How is a - (b - c) = c - (b - a) called?
$begingroup$
What is the following property of a subtraction operator called?
$a - (b - c) = c - (b - a)$
Having a name for this property is important for algebraic structures defined with a subtraction operator but no addition operator. And these are important for measures such as temperatures that can be subtracted from each other but not added to each other. Physicists call these intensive measures, while statisticians call these intensive variables.
For example, let’s call ordinal a type $mathfrak{O}$ which terms can be subtracted but not added. The best way to define the subtraction operator for this type is:
$- : mathfrak{O} times mathfrak{O} longrightarrow mathfrak{O} \ a - (b - c) = c - (b - a)$
Interestingly, the codomain of the subtraction operator should actually be an extended type called cardinal that does support addition, with addition defined as:
$+ : mathfrak{C} times mathfrak{C} longrightarrow mathfrak{C} \ a + b = c Longleftrightarrow c - a = b$
Therefore, the codomain of the subtraction operator should be $mathfrak{O}$, not $mathfrak{C}$.
$- : mathfrak{O} times mathfrak{O} longrightarrow mathfrak{C}$
But, when the domain of the subtraction operator is $mathfrak{O} times mathfrak{C}$, then its codomain remains $mathfrak{O}$. In other words, you can subtract a temperature from a temperature, and this gives you a temperature delta. But you can also subtract a temperature delta from a temperature, and this gives you a temperature.
From there, it becomes obvious that you can add a temperature delta to a temperature, which gives you a temperature, while adding two temperature deltas to each other gives a temperature delta. In other words, there is no commutative addition operator for the ordinal type $mathfrak{O}$, but there is something that comes pretty close (hence the confusion for most people).
If you define the ordinal type $mathfrak{O}$ as a coinductive type with successor as coinductive operator, an equivalence relation, and a total order relation, you can define a pair of forward and backward operators once you have added the subtraction operator and the cardinal type $mathfrak{C}$, with $mathfrak{O} times mathfrak{C}$ as domain and $mathfrak{O}$ as codomain.
The forward operator is equivalent to performing multiple successor operations, while the backward operator is equivalent to performing multiple predecessor operations. And with these, you have operators that let you “add” and “subtract” temperature deltas to and from temperatures, without letting you add two temperatures together.
This line of thinking has some interesting consequences. For example, it suggests a different definition for the arithmetic mean. Indeed, if one defines the arithmetic mean as a sum divided by a count, how does one gets the average of two temperatures since they cannot be summed?
In order to work around this problem, one has to define the subdivision operator on an extension of the cardinal type $mathfrak{C}$. We will call it the fractional type $mathfrak{F}$. From there, the arithmetic mean of two ordinal terms (like two temperatures) is defined as:
$a, b : mathfrak{C} mid a < b \ mean(a, b) = forward(a, (b - a) / 2)$
I should also mention that all this should apply to both sets or types, but I do all my work with types instead of sets, just because non-mathematicians are more comfortable with them, and because it lends itself nicely to the approach of building a hierarchy of type whereby an extended type is defined from a primitive type by simply adding an extra relation or operation.
I do not know how far this could go, but I am putting my ideas together on this notebook.
arithmetic
edited Jan 10 at 19:09
Namaste
1
asked Dec 31 '18 at 1:47
ismaelismael
281216
$endgroup$
|
show 8 more comments
$begingroup$
What is the following property of a subtraction operator called?
$a - (b - c) = c - (b - a)$
Having a name for this property is important for algebraic structures defined with a subtraction operator but no addition operator. And these are important for measures such as temperatures that can be subtracted from each other but not added to each other. Physicists call these intensive measures, while statisticians call these intensive variables.
For example, let’s call ordinal a type $mathfrak{O}$ which terms can be subtracted but not added. The best way to define the subtraction operator for this type is:
$- : mathfrak{O} times mathfrak{O} longrightarrow mathfrak{O} \ a - (b - c) = c - (b - a)$
Interestingly, the codomain of the subtraction operator should actually be an extended type called cardinal that does support addition, with addition defined as:
$+ : mathfrak{C} times mathfrak{C} longrightarrow mathfrak{C} \ a + b = c Longleftrightarrow c - a = b$
Therefore, the codomain of the subtraction operator should be $mathfrak{O}$, not $mathfrak{C}$.
