moving in the right, then its velocity will be positive, and if it's moving in the left direction, then its velocity will be negative.



                                      enter image description here



                                      Imagine now that you have a video of the above scene, and time is positive if you play the video normally but is negative if you play it backwards. We also know the following: $$rm Velocity=dfrac{rm Distance}{rm Time}.$$
                                      Solving for distance we get: $$rm Velocitytimes{rm Time}={rm Distance}.$$
                                      Here the important part comes, if the car is moving in the $+$ direction and the time the video is played is positive, i.e. the video is played normally, then you'll see that the car moves along the $+$ direction and you'll calculate that it moves "a positive distance". Therefore the following holds: $$rm positivetimes positive=positive.$$



                                      If the car moves along the $-$ direction and the time the video is played is positive, i.e. the video is played normally, then you'll see the car going along the $-$ direction, you'll then calculate that it moves "a negative distance". Thus: $$rm negativetimes positive=negative.$$



                                      If however, the car moves along the $+$ direction but the time the video is played is negative, i.e. the video is played backwards, then you'll see that the car is moving in the $-$ direction, and you'll calculate that it moves "a negative distance". Thus: $$rm positivetimes negative=negative.$$



                                      If the car moves along the $-$ direction but the time the video is played is negative, i.e. the video is played backwards, you'll see that the car moves along the $+$ direction! Thus it moves "a positive distance". And therefore: $$color{grey}{boxed{color{white}{underline{overline{color{black}{displaystylerm, negativetimes negative=positive.,}}}}}}$$ As you have seen, it takes only a little bit of imagination for it to make sense.



                                      I hope this helps.
                                      Best wishes, $mathcal H$akim.






                                      share|cite|improve this answer











                                      $endgroup$









                                      • 6




                                        $begingroup$
                                        The material you quoted contains the interesting statement "Whatever explanations they offered could not overcome my strong sense that multiplying intensifies something". I wonder whether anyone tried the very short answer "yes, but negative intensification is attenuation"?
                                        $endgroup$
                                        – Andreas Blass
                                        Jun 29 '14 at 21:43










                                      • $begingroup$
                                        @AndreasBlass Or in another form: "negative multiplication is reverse intensification".
                                        $endgroup$
                                        – Hakim
                                        Feb 22 '15 at 19:56










                                      • $begingroup$
                                        @Hakim, Doesn't Velocity×Time equal displacement instead of distance?
                                        $endgroup$
                                        – Pacerier
                                        Feb 2 '18 at 7:39
















                                      28












                                      $begingroup$

                                      This is really one of those important questions that leads many people to say "Math sucks!". In fact, for many students, mathematics stopped making sense somewhere along the way. Either slowly or dramatically, they gave up on the field as hopelessly baffling and difficult, and they grew up to be adults who — confident that others share their experience — nonchalantly announce, “Math was just not for me
                                      or “I was never good at it.
                                      Usually the process is gradual, but for Ruth
                                      McNeill, the turning point was clearly defined. In an
                                      article in the Journal of Mathematical Behavior, she
                                      described how it happened:




                                      What did me in was the idea that a negative number
                                      times a negative number comes out to a positive
                                      number. This seemed (and still seems) inherently
                                      unlikely — counter intuitive, as mathematicians say.
                                      I wrestled with the idea for what I imagine to be
                                      several weeks, trying to get a sensible explanation
                                      from my teacher, my classmates, my parents, any-
                                      body. Whatever explanations they offered could not
                                      overcome my strong sense that multiplying intensifies something, and thus two negative numbers
                                      multiplied together should properly produce a very
                                      negative result. I have since been offered a moderately convincing explanation that features a film of a
                                      swimming pool being drained that gets run back-
                                      wards through the projector. At the time, however,
                                      nothing convinced me. The most commonsense of
                                      all school subjects had abandoned common sense;
                                      I was indignant and baffled.


                                      Meanwhile, the curriculum kept rolling on, and I
                                      could see that I couldn’t stay behind, stuck on nega-
                                      tive times negative. I would have to pay attention to
                                      the next topic, and the only practical course open to
                                      me was to pretend to agree that negative times nega-
                                      tive equals positive. The book and the teacher and
                                      the general consensus of the algebra survivors of so-
                                      ciety were clearly more powerful than I was. I capitu-
                                      lated. I did the rest of algebra, and geometry, and
                                      trigonometry; I did them in the advanced sections,
                                      and I often had that nice sense of “aha!” when I
                                      could suddenly see how a proof was going to come
                                      out. Underneath, however, a kind of resentment and
                                      betrayal lurked, and I was not surprised or dismayed
                                      by any further foolishness my math teachers had up
                                      their sleeves.... Intellectually, I was disengaged, and
                                      when math was no longer required, I took German
                                      instead.






                                      I will show in this answer that: negative $times$ negative $=$ positive, is in fact not counter-intuitive at all! There are many ways that we can use to show that result, but I'd like to show my personal way of thinking about the latter.



                                      Let's imagine we're sitting near a road, and there is a car that is moving with a constant speed. We also have a clock, and so we can measure time.



                                      Before going any further, we should first specify some assumptions like if the car is moving in the right, then its velocity will be positive, and if it's moving in the left direction, then its velocity will be negative.



                                      enter image description here



                                      Imagine now that you have a video of the above scene, and time is positive if you play the video normally but is negative if you play it backwards. We also know the following: $$rm Velocity=dfrac{rm Distance}{rm Time}.$$
                                      Solving for distance we get: $$rm Velocitytimes{rm Time}={rm Distance}.$$
                                      Here the important part comes, if the car is moving in the $+$ direction and the time the video is played is positive, i.e. the video is played normally, then you'll see that the car moves along the $+$ direction and you'll calculate that it moves "a positive distance". Therefore the following holds: $$rm positivetimes positive=positive.$$



                                      If the car moves along the $-$ direction and the time the video is played is positive, i.e. the video is played normally, then you'll see the car going along the $-$ direction, you'll then calculate that it moves "a negative distance". Thus: $$rm negativetimes positive=negative.$$



                                      If however, the car moves along the $+$ direction but the time the video is played is negative, i.e. the video is played backwards, then you'll see that the car is moving in the $-$ direction, and you'll calculate that it moves "a negative distance". Thus: $$rm positivetimes negative=negative.$$



                                      If the car moves along the $-$ direction but the time the video is played is negative, i.e. the video is played backwards, you'll see that the car moves along the $+$ direction! Thus it moves "a positive distance". And therefore: $$color{grey}{boxed{color{white}{underline{overline{color{black}{displaystylerm, negativetimes negative=positive.,}}}}}}$$ As you have seen, it takes only a little bit of imagination for it to make sense.



                                      I hope this helps.
                                      Best wishes, $mathcal H$akim.






                                      share|cite|improve this answer











                                      $endgroup$









                                      • 6




                                        $begingroup$
                                        The material you quoted contains the interesting statement "Whatever explanations they offered could not overcome my strong sense that multiplying intensifies something". I wonder whether anyone tried the very short answer "yes, but negative intensification is attenuation"?
                                        $endgroup$
                                        – Andreas Blass
                                        Jun 29 '14 at 21:43










                                      • $begingroup$
                                        @AndreasBlass Or in another form: "negative multiplication is reverse intensification".
                                        $endgroup$
                                        – Hakim
                                        Feb 22 '15 at 19:56










                                      • $begingroup$
                                        @Hakim, Doesn't Velocity×Time equal displacement instead of distance?
                                        $endgroup$
                                        – Pacerier
                                        Feb 2 '18 at 7:39














                                      28












                                      28








                                      28





                                      $begingroup$

                                      This is really one of those important questions that leads many people to say "Math sucks!". In fact, for many students, mathematics stopped making sense somewhere along the way. Either slowly or dramatically, they gave up on the field as hopelessly baffling and difficult, and they grew up to be adults who — confident that others share their experience — nonchalantly announce, “Math was just not for me
                                      or “I was never good at it.
                                      Usually the process is gradual, but for Ruth
                                      McNeill, the turning point was clearly defined. In an
                                      article in the Journal of Mathematical Behavior, she
                                      described how it happened:




                                      What did me in was the idea that a negative number
                                      times a negative number comes out to a positive
                                      number. This seemed (and still seems) inherently
                                      unlikely — counter intuitive, as mathematicians say.
                                      I wrestled with the idea for what I imagine to be
                                      several weeks, trying to get a sensible explanation
                                      from my teacher, my classmates, my parents, any-
                                      body. Whatever explanations they offered could not
                                      overcome my strong sense that multiplying intensifies something, and thus two negative numbers
                                      multiplied together should properly produce a very
                                      negative result. I have since been offered a moderately convincing explanation that features a film of a
                                      swimming pool being drained that gets run back-
                                      wards through the projector. At the time, however,
                                      nothing convinced me. The most commonsense of
                                      all school subjects had abandoned common sense;
                                      I was indignant and baffled.


                                      Meanwhile, the curriculum kept rolling on, and I
                                      could see that I couldn’t stay behind, stuck on nega-
                                      tive times negative. I would have to pay attention to
                                      the next topic, and the only practical course open to
                                      me was to pretend to agree that negative times nega-
                                      tive equals positive. The book and the teacher and
                                      the general consensus of the algebra survivors of so-
                                      ciety were clearly more powerful than I was. I capitu-
                                      lated. I did the rest of algebra, and geometry, and
                                      trigonometry; I did them in the advanced sections,
                                      and I often had that nice sense of “aha!” when I
                                      could suddenly see how a proof was going to come
                                      out. Underneath, however, a kind of resentment and
                                      betrayal lurked, and I was not surprised or dismayed
                                      by any further foolishness my math teachers had up
                                      their sleeves.... Intellectually, I was disengaged, and
                                      when math was no longer required, I took German
                                      instead.






                                      I will show in this answer that: negative $times$ negative $=$ positive, is in fact not counter-intuitive at all! There are many ways that we can use to show that result, but I'd like to show my personal way of thinking about the latter.



                                      Let's imagine we're sitting near a road, and there is a car that is moving with a constant speed. We also have a clock, and so we can measure time.



                                      Before going any further, we should first specify some assumptions like if the car is moving in the right, then its velocity will be positive, and if it's moving in the left direction, then its velocity will be negative.



                                      enter image description here



                                      Imagine now that you have a video of the above scene, and time is positive if you play the video normally but is negative if you play it backwards. We also know the following: $$rm Velocity=dfrac{rm Distance}{rm Time}.$$
                                      Solving for distance we get: $$rm Velocitytimes{rm Time}={rm Distance}.$$
                                      Here the important part comes, if the car is moving in the $+$ direction and the time the video is played is positive, i.e. the video is played normally, then you'll see that the car moves along the $+$ direction and you'll calculate that it moves "a positive distance". Therefore the following holds: $$rm positivetimes positive=positive.$$



                                      If the car moves along the $-$ direction and the time the video is played is positive, i.e. the video is played normally, then you'll see the car going along the $-$ direction, you'll then calculate that it moves "a negative distance". Thus: $$rm negativetimes positive=negative.$$



                                      If however, the car moves along the $+$ direction but the time the video is played is negative, i.e. the video is played backwards, then you'll see that the car is moving in the $-$ direction, and you'll calculate that it moves "a negative distance". Thus: $$rm positivetimes negative=negative.$$



                                      If the car moves along the $-$ direction but the time the video is played is negative, i.e. the video is played backwards, you'll see that the car moves along the $+$ direction! Thus it moves "a positive distance". And therefore: $$color{grey}{boxed{color{white}{underline{overline{color{black}{displaystylerm, negativetimes negative=positive.,}}}}}}$$ As you have seen, it takes only a little bit of imagination for it to make sense.



                                      I hope this helps.
                                      Best wishes, $mathcal H$akim.






                                      share|cite|improve this answer











                                      $endgroup$



                                      This is really one of those important questions that leads many people to say "Math sucks!". In fact, for many students, mathematics stopped making sense somewhere along the way. Either slowly or dramatically, they gave up on the field as hopelessly baffling and difficult, and they grew up to be adults who — confident that others share their experience — nonchalantly announce, “Math was just not for me
                                      or “I was never good at it.
                                      Usually the process is gradual, but for Ruth
                                      McNeill, the turning point was clearly defined. In an
                                      article in the Journal of Mathematical Behavior, she
                                      described how it happened:




                                      What did me in was the idea that a negative number
                                      times a negative number comes out to a positive
                                      number. This seemed (and still seems) inherently
                                      unlikely — counter intuitive, as mathematicians say.
                                      I wrestled with the idea for what I imagine to be
                                      several weeks, trying to get a sensible explanation
                                      from my teacher, my classmates, my parents, any-
                                      body. Whatever explanations they offered could not
                                      overcome my strong sense that multiplying intensifies something, and thus two negative numbers
                                      multiplied together should properly produce a very
                                      negative result. I have since been offered a moderately convincing explanation that features a film of a
                                      swimming pool being drained that gets run back-
                                      wards through the projector. At the time, however,
                                      nothing convinced me. The most commonsense of
                                      all school subjects had abandoned common sense;
                                      I was indignant and baffled.


                                      Meanwhile, the curriculum kept rolling on, and I
                                      could see that I couldn’t stay behind, stuck on nega-
                                      tive times negative. I would have to pay attention to
                                      the next topic, and the only practical course open to
                                      me was to pretend to agree that negative times nega-
                                      tive equals positive. The book and the teacher and
                                      the general consensus of the algebra survivors of so-
                                      ciety were clearly more powerful than I was. I capitu-
                                      lated. I did the rest of algebra, and geometry, and
                                      trigonometry; I did them in the advanced sections,
                                      and I often had that nice sense of “aha!” when I
                                      could suddenly see how a proof was going to come
                                      out. Underneath, however, a kind of resentment and
                                      betrayal lurked, and I was not surprised or dismayed
                                      by any further foolishness my math teachers had up
                                      their sleeves.... Intellectually, I was disengaged, and
                                      when math was no longer required, I took German
                                      instead.






                                      I will show in this answer that: negative $times$ negative $=$ positive, is in fact not counter-intuitive at all! There are many ways that we can use to show that result, but I'd like to show my personal way of thinking about the latter.



                                      Let's imagine we're sitting near a road, and there is a car that is moving with a constant speed. We also have a clock, and so we can measure time.



                                      Before going any further, we should first specify some assumptions like if the car is moving in the right, then its velocity will be positive, and if it's moving in the left direction, then its velocity will be negative.



                                      enter image description here



                                      Imagine now that you have a video of the above scene, and time is positive if you play the video normally but is negative if you play it backwards. We also know the following: $$rm Velocity=dfrac{rm Distance}{rm Time}.$$
                                      Solving for distance we get: $$rm Velocitytimes{rm Time}={rm Distance}.$$
                                      Here the important part comes, if the car is moving in the $+$ direction and the time the video is played is positive, i.e. the video is played normally, then you'll see that the car moves along the $+$ direction and you'll calculate that it moves "a positive distance". Therefore the following holds: $$rm positivetimes positive=positive.$$



                                      If the car moves along the $-$ direction and the time the video is played is positive, i.e. the video is played normally, then you'll see the car going along the $-$ direction, you'll then calculate that it moves "a negative distance". Thus: $$rm negativetimes positive=negative.$$



                                      If however, the car moves along the $+$ direction but the time the video is played is negative, i.e. the video is played backwards, then you'll see that the car is moving in the $-$ direction, and you'll calculate that it moves "a negative distance". Thus: $$rm positivetimes negative=negative.$$



                                      If the car moves along the $-$ direction but the time the video is played is negative, i.e. the video is played backwards, you'll see that the car moves along the $+$ direction! Thus it moves "a positive distance". And therefore: $$color{grey}{boxed{color{white}{underline{overline{color{black}{displaystylerm, negativetimes negative=positive.,}}}}}}$$ As you have seen, it takes only a little bit of imagination for it to make sense.



                                      I hope this helps.
                                      Best wishes, $mathcal H$akim.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Apr 14 '14 at 22:45

























                                      answered Mar 8 '14 at 16:21









                                      HakimHakim

                                      9,21953157




                                      9,21953157








                                      • 6




                                        $begingroup$
                                        The material you quoted contains the interesting statement "Whatever explanations they offered could not overcome my strong sense that multiplying intensifies something". I wonder whether anyone tried the very short answer "yes, but negative intensification is attenuation"?
                                        $endgroup$
                                        – Andreas Blass
                                        Jun 29 '14 at 21:43










                                      • $begingroup$
                                        @AndreasBlass Or in another form: "negative multiplication is reverse intensification".
                                        $endgroup$
                                        – Hakim
                                        Feb 22 '15 at 19:56










                                      • $begingroup$
                                        @Hakim, Doesn't Velocity×Time equal displacement instead of distance?
                                        $endgroup$
                                        – Pacerier
                                        Feb 2 '18 at 7:39














                                      • 6




                                        $begingroup$
                                        The material you quoted contains the interesting statement "Whatever explanations they offered could not overcome my strong sense that multiplying intensifies something". I wonder whether anyone tried the very short answer "yes, but negative intensification is attenuation"?
                                        $endgroup$
                                        – Andreas Blass
                                        Jun 29 '14 at 21:43










                                      • $begingroup$
                                        @AndreasBlass Or in another form: "negative multiplication is reverse intensification".
                                        $endgroup$
                                        – Hakim
                                        Feb 22 '15 at 19:56










                                      • $begingroup$
                                        @Hakim, Doesn't Velocity×Time equal displacement instead of distance?
                                        $endgroup$
                                        – Pacerier
                                        Feb 2 '18 at 7:39








                                      6




                                      6




                                      $begingroup$
                                      The material you quoted contains the interesting statement "Whatever explanations they offered could not overcome my strong sense that multiplying intensifies something". I wonder whether anyone tried the very short answer "yes, but negative intensification is attenuation"?
                                      $endgroup$
                                      – Andreas Blass
                                      Jun 29 '14 at 21:43




                                      $begingroup$
                                      The material you quoted contains the interesting statement "Whatever explanations they offered could not overcome my strong sense that multiplying intensifies something". I wonder whether anyone tried the very short answer "yes, but negative intensification is attenuation"?
                                      $endgroup$
                                      – Andreas Blass
                                      Jun 29 '14 at 21:43












                                      $begingroup$
                                      @AndreasBlass Or in another form: "negative multiplication is reverse intensification".
                                      $endgroup$
                                      – Hakim
                                      Feb 22 '15 at 19:56




                                      $begingroup$
                                      @AndreasBlass Or in another form: "negative multiplication is reverse intensification".
                                      $endgroup$
                                      – Hakim
                                      Feb 22 '15 at 19:56












                                      $begingroup$
                                      @Hakim, Doesn't Velocity×Time equal displacement instead of distance?
                                      $endgroup$
                                      – Pacerier
                                      Feb 2 '18 at 7:39




                                      $begingroup$
                                      @Hakim, Doesn't Velocity×Time equal displacement instead of distance?
                                      $endgroup$
                                      – Pacerier
                                      Feb 2 '18 at 7:39











                                      22












                                      $begingroup$

                                      The elementary intuition behind the product of two negatives can be thought of as follows. You have a bank account. You pay 3 bills for 40 dollars each, $3 cdot (-40) = -120$ is added to your account.



                                      The opposite of being billed would be billing someone else.



                                      So, if you bill 3 people for 40 dollars each, $(-3) cdot (-40)$ is added to your account. This value should be positive since it results in you receiving money.






                                      share|cite|improve this answer











                                      $endgroup$


















                                        22












                                        $begingroup$

                                        The elementary intuition behind the product of two negatives can be thought of as follows. You have a bank account. You pay 3 bills for 40 dollars each, $3 cdot (-40) = -120$ is added to your account.



                                        The opposite of being billed would be billing someone else.



                                        So, if you bill 3 people for 40 dollars each, $(-3) cdot (-40)$ is added to your account. This value should be positive since it results in you receiving money.






                                        share|cite|improve this answer











                                        $endgroup$
















                                          22












                                          22








                                          22





                                          $begingroup$

                                          The elementary intuition behind the product of two negatives can be thought of as follows. You have a bank account. You pay 3 bills for 40 dollars each, $3 cdot (-40) = -120$ is added to your account.



                                          The opposite of being billed would be billing someone else.



                                          So, if you bill 3 people for 40 dollars each, $(-3) cdot (-40)$ is added to your account. This value should be positive since it results in you receiving money.






                                          share|cite|improve this answer











                                          $endgroup$



                                          The elementary intuition behind the product of two negatives can be thought of as follows. You have a bank account. You pay 3 bills for 40 dollars each, $3 cdot (-40) = -120$ is added to your account.



                                          The opposite of being billed would be billing someone else.



                                          So, if you bill 3 people for 40 dollars each, $(-3) cdot (-40)$ is added to your account. This value should be positive since it results in you receiving money.







                                          share|cite|improve this answer














                                          share|cite|improve this answer



                                          share|cite|improve this answer








                                          edited Aug 31 '11 at 7:25









                                          Mateen Ulhaq

                                          73821228




                                          73821228










                                          answered Nov 12 '10 at 3:25









                                          futurebirdfuturebird

                                          3,60412752




                                          3,60412752























                                              20












                                              $begingroup$

                                              Here's a proof. First, for all $x$, $xcdot 0=xcdot(0+0)=xcdot 0 +xcdot 0$. Subtracting $xcdot0$ from each side, $xcdot0=0$. Now, for all $x$ and $y$, $0=xcdot0=xcdot(-y+y)=xcdot(-y)+xcdot y$. Subtracting $xcdot y$ from both sides, $xcdot(-y)=-(xcdot y)$. Applying this twice along with the identity $-(-a)=a$, $(-x)cdot(-y)=-(-x)cdot y=-(-(xcdot y))=xcdot y$.



                                              Your proof implicitly uses the fact that $-xy=(-x)y$, and assumes that there are only two possibilities, $xy$ or $-xy$, then shows that the latter is impossible. These seem like plausible assumptions, but I tried to be very careful in my proof above (thus using $-(xcdot y)$ rather than simply $-xy$ to not be confused with $(-x)cdot y$).



                                              I only have a vague intuitive notion that I probably can't explain well, but I sometimes think of a negative number like $-5$ as being "$5$ in the other direction", and so multiplying by $-5$ means "multiply by $5$ and switch direction", i.e., sign. This means if you multiply $-5$ by a negative number, you should switch its direction back to positive.






                                              share|cite|improve this answer











                                              $endgroup$


















                                                20












                                                $begingroup$

                                                Here's a proof. First, for all $x$, $xcdot 0=xcdot(0+0)=xcdot 0 +xcdot 0$. Subtracting $xcdot0$ from each side, $xcdot0=0$. Now, for all $x$ and $y$, $0=xcdot0=xcdot(-y+y)=xcdot(-y)+xcdot y$. Subtracting $xcdot y$ from both sides, $xcdot(-y)=-(xcdot y)$. Applying this twice along with the identity $-(-a)=a$, $(-x)cdot(-y)=-(-x)cdot y=-(-(xcdot y))=xcdot y$.



                                                Your proof implicitly uses the fact that $-xy=(-x)y$, and assumes that there are only two possibilities, $xy$ or $-xy$, then shows that the latter is impossible. These seem like plausible assumptions, but I tried to be very careful in my proof above (thus using $-(xcdot y)$ rather than simply $-xy$ to not be confused with $(-x)cdot y$).



                                                I only have a vague intuitive notion that I probably can't explain well, but I sometimes think of a negative number like $-5$ as being "$5$ in the other direction", and so multiplying by $-5$ means "multiply by $5$ and switch direction", i.e., sign. This means if you multiply $-5$ by a negative number, you should switch its direction back to positive.






                                                share|cite|improve this answer











                                                $endgroup$
















                                                  20












                                                  20








                                                  20





                                                  $begingroup$

                                                  Here's a proof. First, for all $x$, $xcdot 0=xcdot(0+0)=xcdot 0 +xcdot 0$. Subtracting $xcdot0$ from each side, $xcdot0=0$. Now, for all $x$ and $y$, $0=xcdot0=xcdot(-y+y)=xcdot(-y)+xcdot y$. Subtracting $xcdot y$ from both sides, $xcdot(-y)=-(xcdot y)$. Applying this twice along with the identity $-(-a)=a$, $(-x)cdot(-y)=-(-x)cdot y=-(-(xcdot y))=xcdot y$.



                                                  Your proof implicitly uses the fact that $-xy=(-x)y$, and assumes that there are only two possibilities, $xy$ or $-xy$, then shows that the latter is impossible. These seem like plausible assumptions, but I tried to be very careful in my proof above (thus using $-(xcdot y)$ rather than simply $-xy$ to not be confused with $(-x)cdot y$).



