Incredibly basic notation question: Is $-1^2$ 1 or -1? How about $-x^2$ where $x$ is $1$?
$begingroup$
Notation question: Is $-1^2$ equal to $1$ (because it's $(-1)^2$, as seems obvious) or $-1$ (because it's $-(1^2)$). E.g., is the $-1$ a single thing (token or whatever) in the equation? Similarly, let $x$ be $1$. Is $-x^2$ $1$ or $-1$?
I believe $-1^2$ is $1$ because "$-1$" is a single thing (a negative number), but that $-x^2$ where $x$ is $1$ is $-1$ because it's $-(x^2)$, the "$-x$" is not a single thing and exponeniation happens before negation.
Am I correct or incorrect? Or correct, but for the wrong reason?
This may seem incredibly odd to ask, but I'm suddenly unsure of something I've been sure of for more than four decades. Why would I question it? Because in JavaScript, -1 is not a numeric literal, it's a unary - operator followed by a numeric literal for the number 1, and this had pretty major ramifications for adding an exponentiation operator to the language, because the unary - has very high precedence. But I'm sure...ish that $-1$ in math notation is a literal number (and that exponentiation happens before negation).
notation
asked Dec 5 '18 at 15:29
T.J. CrowderT.J. Crowder
1185
$endgroup$
add a comment |
$begingroup$
Notation question: Is $-1^2$ equal to $1$ (because it's $(-1)^2$, as seems obvious) or $-1$ (because it's $-(1^2)$). E.g., is the $-1$ a single thing (token or whatever) in the equation? Similarly, let $x$ be $1$. Is $-x^2$ $1$ or $-1$?
I believe $-1^2$ is $1$ because "$-1$" is a single thing (a negative number), but that $-x^2$ where $x$ is $1$ is $-1$ because it's $-(x^2)$, the "$-x$" is not a single thing and exponeniation happens before negation.
Am I correct or incorrect? Or correct, but for the wrong reason?
This may seem incredibly odd to ask, but I'm suddenly unsure of something I've been sure of for more than four decades. Why would I question it? Because in JavaScript, -1 is not a numeric literal, it's a unary - operator followed by a numeric literal for the number 1, and this had pretty major ramifications for adding an exponentiation operator to the language, because the unary - has very high precedence. But I'm sure...ish that $-1$ in math notation is a literal number (and that exponentiation happens before negation).
notation
asked Dec 5 '18 at 15:29
T.J. CrowderT.J. Crowder
1185
$endgroup$
3
$begingroup$
This is purely anecdotal, but I've always read $-1^2$ as $-(1^2)$.
$endgroup$
– axblount
Dec 5 '18 at 15:33
3
$begingroup$
In analysis (such as power series expansions of trigonometric functions) you will see a lot of expressions involving $(-1)^n$. It is invariably always written this way, because $-1^n$ would mean something different (and trivial).
$endgroup$
– Henning Makholm
Dec 5 '18 at 15:40
add a comment |
$begingroup$
Notation question: Is $-1^2$ equal to $1$ (because it's $(-1)^2$, as seems obvious) or $-1$ (because it's $-(1^2)$). E.g., is the $-1$ a single thing (token or whatever) in the equation? Similarly, let $x$ be $1$. Is $-x^2$ $1$ or $-1$?
I believe $-1^2$ is $1$ because "$-1$" is a single thing (a negative number), but that $-x^2$ where $x$ is $1$ is $-1$ because it's $-(x^2)$, the "$-x$" is not a single thing and exponeniation happens before negation.
Am I correct or incorrect? Or correct, but for the wrong reason?
This may seem incredibly odd to ask, but I'm suddenly unsure of something I've been sure of for more than four decades. Why would I question it? Because in JavaScript, -1 is not a numeric literal, it's a unary - operator followed by a numeric literal for the number 1, and this had pretty major ramifications for adding an exponentiation operator to the language, because the unary - has very high precedence. But I'm sure...ish that $-1$ in math notation is a literal number (and that exponentiation happens before negation).
notation
asked Dec 5 '18 at 15:29
T.J. CrowderT.J. Crowder
1185
$endgroup$
Notation question: Is $-1^2$ equal to $1$ (because it's $(-1)^2$, as seems obvious) or $-1$ (because it's $-(1^2)$). E.g., is the $-1$ a single thing (token or whatever) in the equation? Similarly, let $x$ be $1$. Is $-x^2$ $1$ or $-1$?