$- : mathfrak{O} times mathfrak{O} longrightarrow mathfrak{C}$
But, when the domain of the subtraction operator is $mathfrak{O} times mathfrak{C}$, then its codomain remains $mathfrak{O}$. In other words, you can subtract a temperature from a temperature, and this gives you a temperature delta. But you can also subtract a temperature delta from a temperature, and this gives you a temperature.
From there, it becomes obvious that you can add a temperature delta to a temperature, which gives you a temperature, while adding two temperature deltas to each other gives a temperature delta. In other words, there is no commutative addition operator for the ordinal type $mathfrak{O}$, but there is something that comes pretty close (hence the confusion for most people).
If you define the ordinal type $mathfrak{O}$ as a coinductive type with successor as coinductive operator, an equivalence relation, and a total order relation, you can define a pair of forward and backward operators once you have added the subtraction operator and the cardinal type $mathfrak{C}$, with $mathfrak{O} times mathfrak{C}$ as domain and $mathfrak{O}$ as codomain.
The forward operator is equivalent to performing multiple successor operations, while the backward operator is equivalent to performing multiple predecessor operations. And with these, you have operators that let you “add” and “subtract” temperature deltas to and from temperatures, without letting you add two temperatures together.
This line of thinking has some interesting consequences. For example, it suggests a different definition for the arithmetic mean. Indeed, if one defines the arithmetic mean as a sum divided by a count, how does one gets the average of two temperatures since they cannot be summed?
In order to work around this problem, one has to define the subdivision operator on an extension of the cardinal type $mathfrak{C}$. We will call it the fractional type $mathfrak{F}$. From there, the arithmetic mean of two ordinal terms (like two temperatures) is defined as:
$a, b : mathfrak{C} mid a < b \ mean(a, b) = forward(a, (b - a) / 2)$
I should also mention that all this should apply to both sets or types, but I do all my work with types instead of sets, just because non-mathematicians are more comfortable with them, and because it lends itself nicely to the approach of building a hierarchy of type whereby an extended type is defined from a primitive type by simply adding an extra relation or operation.
I do not know how far this could go, but I am putting my ideas together on this notebook.
arithmetic
edited Jan 10 at 19:09
Namaste
1
asked Dec 31 '18 at 1:47
ismaelismael
281216
$endgroup$
1
$begingroup$
This is a composition of properties (conmutative and associative)
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 1:53
3
$begingroup$
Why would it have a name?
$endgroup$
– fleablood
Dec 31 '18 at 1:59
1
$begingroup$
Even if you only have interval variables, you have a category error. Take "time" for instance. You can take the difference between two clock times, which gives you a duration. A duration is not a time. You can add a duration to a time and get another time. Adding two times is nonsensical. A time doesn't even have a well-defined unit of measure.
$endgroup$
– obscurans
Dec 31 '18 at 3:46
2
$begingroup$
In the context of affine geometry, the difference of two points in an affine space is a vector in the underlying vector space. In the context of electricity voltage is determined by a difference in electrical potential. In both these cases, however, you can't take take the difference of two elements and then take the difference of that with a third element. That is, $(a-b)-c$ makes no sense in these contexts.
$endgroup$
– Somos
Dec 31 '18 at 6:09
2
$begingroup$
You can, however, define $a-(b-c)$ which is equal to $c-(b-a)$, which is the equivalent form of $a+(b-c)=b+(a-c)$. The two operations that are defined are 1. subtraction of two points, forming a vector, and 2. adding a vector to a point, getting another point. Vectors have a zero so they can be scaled by $-1$.
$endgroup$
– obscurans
Dec 31 '18 at 7:10
|
show 8 more comments
$begingroup$
What is the following property of a subtraction operator called?
$a - (b - c) = c - (b - a)$
Having a name for this property is important for algebraic structures defined with a subtraction operator but no addition operator. And these are important for measures such as temperatures that can be subtracted from each other but not added to each other. Physicists call these intensive measures, while statisticians call these intensive variables.