                                                  I only have a vague intuitive notion that I probably can't explain well, but I sometimes think of a negative number like $-5$ as being "$5$ in the other direction", and so multiplying by $-5$ means "multiply by $5$ and switch direction", i.e., sign. This means if you multiply $-5$ by a negative number, you should switch its direction back to positive.






                                                  share|cite|improve this answer











                                                  $endgroup$



                                                  Here's a proof. First, for all $x$, $xcdot 0=xcdot(0+0)=xcdot 0 +xcdot 0$. Subtracting $xcdot0$ from each side, $xcdot0=0$. Now, for all $x$ and $y$, $0=xcdot0=xcdot(-y+y)=xcdot(-y)+xcdot y$. Subtracting $xcdot y$ from both sides, $xcdot(-y)=-(xcdot y)$. Applying this twice along with the identity $-(-a)=a$, $(-x)cdot(-y)=-(-x)cdot y=-(-(xcdot y))=xcdot y$.



                                                  Your proof implicitly uses the fact that $-xy=(-x)y$, and assumes that there are only two possibilities, $xy$ or $-xy$, then shows that the latter is impossible. These seem like plausible assumptions, but I tried to be very careful in my proof above (thus using $-(xcdot y)$ rather than simply $-xy$ to not be confused with $(-x)cdot y$).



                                                  I only have a vague intuitive notion that I probably can't explain well, but I sometimes think of a negative number like $-5$ as being "$5$ in the other direction", and so multiplying by $-5$ means "multiply by $5$ and switch direction", i.e., sign. This means if you multiply $-5$ by a negative number, you should switch its direction back to positive.







                                                  share|cite|improve this answer














                                                  share|cite|improve this answer



                                                  share|cite|improve this answer








                                                  answered Nov 12 '10 at 2:36


























                                                  community wiki





                                                  Jonas Meyer
























                                                      19












                                                      $begingroup$

                                                      I have always viewed negative numbers as a "flip" on the number line.



                                                      For example, -2 is the same as 2, but flip-mirrored to the other side of the zero.



                                                      Multiplication then works as follows:



                                                      2 x 3 has no flips, so it's just 2x3 = 6.



                                                      -2 x 3 has one flip, so you start with 2x3 = 6 but with one flip so it becomes -6 instead.



                                                      2 x -3 is the same, only one flip, so it's 2x3 = 6 but flipped to -6.



                                                      -2 x -3 has two flips, so you start with 2x3 = 6. When you apply the two flips it gets you back to where you started because you flip to negative and then flip back. So -3 x -2 = 6.






                                                      share|cite|improve this answer











                                                      $endgroup$


















                                                        19












                                                        $begingroup$

                                                        I have always viewed negative numbers as a "flip" on the number line.



                                                        For example, -2 is the same as 2, but flip-mirrored to the other side of the zero.



                                                        Multiplication then works as follows:



                                                        2 x 3 has no flips, so it's just 2x3 = 6.



                                                        -2 x 3 has one flip, so you start with 2x3 = 6 but with one flip so it becomes -6 instead.



                                                        2 x -3 is the same, only one flip, so it's 2x3 = 6 but flipped to -6.



                                                        -2 x -3 has two flips, so you start with 2x3 = 6. When you apply the two flips it gets you back to where you started because you flip to negative and then flip back. So -3 x -2 = 6.






                                                        share|cite|improve this answer











                                                        $endgroup$
















                                                          19












                                                          19








                                                          19





                                                          $begingroup$

                                                          I have always viewed negative numbers as a "flip" on the number line.



                                                          For example, -2 is the same as 2, but flip-mirrored to the other side of the zero.



                                                          Multiplication then works as follows:



                                                          2 x 3 has no flips, so it's just 2x3 = 6.



                                                          -2 x 3 has one flip, so you start with 2x3 = 6 but with one flip so it becomes -6 instead.



                                                          2 x -3 is the same, only one flip, so it's 2x3 = 6 but flipped to -6.



                                                          -2 x -3 has two flips, so you start with 2x3 = 6. When you apply the two flips it gets you back to where you started because you flip to negative and then flip back. So -3 x -2 = 6.






                                                          share|cite|improve this answer











                                                          $endgroup$



                                                          I have always viewed negative numbers as a "flip" on the number line.



                                                          For example, -2 is the same as 2, but flip-mirrored to the other side of the zero.



                                                          Multiplication then works as follows:



                                                          2 x 3 has no flips, so it's just 2x3 = 6.



                                                          -2 x 3 has one flip, so you start with 2x3 = 6 but with one flip so it becomes -6 instead.



                                                          2 x -3 is the same, only one flip, so it's 2x3 = 6 but flipped to -6.



                                                          -2 x -3 has two flips, so you start with 2x3 = 6. When you apply the two flips it gets you back to where you started because you flip to negative and then flip back. So -3 x -2 = 6.







                                                          share|cite|improve this answer














                                                          share|cite|improve this answer



                                                          share|cite|improve this answer








                                                          edited Nov 9 '15 at 6:57

























                                                          answered Nov 9 '15 at 5:59







                                                          user34016






























                                                              12












                                                              $begingroup$

                                                              I think the x and y get in the way a bit; you can see the crucial steps using just 1 and -1. What you've really shown is that (-1)(-1)=-1 leads to a contradiction. If we divide by -1, we get -1=1, which is not true!



                                                              Getting the right answer, (-1)(-1)=1, uses a couple more steps: First, you must agree that (1)+(-1)=0, (1)(-1)=-1, and (0)(-1)=0.



                                                              Now, we multiply the first equation by (-1) and use the distributive property to get (-1)(-1)+(-1)(1)=(-1)(0). Now, we simplify the parts we know to get (-1)(-1)+(-1)=0. Solve for (-1)(-1), and you get (-1)(-1)=1.



                                                              So, we must have (-1)(-1)=1 if we accept basic rules of arithmetic: 0 is the additive identity, 1 is the multiplicative identity, -1 is the additive inverse of 1, and multiplication distributes over addition.



                                                              One physical explanation people often like for negative*negative=positive is multiplying rates. You can film someone walking forwards (positive rate) or walking backwards (negative rate). Now, play the film back, but in reverse (another negative rate). What do you see if you play a film backwards of someone walking backwards? You see the person walking forwards, because negative*negative=positive!






                                                              share|cite|improve this answer









                                                              $endgroup$


















                                                                12












                                                                $begingroup$

                                                                I think the x and y get in the way a bit; you can see the crucial steps using just 1 and -1. What you've really shown is that (-1)(-1)=-1 leads to a contradiction. If we divide by -1, we get -1=1, which is not true!



                                                                Getting the right answer, (-1)(-1)=1, uses a couple more steps: First, you must agree that (1)+(-1)=0, (1)(-1)=-1, and (0)(-1)=0.



                                                                Now, we multiply the first equation by (-1) and use the distributive property to get (-1)(-1)+(-1)(1)=(-1)(0). Now, we simplify the parts we know to get (-1)(-1)+(-1)=0. Solve for (-1)(-1), and you get (-1)(-1)=1.



                                                                So, we must have (-1)(-1)=1 if we accept basic rules of arithmetic: 0 is the additive identity, 1 is the multiplicative identity, -1 is the additive inverse of 1, and multiplication distributes over addition.



                                                                One physical explanation people often like for negative*negative=positive is multiplying rates. You can film someone walking forwards (positive rate) or walking backwards (negative rate). Now, play the film back, but in reverse (another negative rate). What do you see if you play a film backwards of someone walking backwards? You see the person walking forwards, because negative*negative=positive!






                                                                share|cite|improve this answer









                                                                $endgroup$
















                                                                  12












                                                                  12








                                                                  12





                                                                  $begingroup$

                                                                  I think the x and y get in the way a bit; you can see the crucial steps using just 1 and -1. What you've really shown is that (-1)(-1)=-1 leads to a contradiction. If we divide by -1, we get -1=1, which is not true!



                                                                  Getting the right answer, (-1)(-1)=1, uses a couple more steps: First, you must agree that (1)+(-1)=0, (1)(-1)=-1, and (0)(-1)=0.



                                                                  Now, we multiply the first equation by (-1) and use the distributive property to get (-1)(-1)+(-1)(1)=(-1)(0). Now, we simplify the parts we know to get (-1)(-1)+(-1)=0. Solve for (-1)(-1), and you get (-1)(-1)=1.



                                                                  So, we must have (-1)(-1)=1 if we accept basic rules of arithmetic: 0 is the additive identity, 1 is the multiplicative identity, -1 is the additive inverse of 1, and multiplication distributes over addition.



                                                                  One physical explanation people often like for negative*negative=positive is multiplying rates. You can film someone walking forwards (positive rate) or walking backwards (negative rate). Now, play the film back, but in reverse (another negative rate). What do you see if you play a film backwards of someone walking backwards? You see the person walking forwards, because negative*negative=positive!






                                                                  share|cite|improve this answer









                                                                  $endgroup$



                                                                  I think the x and y get in the way a bit; you can see the crucial steps using just 1 and -1. What you've really shown is that (-1)(-1)=-1 leads to a contradiction. If we divide by -1, we get -1=1, which is not true!



                                                                  Getting the right answer, (-1)(-1)=1, uses a couple more steps: First, you must agree that (1)+(-1)=0, (1)(-1)=-1, and (0)(-1)=0.



                                                                  Now, we multiply the first equation by (-1) and use the distributive property to get (-1)(-1)+(-1)(1)=(-1)(0). Now, we simplify the parts we know to get (-1)(-1)+(-1)=0. Solve for (-1)(-1), and you get (-1)(-1)=1.



                                                                  So, we must have (-1)(-1)=1 if we accept basic rules of arithmetic: 0 is the additive identity, 1 is the multiplicative identity, -1 is the additive inverse of 1, and multiplication distributes over addition.



                                                                  One physical explanation people often like for negative*negative=positive is multiplying rates. You can film someone walking forwards (positive rate) or walking backwards (negative rate). Now, play the film back, but in reverse (another negative rate). What do you see if you play a film backwards of someone walking backwards? You see the person walking forwards, because negative*negative=positive!







                                                                  share|cite|improve this answer












                                                                  share|cite|improve this answer



                                                                  share|cite|improve this answer










                                                                  answered Nov 12 '10 at 2:37









                                                                  Jonas KibelbekJonas Kibelbek

                                                                  5,24922330




                                                                  5,24922330























                                                                      12












                                                                      $begingroup$

                                                                      Quite a good explanation is that one wants the distributive law to work in general with positive quantities when you add (smaller) negative ones:
                                                                      If $x>age0$ and $y>bge0$ then
                                                                      $$
                                                                      (x-a)(y-b)=(x+(-a))(y+(-b))=xy+(-a)y+x(-b)+(-a)(-b)
                                                                      $$
                                                                      For this to always work, one needs $(-a)y=-(ay)$ in case $b=0$, $x(-b)=-xb$ in case $a=0$, and $(-a)(-b)=ab$.






                                                                      share|cite|improve this answer









                                                                      $endgroup$


















                                                                        12












                                                                        $begingroup$

                                                                        Quite a good explanation is that one wants the distributive law to work in general with positive quantities when you add (smaller) negative ones:
                                                                        If $x>age0$ and $y>bge0$ then
                                                                        $$
                                                                        (x-a)(y-b)=(x+(-a))(y+(-b))=xy+(-a)y+x(-b)+(-a)(-b)
                                                                        $$
                                                                        For this to always work, one needs $(-a)y=-(ay)$ in case $b=0$, $x(-b)=-xb$ in case $a=0$, and $(-a)(-b)=ab$.






                                                                        share|cite|improve this answer









                                                                        $endgroup$
















                                                                          12












                                                                          12








                                                                          12





                                                                          $begingroup$

                                                                          Quite a good explanation is that one wants the distributive law to work in general with positive quantities when you add (smaller) negative ones:
                                                                          If $x>age0$ and $y>bge0$ then
                                                                          $$
                                                                          (x-a)(y-b)=(x+(-a))(y+(-b))=xy+(-a)y+x(-b)+(-a)(-b)
                                                                          $$
                                                                          For this to always work, one needs $(-a)y=-(ay)$ in case $b=0$, $x(-b)=-xb$ in case $a=0$, and $(-a)(-b)=ab$.






                                                                          share|cite|improve this answer









                                                                          $endgroup$



                                                                          Quite a good explanation is that one wants the distributive law to work in general with positive quantities when you add (smaller) negative ones:
                                                                          If $x>age0$ and $y>bge0$ then
                                                                          $$
                                                                          (x-a)(y-b)=(x+(-a))(y+(-b))=xy+(-a)y+x(-b)+(-a)(-b)
                                                                          $$
                                                                          For this to always work, one needs $(-a)y=-(ay)$ in case $b=0$, $x(-b)=-xb$ in case $a=0$, and $(-a)(-b)=ab$.







                                                                          share|cite|improve this answer












                                                                          share|cite|improve this answer



                                                                          share|cite|improve this answer










                                                                          answered Nov 12 '10 at 3:47









                                                                          Bob PegoBob Pego

                                                                          3,7572016




                                                                          3,7572016























                                                                              12












                                                                              $begingroup$

                                                                              One thing that must be understood is that this law cannot be proven in the same way that the laws of positive rational and integral arithmetic can be. The reason for this is that negatives lack any "external" (external to mathematics, ie. pre-axiomatic, intuitive, conceptuel, empirical, physical, etc.) definition.



                                                                              For example. Without even getting into the Peano axioms, I can prove that, where $a$ and $b$ are positive integers, $ab=ba$. Indeed, $ab$ is just the process of taking $a$ sets of $b$. Take one element from each of these sets, thus forming a set of $a$ elements. Repeat this $b$ times: you will clearly use up exactly all of the elements and obtain $b$ sets of $a$ elements, in other words, $ba$. Similar informal (but entirely convincing, reasonable, and I would say irrefutable) reasoning can be used to demonstrate the rules for manipulating positive fractions, say.



                                                                              Notice that in the above paragraph I used the fact that both positive integers and positive integer multiplication have pre-axiomatic, "physical" definitions.



                                                                              Ask someone why the product of two negatives is positive, and the best they can do is explain, not prove. "Well, negative kind of means 'opposite', so doing the opposite twice means doing the usual, ie positive" does not constitute a proof, but merely an explanation serving to make the accepted mathematical axiom less surprising. Another common one begins with "we would like the usual properties of arithmetic to hold, so assume they do...", but then it remains to be explained why it's so important that the usual laws of arithmetic hold. Euler himself, in an early chapter of his textbook on algebra, gave the following supremely questionable justification. After justifying $(-a)b=-(ab)$ by analogies with debts, he writes:




                                                                              It remains to resolve the case in which - is multiplied by -; or, for example, -a by -b. It is evident, at first sight, with regard to the letters, that the product will be ab; but it is doubtful whether the sign + or the sign - is to be placed before it, all we know is, that it must be one or the other of these signs. Now, I say that it cannot be the sign -: for -a by +b gives -ab, and -a by -b cannot produce the same result as -a by +b...




                                                                              With no disrespect to Euler (especially consdiering this was intended as an introductory textbook), I think we can agree that this is a pretty philosophically dubious argument.



                                                                              The reason it is impossible is because there is no pre-axiomatic definition for what a negative number or negative multiplication really is. Oh, you could probably come up with one involving opposite "directions", and notions of symmetry, but it would be quite artificial and not at all obviously "the best" definition. In my opinion, negatives are ultimately best understood as purely abstract objects. It so happens - and this is quite myseterious - that these utterly abstract laws of calculation lead to physically meaningful results. This was nicely expressed in 1778 by the mathematician John Playfair, when addressing the then controversial issues of negative and complex numbers:




                                                                              Here then is a paradox which remains to be explained. If the operations of this imaginary arithmetic are unintelligible, why are they not altogether useless? Is investigation an art so mechanical, that it may be conducted by certain manual operations? Or is truth so easily discovered, that intelligence is not necessary to give success to our researches?




                                                                              Quoted in Negative Math: How Mathematical Rules Can be Positively Bent by Alberto A. Martínez.



                                                                              One way of approching the problem is with the idea that negative numbers are a different name for subtraction. The differences between subtraction and addition force us, if we reject negatives, to create many different rules covering all the different possibilities ($a - b$, $b - a$, and $a + b$, and if the particular theorem or problem involves more than two variables, the difficulty is compounded further...). The idea of negatives could be described as the insight that rather than having two operations and one type of number, we can have one operation and two types of number. Indeed, if you start with some perfectly physically meaningful axioms about subtraction, you will find that the $(-1)(-1)=1$ law seems to be implicit within them. Hint: starting from the very reasonable axioms $a(b-c)=ab-ac , a - (b - c) = a - b + c$, consider the product $(a-b)(c-d)$.



                                                                              But even that explanation doesn't altogether satisfy me. I've become convinced that my education cheated me on how deep an idea negative numbers are, and I expect to remain puzzled by them for many years. Anyway, I hope some of the above is useful to someone.






                                                                              share|cite|improve this answer











                                                                              $endgroup$









                                                                              • 1




                                                                                $begingroup$
                                                                                This $(-a)(-b)=ab$ stuff started when we gave up on distinguishing multiplier and multiplied in a multiplication. I can have three unities of apples, but I can't have apple unities of threes. In this sense, it's possible to have 3 unities of an 1D vector backwards $3 times -1$ but not -1 unities of a vector forwards $-1 times 3$, unless we define that a negative multiplier change the sense of the vector. I think a negative-free algebra can be created by distinguish three kind of numbers: scalar (no sign), up and down.
                                                                                $endgroup$
                                                                                – Leonardo Castro
                                                                                Sep 9 '16 at 12:22












                                                                              • $begingroup$
                                                                                If you add an up and a down, you cancel out the amount they have in common keep the remaining up or down part. This would the subtraction of this algebra. Multiplying by $-1$ should be replaced by a simple inversion operator. Actually, $+1$ and $-1$ as multipliers are usually taken as the identity and inversion operators. In that way, they form a simple identity-inversion group and so $-1 times -1 = 1$ since two inversions are equivalent to identity. I don't think, however, that algebra should necessarily be that way.
                                                                                $endgroup$
                                                                                – Leonardo Castro
                                                                                Sep 9 '16 at 12:33










                                                                              • $begingroup$
                                                                                I used up and down just to avoid all the assumptions about positive and negative, but they could be right and left, yin and yang, whatever... I wonder if yin times yin is yang or the other way around.
                                                                                $endgroup$
                                                                                – Leonardo Castro
                                                                                Sep 9 '16 at 12:35


















                                                                              12












                                                                              $begingroup$

                                                                              One thing that must be understood is that this law cannot be proven in the same way that the laws of positive rational and integral arithmetic can be. The reason for this is that negatives lack any "external" (external to mathematics, ie. pre-axiomatic, intuitive, conceptuel, empirical, physical, etc.) definition.



                                                                              For example. Without even getting into the Peano axioms, I can prove that, where $a$ and $b$ are positive integers, $ab=ba$. Indeed, $ab$ is just the process of taking $a$ sets of $b$. Take one element from each of these sets, thus forming a set of $a$ elements. Repeat this $b$ times: you will clearly use up exactly all of the elements and obtain $b$ sets of $a$ elements, in other words, $ba$. Similar informal (but entirely convincing, reasonable, and I would say irrefutable) reasoning can be used to demonstrate the rules for manipulating positive fractions, say.



                                                                              Notice that in the above paragraph I used the fact that both positive integers and positive integer multiplication have pre-axiomatic, "physical" definitions.



                                                                              Ask someone why the product of two negatives is positive, and the best they can do is explain, not prove. "Well, negative kind of means 'opposite', so doing the opposite twice means doing the usual, ie positive" does not constitute a proof, but merely an explanation serving to make the accepted mathematical axiom less surprising. Another common one begins with "we would like the usual properties of arithmetic to hold, so assume they do...", but then it remains to be explained why it's so important that the usual laws of arithmetic hold. Euler himself, in an early chapter of his textbook on algebra, gave the following supremely questionable justification. After justifying $(-a)b=-(ab)$ by analogies with debts, he writes:




                                                                              It remains to resolve the case in which - is multiplied by -; or, for example, -a by -b. It is evident, at first sight, with regard to the letters, that the product will be ab; but it is doubtful whether the sign + or the sign - is to be placed before it, all we know is, that it must be one or the other of these signs. Now, I say that it cannot be the sign -: for -a by +b gives -ab, and -a by -b cannot produce the same result as -a by +b...




                                                                              With no disrespect to Euler (especially consdiering this was intended as an introductory textbook), I think we can agree that this is a pretty philosophically dubious argument.



                                                                              The reason it is impossible is because there is no pre-axiomatic definition for what a negative number or negative multiplication really is. Oh, you could probably come up with one involving opposite "directions", and notions of symmetry, but it would be quite artificial and not at all obviously "the best" definition. In my opinion, negatives are ultimately best understood as purely abstract objects. It so happens - and this is quite myseterious - that these utterly abstract laws of calculation lead to physically meaningful results. This was nicely expressed in 1778 by the mathematician John Playfair, when addressing the then controversial issues of negative and complex numbers:




                                                                              Here then is a paradox which remains to be explained. If the operations of this imaginary arithmetic are unintelligible, why are they not altogether useless? Is investigation an art so mechanical, that it may be conducted by certain manual operations? Or is truth so easily discovered, that intelligence is not necessary to give success to our researches?




                                                                              Quoted in Negative Math: How Mathematical Rules Can be Positively Bent by Alberto A. Martínez.



                                                                              One way of approching the problem is with the idea that negative numbers are a different name for subtraction. The differences between subtraction and addition force us, if we reject negatives, to create many different rules covering all the different possibilities ($a - b$, $b - a$, and $a + b$, and if the particular theorem or problem involves more than two variables, the difficulty is compounded further...). The idea of negatives could be described as the insight that rather than having two operations and one type of number, we can have one operation and two types of number. Indeed, if you start with some perfectly physically meaningful axioms about subtraction, you will find that the $(-1)(-1)=1$ law seems to be implicit within them. Hint: starting from the very reasonable axioms $a(b-c)=ab-ac , a - (b - c) = a - b + c$, consider the product $(a-b)(c-d)$.



                                                                              But even that explanation doesn't altogether satisfy me. I've become convinced that my education cheated me on how deep an idea negative numbers are, and I expect to remain puzzled by them for many years. Anyway, I hope some of the above is useful to someone.






                                                                              share|cite|improve this answer











                                                                              $endgroup$









                                                                              • 1




                                                                                $begingroup$
                                                                                This $(-a)(-b)=ab$ stuff started when we gave up on distinguishing multiplier and multiplied in a multiplication. I can have three unities of apples, but I can't have apple unities of threes. In this sense, it's possible to have 3 unities of an 1D vector backwards $3 times -1$ but not -1 unities of a vector forwards $-1 times 3$, unless we define that a negative multiplier change the sense of the vector. I think a negative-free algebra can be created by distinguish three kind of numbers: scalar (no sign), up and down.
                                                                                $endgroup$
                                                                                – Leonardo Castro
                                                                                Sep 9 '16 at 12:22












                                                                              • $begingroup$
                                                                                If you add an up and a down, you cancel out the amount they have in common keep the remaining up or down part. This would the subtraction of this algebra. Multiplying by $-1$ should be replaced by a simple inversion operator. Actually, $+1$ and $-1$ as multipliers are usually taken as the identity and inversion operators. In that way, they form a simple identity-inversion group and so $-1 times -1 = 1$ since two inversions are equivalent to identity. I don't think, however, that algebra should necessarily be that way.
                                                                                $endgroup$
                                                                                – Leonardo Castro
                                                                                Sep 9 '16 at 12:33










                                                                              • $begingroup$
                                                                                I used up and down just to avoid all the assumptions about positive and negative, but they could be right and left, yin and yang, whatever... I wonder if yin times yin is yang or the other way around.
                                                                                $endgroup$
                                                                                – Leonardo Castro
                                                                                Sep 9 '16 at 12:35
















                                                                              12












                                                                              12








                                                                              12





                                                                              $begingroup$

                                                                              One thing that must be understood is that this law cannot be proven in the same way that the laws of positive rational and integral arithmetic can be. The reason for this is that negatives lack any "external" (external to mathematics, ie. pre-axiomatic, intuitive, conceptuel, empirical, physical, etc.) definition.



                                                                              For example. Without even getting into the Peano axioms, I can prove that, where $a$ and $b$ are positive integers, $ab=ba$. Indeed, $ab$ is just the process of taking $a$ sets of $b$. Take one element from each of these sets, thus forming a set of $a$ elements. Repeat this $b$ times: you will clearly use up exactly all of the elements and obtain $b$ sets of $a$ elements, in other words, $ba$. Similar informal (but entirely convincing, reasonable, and I would say irrefutable) reasoning can be used to demonstrate the rules for manipulating positive fractions, say.



                                                                              Notice that in the above paragraph I used the fact that both positive integers and positive integer multiplication have pre-axiomatic, "physical" definitions.