I believe $-1^2$ is $1$ because "$-1$" is a single thing (a negative number), but that $-x^2$ where $x$ is $1$ is $-1$ because it's $-(x^2)$, the "$-x$" is not a single thing and exponeniation happens before negation.
Am I correct or incorrect? Or correct, but for the wrong reason?
This may seem incredibly odd to ask, but I'm suddenly unsure of something I've been sure of for more than four decades. Why would I question it? Because in JavaScript, -1 is not a numeric literal, it's a unary - operator followed by a numeric literal for the number 1, and this had pretty major ramifications for adding an exponentiation operator to the language, because the unary - has very high precedence. But I'm sure...ish that $-1$ in math notation is a literal number (and that exponentiation happens before negation).
notation
notation
asked Dec 5 '18 at 15:29
T.J. CrowderT.J. Crowder
1185
asked Dec 5 '18 at 15:29
T.J. CrowderT.J. Crowder
1185
asked Dec 5 '18 at 15:29
T.J. CrowderT.J. Crowder
1185
asked Dec 5 '18 at 15:29
T.J. CrowderT.J. Crowder
1185
asked Dec 5 '18 at 15:29
T.J. CrowderT.J. Crowder
1185
1185
3
$begingroup$
This is purely anecdotal, but I've always read $-1^2$ as $-(1^2)$.
$endgroup$
– axblount
Dec 5 '18 at 15:33
3
$begingroup$
In analysis (such as power series expansions of trigonometric functions) you will see a lot of expressions involving $(-1)^n$. It is invariably always written this way, because $-1^n$ would mean something different (and trivial).
$endgroup$
– Henning Makholm
Dec 5 '18 at 15:40
add a comment |
3
$begingroup$
This is purely anecdotal, but I've always read $-1^2$ as $-(1^2)$.
$endgroup$
– axblount
Dec 5 '18 at 15:33
3
$begingroup$
In analysis (such as power series expansions of trigonometric functions) you will see a lot of expressions involving $(-1)^n$. It is invariably always written this way, because $-1^n$ would mean something different (and trivial).
$endgroup$
– Henning Makholm
Dec 5 '18 at 15:40
3
3
$begingroup$
This is purely anecdotal, but I've always read $-1^2$ as $-(1^2)$.
$endgroup$
– axblount
Dec 5 '18 at 15:33
$begingroup$
This is purely anecdotal, but I've always read $-1^2$ as $-(1^2)$.
$endgroup$
– axblount
Dec 5 '18 at 15:33
3
3
$begingroup$
In analysis (such as power series expansions of trigonometric functions) you will see a lot of expressions involving $(-1)^n$. It is invariably always written this way, because $-1^n$ would mean something different (and trivial).
$endgroup$
– Henning Makholm
Dec 5 '18 at 15:40
$begingroup$
In analysis (such as power series expansions of trigonometric functions) you will see a lot of expressions involving $(-1)^n$. It is invariably always written this way, because $-1^n$ would mean something different (and trivial).
$endgroup$
– Henning Makholm
Dec 5 '18 at 15:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The exponent binds first, so $-1^2 = -1cdot 1 = -1.$
answered Dec 5 '18 at 15:31
ncmathsadistncmathsadist
42.4k260103
$endgroup$
$begingroup$
... wow .... I'm glad I asked. I'd gotten myself into a very confused state.
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:32
add a comment |
$begingroup$
You’re confusing $(-1)^2$ with $-1^2$. Recall that exponents have the priority, so you get
$$-1^2 = -(1)^2 = -1cdot 1 = -1$$
On the other hand, for $(-1)^2$, you get
$$(-1)^2 = (-1)cdot(-1) = +1$$
The same idea applies to all negative numbers raised to an even power. If the whole negative number is in parentheses, your answer becomes positive, but if the number is not within parentheses, the exponent comes first so the answer becomes negative.
answered Dec 5 '18 at 15:35
KM101KM101
5,9361523
$endgroup$
2
$begingroup$
I remembered that exponents had priority. My confusion was whether $-1$ is a single token (as we'd put it in programming), a number; or two tokens, negation and a number. I got myself in a right confused state over it. But of course, for years I've been reading things like $-2^{32} - 1$ correctly. Off day. :-)
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:43
$begingroup$
(sigh... $-2^{32} + 1$ I mean...)
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:55
2
$begingroup$
@T.J.Crowder - that confuses a lot of people! Thinking about it as a single token seems most natural at first, but you figured out in the question exactly why we don't do that: because we want $-x^2$ and $x^2$ to be different.