                                                                              Ask someone why the product of two negatives is positive, and the best they can do is explain, not prove. "Well, negative kind of means 'opposite', so doing the opposite twice means doing the usual, ie positive" does not constitute a proof, but merely an explanation serving to make the accepted mathematical axiom less surprising. Another common one begins with "we would like the usual properties of arithmetic to hold, so assume they do...", but then it remains to be explained why it's so important that the usual laws of arithmetic hold. Euler himself, in an early chapter of his textbook on algebra, gave the following supremely questionable justification. After justifying $(-a)b=-(ab)$ by analogies with debts, he writes:




                                                                              It remains to resolve the case in which - is multiplied by -; or, for example, -a by -b. It is evident, at first sight, with regard to the letters, that the product will be ab; but it is doubtful whether the sign + or the sign - is to be placed before it, all we know is, that it must be one or the other of these signs. Now, I say that it cannot be the sign -: for -a by +b gives -ab, and -a by -b cannot produce the same result as -a by +b...




                                                                              With no disrespect to Euler (especially consdiering this was intended as an introductory textbook), I think we can agree that this is a pretty philosophically dubious argument.



                                                                              The reason it is impossible is because there is no pre-axiomatic definition for what a negative number or negative multiplication really is. Oh, you could probably come up with one involving opposite "directions", and notions of symmetry, but it would be quite artificial and not at all obviously "the best" definition. In my opinion, negatives are ultimately best understood as purely abstract objects. It so happens - and this is quite myseterious - that these utterly abstract laws of calculation lead to physically meaningful results. This was nicely expressed in 1778 by the mathematician John Playfair, when addressing the then controversial issues of negative and complex numbers:




                                                                              Here then is a paradox which remains to be explained. If the operations of this imaginary arithmetic are unintelligible, why are they not altogether useless? Is investigation an art so mechanical, that it may be conducted by certain manual operations? Or is truth so easily discovered, that intelligence is not necessary to give success to our researches?




                                                                              Quoted in Negative Math: How Mathematical Rules Can be Positively Bent by Alberto A. Martínez.



                                                                              One way of approching the problem is with the idea that negative numbers are a different name for subtraction. The differences between subtraction and addition force us, if we reject negatives, to create many different rules covering all the different possibilities ($a - b$, $b - a$, and $a + b$, and if the particular theorem or problem involves more than two variables, the difficulty is compounded further...). The idea of negatives could be described as the insight that rather than having two operations and one type of number, we can have one operation and two types of number. Indeed, if you start with some perfectly physically meaningful axioms about subtraction, you will find that the $(-1)(-1)=1$ law seems to be implicit within them. Hint: starting from the very reasonable axioms $a(b-c)=ab-ac , a - (b - c) = a - b + c$, consider the product $(a-b)(c-d)$.



                                                                              But even that explanation doesn't altogether satisfy me. I've become convinced that my education cheated me on how deep an idea negative numbers are, and I expect to remain puzzled by them for many years. Anyway, I hope some of the above is useful to someone.






                                                                              share|cite|improve this answer











                                                                              $endgroup$



                                                                              One thing that must be understood is that this law cannot be proven in the same way that the laws of positive rational and integral arithmetic can be. The reason for this is that negatives lack any "external" (external to mathematics, ie. pre-axiomatic, intuitive, conceptuel, empirical, physical, etc.) definition.



                                                                              For example. Without even getting into the Peano axioms, I can prove that, where $a$ and $b$ are positive integers, $ab=ba$. Indeed, $ab$ is just the process of taking $a$ sets of $b$. Take one element from each of these sets, thus forming a set of $a$ elements. Repeat this $b$ times: you will clearly use up exactly all of the elements and obtain $b$ sets of $a$ elements, in other words, $ba$. Similar informal (but entirely convincing, reasonable, and I would say irrefutable) reasoning can be used to demonstrate the rules for manipulating positive fractions, say.



                                                                              Notice that in the above paragraph I used the fact that both positive integers and positive integer multiplication have pre-axiomatic, "physical" definitions.



                                                                              Ask someone why the product of two negatives is positive, and the best they can do is explain, not prove. "Well, negative kind of means 'opposite', so doing the opposite twice means doing the usual, ie positive" does not constitute a proof, but merely an explanation serving to make the accepted mathematical axiom less surprising. Another common one begins with "we would like the usual properties of arithmetic to hold, so assume they do...", but then it remains to be explained why it's so important that the usual laws of arithmetic hold. Euler himself, in an early chapter of his textbook on algebra, gave the following supremely questionable justification. After justifying $(-a)b=-(ab)$ by analogies with debts, he writes:




                                                                              It remains to resolve the case in which - is multiplied by -; or, for example, -a by -b. It is evident, at first sight, with regard to the letters, that the product will be ab; but it is doubtful whether the sign + or the sign - is to be placed before it, all we know is, that it must be one or the other of these signs. Now, I say that it cannot be the sign -: for -a by +b gives -ab, and -a by -b cannot produce the same result as -a by +b...




                                                                              With no disrespect to Euler (especially consdiering this was intended as an introductory textbook), I think we can agree that this is a pretty philosophically dubious argument.



                                                                              The reason it is impossible is because there is no pre-axiomatic definition for what a negative number or negative multiplication really is. Oh, you could probably come up with one involving opposite "directions", and notions of symmetry, but it would be quite artificial and not at all obviously "the best" definition. In my opinion, negatives are ultimately best understood as purely abstract objects. It so happens - and this is quite myseterious - that these utterly abstract laws of calculation lead to physically meaningful results. This was nicely expressed in 1778 by the mathematician John Playfair, when addressing the then controversial issues of negative and complex numbers:




                                                                              Here then is a paradox which remains to be explained. If the operations of this imaginary arithmetic are unintelligible, why are they not altogether useless? Is investigation an art so mechanical, that it may be conducted by certain manual operations? Or is truth so easily discovered, that intelligence is not necessary to give success to our researches?




                                                                              Quoted in Negative Math: How Mathematical Rules Can be Positively Bent by Alberto A. Martínez.



                                                                              One way of approching the problem is with the idea that negative numbers are a different name for subtraction. The differences between subtraction and addition force us, if we reject negatives, to create many different rules covering all the different possibilities ($a - b$, $b - a$, and $a + b$, and if the particular theorem or problem involves more than two variables, the difficulty is compounded further...). The idea of negatives could be described as the insight that rather than having two operations and one type of number, we can have one operation and two types of number. Indeed, if you start with some perfectly physically meaningful axioms about subtraction, you will find that the $(-1)(-1)=1$ law seems to be implicit within them. Hint: starting from the very reasonable axioms $a(b-c)=ab-ac , a - (b - c) = a - b + c$, consider the product $(a-b)(c-d)$.



                                                                              But even that explanation doesn't altogether satisfy me. I've become convinced that my education cheated me on how deep an idea negative numbers are, and I expect to remain puzzled by them for many years. Anyway, I hope some of the above is useful to someone.







                                                                              share|cite|improve this answer














                                                                              share|cite|improve this answer



                                                                              share|cite|improve this answer








                                                                              edited Nov 20 '14 at 0:19

























                                                                              answered May 2 '13 at 13:44









                                                                              Jack MJack M

                                                                              18.9k33882




                                                                              18.9k33882








                                                                              • 1




                                                                                $begingroup$
                                                                                This $(-a)(-b)=ab$ stuff started when we gave up on distinguishing multiplier and multiplied in a multiplication. I can have three unities of apples, but I can't have apple unities of threes. In this sense, it's possible to have 3 unities of an 1D vector backwards $3 times -1$ but not -1 unities of a vector forwards $-1 times 3$, unless we define that a negative multiplier change the sense of the vector. I think a negative-free algebra can be created by distinguish three kind of numbers: scalar (no sign), up and down.
                                                                                $endgroup$
                                                                                – Leonardo Castro
                                                                                Sep 9 '16 at 12:22












                                                                              • $begingroup$
                                                                                If you add an up and a down, you cancel out the amount they have in common keep the remaining up or down part. This would the subtraction of this algebra. Multiplying by $-1$ should be replaced by a simple inversion operator. Actually, $+1$ and $-1$ as multipliers are usually taken as the identity and inversion operators. In that way, they form a simple identity-inversion group and so $-1 times -1 = 1$ since two inversions are equivalent to identity. I don't think, however, that algebra should necessarily be that way.
                                                                                $endgroup$
                                                                                – Leonardo Castro
                                                                                Sep 9 '16 at 12:33










                                                                              • $begingroup$
                                                                                I used up and down just to avoid all the assumptions about positive and negative, but they could be right and left, yin and yang, whatever... I wonder if yin times yin is yang or the other way around.
                                                                                $endgroup$
                                                                                – Leonardo Castro
                                                                                Sep 9 '16 at 12:35
















                                                                              • 1




                                                                                $begingroup$
                                                                                This $(-a)(-b)=ab$ stuff started when we gave up on distinguishing multiplier and multiplied in a multiplication. I can have three unities of apples, but I can't have apple unities of threes. In this sense, it's possible to have 3 unities of an 1D vector backwards $3 times -1$ but not -1 unities of a vector forwards $-1 times 3$, unless we define that a negative multiplier change the sense of the vector. I think a negative-free algebra can be created by distinguish three kind of numbers: scalar (no sign), up and down.
                                                                                $endgroup$
                                                                                – Leonardo Castro
                                                                                Sep 9 '16 at 12:22












                                                                              • $begingroup$
                                                                                If you add an up and a down, you cancel out the amount they have in common keep the remaining up or down part. This would the subtraction of this algebra. Multiplying by $-1$ should be replaced by a simple inversion operator. Actually, $+1$ and $-1$ as multipliers are usually taken as the identity and inversion operators. In that way, they form a simple identity-inversion group and so $-1 times -1 = 1$ since two inversions are equivalent to identity. I don't think, however, that algebra should necessarily be that way.
                                                                                $endgroup$
                                                                                – Leonardo Castro
                                                                                Sep 9 '16 at 12:33










                                                                              • $begingroup$
                                                                                I used up and down just to avoid all the assumptions about positive and negative, but they could be right and left, yin and yang, whatever... I wonder if yin times yin is yang or the other way around.
                                                                                $endgroup$
                                                                                – Leonardo Castro
                                                                                Sep 9 '16 at 12:35










                                                                              1




                                                                              1




                                                                              $begingroup$
                                                                              This $(-a)(-b)=ab$ stuff started when we gave up on distinguishing multiplier and multiplied in a multiplication. I can have three unities of apples, but I can't have apple unities of threes. In this sense, it's possible to have 3 unities of an 1D vector backwards $3 times -1$ but not -1 unities of a vector forwards $-1 times 3$, unless we define that a negative multiplier change the sense of the vector. I think a negative-free algebra can be created by distinguish three kind of numbers: scalar (no sign), up and down.
                                                                              $endgroup$
                                                                              – Leonardo Castro
                                                                              Sep 9 '16 at 12:22






                                                                              $begingroup$
                                                                              This $(-a)(-b)=ab$ stuff started when we gave up on distinguishing multiplier and multiplied in a multiplication. I can have three unities of apples, but I can't have apple unities of threes. In this sense, it's possible to have 3 unities of an 1D vector backwards $3 times -1$ but not -1 unities of a vector forwards $-1 times 3$, unless we define that a negative multiplier change the sense of the vector. I think a negative-free algebra can be created by distinguish three kind of numbers: scalar (no sign), up and down.
                                                                              $endgroup$
                                                                              – Leonardo Castro
                                                                              Sep 9 '16 at 12:22














                                                                              $begingroup$
                                                                              If you add an up and a down, you cancel out the amount they have in common keep the remaining up or down part. This would the subtraction of this algebra. Multiplying by $-1$ should be replaced by a simple inversion operator. Actually, $+1$ and $-1$ as multipliers are usually taken as the identity and inversion operators. In that way, they form a simple identity-inversion group and so $-1 times -1 = 1$ since two inversions are equivalent to identity. I don't think, however, that algebra should necessarily be that way.
                                                                              $endgroup$
                                                                              – Leonardo Castro
                                                                              Sep 9 '16 at 12:33




                                                                              $begingroup$
                                                                              If you add an up and a down, you cancel out the amount they have in common keep the remaining up or down part. This would the subtraction of this algebra. Multiplying by $-1$ should be replaced by a simple inversion operator. Actually, $+1$ and $-1$ as multipliers are usually taken as the identity and inversion operators. In that way, they form a simple identity-inversion group and so $-1 times -1 = 1$ since two inversions are equivalent to identity. I don't think, however, that algebra should necessarily be that way.
                                                                              $endgroup$
                                                                              – Leonardo Castro
                                                                              Sep 9 '16 at 12:33












                                                                              $begingroup$
                                                                              I used up and down just to avoid all the assumptions about positive and negative, but they could be right and left, yin and yang, whatever... I wonder if yin times yin is yang or the other way around.
                                                                              $endgroup$
                                                                              – Leonardo Castro
                                                                              Sep 9 '16 at 12:35






                                                                              $begingroup$
                                                                              I used up and down just to avoid all the assumptions about positive and negative, but they could be right and left, yin and yang, whatever... I wonder if yin times yin is yang or the other way around.
                                                                              $endgroup$
                                                                              – Leonardo Castro
                                                                              Sep 9 '16 at 12:35













                                                                              12












                                                                              $begingroup$

                                                                              Extend reals to the complex plane. Multiplication by $-1$ is a rotation by $pi$ radians. When you multiply by two negatives, you rotate by $2pi$. :-)






                                                                              share|cite|improve this answer











                                                                              $endgroup$


















                                                                                12












                                                                                $begingroup$

                                                                                Extend reals to the complex plane. Multiplication by $-1$ is a rotation by $pi$ radians. When you multiply by two negatives, you rotate by $2pi$. :-)






                                                                                share|cite|improve this answer











                                                                                $endgroup$
















                                                                                  12












                                                                                  12








                                                                                  12





                                                                                  $begingroup$

                                                                                  Extend reals to the complex plane. Multiplication by $-1$ is a rotation by $pi$ radians. When you multiply by two negatives, you rotate by $2pi$. :-)






                                                                                  share|cite|improve this answer











                                                                                  $endgroup$



                                                                                  Extend reals to the complex plane. Multiplication by $-1$ is a rotation by $pi$ radians. When you multiply by two negatives, you rotate by $2pi$. :-)







                                                                                  share|cite|improve this answer














                                                                                  share|cite|improve this answer



                                                                                  share|cite|improve this answer








                                                                                  edited Oct 24 '15 at 19:36









                                                                                  akukas

                                                                                  150111




                                                                                  150111










                                                                                  answered Apr 14 '14 at 23:26









                                                                                  JasonJason

                                                                                  1,367610




                                                                                  1,367610























                                                                                      10












                                                                                      $begingroup$

                                                                                      Why don't we tell a story! The dastardly Dalton gang is on the loose but Al Catchem is hot on their trail and nearly catches them at their latest heist. When pulling out of the parking lot, there are a few possibilities:




                                                                                      • Al is quick, up to the task and catches the Daltons,

                                                                                      • Al's car stalls and the Daltons get away,

                                                                                      • The Dalton's are tipped off early and escape, or

                                                                                      • The Dalton's car stalls, they are captured and the city is saved.


                                                                                      In summary:




                                                                                      • positive $times$ positive: If a good thing happens to a good person, that's good! :)

                                                                                      • negative $times$ positive: If a bad thing happends to a good person, that's bad. :(

                                                                                      • positive $times$ negative: If a good thing happens to a bad person, that's also bad. :(

                                                                                      • negative $times$ negative: But, if a bad thing happens to a bad person, that's good!! :)






                                                                                      share|cite|improve this answer











                                                                                      $endgroup$









                                                                                      • 22




                                                                                        $begingroup$
                                                                                        While I can understand the sentiment, I'm not sure this is the kind of morality we should be teaching 8-year-old kids.
                                                                                        $endgroup$
                                                                                        – MichaelS
                                                                                        Nov 9 '15 at 3:32






                                                                                      • 6




                                                                                        $begingroup$
                                                                                        Downvoted because this isn't really a mathematical explanation.
                                                                                        $endgroup$
                                                                                        – Daniel R. Collins
                                                                                        Nov 9 '15 at 3:39






                                                                                      • 7




                                                                                        $begingroup$
                                                                                        @DanielR.Collins That is funny - the question is how to explain the idea to an 8 year old. :)
                                                                                        $endgroup$
                                                                                        – Mark McClure
                                                                                        Nov 9 '15 at 3:40






                                                                                      • 2




                                                                                        $begingroup$
                                                                                        @MichaelS You are, of course, correct. When I've actually done this, which I have many times, I've tried to tell a story with it that (hopefully) makes the morality reasonably clear. Turns out, you can have great fun with it!
                                                                                        $endgroup$
                                                                                        – Mark McClure
                                                                                        Nov 9 '15 at 4:10






                                                                                      • 4




                                                                                        $begingroup$
                                                                                        Why is this immoral? From a naturalistic philosophical viewpoint, this is arguably the most moral, since babies have been shown to prefer "good" characters over "evil" ones.
                                                                                        $endgroup$
                                                                                        – March Ho
                                                                                        Nov 9 '15 at 13:48
















                                                                                      10












                                                                                      $begingroup$

                                                                                      Why don't we tell a story! The dastardly Dalton gang is on the loose but Al Catchem is hot on their trail and nearly catches them at their latest heist. When pulling out of the parking lot, there are a few possibilities:




                                                                                      • Al is quick, up to the task and catches the Daltons,

                                                                                      • Al's car stalls and the Daltons get away,

                                                                                      • The Dalton's are tipped off early and escape, or

                                                                                      • The Dalton's car stalls, they are captured and the city is saved.


                                                                                      In summary:




                                                                                      • positive $times$ positive: If a good thing happens to a good person, that's good! :)

                                                                                      • negative $times$ positive: If a bad thing happends to a good person, that's bad. :(

                                                                                      • positive $times$ negative: If a good thing happens to a bad person, that's also bad. :(

                                                                                      • negative $times$ negative: But, if a bad thing happens to a bad person, that's good!! :)






                                                                                      share|cite|improve this answer











                                                                                      $endgroup$









                                                                                      • 22




                                                                                        $begingroup$
                                                                                        While I can understand the sentiment, I'm not sure this is the kind of morality we should be teaching 8-year-old kids.
                                                                                        $endgroup$
                                                                                        – MichaelS
                                                                                        Nov 9 '15 at 3:32






                                                                                      • 6




                                                                                        $begingroup$
                                                                                        Downvoted because this isn't really a mathematical explanation.
                                                                                        $endgroup$
                                                                                        – Daniel R. Collins
                                                                                        Nov 9 '15 at 3:39






                                                                                      • 7




                                                                                        $begingroup$
                                                                                        @DanielR.Collins That is funny - the question is how to explain the idea to an 8 year old. :)
                                                                                        $endgroup$
                                                                                        – Mark McClure
                                                                                        Nov 9 '15 at 3:40






                                                                                      • 2




                                                                                        $begingroup$
                                                                                        @MichaelS You are, of course, correct. When I've actually done this, which I have many times, I've tried to tell a story with it that (hopefully) makes the morality reasonably clear. Turns out, you can have great fun with it!
                                                                                        $endgroup$
                                                                                        – Mark McClure
                                                                                        Nov 9 '15 at 4:10






                                                                                      • 4




                                                                                        $begingroup$
                                                                                        Why is this immoral? From a naturalistic philosophical viewpoint, this is arguably the most moral, since babies have been shown to prefer "good" characters over "evil" ones.
                                                                                        $endgroup$
                                                                                        – March Ho
                                                                                        Nov 9 '15 at 13:48














                                                                                      10












                                                                                      10








                                                                                      10





                                                                                      $begingroup$

                                                                                      Why don't we tell a story! The dastardly Dalton gang is on the loose but Al Catchem is hot on their trail and nearly catches them at their latest heist. When pulling out of the parking lot, there are a few possibilities:




                                                                                      • Al is quick, up to the task and catches the Daltons,

                                                                                      • Al's car stalls and the Daltons get away,

                                                                                      • The Dalton's are tipped off early and escape, or

                                                                                      • The Dalton's car stalls, they are captured and the city is saved.


                                                                                      In summary:




                                                                                      • positive $times$ positive: If a good thing happens to a good person, that's good! :)

                                                                                      • negative $times$ positive: If a bad thing happends to a good person, that's bad. :(

                                                                                      • positive $times$ negative: If a good thing happens to a bad person, that's also bad. :(

                                                                                      • negative $times$ negative: But, if a bad thing happens to a bad person, that's good!! :)






                                                                                      share|cite|improve this answer











                                                                                      $endgroup$



                                                                                      Why don't we tell a story! The dastardly Dalton gang is on the loose but Al Catchem is hot on their trail and nearly catches them at their latest heist. When pulling out of the parking lot, there are a few possibilities:




                                                                                      • Al is quick, up to the task and catches the Daltons,

                                                                                      • Al's car stalls and the Daltons get away,

                                                                                      • The Dalton's are tipped off early and escape, or

                                                                                      • The Dalton's car stalls, they are captured and the city is saved.


                                                                                      In summary:




                                                                                      • positive $times$ positive: If a good thing happens to a good person, that's good! :)

                                                                                      • negative $times$ positive: If a bad thing happends to a good person, that's bad. :(

                                                                                      • positive $times$ negative: If a good thing happens to a bad person, that's also bad. :(

                                                                                      • negative $times$ negative: But, if a bad thing happens to a bad person, that's good!! :)







                                                                                      share|cite|improve this answer














                                                                                      share|cite|improve this answer



                                                                                      share|cite|improve this answer








                                                                                      edited Nov 9 '15 at 6:30

























                                                                                      answered Nov 9 '15 at 3:25









                                                                                      Mark McClureMark McClure

                                                                                      23.9k34472




                                                                                      23.9k34472








                                                                                      • 22




                                                                                        $begingroup$
                                                                                        While I can understand the sentiment, I'm not sure this is the kind of morality we should be teaching 8-year-old kids.
                                                                                        $endgroup$
                                                                                        – MichaelS
                                                                                        Nov 9 '15 at 3:32






                                                                                      • 6




                                                                                        $begingroup$
                                                                                        Downvoted because this isn't really a mathematical explanation.
                                                                                        $endgroup$
                                                                                        – Daniel R. Collins
                                                                                        Nov 9 '15 at 3:39






                                                                                      • 7




                                                                                        $begingroup$
                                                                                        @DanielR.Collins That is funny - the question is how to explain the idea to an 8 year old. :)
                                                                                        $endgroup$
                                                                                        – Mark McClure
                                                                                        Nov 9 '15 at 3:40






                                                                                      • 2




                                                                                        $begingroup$
                                                                                        @MichaelS You are, of course, correct. When I've actually done this, which I have many times, I've tried to tell a story with it that (hopefully) makes the morality reasonably clear. Turns out, you can have great fun with it!
                                                                                        $endgroup$
                                                                                        – Mark McClure
                                                                                        Nov 9 '15 at 4:10






                                                                                      • 4




                                                                                        $begingroup$
                                                                                        Why is this immoral? From a naturalistic philosophical viewpoint, this is arguably the most moral, since babies have been shown to prefer "good" characters over "evil" ones.
                                                                                        $endgroup$
                                                                                        – March Ho
                                                                                        Nov 9 '15 at 13:48














                                                                                      • 22




                                                                                        $begingroup$
                                                                                        While I can understand the sentiment, I'm not sure this is the kind of morality we should be teaching 8-year-old kids.
                                                                                        $endgroup$
                                                                                        – MichaelS
                                                                                        Nov 9 '15 at 3:32






                                                                                      • 6




                                                                                        $begingroup$
                                                                                        Downvoted because this isn't really a mathematical explanation.
                                                                                        $endgroup$
                                                                                        – Daniel R. Collins
                                                                                        Nov 9 '15 at 3:39






                                                                                      • 7




                                                                                        $begingroup$
                                                                                        @DanielR.Collins That is funny - the question is how to explain the idea to an 8 year old. :)
                                                                                        $endgroup$
                                                                                        – Mark McClure
                                                                                        Nov 9 '15 at 3:40






                                                                                      • 2




                                                                                        $begingroup$
                                                                                        @MichaelS You are, of course, correct. When I've actually done this, which I have many times, I've tried to tell a story with it that (hopefully) makes the morality reasonably clear. Turns out, you can have great fun with it!
                                                                                        $endgroup$
                                                                                        – Mark McClure
                                                                                        Nov 9 '15 at 4:10






                                                                                      • 4




                                                                                        $begingroup$
                                                                                        Why is this immoral? From a naturalistic philosophical viewpoint, this is arguably the most moral, since babies have been shown to prefer "good" characters over "evil" ones.
                                                                                        $endgroup$
                                                                                        – March Ho
                                                                                        Nov 9 '15 at 13:48








                                                                                      22




                                                                                      22




                                                                                      $begingroup$
                                                                                      While I can understand the sentiment, I'm not sure this is the kind of morality we should be teaching 8-year-old kids.
                                                                                      $endgroup$
                                                                                      – MichaelS
                                                                                      Nov 9 '15 at 3:32




                                                                                      $begingroup$
                                                                                      While I can understand the sentiment, I'm not sure this is the kind of morality we should be teaching 8-year-old kids.
                                                                                      $endgroup$
                                                                                      – MichaelS
                                                                                      Nov 9 '15 at 3:32




                                                                                      6




                                                                                      6




                                                                                      $begingroup$
                                                                                      Downvoted because this isn't really a mathematical explanation.
                                                                                      $endgroup$
                                                                                      – Daniel R. Collins
                                                                                      Nov 9 '15 at 3:39




                                                                                      $begingroup$
                                                                                      Downvoted because this isn't really a mathematical explanation.
                                                                                      $endgroup$
                                                                                      – Daniel R. Collins
                                                                                      Nov 9 '15 at 3:39




                                                                                      7




                                                                                      7




                                                                                      $begingroup$
                                                                                      @DanielR.Collins That is funny - the question is how to explain the idea to an 8 year old. :)
                                                                                      $endgroup$
                                                                                      – Mark McClure
                                                                                      Nov 9 '15 at 3:40




                                                                                      $begingroup$
                                                                                      @DanielR.Collins That is funny - the question is how to explain the idea to an 8 year old. :)
                                                                                      $endgroup$
                                                                                      – Mark McClure
                                                                                      Nov 9 '15 at 3:40




                                                                                      2




                                                                                      2




                                                                                      $begingroup$
                                                                                      @MichaelS You are, of course, correct. When I've actually done this, which I have many times, I've tried to tell a story with it that (hopefully) makes the morality reasonably clear. Turns out, you can have great fun with it!
                                                                                      $endgroup$
                                                                                      – Mark McClure
                                                                                      Nov 9 '15 at 4:10




                                                                                      $begingroup$
                                                                                      @MichaelS You are, of course, correct. When I've actually done this, which I have many times, I've tried to tell a story with it that (hopefully) makes the morality reasonably clear. Turns out, you can have great fun with it!
                                                                                      $endgroup$
                                                                                      – Mark McClure
                                                                                      Nov 9 '15 at 4:10




                                                                                      4




                                                                                      4




                                                                                      $begingroup$
                                                                                      Why is this immoral? From a naturalistic philosophical viewpoint, this is arguably the most moral, since babies have been shown to prefer "good" characters over "evil" ones.
                                                                                      $endgroup$
                                                                                      – March Ho
                                                                                      Nov 9 '15 at 13:48




                                                                                      $begingroup$
                                                                                      Why is this immoral? From a naturalistic philosophical viewpoint, this is arguably the most moral, since babies have been shown to prefer "good" characters over "evil" ones.
                                                                                      $endgroup$
                                                                                      – March Ho
                                                                                      Nov 9 '15 at 13:48











                                                                                      9












                                                                                      $begingroup$

                                                                                      I prefer the explanation by my favorite mathematician , V. I. Arnold (physicist really, since in his own words, "mathematics is a part of physics" and "an experimental science"). I believe it's the most natural (yet totally mathematical) explanation of a basic notion like multiplication of negative numbers.



                                                                                      This is an excerpt from Arnold's wonderful memoir "Yesterday and Long Ago" (3d ed., available in English from Springer), full of world history, drama and ingenious storytelling. The 2007 translation into English, I believe, is not of best quality, but it's the only one so far.



                                                                                      from the short story The Arnold Family




                                                                                      I faced a real difficulty with school mathematics several years after
                                                                                      the multiplication table: it was necessary to leam that “minus
                                                                                      multiplied by minus is plus” I wanted to know the proof of this rule;
                                                                                      I have never been able to leam by heart what is not properly
                                                                                      understood. I asked my father to explain the reason why (—1) • (—2) =
                                                                                      (+2). He, being a student of great algebraists, S. O. Shatunovsky and
                                                                                      E. Noether, gave the following “scientific explanation”: “The point
                                                                                      is,” he said: “that numbers form a field such that the distributive
                                                                                      law (x+y)z=xz+yz holds. And if the product of minus by minus had not
                                                                                      been plus, this law would be broken”.



                                                                                      However, for me this “deductive” (actually juridical) explanation did
                                                                                      not prove anything - what of it! One can study any axioms! Since
                                                                                      that day I have preserved the healthy aversion of a naturalist to the
                                                                                      axiomatic method with its non-motivated axioms.



                                                                                      The axiomophile Rene Descartes stated that “neither experimental tests
                                                                                      that axioms reflect a reality, nor comparison of theoretical results
                                                                                      with reality should be a part of science” (why should results
                                                                                      correspond to reality if the initial principles do not correspond to
                                                                                      it?).



                                                                                      Another thesis of Descartes’ theory and methods of education is even
                                                                                      more peculiar and contemporary: “It is necessary to forbid all other
                                                                                      methods of teaching except mine because only this method is
                                                                                      politically correct
                                                                                      : with my purely deductive method any dull
                                                                                      student can be taught as successfully as the most gifted one
                                                                                      , while
                                                                                      with other methods imagination and even drawings are used
                                                                                      unavoidably, and for this reason geniuses advance faster
                                                                                      than dunces
                                                                                      ”.



                                                                                      Contrary to the deductive theories of my father and Descartes, as a
                                                                                      ten year old, I started thinking about a naturally-scientific sense
                                                                                      of the rule of signs
                                                                                      , and I have come to the following conclusion. A
                                                                                      real (positive or negative) number is a vector on the axis of
                                                                                      coordinates (if a number is positive the corresponding vector is
                                                                                      positively directed along this axis).



                                                                                      A product of two numbers is an area of a rectangle whose sides
                                                                                      correspond to these numbers (one vector is along one axis and the
                                                                                      other is along a perpendicular axis in the plane). A rectangle, given
                                                                                      by an ordered pair of vectors, possesses, as a part of the plane, a
                                                                                      definite orientation (rotation from one vector to another can be
                                                                                      clockwise or anti-clockwise). The anti-clockwise rotation is
                                                                                      customarily considered positive and the clockwise rotation is then
                                                                                      negative. And lastly, the area of a parallelogram (for example, a
                                                                                      rectangle) generated by the two vectors x and у (taken in a
                                                                                      definite order) emanating from one point in the plane is considered
                                                                                      to be positive
                                                                                      if the pair of vectors (taken in this order) defines
                                                                                      positive orientation of the plane (and negative if the rotation from the direction of the vector x to the direction of the vector у is
                                                                                      negative).



                                                                                      Thus, the rule of signs is not an axiom taken out of the blue, but
                                                                                      becomes a natural property of orientation which is easily verified
                                                                                      experimentally.




                                                                                      from the short story Axiomatic Method




                                                                                      My first trouble in school was caused by the rule for multiplication
                                                                                      of negative numbers, and I asked my father to explain this peculiar
                                                                                      rule.



                                                                                      He, as a faithful student of Emmy Noether (and consequently of Hilbert
                                                                                      and Dedekind) started explaining to his eleven-year-old son the
                                                                                      principles of axiomatic science: a definition is chosen such that the
                                                                                      distributive identity a(b+c)=ab+ac holds.



                                                                                      The axiomatic method requires that one should accept any axiom with a
                                                                                      hope that its corollaries would be fruitful (probably this could be
                                                                                      understood by the age of thirty when it would be possible to read and
                                                                                      appreciate “Anna Karenina”). My father did not say a word either about
                                                                                      the oriented area of a rectangular or about any non-mathematicai
                                                                                      interpretation of signs and products.



                                                                                      This “algebraic” explanation was not able to shake either my hearty
                                                                                      love for my father or a deep respect of his science. But since that
                                                                                      time I have disliked the axiomatic method with its non-motivated
                                                                                      definitions. Probably it was for this reason that by this time I got
                                                                                      used to talking with non-algebraists (like L. I. Mandel’shtam, I. E.
                                                                                      Tamm, P. S. Novikov, E. L. Feinberg, M. A. Leontovich, and A. G.
                                                                                      Gurvich) who treated an ignorant interlocutor with full respect and
                                                                                      tried to explain non-trivial ideas and facts of various sciences such
                                                                                      as physics and biology, astronomy and radiolocation.



                                                                                      Negative numbers I came to understand a year later while deriving an
                                                                                      “equation of time”, which takes into account a correction for the
                                                                                      length of a day corresponding to the time of year. It is not possible
                                                                                      to explain to algebraists that their axiomatic method is mostly
                                                                                      useless for students.



                                                                                      One should ask children: at what time will high tide be tomorrow if
                                                                                      today it is at 3 pm? This is a feasible problem, and it helps children
                                                                                      to understand negative numbers better than algebraic rules do. Once I
                                                                                      read from an ancient author (probably from Herodotus) that the tides
                                                                                      "always occur three and nine o'clock". To understand that the monthly
                                                                                      rotation of the Moon about the Earth affects the tide timetable, there
                                                                                      is no need to live near an ocean. Here, not in axioms, is laid true
                                                                                      mathematics.







                                                                                      share|cite|improve this answer









                                                                                      $endgroup$













                                                                                      • $begingroup$
                                                                                        2 very good and insightful anecdotes-although I wouldn't really call either of them PROOFS, would anyone here?They're more intuitive explanations why it's so.Of course, the natural next question is:Can either of these arguments be precisely reformulated as proofs?
                                                                                        $endgroup$
                                                                                        – Mathemagician1234
                                                                                        Jun 13 '15 at 3:21








                                                                                      • 1




                                                                                        $begingroup$
                                                                                        @Mathemagician1234 you know, I agree with Arnold on the idea that explanations or intuition like this are much much more powerful than any proofs or definitions, because they are real and without artificial restrictions of formal systems. Using them, one can come up with a correct system of definitions in no time, if they have to. However, having the definitions in the first place does not add anything to one's true understanding of the subject.
                                                                                        $endgroup$
                                                                                        – level1807
                                                                                        Jun 13 '15 at 7:36










                                                                                      • $begingroup$
                                                                                        With the interpretation in terms of oriented areas we would have $xy=-yx$ which is kind of problematic.
                                                                                        $endgroup$
                                                                                        – Sergio
                                                                                        Feb 2 '18 at 22:55
















                                                                                      9












                                                                                      $begingroup$

                                                                                      I prefer the explanation by my favorite mathematician , V. I. Arnold (physicist really, since in his own words, "mathematics is a part of physics" and "an experimental science"). I believe it's the most natural (yet totally mathematical) explanation of a basic notion like multiplication of negative numbers.



                                                                                      This is an excerpt from Arnold's wonderful memoir "Yesterday and Long Ago" (3d ed., available in English from Springer), full of world history, drama and ingenious storytelling. The 2007 translation into English, I believe, is not of best quality, but it's the only one so far.



                                                                                      from the short story The Arnold Family




                                                                                      I faced a real difficulty with school mathematics several years after
                                                                                      the multiplication table: it was necessary to leam that “minus
                                                                                      multiplied by minus is plus” I wanted to know the proof of this rule;
                                                                                      I have never been able to leam by heart what is not properly
                                                                                      understood. I asked my father to explain the reason why (—1) • (—2) =
                                                                                      (+2). He, being a student of great algebraists, S. O. Shatunovsky and
                                                                                      E. Noether, gave the following “scientific explanation”: “The point
                                                                                      is,” he said: “that numbers form a field such that the distributive
                                                                                      law (x+y)z=xz+yz holds. And if the product of minus by minus had not
                                                                                      been plus, this law would be broken”.



                                                                                      However, for me this “deductive” (actually juridical) explanation did
                                                                                      not prove anything - what of it! One can study any axioms! Since
                                                                                      that day I have preserved the healthy aversion of a naturalist to the
                                                                                      axiomatic method with its non-motivated axioms.



                                                                                      The axiomophile Rene Descartes stated that “neither experimental tests
                                                                                      that axioms reflect a reality, nor comparison of theoretical results
                                                                                      with reality should be a part of science” (why should results
                                                                                      correspond to reality if the initial principles do not correspond to
                                                                                      it?).



                                                                                      Another thesis of Descartes’ theory and methods of education is even
                                                                                      more peculiar and contemporary: “It is necessary to forbid all other
                                                                                      methods of teaching except mine because only this method is
                                                                                      politically correct
                                                                                      : with my purely deductive method any dull
                                                                                      student can be taught as successfully as the most gifted one
                                                                                      , while
                                                                                      with other methods imagination and even drawings are used
                                                                                      unavoidably, and for this reason geniuses advance faster
                                                                                      than dunces
                                                                                      ”.



                                                                                      Contrary to the deductive theories of my father and Descartes, as a
                                                                                      ten year old, I started thinking about a naturally-scientific sense
                                                                                      of the rule of signs
                                                                                      , and I have come to the following conclusion. A
                                                                                      real (positive or negative) number is a vector on the axis of
                                                                                      coordinates (if a number is positive the corresponding vector is
                                                                                      positively directed along this axis).



                                                                                      A product of two numbers is an area of a rectangle whose sides
                                                                                      correspond to these numbers (one vector is along one axis and the
                                                                                      other is along a perpendicular axis in the plane). A rectangle, given
                                                                                      by an ordered pair of vectors, possesses, as a part of the plane, a
                                                                                      definite orientation (rotation from one vector to another can be
                                                                                      clockwise or anti-clockwise). The anti-clockwise rotation is
                                                                                      customarily considered positive and the clockwise rotation is then
                                                                                      negative. And lastly, the area of a parallelogram (for example, a
                                                                                      rectangle) generated by the two vectors x and у (taken in a
                                                                                      definite order) emanating from one point in the plane is considered
                                                                                      to be positive
                                                                                      if the pair of vectors (taken in this order) defines
                                                                                      positive orientation of the plane (and negative if the rotation from the direction of the vector x to the direction of the vector у is
                                                                                      negative).



                                                                                      Thus, the rule of signs is not an axiom taken out of the blue, but
                                                                                      becomes a natural property of orientation which is easily verified
                                                                                      experimentally.




                                                                                      from the short story Axiomatic Method




                                                                                      My first trouble in school was caused by the rule for multiplication
                                                                                      of negative numbers, and I asked my father to explain this peculiar
                                                                                      rule.



                                                                                      He, as a faithful student of Emmy Noether (and consequently of Hilbert
                                                                                      and Dedekind) started explaining to his eleven-year-old son the
                                                                                      principles of axiomatic science: a definition is chosen such that the
                                                                                      distributive identity a(b+c)=ab+ac holds.



                                                                                      The axiomatic method requires that one should accept any axiom with a
                                                                                      hope that its corollaries would be fruitful (probably this could be
                                                                                      understood by the age of thirty when it would be possible to read and
                                                                                      appreciate “Anna Karenina”). My father did not say a word either about
                                                                                      the oriented area of a rectangular or about any non-mathematicai
                                                                                      interpretation of signs and products.



                                                                                      This “algebraic” explanation was not able to shake either my hearty
                                                                                      love for my father or a deep respect of his science. But since that
                                                                                      time I have disliked the axiomatic method with its non-motivated
                                                                                      definitions. Probably it was for this reason that by this time I got
                                                                                      used to talking with non-algebraists (like L. I. Mandel’shtam, I. E.
                                                                                      Tamm, P. S. Novikov, E. L. Feinberg, M. A. Leontovich, and A. G.
                                                                                      Gurvich) who treated an ignorant interlocutor with full respect and
                                                                                      tried to explain non-trivial ideas and facts of various sciences such
                                                                                      as physics and biology, astronomy and radiolocation.



                                                                                      Negative numbers I came to understand a year later while deriving an
                                                                                      “equation of time”, which takes into account a correction for the
                                                                                      length of a day corresponding to the time of year. It is not possible
                                                                                      to explain to algebraists that their axiomatic method is mostly
                                                                                      useless for students.



                                                                                      One should ask children: at what time will high tide be tomorrow if
                                                                                      today it is at 3 pm? This is a feasible problem, and it helps children
                                                                                      to understand negative numbers better than algebraic rules do. Once I
                                                                                      read from an ancient author (probably from Herodotus) that the tides
                                                                                      "always occur three and nine o'clock". To understand that the monthly
                                                                                      rotation of the Moon about the Earth affects the tide timetable, there
                                                                                      is no need to live near an ocean. Here, not in axioms, is laid true
                                                                                      mathematics.







                                                                                      share|cite|improve this answer









                                                                                      $endgroup$













                                                                                      • $begingroup$
                                                                                        2 very good and insightful anecdotes-although I wouldn't really call either of them PROOFS, would anyone here?They're more intuitive explanations why it's so.Of course, the natural next question is:Can either of these arguments be precisely reformulated as proofs?
                                                                                        $endgroup$
                                                                                        – Mathemagician1234
                                                                                        Jun 13 '15 at 3:21








                                                                                      • 1




                                                                                        $begingroup$
                                                                                        @Mathemagician1234 you know, I agree with Arnold on the idea that explanations or intuition like this are much much more powerful than any proofs or definitions, because they are real and without artificial restrictions of formal systems. Using them, one can come up with a correct system of definitions in no time, if they have to. However, having the definitions in the first place does not add anything to one's true understanding of the subject.
                                                                                        $endgroup$
                                                                                        – level1807
                                                                                        Jun 13 '15 at 7:36










                                                                                      • $begingroup$
                                                                                        With the interpretation in terms of oriented areas we would have $xy=-yx$ which is kind of problematic.
                                                                                        $endgroup$
                                                                                        – Sergio
                                                                                        Feb 2 '18 at 22:55














                                                                                      9












                                                                                      9








                                                                                      9





                                                                                      $begingroup$

                                                                                      I prefer the explanation by my favorite mathematician , V. I. Arnold (physicist really, since in his own words, "mathematics is a part of physics" and "an experimental science"). I believe it's the most natural (yet totally mathematical) explanation of a basic notion like multiplication of negative numbers.



                                                                                      This is an excerpt from Arnold's wonderful memoir "Yesterday and Long Ago" (3d ed., available in English from Springer), full of world history, drama and ingenious storytelling. The 2007 translation into English, I believe, is not of best quality, but it's the only one so far.



                                                                                      from the short story The Arnold Family




                                                                                      I faced a real difficulty with school mathematics several years after
                                                                                      the multiplication table: it was necessary to leam that “minus
                                                                                      multiplied by minus is plus” I wanted to know the proof of this rule;
                                                                                      I have never been able to leam by heart what is not properly
                                                                                      understood. I asked my father to explain the reason why (—1) • (—2) =
                                                                                      (+2). He, being a student of great algebraists, S. O. Shatunovsky and
                                                                                      E. Noether, gave the following “scientific explanation”: “The point
                                                                                      is,” he said: “that numbers form a field such that the distributive
                                                                                      law (x+y)z=xz+yz holds. And if the product of minus by minus had not
                                                                                      been plus, this law would be broken”.



                                                                                      However, for me this “deductive” (actually juridical) explanation did
                                                                                      not prove anything - what of it! One can study any axioms! Since
                                                                                      that day I have preserved the healthy aversion of a naturalist to the
                                                                                      axiomatic method with its non-motivated axioms.



                                                                                      The axiomophile Rene Descartes stated that “neither experimental tests
                                                                                      that axioms reflect a reality, nor comparison of theoretical results
                                                                                      with reality should be a part of science” (why should results
                                                                                      correspond to reality if the initial principles do not correspond to
                                                                                      it?).



                                                                                      Another thesis of Descartes’ theory and methods of education is even
                                                                                      more peculiar and contemporary: “It is necessary to forbid all other
                                                                                      methods of teaching except mine because only this method is
                                                                                      politically correct
                                                                                      : with my purely deductive method any dull
                                                                                      student can be taught as successfully as the most gifted one
                                                                                      , while
                                                                                      with other methods imagination and even drawings are used
                                                                                      unavoidably, and for this reason geniuses advance faster
                                                                                      than dunces
                                                                                      ”.



                                                                                      Contrary to the deductive theories of my father and Descartes, as a
                                                                                      ten year old, I started thinking about a naturally-scientific sense
                                                                                      of the rule of signs
                                                                                      , and I have come to the following conclusion. A
                                                                                      real (positive or negative) number is a vector on the axis of
                                                                                      coordinates (if a number is positive the corresponding vector is
                                                                                      positively directed along this axis).



                                                                                      A product of two numbers is an area of a rectangle whose sides
                                                                                      correspond to these numbers (one vector is along one axis and the
                                                                                      other is along a perpendicular axis in the plane). A rectangle, given
                                                                                      by an ordered pair of vectors, possesses, as a part of the plane, a
                                                                                      definite orientation (rotation from one vector to another can be
                                                                                      clockwise or anti-clockwise). The anti-clockwise rotation is
                                                                                      customarily considered positive and the clockwise rotation is then
                                                                                      negative. And lastly, the area of a parallelogram (for example, a
                                                                                      rectangle) generated by the two vectors x and у (taken in a
                                                                                      definite order) emanating from one point in the plane is considered
                                                                                      to be positive
                                                                                      if the pair of vectors (taken in this order) defines
                                                                                      positive orientation of the plane (and negative if the rotation from the direction of the vector x to the direction of the vector у is
                                                                                      negative).



                                                                                      Thus, the rule of signs is not an axiom taken out of the blue, but
                                                                                      becomes a natural property of orientation which is easily verified
                                                                                      experimentally.




                                                                                      from the short story Axiomatic Method




                                                                                      My first trouble in school was caused by the rule for multiplication
                                                                                      of negative numbers, and I asked my father to explain this peculiar
                                                                                      rule.



                                                                                      He, as a faithful student of Emmy Noether (and consequently of Hilbert
                                                                                      and Dedekind) started explaining to his eleven-year-old son the
                                                                                      principles of axiomatic science: a definition is chosen such that the
                                                                                      distributive identity a(b+c)=ab+ac holds.



                                                                                      The axiomatic method requires that one should accept any axiom with a
                                                                                      hope that its corollaries would be fruitful (probably this could be
                                                                                      understood by the age of thirty when it would be possible to read and
                                                                                      appreciate “Anna Karenina”). My father did not say a word either about
                                                                                      the oriented area of a rectangular or about any non-mathematicai
                                                                                      interpretation of signs and products.



                                                                                      This “algebraic” explanation was not able to shake either my hearty
                                                                                      love for my father or a deep respect of his science. But since that
                                                                                      time I have disliked the axiomatic method with its non-motivated
                                                                                      definitions. Probably it was for this reason that by this time I got
                                                                                      used to talking with non-algebraists (like L. I. Mandel’shtam, I. E.
                                                                                      Tamm, P. S. Novikov, E. L. Feinberg, M. A. Leontovich, and A. G.
                                                                                      Gurvich) who treated an ignorant interlocutor with full respect and
                                                                                      tried to explain non-trivial ideas and facts of various sciences such
                                                                                      as physics and biology, astronomy and radiolocation.



                                                                                      Negative numbers I came to understand a year later while deriving an
                                                                                      “equation of time”, which takes into account a correction for the
                                                                                      length of a day corresponding to the time of year. It is not possible
                                                                                      to explain to algebraists that their axiomatic method is mostly
                                                                                      useless for students.



                                                                                      One should ask children: at what time will high tide be tomorrow if
                                                                                      today it is at 3 pm? This is a feasible problem, and it helps children
                                                                                      to understand negative numbers better than algebraic rules do. Once I
                                                                                      read from an ancient author (probably from Herodotus) that the tides
                                                                                      "always occur three and nine o'clock". To understand that the monthly
                                                                                      rotation of the Moon about the Earth affects the tide timetable, there
                                                                                      is no need to live near an ocean. Here, not in axioms, is laid true
                                                                                      mathematics.







                                                                                      share|cite|improve this answer









                                                                                      $endgroup$



                                                                                      I prefer the explanation by my favorite mathematician , V. I. Arnold (physicist really, since in his own words, "mathematics is a part of physics" and "an experimental science"). I believe it's the most natural (yet totally mathematical) explanation of a basic notion like multiplication of negative numbers.



                                                                                      This is an excerpt from Arnold's wonderful memoir "Yesterday and Long Ago" (3d ed., available in English from Springer), full of world history, drama and ingenious storytelling. The 2007 translation into English, I believe, is not of best quality, but it's the only one so far.