$endgroup$
– Deusovi
Dec 5 '18 at 17:53
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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votes
$begingroup$
The exponent binds first, so $-1^2 = -1cdot 1 = -1.$
answered Dec 5 '18 at 15:31
ncmathsadistncmathsadist
42.4k260103
$endgroup$
$begingroup$
... wow .... I'm glad I asked. I'd gotten myself into a very confused state.
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:32
add a comment |
$begingroup$
The exponent binds first, so $-1^2 = -1cdot 1 = -1.$
answered Dec 5 '18 at 15:31
ncmathsadistncmathsadist
42.4k260103
$endgroup$
$begingroup$
... wow .... I'm glad I asked. I'd gotten myself into a very confused state.
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:32
add a comment |
$begingroup$
The exponent binds first, so $-1^2 = -1cdot 1 = -1.$
answered Dec 5 '18 at 15:31
ncmathsadistncmathsadist
42.4k260103
$endgroup$
The exponent binds first, so $-1^2 = -1cdot 1 = -1.$
answered Dec 5 '18 at 15:31
ncmathsadistncmathsadist
42.4k260103
answered Dec 5 '18 at 15:31
ncmathsadistncmathsadist
42.4k260103
answered Dec 5 '18 at 15:31
ncmathsadistncmathsadist
42.4k260103
answered Dec 5 '18 at 15:31
ncmathsadistncmathsadist
42.4k260103
42.4k260103
$begingroup$
... wow .... I'm glad I asked. I'd gotten myself into a very confused state.
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:32
add a comment |
$begingroup$
... wow .... I'm glad I asked. I'd gotten myself into a very confused state.
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:32
$begingroup$
... wow .... I'm glad I asked. I'd gotten myself into a very confused state.
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:32
$begingroup$
... wow .... I'm glad I asked. I'd gotten myself into a very confused state.
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:32
add a comment |
$begingroup$
You’re confusing $(-1)^2$ with $-1^2$. Recall that exponents have the priority, so you get
$$-1^2 = -(1)^2 = -1cdot 1 = -1$$
On the other hand, for $(-1)^2$, you get
$$(-1)^2 = (-1)cdot(-1) = +1$$
The same idea applies to all negative numbers raised to an even power. If the whole negative number is in parentheses, your answer becomes positive, but if the number is not within parentheses, the exponent comes first so the answer becomes negative.
answered Dec 5 '18 at 15:35
KM101KM101
5,9361523
$endgroup$
2
$begingroup$
I remembered that exponents had priority. My confusion was whether $-1$ is a single token (as we'd put it in programming), a number; or two tokens, negation and a number. I got myself in a right confused state over it. But of course, for years I've been reading things like $-2^{32} - 1$ correctly. Off day. :-)
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:43
$begingroup$
(sigh... $-2^{32} + 1$ I mean...)
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:55
2
$begingroup$
@T.J.Crowder - that confuses a lot of people! Thinking about it as a single token seems most natural at first, but you figured out in the question exactly why we don't do that: because we want $-x^2$ and $x^2$ to be different.
$endgroup$
– Deusovi
Dec 5 '18 at 17:53
add a comment |
$begingroup$
You’re confusing $(-1)^2$ with $-1^2$. Recall that exponents have the priority, so you get
$$-1^2 = -(1)^2 = -1cdot 1 = -1$$
On the other hand, for $(-1)^2$, you get
$$(-1)^2 = (-1)cdot(-1) = +1$$
The same idea applies to all negative numbers raised to an even power. If the whole negative number is in parentheses, your answer becomes positive, but if the number is not within parentheses, the exponent comes first so the answer becomes negative.
answered Dec 5 '18 at 15:35
KM101KM101
5,9361523
$endgroup$
2
$begingroup$
I remembered that exponents had priority. My confusion was whether $-1$ is a single token (as we'd put it in programming), a number; or two tokens, negation and a number. I got myself in a right confused state over it. But of course, for years I've been reading things like $-2^{32} - 1$ correctly. Off day. :-)
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:43
$begingroup$
(sigh... $-2^{32} + 1$ I mean...)
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:55
2
$begingroup$
@T.J.Crowder - that confuses a lot of people! Thinking about it as a single token seems most natural at first, but you figured out in the question exactly why we don't do that: because we want $-x^2$ and $x^2$ to be different.