                                                                                      from the short story The Arnold Family




                                                                                      I faced a real difficulty with school mathematics several years after
                                                                                      the multiplication table: it was necessary to leam that “minus
                                                                                      multiplied by minus is plus” I wanted to know the proof of this rule;
                                                                                      I have never been able to leam by heart what is not properly
                                                                                      understood. I asked my father to explain the reason why (—1) • (—2) =
                                                                                      (+2). He, being a student of great algebraists, S. O. Shatunovsky and
                                                                                      E. Noether, gave the following “scientific explanation”: “The point
                                                                                      is,” he said: “that numbers form a field such that the distributive
                                                                                      law (x+y)z=xz+yz holds. And if the product of minus by minus had not
                                                                                      been plus, this law would be broken”.



                                                                                      However, for me this “deductive” (actually juridical) explanation did
                                                                                      not prove anything - what of it! One can study any axioms! Since
                                                                                      that day I have preserved the healthy aversion of a naturalist to the
                                                                                      axiomatic method with its non-motivated axioms.



                                                                                      The axiomophile Rene Descartes stated that “neither experimental tests
                                                                                      that axioms reflect a reality, nor comparison of theoretical results
                                                                                      with reality should be a part of science” (why should results
                                                                                      correspond to reality if the initial principles do not correspond to
                                                                                      it?).



                                                                                      Another thesis of Descartes’ theory and methods of education is even
                                                                                      more peculiar and contemporary: “It is necessary to forbid all other
                                                                                      methods of teaching except mine because only this method is
                                                                                      politically correct
                                                                                      : with my purely deductive method any dull
                                                                                      student can be taught as successfully as the most gifted one
                                                                                      , while
                                                                                      with other methods imagination and even drawings are used
                                                                                      unavoidably, and for this reason geniuses advance faster
                                                                                      than dunces
                                                                                      ”.



                                                                                      Contrary to the deductive theories of my father and Descartes, as a
                                                                                      ten year old, I started thinking about a naturally-scientific sense
                                                                                      of the rule of signs
                                                                                      , and I have come to the following conclusion. A
                                                                                      real (positive or negative) number is a vector on the axis of
                                                                                      coordinates (if a number is positive the corresponding vector is
                                                                                      positively directed along this axis).



                                                                                      A product of two numbers is an area of a rectangle whose sides
                                                                                      correspond to these numbers (one vector is along one axis and the
                                                                                      other is along a perpendicular axis in the plane). A rectangle, given
                                                                                      by an ordered pair of vectors, possesses, as a part of the plane, a
                                                                                      definite orientation (rotation from one vector to another can be
                                                                                      clockwise or anti-clockwise). The anti-clockwise rotation is
                                                                                      customarily considered positive and the clockwise rotation is then
                                                                                      negative. And lastly, the area of a parallelogram (for example, a
                                                                                      rectangle) generated by the two vectors x and у (taken in a
                                                                                      definite order) emanating from one point in the plane is considered
                                                                                      to be positive
                                                                                      if the pair of vectors (taken in this order) defines
                                                                                      positive orientation of the plane (and negative if the rotation from the direction of the vector x to the direction of the vector у is
                                                                                      negative).



                                                                                      Thus, the rule of signs is not an axiom taken out of the blue, but
                                                                                      becomes a natural property of orientation which is easily verified
                                                                                      experimentally.




                                                                                      from the short story Axiomatic Method




                                                                                      My first trouble in school was caused by the rule for multiplication
                                                                                      of negative numbers, and I asked my father to explain this peculiar
                                                                                      rule.



                                                                                      He, as a faithful student of Emmy Noether (and consequently of Hilbert
                                                                                      and Dedekind) started explaining to his eleven-year-old son the
                                                                                      principles of axiomatic science: a definition is chosen such that the
                                                                                      distributive identity a(b+c)=ab+ac holds.



                                                                                      The axiomatic method requires that one should accept any axiom with a
                                                                                      hope that its corollaries would be fruitful (probably this could be
                                                                                      understood by the age of thirty when it would be possible to read and
                                                                                      appreciate “Anna Karenina”). My father did not say a word either about
                                                                                      the oriented area of a rectangular or about any non-mathematicai
                                                                                      interpretation of signs and products.



                                                                                      This “algebraic” explanation was not able to shake either my hearty
                                                                                      love for my father or a deep respect of his science. But since that
                                                                                      time I have disliked the axiomatic method with its non-motivated
                                                                                      definitions. Probably it was for this reason that by this time I got
                                                                                      used to talking with non-algebraists (like L. I. Mandel’shtam, I. E.
                                                                                      Tamm, P. S. Novikov, E. L. Feinberg, M. A. Leontovich, and A. G.
                                                                                      Gurvich) who treated an ignorant interlocutor with full respect and
                                                                                      tried to explain non-trivial ideas and facts of various sciences such
                                                                                      as physics and biology, astronomy and radiolocation.



                                                                                      Negative numbers I came to understand a year later while deriving an
                                                                                      “equation of time”, which takes into account a correction for the
                                                                                      length of a day corresponding to the time of year. It is not possible
                                                                                      to explain to algebraists that their axiomatic method is mostly
                                                                                      useless for students.



                                                                                      One should ask children: at what time will high tide be tomorrow if
                                                                                      today it is at 3 pm? This is a feasible problem, and it helps children
                                                                                      to understand negative numbers better than algebraic rules do. Once I
                                                                                      read from an ancient author (probably from Herodotus) that the tides
                                                                                      "always occur three and nine o'clock". To understand that the monthly
                                                                                      rotation of the Moon about the Earth affects the tide timetable, there
                                                                                      is no need to live near an ocean. Here, not in axioms, is laid true
                                                                                      mathematics.








                                                                                      share|cite|improve this answer












                                                                                      share|cite|improve this answer



                                                                                      share|cite|improve this answer










                                                                                      answered May 11 '15 at 15:53









                                                                                      level1807level1807

                                                                                      817815




                                                                                      817815












                                                                                      • $begingroup$
                                                                                        2 very good and insightful anecdotes-although I wouldn't really call either of them PROOFS, would anyone here?They're more intuitive explanations why it's so.Of course, the natural next question is:Can either of these arguments be precisely reformulated as proofs?
                                                                                        $endgroup$
                                                                                        – Mathemagician1234
                                                                                        Jun 13 '15 at 3:21








                                                                                      • 1




                                                                                        $begingroup$
                                                                                        @Mathemagician1234 you know, I agree with Arnold on the idea that explanations or intuition like this are much much more powerful than any proofs or definitions, because they are real and without artificial restrictions of formal systems. Using them, one can come up with a correct system of definitions in no time, if they have to. However, having the definitions in the first place does not add anything to one's true understanding of the subject.
                                                                                        $endgroup$
                                                                                        – level1807
                                                                                        Jun 13 '15 at 7:36










                                                                                      • $begingroup$
                                                                                        With the interpretation in terms of oriented areas we would have $xy=-yx$ which is kind of problematic.
                                                                                        $endgroup$
                                                                                        – Sergio
                                                                                        Feb 2 '18 at 22:55


















                                                                                      • $begingroup$
                                                                                        2 very good and insightful anecdotes-although I wouldn't really call either of them PROOFS, would anyone here?They're more intuitive explanations why it's so.Of course, the natural next question is:Can either of these arguments be precisely reformulated as proofs?
                                                                                        $endgroup$
                                                                                        – Mathemagician1234
                                                                                        Jun 13 '15 at 3:21








                                                                                      • 1




                                                                                        $begingroup$
                                                                                        @Mathemagician1234 you know, I agree with Arnold on the idea that explanations or intuition like this are much much more powerful than any proofs or definitions, because they are real and without artificial restrictions of formal systems. Using them, one can come up with a correct system of definitions in no time, if they have to. However, having the definitions in the first place does not add anything to one's true understanding of the subject.
                                                                                        $endgroup$
                                                                                        – level1807
                                                                                        Jun 13 '15 at 7:36










                                                                                      • $begingroup$
                                                                                        With the interpretation in terms of oriented areas we would have $xy=-yx$ which is kind of problematic.
                                                                                        $endgroup$
                                                                                        – Sergio
                                                                                        Feb 2 '18 at 22:55
















                                                                                      $begingroup$
                                                                                      2 very good and insightful anecdotes-although I wouldn't really call either of them PROOFS, would anyone here?They're more intuitive explanations why it's so.Of course, the natural next question is:Can either of these arguments be precisely reformulated as proofs?
                                                                                      $endgroup$
                                                                                      – Mathemagician1234
                                                                                      Jun 13 '15 at 3:21






                                                                                      $begingroup$
                                                                                      2 very good and insightful anecdotes-although I wouldn't really call either of them PROOFS, would anyone here?They're more intuitive explanations why it's so.Of course, the natural next question is:Can either of these arguments be precisely reformulated as proofs?
                                                                                      $endgroup$
                                                                                      – Mathemagician1234
                                                                                      Jun 13 '15 at 3:21






                                                                                      1




                                                                                      1




                                                                                      $begingroup$
                                                                                      @Mathemagician1234 you know, I agree with Arnold on the idea that explanations or intuition like this are much much more powerful than any proofs or definitions, because they are real and without artificial restrictions of formal systems. Using them, one can come up with a correct system of definitions in no time, if they have to. However, having the definitions in the first place does not add anything to one's true understanding of the subject.
                                                                                      $endgroup$
                                                                                      – level1807
                                                                                      Jun 13 '15 at 7:36




                                                                                      $begingroup$
                                                                                      @Mathemagician1234 you know, I agree with Arnold on the idea that explanations or intuition like this are much much more powerful than any proofs or definitions, because they are real and without artificial restrictions of formal systems. Using them, one can come up with a correct system of definitions in no time, if they have to. However, having the definitions in the first place does not add anything to one's true understanding of the subject.
                                                                                      $endgroup$
                                                                                      – level1807
                                                                                      Jun 13 '15 at 7:36












                                                                                      $begingroup$
                                                                                      With the interpretation in terms of oriented areas we would have $xy=-yx$ which is kind of problematic.
                                                                                      $endgroup$
                                                                                      – Sergio
                                                                                      Feb 2 '18 at 22:55




                                                                                      $begingroup$
                                                                                      With the interpretation in terms of oriented areas we would have $xy=-yx$ which is kind of problematic.
                                                                                      $endgroup$
                                                                                      – Sergio
                                                                                      Feb 2 '18 at 22:55











                                                                                      8












                                                                                      $begingroup$

                                                                                      Perhaps some intuition can be gained by plotting each number's position on the number line. Taking the inverse of any number is visualized by taking the mirror-image of the original plot. So the inverse of a positive number (a point to the right of zero) is a negative number (a point to the left of zero, at the same distance from zero). Likewise, the inverse of a negative number is a positive number. If we agree that multiplying a number by -1 is the same as finding the inverse, then we can see that the product of two negatives must be a positive, because the mirror-image of a mirror-image is the original image.






                                                                                      share|cite|improve this answer









                                                                                      $endgroup$


















                                                                                        8












                                                                                        $begingroup$

                                                                                        Perhaps some intuition can be gained by plotting each number's position on the number line. Taking the inverse of any number is visualized by taking the mirror-image of the original plot. So the inverse of a positive number (a point to the right of zero) is a negative number (a point to the left of zero, at the same distance from zero). Likewise, the inverse of a negative number is a positive number. If we agree that multiplying a number by -1 is the same as finding the inverse, then we can see that the product of two negatives must be a positive, because the mirror-image of a mirror-image is the original image.






                                                                                        share|cite|improve this answer









                                                                                        $endgroup$
















                                                                                          8












                                                                                          8








                                                                                          8





                                                                                          $begingroup$

                                                                                          Perhaps some intuition can be gained by plotting each number's position on the number line. Taking the inverse of any number is visualized by taking the mirror-image of the original plot. So the inverse of a positive number (a point to the right of zero) is a negative number (a point to the left of zero, at the same distance from zero). Likewise, the inverse of a negative number is a positive number. If we agree that multiplying a number by -1 is the same as finding the inverse, then we can see that the product of two negatives must be a positive, because the mirror-image of a mirror-image is the original image.






                                                                                          share|cite|improve this answer









                                                                                          $endgroup$



                                                                                          Perhaps some intuition can be gained by plotting each number's position on the number line. Taking the inverse of any number is visualized by taking the mirror-image of the original plot. So the inverse of a positive number (a point to the right of zero) is a negative number (a point to the left of zero, at the same distance from zero). Likewise, the inverse of a negative number is a positive number. If we agree that multiplying a number by -1 is the same as finding the inverse, then we can see that the product of two negatives must be a positive, because the mirror-image of a mirror-image is the original image.







                                                                                          share|cite|improve this answer












                                                                                          share|cite|improve this answer



                                                                                          share|cite|improve this answer










                                                                                          answered Nov 12 '10 at 21:02









                                                                                          o. nateo. nate

                                                                                          21613




                                                                                          21613























                                                                                              8












                                                                                              $begingroup$

                                                                                              As for the product of two negatives being a positive, simply consider the multiplicative inverse:



                                                                                              $$-acdot -b$$
                                                                                              $$(-1)acdot (-1)b$$
                                                                                              $$(-1)(-1)acdot b$$



                                                                                              Note that $(-1)^{-1} = -1$. Any number times its multiplicative inverse is $1$.



                                                                                              $$(-1)(-1)^{-1}acdot b$$
                                                                                              $$(1)acdot b$$
                                                                                              $$=acdot b$$






                                                                                              share|cite|improve this answer









                                                                                              $endgroup$









                                                                                              • 1




                                                                                                $begingroup$
                                                                                                Great answer! Can you explain how you came up with this? I mean it's pretty self explanatory that a number multiplied by it's multiplicative inverse is 1 but using it here is pretty clever. How did you think to use it here?
                                                                                                $endgroup$
                                                                                                – user500668
                                                                                                Jul 28 '18 at 13:27










                                                                                              • $begingroup$
                                                                                                @user500668 Good question. I used to be clever once upon a time. I believe at the time I was just making use of multiplicative inverses while solving "contest style" math problems.
                                                                                                $endgroup$
                                                                                                – daOnlyBG
                                                                                                Oct 22 '18 at 14:53










                                                                                              • $begingroup$
                                                                                                For what it's worth, I wish whoever gave me a "-1" score would have the decency to justify their action.
                                                                                                $endgroup$
                                                                                                – daOnlyBG
                                                                                                Oct 22 '18 at 14:54
















                                                                                              8












                                                                                              $begingroup$

                                                                                              As for the product of two negatives being a positive, simply consider the multiplicative inverse:



                                                                                              $$-acdot -b$$
                                                                                              $$(-1)acdot (-1)b$$
                                                                                              $$(-1)(-1)acdot b$$



                                                                                              Note that $(-1)^{-1} = -1$. Any number times its multiplicative inverse is $1$.



                                                                                              $$(-1)(-1)^{-1}acdot b$$
                                                                                              $$(1)acdot b$$
                                                                                              $$=acdot b$$






                                                                                              share|cite|improve this answer









                                                                                              $endgroup$









                                                                                              • 1




                                                                                                $begingroup$
                                                                                                Great answer! Can you explain how you came up with this? I mean it's pretty self explanatory that a number multiplied by it's multiplicative inverse is 1 but using it here is pretty clever. How did you think to use it here?
                                                                                                $endgroup$
                                                                                                – user500668
                                                                                                Jul 28 '18 at 13:27










                                                                                              • $begingroup$
                                                                                                @user500668 Good question. I used to be clever once upon a time. I believe at the time I was just making use of multiplicative inverses while solving "contest style" math problems.
                                                                                                $endgroup$
                                                                                                – daOnlyBG
                                                                                                Oct 22 '18 at 14:53










                                                                                              • $begingroup$
                                                                                                For what it's worth, I wish whoever gave me a "-1" score would have the decency to justify their action.
                                                                                                $endgroup$
                                                                                                – daOnlyBG
                                                                                                Oct 22 '18 at 14:54














                                                                                              8












                                                                                              8








                                                                                              8





                                                                                              $begingroup$

                                                                                              As for the product of two negatives being a positive, simply consider the multiplicative inverse:



                                                                                              $$-acdot -b$$
                                                                                              $$(-1)acdot (-1)b$$
                                                                                              $$(-1)(-1)acdot b$$



                                                                                              Note that $(-1)^{-1} = -1$. Any number times its multiplicative inverse is $1$.



                                                                                              $$(-1)(-1)^{-1}acdot b$$
                                                                                              $$(1)acdot b$$
                                                                                              $$=acdot b$$






                                                                                              share|cite|improve this answer









                                                                                              $endgroup$



                                                                                              As for the product of two negatives being a positive, simply consider the multiplicative inverse:



                                                                                              $$-acdot -b$$
                                                                                              $$(-1)acdot (-1)b$$
                                                                                              $$(-1)(-1)acdot b$$



                                                                                              Note that $(-1)^{-1} = -1$. Any number times its multiplicative inverse is $1$.



                                                                                              $$(-1)(-1)^{-1}acdot b$$
                                                                                              $$(1)acdot b$$
                                                                                              $$=acdot b$$







                                                                                              share|cite|improve this answer












                                                                                              share|cite|improve this answer



                                                                                              share|cite|improve this answer










                                                                                              answered Nov 19 '14 at 23:59









                                                                                              daOnlyBGdaOnlyBG

                                                                                              2,31371734




                                                                                              2,31371734








                                                                                              • 1




                                                                                                $begingroup$
                                                                                                Great answer! Can you explain how you came up with this? I mean it's pretty self explanatory that a number multiplied by it's multiplicative inverse is 1 but using it here is pretty clever. How did you think to use it here?
                                                                                                $endgroup$
                                                                                                – user500668
                                                                                                Jul 28 '18 at 13:27










                                                                                              • $begingroup$
                                                                                                @user500668 Good question. I used to be clever once upon a time. I believe at the time I was just making use of multiplicative inverses while solving "contest style" math problems.
                                                                                                $endgroup$
                                                                                                – daOnlyBG
                                                                                                Oct 22 '18 at 14:53










                                                                                              • $begingroup$
                                                                                                For what it's worth, I wish whoever gave me a "-1" score would have the decency to justify their action.
                                                                                                $endgroup$
                                                                                                – daOnlyBG
                                                                                                Oct 22 '18 at 14:54














                                                                                              • 1




                                                                                                $begingroup$
                                                                                                Great answer! Can you explain how you came up with this? I mean it's pretty self explanatory that a number multiplied by it's multiplicative inverse is 1 but using it here is pretty clever. How did you think to use it here?
                                                                                                $endgroup$
                                                                                                – user500668
                                                                                                Jul 28 '18 at 13:27










                                                                                              • $begingroup$
                                                                                                @user500668 Good question. I used to be clever once upon a time. I believe at the time I was just making use of multiplicative inverses while solving "contest style" math problems.
                                                                                                $endgroup$
                                                                                                – daOnlyBG
                                                                                                Oct 22 '18 at 14:53










                                                                                              • $begingroup$
                                                                                                For what it's worth, I wish whoever gave me a "-1" score would have the decency to justify their action.
                                                                                                $endgroup$
                                                                                                – daOnlyBG
                                                                                                Oct 22 '18 at 14:54








                                                                                              1




                                                                                              1




                                                                                              $begingroup$
                                                                                              Great answer! Can you explain how you came up with this? I mean it's pretty self explanatory that a number multiplied by it's multiplicative inverse is 1 but using it here is pretty clever. How did you think to use it here?
                                                                                              $endgroup$
                                                                                              – user500668
                                                                                              Jul 28 '18 at 13:27




                                                                                              $begingroup$
                                                                                              Great answer! Can you explain how you came up with this? I mean it's pretty self explanatory that a number multiplied by it's multiplicative inverse is 1 but using it here is pretty clever. How did you think to use it here?
                                                                                              $endgroup$
                                                                                              – user500668
                                                                                              Jul 28 '18 at 13:27












                                                                                              $begingroup$
                                                                                              @user500668 Good question. I used to be clever once upon a time. I believe at the time I was just making use of multiplicative inverses while solving "contest style" math problems.
                                                                                              $endgroup$
                                                                                              – daOnlyBG
                                                                                              Oct 22 '18 at 14:53




                                                                                              $begingroup$
                                                                                              @user500668 Good question. I used to be clever once upon a time. I believe at the time I was just making use of multiplicative inverses while solving "contest style" math problems.
                                                                                              $endgroup$
                                                                                              – daOnlyBG
                                                                                              Oct 22 '18 at 14:53












                                                                                              $begingroup$
                                                                                              For what it's worth, I wish whoever gave me a "-1" score would have the decency to justify their action.
                                                                                              $endgroup$
                                                                                              – daOnlyBG
                                                                                              Oct 22 '18 at 14:54




                                                                                              $begingroup$
                                                                                              For what it's worth, I wish whoever gave me a "-1" score would have the decency to justify their action.
                                                                                              $endgroup$
                                                                                              – daOnlyBG
                                                                                              Oct 22 '18 at 14:54











                                                                                              7












                                                                                              $begingroup$

                                                                                              A bit late, but I have written a formal version of Jonas' proof here. The "applying this twice" was the only tricky part.



                                                                                              See Lemma 7 starting at line 224 in http://www.dcproof.com/FieldLemmas.htm






                                                                                              share|cite|improve this answer











                                                                                              $endgroup$


















                                                                                                7












                                                                                                $begingroup$

                                                                                                A bit late, but I have written a formal version of Jonas' proof here. The "applying this twice" was the only tricky part.



                                                                                                See Lemma 7 starting at line 224 in http://www.dcproof.com/FieldLemmas.htm






                                                                                                share|cite|improve this answer











                                                                                                $endgroup$
















                                                                                                  7












                                                                                                  7








                                                                                                  7





                                                                                                  $begingroup$

                                                                                                  A bit late, but I have written a formal version of Jonas' proof here. The "applying this twice" was the only tricky part.



                                                                                                  See Lemma 7 starting at line 224 in http://www.dcproof.com/FieldLemmas.htm






                                                                                                  share|cite|improve this answer











                                                                                                  $endgroup$



                                                                                                  A bit late, but I have written a formal version of Jonas' proof here. The "applying this twice" was the only tricky part.



                                                                                                  See Lemma 7 starting at line 224 in http://www.dcproof.com/FieldLemmas.htm







                                                                                                  share|cite|improve this answer














                                                                                                  share|cite|improve this answer



                                                                                                  share|cite|improve this answer








                                                                                                  edited Jun 5 '12 at 4:40

























                                                                                                  answered Jun 5 '12 at 4:29









                                                                                                  Dan ChristensenDan Christensen

                                                                                                  8,74321835




                                                                                                  8,74321835























                                                                                                      7












                                                                                                      $begingroup$

                                                                                                      I would explain it by number patterns.



                                                                                                      First, to establish that a positive times a negative is negative: $3 times 2 = 6, 3 times 1 = 3, 3 times 0 = 0$. Notice in each case, as we reduce the second factor by 1, the product is being reduced by 3. So for consistency the next product in the pattern must be $0 - 3 = -3$. Therefore we have $3 times (-1) = -3, 3 times (-2) = -6$, and likewise a negative for any other other positive times a negative.



                                                                                                      Second, to establish that a negative times a negative is positive: we now know that $3 times (-2) = -6, 2 times (-2) = -4, 1 times (-2) = -2, 0 times (-2) = 0$. Notice in each case, as we reduce the first multiplier by 1, the product is being increased by 2. So for consistency the next product in the pattern must be $0 + 2 = 2$. Therefore we have $(-1) times (-2) = 2, (-2) times (-2) = 4$, and likewise a positive for any other other negative times a negative.






                                                                                                      share|cite|improve this answer









                                                                                                      $endgroup$













                                                                                                      • $begingroup$
                                                                                                        I tried explaining this to him and he looked even more confused than when I narrated the other examples, I just think this is a bit too complicated for someone like him to understand yet.
                                                                                                        $endgroup$
                                                                                                        – yuritsuki
                                                                                                        Nov 9 '15 at 7:20










                                                                                                      • $begingroup$
                                                                                                        Funny - this is the second argument that I typically present to elementary teachers and their students after my cops and robbers story! Note that it has been presented on this site before as well.
                                                                                                        $endgroup$
                                                                                                        – Mark McClure
                                                                                                        Nov 9 '15 at 15:16
















                                                                                                      7












                                                                                                      $begingroup$

                                                                                                      I would explain it by number patterns.



                                                                                                      First, to establish that a positive times a negative is negative: $3 times 2 = 6, 3 times 1 = 3, 3 times 0 = 0$. Notice in each case, as we reduce the second factor by 1, the product is being reduced by 3. So for consistency the next product in the pattern must be $0 - 3 = -3$. Therefore we have $3 times (-1) = -3, 3 times (-2) = -6$, and likewise a negative for any other other positive times a negative.



                                                                                                      Second, to establish that a negative times a negative is positive: we now know that $3 times (-2) = -6, 2 times (-2) = -4, 1 times (-2) = -2, 0 times (-2) = 0$. Notice in each case, as we reduce the first multiplier by 1, the product is being increased by 2. So for consistency the next product in the pattern must be $0 + 2 = 2$. Therefore we have $(-1) times (-2) = 2, (-2) times (-2) = 4$, and likewise a positive for any other other negative times a negative.






                                                                                                      share|cite|improve this answer









                                                                                                      $endgroup$













                                                                                                      • $begingroup$
                                                                                                        I tried explaining this to him and he looked even more confused than when I narrated the other examples, I just think this is a bit too complicated for someone like him to understand yet.
                                                                                                        $endgroup$
                                                                                                        – yuritsuki
                                                                                                        Nov 9 '15 at 7:20










                                                                                                      • $begingroup$
                                                                                                        Funny - this is the second argument that I typically present to elementary teachers and their students after my cops and robbers story! Note that it has been presented on this site before as well.
                                                                                                        $endgroup$
                                                                                                        – Mark McClure
                                                                                                        Nov 9 '15 at 15:16














                                                                                                      7












                                                                                                      7








                                                                                                      7





                                                                                                      $begingroup$

                                                                                                      I would explain it by number patterns.



                                                                                                      First, to establish that a positive times a negative is negative: $3 times 2 = 6, 3 times 1 = 3, 3 times 0 = 0$. Notice in each case, as we reduce the second factor by 1, the product is being reduced by 3. So for consistency the next product in the pattern must be $0 - 3 = -3$. Therefore we have $3 times (-1) = -3, 3 times (-2) = -6$, and likewise a negative for any other other positive times a negative.



                                                                                                      Second, to establish that a negative times a negative is positive: we now know that $3 times (-2) = -6, 2 times (-2) = -4, 1 times (-2) = -2, 0 times (-2) = 0$. Notice in each case, as we reduce the first multiplier by 1, the product is being increased by 2. So for consistency the next product in the pattern must be $0 + 2 = 2$. Therefore we have $(-1) times (-2) = 2, (-2) times (-2) = 4$, and likewise a positive for any other other negative times a negative.






                                                                                                      share|cite|improve this answer









                                                                                                      $endgroup$



                                                                                                      I would explain it by number patterns.



                                                                                                      First, to establish that a positive times a negative is negative: $3 times 2 = 6, 3 times 1 = 3, 3 times 0 = 0$. Notice in each case, as we reduce the second factor by 1, the product is being reduced by 3. So for consistency the next product in the pattern must be $0 - 3 = -3$. Therefore we have $3 times (-1) = -3, 3 times (-2) = -6$, and likewise a negative for any other other positive times a negative.



                                                                                                      Second, to establish that a negative times a negative is positive: we now know that $3 times (-2) = -6, 2 times (-2) = -4, 1 times (-2) = -2, 0 times (-2) = 0$. Notice in each case, as we reduce the first multiplier by 1, the product is being increased by 2. So for consistency the next product in the pattern must be $0 + 2 = 2$. Therefore we have $(-1) times (-2) = 2, (-2) times (-2) = 4$, and likewise a positive for any other other negative times a negative.







                                                                                                      share|cite|improve this answer












                                                                                                      share|cite|improve this answer



                                                                                                      share|cite|improve this answer










                                                                                                      answered Nov 9 '15 at 3:36









                                                                                                      Daniel R. CollinsDaniel R. Collins

                                                                                                      5,9251534




                                                                                                      5,9251534












                                                                                                      • $begingroup$
                                                                                                        I tried explaining this to him and he looked even more confused than when I narrated the other examples, I just think this is a bit too complicated for someone like him to understand yet.
                                                                                                        $endgroup$
                                                                                                        – yuritsuki
                                                                                                        Nov 9 '15 at 7:20










                                                                                                      • $begingroup$
                                                                                                        Funny - this is the second argument that I typically present to elementary teachers and their students after my cops and robbers story! Note that it has been presented on this site before as well.
                                                                                                        $endgroup$
                                                                                                        – Mark McClure
                                                                                                        Nov 9 '15 at 15:16


















                                                                                                      • $begingroup$
                                                                                                        I tried explaining this to him and he looked even more confused than when I narrated the other examples, I just think this is a bit too complicated for someone like him to understand yet.
                                                                                                        $endgroup$
                                                                                                        – yuritsuki
                                                                                                        Nov 9 '15 at 7:20










                                                                                                      • $begingroup$
                                                                                                        Funny - this is the second argument that I typically present to elementary teachers and their students after my cops and robbers story! Note that it has been presented on this site before as well.
                                                                                                        $endgroup$
                                                                                                        – Mark McClure
                                                                                                        Nov 9 '15 at 15:16
















                                                                                                      $begingroup$
                                                                                                      I tried explaining this to him and he looked even more confused than when I narrated the other examples, I just think this is a bit too complicated for someone like him to understand yet.
                                                                                                      $endgroup$
                                                                                                      – yuritsuki
                                                                                                      Nov 9 '15 at 7:20




                                                                                                      $begingroup$
                                                                                                      I tried explaining this to him and he looked even more confused than when I narrated the other examples, I just think this is a bit too complicated for someone like him to understand yet.
                                                                                                      $endgroup$
                                                                                                      – yuritsuki
                                                                                                      Nov 9 '15 at 7:20












                                                                                                      $begingroup$
                                                                                                      Funny - this is the second argument that I typically present to elementary teachers and their students after my cops and robbers story! Note that it has been presented on this site before as well.
                                                                                                      $endgroup$
                                                                                                      – Mark McClure
                                                                                                      Nov 9 '15 at 15:16




                                                                                                      $begingroup$
                                                                                                      Funny - this is the second argument that I typically present to elementary teachers and their students after my cops and robbers story! Note that it has been presented on this site before as well.
                                                                                                      $endgroup$
                                                                                                      – Mark McClure
                                                                                                      Nov 9 '15 at 15:16











                                                                                                      7












                                                                                                      $begingroup$

                                                                                                      I would go for the flipping explanation of the negative numbers: multiplying with a negative number flips from positive to negative and from negative to positive.

                                                                                                      Imagine he understands that multiplying with 1 makes no difference, then it's very simple:

                                                                                                      -1 * 1 = -1 can mean two things (for children, the fact that multiplication is commutative is obvious):

                                                                                                      - either it keeps -1 as -1

                                                                                                      - either it flips +1 (positive number) to the negative side



                                                                                                      -1 * (-1) then simply flips it back from the negative to the positive side.



                                                                                                      Good luck






                                                                                                      share|cite|improve this answer











                                                                                                      $endgroup$


















                                                                                                        7












                                                                                                        $begingroup$

                                                                                                        I would go for the flipping explanation of the negative numbers: multiplying with a negative number flips from positive to negative and from negative to positive.

                                                                                                        Imagine he understands that multiplying with 1 makes no difference, then it's very simple:

                                                                                                        -1 * 1 = -1 can mean two things (for children, the fact that multiplication is commutative is obvious):

                                                                                                        - either it keeps -1 as -1

                                                                                                        - either it flips +1 (positive number) to the negative side



                                                                                                        -1 * (-1) then simply flips it back from the negative to the positive side.



                                                                                                        Good luck






                                                                                                        share|cite|improve this answer











                                                                                                        $endgroup$
















                                                                                                          7












                                                                                                          7








                                                                                                          7





                                                                                                          $begingroup$

                                                                                                          I would go for the flipping explanation of the negative numbers: multiplying with a negative number flips from positive to negative and from negative to positive.

                                                                                                          Imagine he understands that multiplying with 1 makes no difference, then it's very simple:

                                                                                                          -1 * 1 = -1 can mean two things (for children, the fact that multiplication is commutative is obvious):

                                                                                                          - either it keeps -1 as -1

                                                                                                          - either it flips +1 (positive number) to the negative side



                                                                                                          -1 * (-1) then simply flips it back from the negative to the positive side.



                                                                                                          Good luck






                                                                                                          share|cite|improve this answer











                                                                                                          $endgroup$



                                                                                                          I would go for the flipping explanation of the negative numbers: multiplying with a negative number flips from positive to negative and from negative to positive.

                                                                                                          Imagine he understands that multiplying with 1 makes no difference, then it's very simple:

                                                                                                          -1 * 1 = -1 can mean two things (for children, the fact that multiplication is commutative is obvious):

                                                                                                          - either it keeps -1 as -1

                                                                                                          - either it flips +1 (positive number) to the negative side



                                                                                                          -1 * (-1) then simply flips it back from the negative to the positive side.



                                                                                                          Good luck







                                                                                                          share|cite|improve this answer














                                                                                                          share|cite|improve this answer



                                                                                                          share|cite|improve this answer








                                                                                                          edited Nov 9 '15 at 19:53









                                                                                                          BLAZE

                                                                                                          6,164112857




                                                                                                          6,164112857










                                                                                                          answered Nov 9 '15 at 8:34









                                                                                                          DominiqueDominique

                                                                                                          31627




                                                                                                          31627























                                                                                                              6












                                                                                                              $begingroup$

                                                                                                              It might be easiest to explain using whole numbers. Suppose $P$ is some positive number. Then $-P$ is negative. Now $-2P$ is $P$ subtracted from $-P$, so is still negative. Subtract another $P$ and you get $-3P$, which is still negative. Similarly for $-4P, -5P$, and so on. So negative times positive is positive. Same idea for positive times negative.



                                                                                                              When it comes to negative times negative, it's a little harder... But how about... $-P$ is negative, so $-(-P)$ is now positive, flipping around $0$. So $-2(-P)$, which is $-(-P)$ added to itself, is still positive. In general adding $-(-P)$ to itself $Q$ times gives $(-Q)(-P)$, which is therefore positive as well.






                                                                                                              share|cite|improve this answer









                                                                                                              $endgroup$


















                                                                                                                6












                                                                                                                $begingroup$

                                                                                                                It might be easiest to explain using whole numbers. Suppose $P$ is some positive number. Then $-P$ is negative. Now $-2P$ is $P$ subtracted from $-P$, so is still negative. Subtract another $P$ and you get $-3P$, which is still negative. Similarly for $-4P, -5P$, and so on. So negative times positive is positive. Same idea for positive times negative.



                                                                                                                When it comes to negative times negative, it's a little harder... But how about... $-P$ is negative, so $-(-P)$ is now positive, flipping around $0$. So $-2(-P)$, which is $-(-P)$ added to itself, is still positive. In general adding $-(-P)$ to itself $Q$ times gives $(-Q)(-P)$, which is therefore positive as well.






                                                                                                                share|cite|improve this answer









                                                                                                                $endgroup$
















                                                                                                                  6












                                                                                                                  6








                                                                                                                  6





                                                                                                                  $begingroup$

                                                                                                                  It might be easiest to explain using whole numbers. Suppose $P$ is some positive number. Then $-P$ is negative. Now $-2P$ is $P$ subtracted from $-P$, so is still negative. Subtract another $P$ and you get $-3P$, which is still negative. Similarly for $-4P, -5P$, and so on. So negative times positive is positive. Same idea for positive times negative.



                                                                                                                  When it comes to negative times negative, it's a little harder... But how about... $-P$ is negative, so $-(-P)$ is now positive, flipping around $0$. So $-2(-P)$, which is $-(-P)$ added to itself, is still positive. In general adding $-(-P)$ to itself $Q$ times gives $(-Q)(-P)$, which is therefore positive as well.






                                                                                                                  share|cite|improve this answer









                                                                                                                  $endgroup$



                                                                                                                  It might be easiest to explain using whole numbers. Suppose $P$ is some positive number. Then $-P$ is negative. Now $-2P$ is $P$ subtracted from $-P$, so is still negative. Subtract another $P$ and you get $-3P$, which is still negative. Similarly for $-4P, -5P$, and so on. So negative times positive is positive. Same idea for positive times negative.



                                                                                                                  When it comes to negative times negative, it's a little harder... But how about... $-P$ is negative, so $-(-P)$ is now positive, flipping around $0$. So $-2(-P)$, which is $-(-P)$ added to itself, is still positive. In general adding $-(-P)$ to itself $Q$ times gives $(-Q)(-P)$, which is therefore positive as well.







                                                                                                                  share|cite|improve this answer












                                                                                                                  share|cite|improve this answer



                                                                                                                  share|cite|improve this answer










                                                                                                                  answered Nov 12 '10 at 21:40









                                                                                                                  ZarraxZarrax

                                                                                                                  35.7k250104




                                                                                                                  35.7k250104























                                                                                                                      6












                                                                                                                      $begingroup$

                                                                                                                      If your son is clear on the concept of money and knows what a credit card is, this might be a good explanation:



                                                                                                                      Imagine you tell your son that you will buy him $color{green}{mathrm{seven}}$ gift vouchers worth £$color{green}{5}$ each and pay for them using your credit card. Explain to your son that you now owe money, and say that it is $7 times -5=-mathrm{£}35$. Not part of your question; but this covers the $ text{positive} times text{negative} = text{negative}$ case.



                                                                                                                      You are having dinner with your best friend when the bill arrives from the credit card company, when your friend sees the bill he generously insists on paying it for you. You now have £$color{blue}{35}$ worth of gift vouchers without having paid anything.



                                                                                                                      So you tell your son that your best friend $color{red}{fbox{took-away}}$ seven $color{red}{fbox{debts}}$ of £$5$ ($color{red}{-7}timescolor{red}{-5}$) and this equals a gain of £$color{blue}{35}$.






                                                                                                                      share|cite|improve this answer











                                                                                                                      $endgroup$









                                                                                                                      • 2




                                                                                                                        $begingroup$
                                                                                                                        (+1) I've seen this taught particularly effectively by drawing the money in a "bag" (I imagine physical cards with + or - numbers on work well for the same reason). The basic premise is that if a bag contains 7, -3, and 5 (total = 9) we see that "taking away" or "crossing out" the -3 makes the new total 12, so "taking away -3 is the same as adding 3". If the bag contains two 6s, among other things, we can easily taking that "minus two lots of six from the bag is the same as minus 12 from the total". Or if the bag contains two -5s, among others, we see "minus two lots of -5 is the same as +10"
                                                                                                                        $endgroup$
                                                                                                                        – Silverfish
                                                                                                                        Nov 9 '15 at 21:17
















                                                                                                                      6












                                                                                                                      $begingroup$

                                                                                                                      If your son is clear on the concept of money and knows what a credit card is, this might be a good explanation:



                                                                                                                      Imagine you tell your son that you will buy him $color{green}{mathrm{seven}}$ gift vouchers worth £$color{green}{5}$ each and pay for them using your credit card. Explain to your son that you now owe money, and say that it is $7 times -5=-mathrm{£}35$. Not part of your question; but this covers the $ text{positive} times text{negative} = text{negative}$ case.



                                                                                                                      You are having dinner with your best friend when the bill arrives from the credit card company, when your friend sees the bill he generously insists on paying it for you. You now have £$color{blue}{35}$ worth of gift vouchers without having paid anything.



                                                                                                                      So you tell your son that your best friend $color{red}{fbox{took-away}}$ seven $color{red}{fbox{debts}}$ of £$5$ ($color{red}{-7}timescolor{red}{-5}$) and this equals a gain of £$color{blue}{35}$.






                                                                                                                      share|cite|improve this answer











                                                                                                                      $endgroup$









                                                                                                                      • 2




                                                                                                                        $begingroup$
                                                                                                                        (+1) I've seen this taught particularly effectively by drawing the money in a "bag" (I imagine physical cards with + or - numbers on work well for the same reason). The basic premise is that if a bag contains 7, -3, and 5 (total = 9) we see that "taking away" or "crossing out" the -3 makes the new total 12, so "taking away -3 is the same as adding 3". If the bag contains two 6s, among other things, we can easily taking that "minus two lots of six from the bag is the same as minus 12 from the total". Or if the bag contains two -5s, among others, we see "minus two lots of -5 is the same as +10"
                                                                                                                        $endgroup$
                                                                                                                        – Silverfish
                                                                                                                        Nov 9 '15 at 21:17














                                                                                                                      6












                                                                                                                      6








                                                                                                                      6





                                                                                                                      $begingroup$

                                                                                                                      If your son is clear on the concept of money and knows what a credit card is, this might be a good explanation:



                                                                                                                      Imagine you tell your son that you will buy him $color{green}{mathrm{seven}}$ gift vouchers worth £$color{green}{5}$ each and pay for them using your credit card. Explain to your son that you now owe money, and say that it is $7 times -5=-mathrm{£}35$. Not part of your question; but this covers the $ text{positive} times text{negative} = text{negative}$ case.



                                                                                                                      You are having dinner with your best friend when the bill arrives from the credit card company, when your friend sees the bill he generously insists on paying it for you. You now have £$color{blue}{35}$ worth of gift vouchers without having paid anything.



                                                                                                                      So you tell your son that your best friend $color{red}{fbox{took-away}}$ seven $color{red}{fbox{debts}}$ of £$5$ ($color{red}{-7}timescolor{red}{-5}$) and this equals a gain of £$color{blue}{35}$.






                                                                                                                      share|cite|improve this answer











                                                                                                                      $endgroup$



                                                                                                                      If your son is clear on the concept of money and knows what a credit card is, this might be a good explanation:



                                                                                                                      Imagine you tell your son that you will buy him $color{green}{mathrm{seven}}$ gift vouchers worth £$color{green}{5}$ each and pay for them using your credit card. Explain to your son that you now owe money, and say that it is $7 times -5=-mathrm{£}35$. Not part of your question; but this covers the $ text{positive} times text{negative} = text{negative}$ case.



                                                                                                                      You are having dinner with your best friend when the bill arrives from the credit card company, when your friend sees the bill he generously insists on paying it for you. You now have £$color{blue}{35}$ worth of gift vouchers without having paid anything.



                                                                                                                      So you tell your son that your best friend $color{red}{fbox{took-away}}$ seven $color{red}{fbox{debts}}$ of £$5$ ($color{red}{-7}timescolor{red}{-5}$) and this equals a gain of £$color{blue}{35}$.







                                                                                                                      share|cite|improve this answer














                                                                                                                      share|cite|improve this answer



                                                                                                                      share|cite|improve this answer








                                                                                                                      edited Nov 10 '15 at 1:16

























                                                                                                                      answered Nov 9 '15 at 9:57









                                                                                                                      BLAZEBLAZE

                                                                                                                      6,164112857




                                                                                                                      6,164112857








                                                                                                                      • 2




                                                                                                                        $begingroup$
                                                                                                                        (+1) I've seen this taught particularly effectively by drawing the money in a "bag" (I imagine physical cards with + or - numbers on work well for the same reason). The basic premise is that if a bag contains 7, -3, and 5 (total = 9) we see that "taking away" or "crossing out" the -3 makes the new total 12, so "taking away -3 is the same as adding 3". If the bag contains two 6s, among other things, we can easily taking that "minus two lots of six from the bag is the same as minus 12 from the total". Or if the bag contains two -5s, among others, we see "minus two lots of -5 is the same as +10"
                                                                                                                        $endgroup$
                                                                                                                        – Silverfish
                                                                                                                        Nov 9 '15 at 21:17














                                                                                                                      • 2




                                                                                                                        $begingroup$
                                                                                                                        (+1) I've seen this taught particularly effectively by drawing the money in a "bag" (I imagine physical cards with + or - numbers on work well for the same reason). The basic premise is that if a bag contains 7, -3, and 5 (total = 9) we see that "taking away" or "crossing out" the -3 makes the new total 12, so "taking away -3 is the same as adding 3". If the bag contains two 6s, among other things, we can easily taking that "minus two lots of six from the bag is the same as minus 12 from the total". Or if the bag contains two -5s, among others, we see "minus two lots of -5 is the same as +10"
                                                                                                                        $endgroup$
                                                                                                                        – Silverfish
                                                                                                                        Nov 9 '15 at 21:17








                                                                                                                      2




                                                                                                                      2




                                                                                                                      $begingroup$
                                                                                                                      (+1) I've seen this taught particularly effectively by drawing the money in a "bag" (I imagine physical cards with + or - numbers on work well for the same reason). The basic premise is that if a bag contains 7, -3, and 5 (total = 9) we see that "taking away" or "crossing out" the -3 makes the new total 12, so "taking away -3 is the same as adding 3". If the bag contains two 6s, among other things, we can easily taking that "minus two lots of six from the bag is the same as minus 12 from the total". Or if the bag contains two -5s, among others, we see "minus two lots of -5 is the same as +10"
                                                                                                                      $endgroup$
                                                                                                                      – Silverfish
                                                                                                                      Nov 9 '15 at 21:17




                                                                                                                      $begingroup$
                                                                                                                      (+1) I've seen this taught particularly effectively by drawing the money in a "bag" (I imagine physical cards with + or - numbers on work well for the same reason). The basic premise is that if a bag contains 7, -3, and 5 (total = 9) we see that "taking away" or "crossing out" the -3 makes the new total 12, so "taking away -3 is the same as adding 3". If the bag contains two 6s, among other things, we can easily taking that "minus two lots of six from the bag is the same as minus 12 from the total". Or if the bag contains two -5s, among others, we see "minus two lots of -5 is the same as +10"
                                                                                                                      $endgroup$
                                                                                                                      – Silverfish
                                                                                                                      Nov 9 '15 at 21:17











                                                                                                                      6












                                                                                                                      $begingroup$

                                                                                                                      Your child may have been introduced to the minus sign by means of the word opposite. This is a great term to use in your conversations. Indeed, $a$ and $-a$ are opposites in that they are additive inverses. On the number line, opposite numbers are mirrored in their distances from zero, which provides a nice visual aid as well. We can use the term to describe arithmetic operations:



                                                                                                                      The opposite of three times five is the opposite of 15.
                                                                                                                      $$-3 times 5 = -15$$



                                                                                                                      The opposite of three times the opposite of five is the opposite of the opposite of fifteen... which is just fifteen.
                                                                                                                      $$-3 times -5 = 15$$



                                                                                                                      Okay, so we can use language to better cultivate our understanding of negative numbers. But what about a physical example? Well, here's a cute one that a friend once told me. It's a bit contrived, but I think it gets the point across. You can turn this into a demonstration too.




                                                                                                                      Suppose an ice cube lowers the temperature of a drink by $1$ degree.



                                                                                                                      Placing three ice cubes in a glass will lower the temperature by $3$ degrees, or
                                                                                                                      $$ 3 times -1 = -3$$



                                                                                                                      Removing two of the ice cubes will raise the temperature by $2$ degrees, or
                                                                                                                      $$ -2 times -1 = 2$$







                                                                                                                      share|cite|improve this answer











                                                                                                                      $endgroup$









                                                                                                                      • 1




                                                                                                                        $begingroup$
                                                                                                                        Can you really warm up a drink by removing ice cubes from it? In reality, an ice cube in a drink absorbs some heat from the drink, and some of the ice may melt and become part of the liquid in the drink, then mix with the other liquid, lowering its temperature. In either case, when you remove the ice, you do not undo either of those effects; the drink remains cooler than before you put the ice in it. So you either have to make some complicated "averaging" argument or you have to rely on a false physical intuition to be "unlearned" later.
                                                                                                                        $endgroup$
                                                                                                                        – David K
                                                                                                                        Nov 9 '15 at 13:49










                                                                                                                      • $begingroup$
                                                                                                                        @DavidK Just use the word "pretend." I doubt the child is ready for a talk about specific heat.
                                                                                                                        $endgroup$
                                                                                                                        – zahbaz
                                                                                                                        Nov 9 '15 at 15:55












                                                                                                                      • $begingroup$
                                                                                                                        What should we pretend? Should we pretend that when you remove some of the ice, the liquid in the glass warms up? That's a "false intuition" of the sort I meant. There are so many analogies that actually work in real life, why not just use one of them instead of making up plausible (to a naive person) but incorrect physics?
                                                                                                                        $endgroup$
                                                                                                                        – David K
                                                                                                                        Nov 9 '15 at 19:23










                                                                                                                      • $begingroup$
                                                                                                                        @DavidK No, we should suppose that the scenario is grossly simplified. Children can understand that sort of disclaimer. The point isn't to teach physics, but to paint a mental image of what opposites mean in terms of multiplication. We can use a metaphorical example as a heuristic, no? I get your point, but I find this example works well enough for pedagogical purposes.. If you want a physical example, then take the ratio of the mass of your ice cube and water to be approx 1:100. Then, yes, one ice cube will (eventually) lower the temperature of the drink by about 1 degree C.
                                                                                                                        $endgroup$
                                                                                                                        – zahbaz
                                                                                                                        Nov 9 '15 at 20:42












                                                                                                                      • $begingroup$
                                                                                                                        Or if you will, replace the word ice cube by weight, glass by buoy, and temperature by height above sea level. It's just an architecture to express opposites.
                                                                                                                        $endgroup$
                                                                                                                        – zahbaz
                                                                                                                        Nov 9 '15 at 21:05
















                                                                                                                      6












                                                                                                                      $begingroup$

                                                                                                                      Your child may have been introduced to the minus sign by means of the word opposite. This is a great term to use in your conversations. Indeed, $a$ and $-a$ are opposites in that they are additive inverses. On the number line, opposite numbers are mirrored in their distances from zero, which provides a nice visual aid as well. We can use the term to describe arithmetic operations:



                                                                                                                      The opposite of three times five is the opposite of 15.
                                                                                                                      $$-3 times 5 = -15$$



                                                                                                                      The opposite of three times the opposite of five is the opposite of the opposite of fifteen... which is just fifteen.
                                                                                                                      $$-3 times -5 = 15$$



                                                                                                                      Okay, so we can use language to better cultivate our understanding of negative numbers. But what about a physical example? Well, here's a cute one that a friend once told me. It's a bit contrived, but I think it gets the point across. You can turn this into a demonstration too.