$endgroup$
– Deusovi
Dec 5 '18 at 17:53
add a comment |
$begingroup$
You’re confusing $(-1)^2$ with $-1^2$. Recall that exponents have the priority, so you get
$$-1^2 = -(1)^2 = -1cdot 1 = -1$$
On the other hand, for $(-1)^2$, you get
$$(-1)^2 = (-1)cdot(-1) = +1$$
The same idea applies to all negative numbers raised to an even power. If the whole negative number is in parentheses, your answer becomes positive, but if the number is not within parentheses, the exponent comes first so the answer becomes negative.
answered Dec 5 '18 at 15:35
KM101KM101
5,9361523
$endgroup$
You’re confusing $(-1)^2$ with $-1^2$. Recall that exponents have the priority, so you get
$$-1^2 = -(1)^2 = -1cdot 1 = -1$$
On the other hand, for $(-1)^2$, you get
$$(-1)^2 = (-1)cdot(-1) = +1$$
The same idea applies to all negative numbers raised to an even power. If the whole negative number is in parentheses, your answer becomes positive, but if the number is not within parentheses, the exponent comes first so the answer becomes negative.
answered Dec 5 '18 at 15:35
KM101KM101
5,9361523
answered Dec 5 '18 at 15:35
KM101KM101
5,9361523
answered Dec 5 '18 at 15:35
KM101KM101
5,9361523
answered Dec 5 '18 at 15:35
KM101KM101
5,9361523
5,9361523
2
$begingroup$
I remembered that exponents had priority. My confusion was whether $-1$ is a single token (as we'd put it in programming), a number; or two tokens, negation and a number. I got myself in a right confused state over it. But of course, for years I've been reading things like $-2^{32} - 1$ correctly. Off day. :-)
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:43
$begingroup$
(sigh... $-2^{32} + 1$ I mean...)
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:55
2
$begingroup$
@T.J.Crowder - that confuses a lot of people! Thinking about it as a single token seems most natural at first, but you figured out in the question exactly why we don't do that: because we want $-x^2$ and $x^2$ to be different.
$endgroup$
– Deusovi
Dec 5 '18 at 17:53
add a comment |
2
$begingroup$
I remembered that exponents had priority. My confusion was whether $-1$ is a single token (as we'd put it in programming), a number; or two tokens, negation and a number. I got myself in a right confused state over it. But of course, for years I've been reading things like $-2^{32} - 1$ correctly. Off day. :-)
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:43
$begingroup$
(sigh... $-2^{32} + 1$ I mean...)
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:55
2
$begingroup$
@T.J.Crowder - that confuses a lot of people! Thinking about it as a single token seems most natural at first, but you figured out in the question exactly why we don't do that: because we want $-x^2$ and $x^2$ to be different.
$endgroup$
– Deusovi
Dec 5 '18 at 17:53
2
2
$begingroup$
I remembered that exponents had priority. My confusion was whether $-1$ is a single token (as we'd put it in programming), a number; or two tokens, negation and a number. I got myself in a right confused state over it. But of course, for years I've been reading things like $-2^{32} - 1$ correctly. Off day. :-)
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:43
$begingroup$
I remembered that exponents had priority. My confusion was whether $-1$ is a single token (as we'd put it in programming), a number; or two tokens, negation and a number. I got myself in a right confused state over it. But of course, for years I've been reading things like $-2^{32} - 1$ correctly. Off day. :-)
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:43
$begingroup$
(sigh... $-2^{32} + 1$ I mean...)
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:55
$begingroup$
(sigh... $-2^{32} + 1$ I mean...)
$endgroup$
– T.J. Crowder
Dec 5 '18 at 15:55
2
2
$begingroup$
@T.J.Crowder - that confuses a lot of people! Thinking about it as a single token seems most natural at first, but you figured out in the question exactly why we don't do that: because we want $-x^2$ and $x^2$ to be different.
$endgroup$
– Deusovi
Dec 5 '18 at 17:53
$begingroup$
@T.J.Crowder - that confuses a lot of people! Thinking about it as a single token seems most natural at first, but you figured out in the question exactly why we don't do that: because we want $-x^2$ and $x^2$ to be different.
$endgroup$
– Deusovi
Dec 5 '18 at 17:53
add a comment |
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$begingroup$
This is purely anecdotal, but I've always read $-1^2$ as $-(1^2)$.
$endgroup$
– axblount
Dec 5 '18 at 15:33
3
$begingroup$
In analysis (such as power series expansions of trigonometric functions) you will see a lot of expressions involving $(-1)^n$. It is invariably always written this way, because $-1^n$ would mean something different (and trivial).
$endgroup$
– Henning Makholm
Dec 5 '18 at 15:40