                                                                                                                      Suppose an ice cube lowers the temperature of a drink by $1$ degree.



                                                                                                                      Placing three ice cubes in a glass will lower the temperature by $3$ degrees, or
                                                                                                                      $$ 3 times -1 = -3$$



                                                                                                                      Removing two of the ice cubes will raise the temperature by $2$ degrees, or
                                                                                                                      $$ -2 times -1 = 2$$







                                                                                                                      share|cite|improve this answer











                                                                                                                      $endgroup$









                                                                                                                      • 1




                                                                                                                        $begingroup$
                                                                                                                        Can you really warm up a drink by removing ice cubes from it? In reality, an ice cube in a drink absorbs some heat from the drink, and some of the ice may melt and become part of the liquid in the drink, then mix with the other liquid, lowering its temperature. In either case, when you remove the ice, you do not undo either of those effects; the drink remains cooler than before you put the ice in it. So you either have to make some complicated "averaging" argument or you have to rely on a false physical intuition to be "unlearned" later.
                                                                                                                        $endgroup$
                                                                                                                        – David K
                                                                                                                        Nov 9 '15 at 13:49










                                                                                                                      • $begingroup$
                                                                                                                        @DavidK Just use the word "pretend." I doubt the child is ready for a talk about specific heat.
                                                                                                                        $endgroup$
                                                                                                                        – zahbaz
                                                                                                                        Nov 9 '15 at 15:55












                                                                                                                      • $begingroup$
                                                                                                                        What should we pretend? Should we pretend that when you remove some of the ice, the liquid in the glass warms up? That's a "false intuition" of the sort I meant. There are so many analogies that actually work in real life, why not just use one of them instead of making up plausible (to a naive person) but incorrect physics?
                                                                                                                        $endgroup$
                                                                                                                        – David K
                                                                                                                        Nov 9 '15 at 19:23










                                                                                                                      • $begingroup$
                                                                                                                        @DavidK No, we should suppose that the scenario is grossly simplified. Children can understand that sort of disclaimer. The point isn't to teach physics, but to paint a mental image of what opposites mean in terms of multiplication. We can use a metaphorical example as a heuristic, no? I get your point, but I find this example works well enough for pedagogical purposes.. If you want a physical example, then take the ratio of the mass of your ice cube and water to be approx 1:100. Then, yes, one ice cube will (eventually) lower the temperature of the drink by about 1 degree C.
                                                                                                                        $endgroup$
                                                                                                                        – zahbaz
                                                                                                                        Nov 9 '15 at 20:42












                                                                                                                      • $begingroup$
                                                                                                                        Or if you will, replace the word ice cube by weight, glass by buoy, and temperature by height above sea level. It's just an architecture to express opposites.
                                                                                                                        $endgroup$
                                                                                                                        – zahbaz
                                                                                                                        Nov 9 '15 at 21:05














                                                                                                                      6












                                                                                                                      6








                                                                                                                      6





                                                                                                                      $begingroup$

                                                                                                                      Your child may have been introduced to the minus sign by means of the word opposite. This is a great term to use in your conversations. Indeed, $a$ and $-a$ are opposites in that they are additive inverses. On the number line, opposite numbers are mirrored in their distances from zero, which provides a nice visual aid as well. We can use the term to describe arithmetic operations:



                                                                                                                      The opposite of three times five is the opposite of 15.
                                                                                                                      $$-3 times 5 = -15$$



                                                                                                                      The opposite of three times the opposite of five is the opposite of the opposite of fifteen... which is just fifteen.
                                                                                                                      $$-3 times -5 = 15$$



                                                                                                                      Okay, so we can use language to better cultivate our understanding of negative numbers. But what about a physical example? Well, here's a cute one that a friend once told me. It's a bit contrived, but I think it gets the point across. You can turn this into a demonstration too.




                                                                                                                      Suppose an ice cube lowers the temperature of a drink by $1$ degree.



                                                                                                                      Placing three ice cubes in a glass will lower the temperature by $3$ degrees, or
                                                                                                                      $$ 3 times -1 = -3$$



                                                                                                                      Removing two of the ice cubes will raise the temperature by $2$ degrees, or
                                                                                                                      $$ -2 times -1 = 2$$







                                                                                                                      share|cite|improve this answer











                                                                                                                      $endgroup$



                                                                                                                      Your child may have been introduced to the minus sign by means of the word opposite. This is a great term to use in your conversations. Indeed, $a$ and $-a$ are opposites in that they are additive inverses. On the number line, opposite numbers are mirrored in their distances from zero, which provides a nice visual aid as well. We can use the term to describe arithmetic operations:



                                                                                                                      The opposite of three times five is the opposite of 15.
                                                                                                                      $$-3 times 5 = -15$$



                                                                                                                      The opposite of three times the opposite of five is the opposite of the opposite of fifteen... which is just fifteen.
                                                                                                                      $$-3 times -5 = 15$$



                                                                                                                      Okay, so we can use language to better cultivate our understanding of negative numbers. But what about a physical example? Well, here's a cute one that a friend once told me. It's a bit contrived, but I think it gets the point across. You can turn this into a demonstration too.




                                                                                                                      Suppose an ice cube lowers the temperature of a drink by $1$ degree.



                                                                                                                      Placing three ice cubes in a glass will lower the temperature by $3$ degrees, or
                                                                                                                      $$ 3 times -1 = -3$$



                                                                                                                      Removing two of the ice cubes will raise the temperature by $2$ degrees, or
                                                                                                                      $$ -2 times -1 = 2$$








                                                                                                                      share|cite|improve this answer














                                                                                                                      share|cite|improve this answer



                                                                                                                      share|cite|improve this answer








                                                                                                                      edited Nov 10 '15 at 4:09

























                                                                                                                      answered Nov 9 '15 at 8:08









                                                                                                                      zahbazzahbaz

                                                                                                                      8,45921938




                                                                                                                      8,45921938








                                                                                                                      • 1




                                                                                                                        $begingroup$
                                                                                                                        Can you really warm up a drink by removing ice cubes from it? In reality, an ice cube in a drink absorbs some heat from the drink, and some of the ice may melt and become part of the liquid in the drink, then mix with the other liquid, lowering its temperature. In either case, when you remove the ice, you do not undo either of those effects; the drink remains cooler than before you put the ice in it. So you either have to make some complicated "averaging" argument or you have to rely on a false physical intuition to be "unlearned" later.
                                                                                                                        $endgroup$
                                                                                                                        – David K
                                                                                                                        Nov 9 '15 at 13:49










                                                                                                                      • $begingroup$
                                                                                                                        @DavidK Just use the word "pretend." I doubt the child is ready for a talk about specific heat.
                                                                                                                        $endgroup$
                                                                                                                        – zahbaz
                                                                                                                        Nov 9 '15 at 15:55












                                                                                                                      • $begingroup$
                                                                                                                        What should we pretend? Should we pretend that when you remove some of the ice, the liquid in the glass warms up? That's a "false intuition" of the sort I meant. There are so many analogies that actually work in real life, why not just use one of them instead of making up plausible (to a naive person) but incorrect physics?
                                                                                                                        $endgroup$
                                                                                                                        – David K
                                                                                                                        Nov 9 '15 at 19:23










                                                                                                                      • $begingroup$
                                                                                                                        @DavidK No, we should suppose that the scenario is grossly simplified. Children can understand that sort of disclaimer. The point isn't to teach physics, but to paint a mental image of what opposites mean in terms of multiplication. We can use a metaphorical example as a heuristic, no? I get your point, but I find this example works well enough for pedagogical purposes.. If you want a physical example, then take the ratio of the mass of your ice cube and water to be approx 1:100. Then, yes, one ice cube will (eventually) lower the temperature of the drink by about 1 degree C.
                                                                                                                        $endgroup$
                                                                                                                        – zahbaz
                                                                                                                        Nov 9 '15 at 20:42












                                                                                                                      • $begingroup$
                                                                                                                        Or if you will, replace the word ice cube by weight, glass by buoy, and temperature by height above sea level. It's just an architecture to express opposites.
                                                                                                                        $endgroup$
                                                                                                                        – zahbaz
                                                                                                                        Nov 9 '15 at 21:05














                                                                                                                      • 1




                                                                                                                        $begingroup$
                                                                                                                        Can you really warm up a drink by removing ice cubes from it? In reality, an ice cube in a drink absorbs some heat from the drink, and some of the ice may melt and become part of the liquid in the drink, then mix with the other liquid, lowering its temperature. In either case, when you remove the ice, you do not undo either of those effects; the drink remains cooler than before you put the ice in it. So you either have to make some complicated "averaging" argument or you have to rely on a false physical intuition to be "unlearned" later.
                                                                                                                        $endgroup$
                                                                                                                        – David K
                                                                                                                        Nov 9 '15 at 13:49










                                                                                                                      • $begingroup$
                                                                                                                        @DavidK Just use the word "pretend." I doubt the child is ready for a talk about specific heat.
                                                                                                                        $endgroup$
                                                                                                                        – zahbaz
                                                                                                                        Nov 9 '15 at 15:55












                                                                                                                      • $begingroup$
                                                                                                                        What should we pretend? Should we pretend that when you remove some of the ice, the liquid in the glass warms up? That's a "false intuition" of the sort I meant. There are so many analogies that actually work in real life, why not just use one of them instead of making up plausible (to a naive person) but incorrect physics?
                                                                                                                        $endgroup$
                                                                                                                        – David K
                                                                                                                        Nov 9 '15 at 19:23










                                                                                                                      • $begingroup$
                                                                                                                        @DavidK No, we should suppose that the scenario is grossly simplified. Children can understand that sort of disclaimer. The point isn't to teach physics, but to paint a mental image of what opposites mean in terms of multiplication. We can use a metaphorical example as a heuristic, no? I get your point, but I find this example works well enough for pedagogical purposes.. If you want a physical example, then take the ratio of the mass of your ice cube and water to be approx 1:100. Then, yes, one ice cube will (eventually) lower the temperature of the drink by about 1 degree C.
                                                                                                                        $endgroup$
                                                                                                                        – zahbaz
                                                                                                                        Nov 9 '15 at 20:42












                                                                                                                      • $begingroup$
                                                                                                                        Or if you will, replace the word ice cube by weight, glass by buoy, and temperature by height above sea level. It's just an architecture to express opposites.
                                                                                                                        $endgroup$
                                                                                                                        – zahbaz
                                                                                                                        Nov 9 '15 at 21:05








                                                                                                                      1




                                                                                                                      1




                                                                                                                      $begingroup$
                                                                                                                      Can you really warm up a drink by removing ice cubes from it? In reality, an ice cube in a drink absorbs some heat from the drink, and some of the ice may melt and become part of the liquid in the drink, then mix with the other liquid, lowering its temperature. In either case, when you remove the ice, you do not undo either of those effects; the drink remains cooler than before you put the ice in it. So you either have to make some complicated "averaging" argument or you have to rely on a false physical intuition to be "unlearned" later.
                                                                                                                      $endgroup$
                                                                                                                      – David K
                                                                                                                      Nov 9 '15 at 13:49




                                                                                                                      $begingroup$
                                                                                                                      Can you really warm up a drink by removing ice cubes from it? In reality, an ice cube in a drink absorbs some heat from the drink, and some of the ice may melt and become part of the liquid in the drink, then mix with the other liquid, lowering its temperature. In either case, when you remove the ice, you do not undo either of those effects; the drink remains cooler than before you put the ice in it. So you either have to make some complicated "averaging" argument or you have to rely on a false physical intuition to be "unlearned" later.
                                                                                                                      $endgroup$
                                                                                                                      – David K
                                                                                                                      Nov 9 '15 at 13:49












                                                                                                                      $begingroup$
                                                                                                                      @DavidK Just use the word "pretend." I doubt the child is ready for a talk about specific heat.
                                                                                                                      $endgroup$
                                                                                                                      – zahbaz
                                                                                                                      Nov 9 '15 at 15:55






                                                                                                                      $begingroup$
                                                                                                                      @DavidK Just use the word "pretend." I doubt the child is ready for a talk about specific heat.
                                                                                                                      $endgroup$
                                                                                                                      – zahbaz
                                                                                                                      Nov 9 '15 at 15:55














                                                                                                                      $begingroup$
                                                                                                                      What should we pretend? Should we pretend that when you remove some of the ice, the liquid in the glass warms up? That's a "false intuition" of the sort I meant. There are so many analogies that actually work in real life, why not just use one of them instead of making up plausible (to a naive person) but incorrect physics?
                                                                                                                      $endgroup$
                                                                                                                      – David K
                                                                                                                      Nov 9 '15 at 19:23




                                                                                                                      $begingroup$
                                                                                                                      What should we pretend? Should we pretend that when you remove some of the ice, the liquid in the glass warms up? That's a "false intuition" of the sort I meant. There are so many analogies that actually work in real life, why not just use one of them instead of making up plausible (to a naive person) but incorrect physics?
                                                                                                                      $endgroup$
                                                                                                                      – David K
                                                                                                                      Nov 9 '15 at 19:23












                                                                                                                      $begingroup$
                                                                                                                      @DavidK No, we should suppose that the scenario is grossly simplified. Children can understand that sort of disclaimer. The point isn't to teach physics, but to paint a mental image of what opposites mean in terms of multiplication. We can use a metaphorical example as a heuristic, no? I get your point, but I find this example works well enough for pedagogical purposes.. If you want a physical example, then take the ratio of the mass of your ice cube and water to be approx 1:100. Then, yes, one ice cube will (eventually) lower the temperature of the drink by about 1 degree C.
                                                                                                                      $endgroup$
                                                                                                                      – zahbaz
                                                                                                                      Nov 9 '15 at 20:42






                                                                                                                      $begingroup$
                                                                                                                      @DavidK No, we should suppose that the scenario is grossly simplified. Children can understand that sort of disclaimer. The point isn't to teach physics, but to paint a mental image of what opposites mean in terms of multiplication. We can use a metaphorical example as a heuristic, no? I get your point, but I find this example works well enough for pedagogical purposes.. If you want a physical example, then take the ratio of the mass of your ice cube and water to be approx 1:100. Then, yes, one ice cube will (eventually) lower the temperature of the drink by about 1 degree C.
                                                                                                                      $endgroup$
                                                                                                                      – zahbaz
                                                                                                                      Nov 9 '15 at 20:42














                                                                                                                      $begingroup$
                                                                                                                      Or if you will, replace the word ice cube by weight, glass by buoy, and temperature by height above sea level. It's just an architecture to express opposites.
                                                                                                                      $endgroup$
                                                                                                                      – zahbaz
                                                                                                                      Nov 9 '15 at 21:05




                                                                                                                      $begingroup$
                                                                                                                      Or if you will, replace the word ice cube by weight, glass by buoy, and temperature by height above sea level. It's just an architecture to express opposites.
                                                                                                                      $endgroup$
                                                                                                                      – zahbaz
                                                                                                                      Nov 9 '15 at 21:05











                                                                                                                      5












                                                                                                                      $begingroup$

                                                                                                                      One way to picture this is to imagine a number line. Then rotate it $180^{circ}$. Each number will now be superimposed over its negative: $-1$ will be where $+1$ was; $+2$ will be where $-2$ was. Rotation of the number line by $180^{circ}$ is the equivalent of multiplying by $-1$.



                                                                                                                      Now do the rotation twice. The number line is unchanged. So, multiplying by $-1$ twice is the same as multiplying by $+1$.



                                                                                                                      This approach has applications with Complex numbers. In these scenarios, the number line is rotated $90^{circ}$ counter clockwise to multiply by $i$.



                                                                                                                      But that's another story.






                                                                                                                      share|cite|improve this answer









                                                                                                                      $endgroup$


















                                                                                                                        5












                                                                                                                        $begingroup$

                                                                                                                        One way to picture this is to imagine a number line. Then rotate it $180^{circ}$. Each number will now be superimposed over its negative: $-1$ will be where $+1$ was; $+2$ will be where $-2$ was. Rotation of the number line by $180^{circ}$ is the equivalent of multiplying by $-1$.



                                                                                                                        Now do the rotation twice. The number line is unchanged. So, multiplying by $-1$ twice is the same as multiplying by $+1$.



                                                                                                                        This approach has applications with Complex numbers. In these scenarios, the number line is rotated $90^{circ}$ counter clockwise to multiply by $i$.



                                                                                                                        But that's another story.






                                                                                                                        share|cite|improve this answer









                                                                                                                        $endgroup$
















                                                                                                                          5












                                                                                                                          5








                                                                                                                          5





                                                                                                                          $begingroup$

                                                                                                                          One way to picture this is to imagine a number line. Then rotate it $180^{circ}$. Each number will now be superimposed over its negative: $-1$ will be where $+1$ was; $+2$ will be where $-2$ was. Rotation of the number line by $180^{circ}$ is the equivalent of multiplying by $-1$.



                                                                                                                          Now do the rotation twice. The number line is unchanged. So, multiplying by $-1$ twice is the same as multiplying by $+1$.



                                                                                                                          This approach has applications with Complex numbers. In these scenarios, the number line is rotated $90^{circ}$ counter clockwise to multiply by $i$.



                                                                                                                          But that's another story.






                                                                                                                          share|cite|improve this answer









                                                                                                                          $endgroup$



                                                                                                                          One way to picture this is to imagine a number line. Then rotate it $180^{circ}$. Each number will now be superimposed over its negative: $-1$ will be where $+1$ was; $+2$ will be where $-2$ was. Rotation of the number line by $180^{circ}$ is the equivalent of multiplying by $-1$.



                                                                                                                          Now do the rotation twice. The number line is unchanged. So, multiplying by $-1$ twice is the same as multiplying by $+1$.



                                                                                                                          This approach has applications with Complex numbers. In these scenarios, the number line is rotated $90^{circ}$ counter clockwise to multiply by $i$.



                                                                                                                          But that's another story.







                                                                                                                          share|cite|improve this answer












                                                                                                                          share|cite|improve this answer



                                                                                                                          share|cite|improve this answer










                                                                                                                          answered Oct 17 '15 at 6:12









                                                                                                                          Adam HrankowskiAdam Hrankowski

                                                                                                                          2,099930




                                                                                                                          2,099930























                                                                                                                              5












                                                                                                                              $begingroup$


                                                                                                                              1. Explain the definition of negative numbers.

                                                                                                                              2. Point out that the definition of $-x$ implies that $-(-x) = x$.

                                                                                                                              3. Explain that $-x = (-1)times x$.

                                                                                                                              4. Point out that (2) and (3) imply that $(-1)times(-1) = 1$.






                                                                                                                              share|cite|improve this answer









                                                                                                                              $endgroup$









                                                                                                                              • 4




                                                                                                                                $begingroup$
                                                                                                                                "Huh??? x is a letter! you can't multiply numbers and letters!"
                                                                                                                                $endgroup$
                                                                                                                                – Paul Sinclair
                                                                                                                                Nov 9 '15 at 4:28






                                                                                                                              • 4




                                                                                                                                $begingroup$
                                                                                                                                @PaulSinclair The children of today are used to using computers, they can easily be taught to think of $x$ as storing some arbitrary number.
                                                                                                                                $endgroup$
                                                                                                                                – Count Iblis
                                                                                                                                Nov 9 '15 at 4:45






                                                                                                                              • 7




                                                                                                                                $begingroup$
                                                                                                                                @CountIbis I wouldnt take that bet. I teach engineers and for many years this has been the assumption. Net result is that i get Bachelors level students that can not do anything with a computer, not even frigging graph a function. Some dont even know what a file is.
                                                                                                                                $endgroup$
                                                                                                                                – joojaa
                                                                                                                                Nov 9 '15 at 5:08












                                                                                                                              • $begingroup$
                                                                                                                                @PaulSinclair Indeed! With kids of that age, a common response is that $x$ means 24. Because $a$ is one, and $b$ is two, and so on, and in that way the more inventive kids discover they can add numbers and letters. It's just that they've made the "wrong" extension into symbols. And explaining that $x$, or indeed $a$ might represent an arbitrary number is surprisingly hard if the kid hasn't developed that kind of abstraction yet. You probably would have more luck with "an empty box that a number can go in", rather than using a letter.
                                                                                                                                $endgroup$
                                                                                                                                – Silverfish
                                                                                                                                Nov 9 '15 at 21:21
















                                                                                                                              5












                                                                                                                              $begingroup$


                                                                                                                              1. Explain the definition of negative numbers.

                                                                                                                              2. Point out that the definition of $-x$ implies that $-(-x) = x$.

                                                                                                                              3. Explain that $-x = (-1)times x$.

                                                                                                                              4. Point out that (2) and (3) imply that $(-1)times(-1) = 1$.






                                                                                                                              share|cite|improve this answer









                                                                                                                              $endgroup$









                                                                                                                              • 4




                                                                                                                                $begingroup$
                                                                                                                                "Huh??? x is a letter! you can't multiply numbers and letters!"
                                                                                                                                $endgroup$
                                                                                                                                – Paul Sinclair
                                                                                                                                Nov 9 '15 at 4:28






                                                                                                                              • 4




                                                                                                                                $begingroup$
                                                                                                                                @PaulSinclair The children of today are used to using computers, they can easily be taught to think of $x$ as storing some arbitrary number.
                                                                                                                                $endgroup$
                                                                                                                                – Count Iblis
                                                                                                                                Nov 9 '15 at 4:45






                                                                                                                              • 7




                                                                                                                                $begingroup$
                                                                                                                                @CountIbis I wouldnt take that bet. I teach engineers and for many years this has been the assumption. Net result is that i get Bachelors level students that can not do anything with a computer, not even frigging graph a function. Some dont even know what a file is.
                                                                                                                                $endgroup$
                                                                                                                                – joojaa
                                                                                                                                Nov 9 '15 at 5:08












                                                                                                                              • $begingroup$
                                                                                                                                @PaulSinclair Indeed! With kids of that age, a common response is that $x$ means 24. Because $a$ is one, and $b$ is two, and so on, and in that way the more inventive kids discover they can add numbers and letters. It's just that they've made the "wrong" extension into symbols. And explaining that $x$, or indeed $a$ might represent an arbitrary number is surprisingly hard if the kid hasn't developed that kind of abstraction yet. You probably would have more luck with "an empty box that a number can go in", rather than using a letter.
                                                                                                                                $endgroup$
                                                                                                                                – Silverfish
                                                                                                                                Nov 9 '15 at 21:21














                                                                                                                              5












                                                                                                                              5








                                                                                                                              5





                                                                                                                              $begingroup$


                                                                                                                              1. Explain the definition of negative numbers.

                                                                                                                              2. Point out that the definition of $-x$ implies that $-(-x) = x$.

                                                                                                                              3. Explain that $-x = (-1)times x$.

                                                                                                                              4. Point out that (2) and (3) imply that $(-1)times(-1) = 1$.






                                                                                                                              share|cite|improve this answer









                                                                                                                              $endgroup$




                                                                                                                              1. Explain the definition of negative numbers.

                                                                                                                              2. Point out that the definition of $-x$ implies that $-(-x) = x$.

                                                                                                                              3. Explain that $-x = (-1)times x$.

                                                                                                                              4. Point out that (2) and (3) imply that $(-1)times(-1) = 1$.







                                                                                                                              share|cite|improve this answer












                                                                                                                              share|cite|improve this answer



                                                                                                                              share|cite|improve this answer










                                                                                                                              answered Nov 9 '15 at 4:02









                                                                                                                              Count IblisCount Iblis

                                                                                                                              8,55221534




                                                                                                                              8,55221534








                                                                                                                              • 4




                                                                                                                                $begingroup$
                                                                                                                                "Huh??? x is a letter! you can't multiply numbers and letters!"
                                                                                                                                $endgroup$
                                                                                                                                – Paul Sinclair
                                                                                                                                Nov 9 '15 at 4:28






                                                                                                                              • 4




                                                                                                                                $begingroup$
                                                                                                                                @PaulSinclair The children of today are used to using computers, they can easily be taught to think of $x$ as storing some arbitrary number.
                                                                                                                                $endgroup$
                                                                                                                                – Count Iblis
                                                                                                                                Nov 9 '15 at 4:45






                                                                                                                              • 7




                                                                                                                                $begingroup$
                                                                                                                                @CountIbis I wouldnt take that bet. I teach engineers and for many years this has been the assumption. Net result is that i get Bachelors level students that can not do anything with a computer, not even frigging graph a function. Some dont even know what a file is.
                                                                                                                                $endgroup$
                                                                                                                                – joojaa
                                                                                                                                Nov 9 '15 at 5:08












                                                                                                                              • $begingroup$
                                                                                                                                @PaulSinclair Indeed! With kids of that age, a common response is that $x$ means 24. Because $a$ is one, and $b$ is two, and so on, and in that way the more inventive kids discover they can add numbers and letters. It's just that they've made the "wrong" extension into symbols. And explaining that $x$, or indeed $a$ might represent an arbitrary number is surprisingly hard if the kid hasn't developed that kind of abstraction yet. You probably would have more luck with "an empty box that a number can go in", rather than using a letter.
                                                                                                                                $endgroup$
                                                                                                                                – Silverfish
                                                                                                                                Nov 9 '15 at 21:21














                                                                                                                              • 4




                                                                                                                                $begingroup$
                                                                                                                                "Huh??? x is a letter! you can't multiply numbers and letters!"
                                                                                                                                $endgroup$
                                                                                                                                – Paul Sinclair
                                                                                                                                Nov 9 '15 at 4:28






                                                                                                                              • 4




                                                                                                                                $begingroup$
                                                                                                                                @PaulSinclair The children of today are used to using computers, they can easily be taught to think of $x$ as storing some arbitrary number.
                                                                                                                                $endgroup$
                                                                                                                                – Count Iblis
                                                                                                                                Nov 9 '15 at 4:45






                                                                                                                              • 7




                                                                                                                                $begingroup$
                                                                                                                                @CountIbis I wouldnt take that bet. I teach engineers and for many years this has been the assumption. Net result is that i get Bachelors level students that can not do anything with a computer, not even frigging graph a function. Some dont even know what a file is.
                                                                                                                                $endgroup$
                                                                                                                                – joojaa
                                                                                                                                Nov 9 '15 at 5:08












                                                                                                                              • $begingroup$
                                                                                                                                @PaulSinclair Indeed! With kids of that age, a common response is that $x$ means 24. Because $a$ is one, and $b$ is two, and so on, and in that way the more inventive kids discover they can add numbers and letters. It's just that they've made the "wrong" extension into symbols. And explaining that $x$, or indeed $a$ might represent an arbitrary number is surprisingly hard if the kid hasn't developed that kind of abstraction yet. You probably would have more luck with "an empty box that a number can go in", rather than using a letter.
                                                                                                                                $endgroup$
                                                                                                                                – Silverfish
                                                                                                                                Nov 9 '15 at 21:21








                                                                                                                              4




                                                                                                                              4




                                                                                                                              $begingroup$
                                                                                                                              "Huh??? x is a letter! you can't multiply numbers and letters!"
                                                                                                                              $endgroup$
                                                                                                                              – Paul Sinclair
                                                                                                                              Nov 9 '15 at 4:28




                                                                                                                              $begingroup$
                                                                                                                              "Huh??? x is a letter! you can't multiply numbers and letters!"
                                                                                                                              $endgroup$
                                                                                                                              – Paul Sinclair
                                                                                                                              Nov 9 '15 at 4:28




                                                                                                                              4




                                                                                                                              4




                                                                                                                              $begingroup$
                                                                                                                              @PaulSinclair The children of today are used to using computers, they can easily be taught to think of $x$ as storing some arbitrary number.
                                                                                                                              $endgroup$
                                                                                                                              – Count Iblis
                                                                                                                              Nov 9 '15 at 4:45




                                                                                                                              $begingroup$
                                                                                                                              @PaulSinclair The children of today are used to using computers, they can easily be taught to think of $x$ as storing some arbitrary number.
                                                                                                                              $endgroup$
                                                                                                                              – Count Iblis
                                                                                                                              Nov 9 '15 at 4:45




                                                                                                                              7




                                                                                                                              7




                                                                                                                              $begingroup$
                                                                                                                              @CountIbis I wouldnt take that bet. I teach engineers and for many years this has been the assumption. Net result is that i get Bachelors level students that can not do anything with a computer, not even frigging graph a function. Some dont even know what a file is.
                                                                                                                              $endgroup$
                                                                                                                              – joojaa
                                                                                                                              Nov 9 '15 at 5:08






                                                                                                                              $begingroup$
                                                                                                                              @CountIbis I wouldnt take that bet. I teach engineers and for many years this has been the assumption. Net result is that i get Bachelors level students that can not do anything with a computer, not even frigging graph a function. Some dont even know what a file is.
                                                                                                                              $endgroup$
                                                                                                                              – joojaa
                                                                                                                              Nov 9 '15 at 5:08














                                                                                                                              $begingroup$
                                                                                                                              @PaulSinclair Indeed! With kids of that age, a common response is that $x$ means 24. Because $a$ is one, and $b$ is two, and so on, and in that way the more inventive kids discover they can add numbers and letters. It's just that they've made the "wrong" extension into symbols. And explaining that $x$, or indeed $a$ might represent an arbitrary number is surprisingly hard if the kid hasn't developed that kind of abstraction yet. You probably would have more luck with "an empty box that a number can go in", rather than using a letter.
                                                                                                                              $endgroup$
                                                                                                                              – Silverfish
                                                                                                                              Nov 9 '15 at 21:21




                                                                                                                              $begingroup$
                                                                                                                              @PaulSinclair Indeed! With kids of that age, a common response is that $x$ means 24. Because $a$ is one, and $b$ is two, and so on, and in that way the more inventive kids discover they can add numbers and letters. It's just that they've made the "wrong" extension into symbols. And explaining that $x$, or indeed $a$ might represent an arbitrary number is surprisingly hard if the kid hasn't developed that kind of abstraction yet. You probably would have more luck with "an empty box that a number can go in", rather than using a letter.
                                                                                                                              $endgroup$
                                                                                                                              – Silverfish
                                                                                                                              Nov 9 '15 at 21:21











                                                                                                                              4












                                                                                                                              $begingroup$

                                                                                                                              Symbology: $$begin{align*}
                                                                                                                              -a times -b &= (-1 times a)times (-1 times b) \
                                                                                                                              &= -1 times a times -1 times b \
                                                                                                                              &= -1 times -1 times a times b \
                                                                                                                              &= (-1 times -1) times (a times b) \
                                                                                                                              &= a times b.
                                                                                                                              end{align*}$$



                                                                                                                              What's going on?:

                                                                                                                              Consider the number line. When I multiply something by $2$, I double its distance from $0$. This happens whether the "something" is positive or negative. When I multiply something by $-2$, I double its distance from $0$ and flip to the other side of the number line. For instance $-2 times 3 : 3 rightarrow 6 rightarrow -6$. If I start with a negative number, it's already on the negative half of the number line and will be flipped to the positive half: $-2 times -3 : -3 rightarrow -6 rightarrow 6$.



                                                                                                                              That is, multiplication by a negative is the same as two steps: multiplying by the thing as if it had no negative, then applying the negative sign. That's what the symbology above says: multiplying by $-a$ is the same as multiplying by $a$ then by $-1$ and similarly for $-b$. But then the two "flip to the other half of the number line"s, the two "$-1$"s, cause two flips. But two flips takes anything away and then right back to itself, so two flips really does nothing. In this way $-1 times -1$ is the same as doing nothing, or is the same as multiplying by $1$. That's why in the symbols above, we can drop the "$-1 times -1$" -- they're the same as multiplying by $1$.






                                                                                                                              share|cite|improve this answer









                                                                                                                              $endgroup$









                                                                                                                              • 1




                                                                                                                                $begingroup$
                                                                                                                                "When I multiply something by −2, I double its distance from 0 and flip to the other side of the number line": I like this answer a lot, but I think to serve as a good explanation for a kid at this level it would help if you could explain why multiplying works on the number line like this (it's safe to assume you can't use the symbology directly with the kid, although you're clearly trying to get across that train of logic). Until you can explain why negative multiplication reflects the number line, we have replaced one "why's does it work like that?" question by another.
                                                                                                                                $endgroup$
                                                                                                                                – Silverfish
                                                                                                                                Nov 9 '15 at 21:14
















                                                                                                                              4












                                                                                                                              $begingroup$

                                                                                                                              Symbology: $$begin{align*}
                                                                                                                              -a times -b &= (-1 times a)times (-1 times b) \
                                                                                                                              &= -1 times a times -1 times b \
                                                                                                                              &= -1 times -1 times a times b \
                                                                                                                              &= (-1 times -1) times (a times b) \
                                                                                                                              &= a times b.
                                                                                                                              end{align*}$$



                                                                                                                              What's going on?:

                                                                                                                              Consider the number line. When I multiply something by $2$, I double its distance from $0$. This happens whether the "something" is positive or negative. When I multiply something by $-2$, I double its distance from $0$ and flip to the other side of the number line. For instance $-2 times 3 : 3 rightarrow 6 rightarrow -6$. If I start with a negative number, it's already on the negative half of the number line and will be flipped to the positive half: $-2 times -3 : -3 rightarrow -6 rightarrow 6$.



                                                                                                                              That is, multiplication by a negative is the same as two steps: multiplying by the thing as if it had no negative, then applying the negative sign. That's what the symbology above says: multiplying by $-a$ is the same as multiplying by $a$ then by $-1$ and similarly for $-b$. But then the two "flip to the other half of the number line"s, the two "$-1$"s, cause two flips. But two flips takes anything away and then right back to itself, so two flips really does nothing. In this way $-1 times -1$ is the same as doing nothing, or is the same as multiplying by $1$. That's why in the symbols above, we can drop the "$-1 times -1$" -- they're the same as multiplying by $1$.






                                                                                                                              share|cite|improve this answer









                                                                                                                              $endgroup$









                                                                                                                              • 1




                                                                                                                                $begingroup$
                                                                                                                                "When I multiply something by −2, I double its distance from 0 and flip to the other side of the number line": I like this answer a lot, but I think to serve as a good explanation for a kid at this level it would help if you could explain why multiplying works on the number line like this (it's safe to assume you can't use the symbology directly with the kid, although you're clearly trying to get across that train of logic). Until you can explain why negative multiplication reflects the number line, we have replaced one "why's does it work like that?" question by another.
                                                                                                                                $endgroup$
                                                                                                                                – Silverfish
                                                                                                                                Nov 9 '15 at 21:14














                                                                                                                              4












                                                                                                                              4








                                                                                                                              4





                                                                                                                              $begingroup$

                                                                                                                              Symbology: $$begin{align*}
                                                                                                                              -a times -b &= (-1 times a)times (-1 times b) \
                                                                                                                              &= -1 times a times -1 times b \
                                                                                                                              &= -1 times -1 times a times b \
                                                                                                                              &= (-1 times -1) times (a times b) \
                                                                                                                              &= a times b.
                                                                                                                              end{align*}$$



                                                                                                                              What's going on?:

                                                                                                                              Consider the number line. When I multiply something by $2$, I double its distance from $0$. This happens whether the "something" is positive or negative. When I multiply something by $-2$, I double its distance from $0$ and flip to the other side of the number line. For instance $-2 times 3 : 3 rightarrow 6 rightarrow -6$. If I start with a negative number, it's already on the negative half of the number line and will be flipped to the positive half: $-2 times -3 : -3 rightarrow -6 rightarrow 6$.



                                                                                                                              That is, multiplication by a negative is the same as two steps: multiplying by the thing as if it had no negative, then applying the negative sign. That's what the symbology above says: multiplying by $-a$ is the same as multiplying by $a$ then by $-1$ and similarly for $-b$. But then the two "flip to the other half of the number line"s, the two "$-1$"s, cause two flips. But two flips takes anything away and then right back to itself, so two flips really does nothing. In this way $-1 times -1$ is the same as doing nothing, or is the same as multiplying by $1$. That's why in the symbols above, we can drop the "$-1 times -1$" -- they're the same as multiplying by $1$.






                                                                                                                              share|cite|improve this answer









                                                                                                                              $endgroup$



                                                                                                                              Symbology: $$begin{align*}
                                                                                                                              -a times -b &= (-1 times a)times (-1 times b) \
                                                                                                                              &= -1 times a times -1 times b \
                                                                                                                              &= -1 times -1 times a times b \
                                                                                                                              &= (-1 times -1) times (a times b) \
                                                                                                                              &= a times b.
                                                                                                                              end{align*}$$



                                                                                                                              What's going on?:

                                                                                                                              Consider the number line. When I multiply something by $2$, I double its distance from $0$. This happens whether the "something" is positive or negative. When I multiply something by $-2$, I double its distance from $0$ and flip to the other side of the number line. For instance $-2 times 3 : 3 rightarrow 6 rightarrow -6$. If I start with a negative number, it's already on the negative half of the number line and will be flipped to the positive half: $-2 times -3 : -3 rightarrow -6 rightarrow 6$.



                                                                                                                              That is, multiplication by a negative is the same as two steps: multiplying by the thing as if it had no negative, then applying the negative sign. That's what the symbology above says: multiplying by $-a$ is the same as multiplying by $a$ then by $-1$ and similarly for $-b$. But then the two "flip to the other half of the number line"s, the two "$-1$"s, cause two flips. But two flips takes anything away and then right back to itself, so two flips really does nothing. In this way $-1 times -1$ is the same as doing nothing, or is the same as multiplying by $1$. That's why in the symbols above, we can drop the "$-1 times -1$" -- they're the same as multiplying by $1$.







                                                                                                                              share|cite|improve this answer












                                                                                                                              share|cite|improve this answer



                                                                                                                              share|cite|improve this answer










                                                                                                                              answered Nov 9 '15 at 3:11









                                                                                                                              Eric TowersEric Towers

                                                                                                                              33.9k22371




                                                                                                                              33.9k22371








                                                                                                                              • 1




                                                                                                                                $begingroup$
                                                                                                                                "When I multiply something by −2, I double its distance from 0 and flip to the other side of the number line": I like this answer a lot, but I think to serve as a good explanation for a kid at this level it would help if you could explain why multiplying works on the number line like this (it's safe to assume you can't use the symbology directly with the kid, although you're clearly trying to get across that train of logic). Until you can explain why negative multiplication reflects the number line, we have replaced one "why's does it work like that?" question by another.
                                                                                                                                $endgroup$
                                                                                                                                – Silverfish
                                                                                                                                Nov 9 '15 at 21:14














                                                                                                                              • 1




                                                                                                                                $begingroup$
                                                                                                                                "When I multiply something by −2, I double its distance from 0 and flip to the other side of the number line": I like this answer a lot, but I think to serve as a good explanation for a kid at this level it would help if you could explain why multiplying works on the number line like this (it's safe to assume you can't use the symbology directly with the kid, although you're clearly trying to get across that train of logic). Until you can explain why negative multiplication reflects the number line, we have replaced one "why's does it work like that?" question by another.
                                                                                                                                $endgroup$
                                                                                                                                – Silverfish
                                                                                                                                Nov 9 '15 at 21:14








                                                                                                                              1




                                                                                                                              1




                                                                                                                              $begingroup$
                                                                                                                              "When I multiply something by −2, I double its distance from 0 and flip to the other side of the number line": I like this answer a lot, but I think to serve as a good explanation for a kid at this level it would help if you could explain why multiplying works on the number line like this (it's safe to assume you can't use the symbology directly with the kid, although you're clearly trying to get across that train of logic). Until you can explain why negative multiplication reflects the number line, we have replaced one "why's does it work like that?" question by another.
                                                                                                                              $endgroup$
                                                                                                                              – Silverfish
                                                                                                                              Nov 9 '15 at 21:14




                                                                                                                              $begingroup$
                                                                                                                              "When I multiply something by −2, I double its distance from 0 and flip to the other side of the number line": I like this answer a lot, but I think to serve as a good explanation for a kid at this level it would help if you could explain why multiplying works on the number line like this (it's safe to assume you can't use the symbology directly with the kid, although you're clearly trying to get across that train of logic). Until you can explain why negative multiplication reflects the number line, we have replaced one "why's does it work like that?" question by another.
                                                                                                                              $endgroup$
                                                                                                                              – Silverfish
                                                                                                                              Nov 9 '15 at 21:14











                                                                                                                              4












                                                                                                                              $begingroup$

                                                                                                                              Well.... this always made sense to me (but I've found it doesn't for others)



                                                                                                                              A) Multiplication is adding a number a bunch of times.



                                                                                                                              1) Negative numbers are numbers that are less than zero. The cancel out positive numbers. They are anti-numbers



                                                                                                                              So)i) positive x positive: add a bunch of positive numbers a positive number of times. Result: A big positive gain.



                                                                                                                              ii) positive x negative: add a bunch of negative anti-numbers. The result is a big amount of potential cancelling. Result: negative.



                                                                                                                              iii) negative x positive: take a bunch of positive numbers and take them away. Result: a loss; negative.



                                                                                                                              iv) negative x negative: take a bunch of anti-numbers and take them away. By taking away the taking away, what's left is putting things back. If you annihilite the annihitation the result is a net gain: Result: positive.






                                                                                                                              share|cite|improve this answer









                                                                                                                              $endgroup$


















                                                                                                                                4












                                                                                                                                $begingroup$

                                                                                                                                Well.... this always made sense to me (but I've found it doesn't for others)



                                                                                                                                A) Multiplication is adding a number a bunch of times.



                                                                                                                                1) Negative numbers are numbers that are less than zero. The cancel out positive numbers. They are anti-numbers



                                                                                                                                So)i) positive x positive: add a bunch of positive numbers a positive number of times. Result: A big positive gain.



                                                                                                                                ii) positive x negative: add a bunch of negative anti-numbers. The result is a big amount of potential cancelling. Result: negative.



                                                                                                                                iii) negative x positive: take a bunch of positive numbers and take them away. Result: a loss; negative.



                                                                                                                                iv) negative x negative: take a bunch of anti-numbers and take them away. By taking away the taking away, what's left is putting things back. If you annihilite the annihitation the result is a net gain: Result: positive.






                                                                                                                                share|cite|improve this answer









                                                                                                                                $endgroup$
















                                                                                                                                  4












                                                                                                                                  4








                                                                                                                                  4





                                                                                                                                  $begingroup$

                                                                                                                                  Well.... this always made sense to me (but I've found it doesn't for others)



                                                                                                                                  A) Multiplication is adding a number a bunch of times.



                                                                                                                                  1) Negative numbers are numbers that are less than zero. The cancel out positive numbers. They are anti-numbers



                                                                                                                                  So)i) positive x positive: add a bunch of positive numbers a positive number of times. Result: A big positive gain.



                                                                                                                                  ii) positive x negative: add a bunch of negative anti-numbers. The result is a big amount of potential cancelling. Result: negative.



                                                                                                                                  iii) negative x positive: take a bunch of positive numbers and take them away. Result: a loss; negative.



                                                                                                                                  iv) negative x negative: take a bunch of anti-numbers and take them away. By taking away the taking away, what's left is putting things back. If you annihilite the annihitation the result is a net gain: Result: positive.






                                                                                                                                  share|cite|improve this answer









                                                                                                                                  $endgroup$



                                                                                                                                  Well.... this always made sense to me (but I've found it doesn't for others)



                                                                                                                                  A) Multiplication is adding a number a bunch of times.



                                                                                                                                  1) Negative numbers are numbers that are less than zero. The cancel out positive numbers. They are anti-numbers



                                                                                                                                  So)i) positive x positive: add a bunch of positive numbers a positive number of times. Result: A big positive gain.



                                                                                                                                  ii) positive x negative: add a bunch of negative anti-numbers. The result is a big amount of potential cancelling. Result: negative.



                                                                                                                                  iii) negative x positive: take a bunch of positive numbers and take them away. Result: a loss; negative.



                                                                                                                                  iv) negative x negative: take a bunch of anti-numbers and take them away. By taking away the taking away, what's left is putting things back. If you annihilite the annihitation the result is a net gain: Result: positive.







                                                                                                                                  share|cite|improve this answer












                                                                                                                                  share|cite|improve this answer



                                                                                                                                  share|cite|improve this answer










                                                                                                                                  answered Nov 9 '15 at 4:31









                                                                                                                                  fleabloodfleablood

                                                                                                                                  1




                                                                                                                                  1























                                                                                                                                      3












                                                                                                                                      $begingroup$

                                                                                                                                      Well, the way I think about it is this. We have the non-negative integers (0,1,2,3,4, etc.).



                                                                                                                                      We introduce the negative numbers and need to define multiplication with negative numbers so that we have internal consistency.



                                                                                                                                      We wish to keep the property that 0*anything = 0, negative or positive.



                                                                                                                                      We also want to keep the distributive property.



                                                                                                                                      In order to keep the above two properties, we're forced to define the product of two negatives as a positive.



                                                                                                                                      0*(-3) = 0



                                                                                                                                      (5 + (-5))*(-3) = 0 (I'm plugging in 5+ (-5) for zero)



                                                                                                                                      5*(-3) + (-5)*(-3) = 0 (distributive property)



                                                                                                                                      add 5*3 to both sides. 5*3 cancels with 5*(-3)



                                                                                                                                      (-5)*(-3) = 5*3






                                                                                                                                      share|cite|improve this answer









                                                                                                                                      $endgroup$


















                                                                                                                                        3












                                                                                                                                        $begingroup$

                                                                                                                                        Well, the way I think about it is this. We have the non-negative integers (0,1,2,3,4, etc.).



                                                                                                                                        We introduce the negative numbers and need to define multiplication with negative numbers so that we have internal consistency.



                                                                                                                                        We wish to keep the property that 0*anything = 0, negative or positive.



                                                                                                                                        We also want to keep the distributive property.



                                                                                                                                        In order to keep the above two properties, we're forced to define the product of two negatives as a positive.



                                                                                                                                        0*(-3) = 0



                                                                                                                                        (5 + (-5))*(-3) = 0 (I'm plugging in 5+ (-5) for zero)



                                                                                                                                        5*(-3) + (-5)*(-3) = 0 (distributive property)



                                                                                                                                        add 5*3 to both sides. 5*3 cancels with 5*(-3)



                                                                                                                                        (-5)*(-3) = 5*3






                                                                                                                                        share|cite|improve this answer









                                                                                                                                        $endgroup$
















                                                                                                                                          3












                                                                                                                                          3








                                                                                                                                          3





                                                                                                                                          $begingroup$

                                                                                                                                          Well, the way I think about it is this. We have the non-negative integers (0,1,2,3,4, etc.).



                                                                                                                                          We introduce the negative numbers and need to define multiplication with negative numbers so that we have internal consistency.



                                                                                                                                          We wish to keep the property that 0*anything = 0, negative or positive.



                                                                                                                                          We also want to keep the distributive property.



                                                                                                                                          In order to keep the above two properties, we're forced to define the product of two negatives as a positive.



                                                                                                                                          0*(-3) = 0



                                                                                                                                          (5 + (-5))*(-3) = 0 (I'm plugging in 5+ (-5) for zero)



                                                                                                                                          5*(-3) + (-5)*(-3) = 0 (distributive property)



                                                                                                                                          add 5*3 to both sides. 5*3 cancels with 5*(-3)



                                                                                                                                          (-5)*(-3) = 5*3






                                                                                                                                          share|cite|improve this answer









                                                                                                                                          $endgroup$



                                                                                                                                          Well, the way I think about it is this. We have the non-negative integers (0,1,2,3,4, etc.).



                                                                                                                                          We introduce the negative numbers and need to define multiplication with negative numbers so that we have internal consistency.



                                                                                                                                          We wish to keep the property that 0*anything = 0, negative or positive.



                                                                                                                                          We also want to keep the distributive property.



                                                                                                                                          In order to keep the above two properties, we're forced to define the product of two negatives as a positive.



                                                                                                                                          0*(-3) = 0



                                                                                                                                          (5 + (-5))*(-3) = 0 (I'm plugging in 5+ (-5) for zero)



                                                                                                                                          5*(-3) + (-5)*(-3) = 0 (distributive property)



                                                                                                                                          add 5*3 to both sides. 5*3 cancels with 5*(-3)



                                                                                                                                          (-5)*(-3) = 5*3







                                                                                                                                          share|cite|improve this answer












                                                                                                                                          share|cite|improve this answer



                                                                                                                                          share|cite|improve this answer










                                                                                                                                          answered Nov 9 '15 at 19:51









                                                                                                                                          Ameet SharmaAmeet Sharma

                                                                                                                                          666517




                                                                                                                                          666517






















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                                                                                                                                              protected by Asaf Karagila Nov 20 '14 at 14:47



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