For example, let’s call ordinal a type $mathfrak{O}$ which terms can be subtracted but not added. The best way to define the subtraction operator for this type is:
$- : mathfrak{O} times mathfrak{O} longrightarrow mathfrak{O} \ a - (b - c) = c - (b - a)$
Interestingly, the codomain of the subtraction operator should actually be an extended type called cardinal that does support addition, with addition defined as:
$+ : mathfrak{C} times mathfrak{C} longrightarrow mathfrak{C} \ a + b = c Longleftrightarrow c - a = b$
Therefore, the codomain of the subtraction operator should be $mathfrak{O}$, not $mathfrak{C}$.
$- : mathfrak{O} times mathfrak{O} longrightarrow mathfrak{C}$
But, when the domain of the subtraction operator is $mathfrak{O} times mathfrak{C}$, then its codomain remains $mathfrak{O}$. In other words, you can subtract a temperature from a temperature, and this gives you a temperature delta. But you can also subtract a temperature delta from a temperature, and this gives you a temperature.
From there, it becomes obvious that you can add a temperature delta to a temperature, which gives you a temperature, while adding two temperature deltas to each other gives a temperature delta. In other words, there is no commutative addition operator for the ordinal type $mathfrak{O}$, but there is something that comes pretty close (hence the confusion for most people).
If you define the ordinal type $mathfrak{O}$ as a coinductive type with successor as coinductive operator, an equivalence relation, and a total order relation, you can define a pair of forward and backward operators once you have added the subtraction operator and the cardinal type $mathfrak{C}$, with $mathfrak{O} times mathfrak{C}$ as domain and $mathfrak{O}$ as codomain.
The forward operator is equivalent to performing multiple successor operations, while the backward operator is equivalent to performing multiple predecessor operations. And with these, you have operators that let you “add” and “subtract” temperature deltas to and from temperatures, without letting you add two temperatures together.
This line of thinking has some interesting consequences. For example, it suggests a different definition for the arithmetic mean. Indeed, if one defines the arithmetic mean as a sum divided by a count, how does one gets the average of two temperatures since they cannot be summed?
In order to work around this problem, one has to define the subdivision operator on an extension of the cardinal type $mathfrak{C}$. We will call it the fractional type $mathfrak{F}$. From there, the arithmetic mean of two ordinal terms (like two temperatures) is defined as:
$a, b : mathfrak{C} mid a < b \ mean(a, b) = forward(a, (b - a) / 2)$
I should also mention that all this should apply to both sets or types, but I do all my work with types instead of sets, just because non-mathematicians are more comfortable with them, and because it lends itself nicely to the approach of building a hierarchy of type whereby an extended type is defined from a primitive type by simply adding an extra relation or operation.
I do not know how far this could go, but I am putting my ideas together on this notebook.
arithmetic
edited Jan 10 at 19:09
Namaste
1
asked Dec 31 '18 at 1:47
ismaelismael
281216
$endgroup$
What is the following property of a subtraction operator called?
$a - (b - c) = c - (b - a)$
Having a name for this property is important for algebraic structures defined with a subtraction operator but no addition operator. And these are important for measures such as temperatures that can be subtracted from each other but not added to each other. Physicists call these intensive measures, while statisticians call these intensive variables.
For example, let’s call ordinal a type $mathfrak{O}$ which terms can be subtracted but not added. The best way to define the subtraction operator for this type is:
$- : mathfrak{O} times mathfrak{O} longrightarrow mathfrak{O} \ a - (b - c) = c - (b - a)$
Interestingly, the codomain of the subtraction operator should actually be an extended type called cardinal that does support addition, with addition defined as:
$+ : mathfrak{C} times mathfrak{C} longrightarrow mathfrak{C} \ a + b = c Longleftrightarrow c - a = b$
Therefore, the codomain of the subtraction operator should be $mathfrak{O}$, not $mathfrak{C}$.
$- : mathfrak{O} times mathfrak{O} longrightarrow mathfrak{C}$
But, when the domain of the subtraction operator is $mathfrak{O} times mathfrak{C}$, then its codomain remains $mathfrak{O}$. In other words, you can subtract a temperature from a temperature, and this gives you a temperature delta. But you can also subtract a temperature delta from a temperature, and this gives you a temperature.
From there, it becomes obvious that you can add a temperature delta to a temperature, which gives you a temperature, while adding two temperature deltas to each other gives a temperature delta. In other words, there is no commutative addition operator for the ordinal type $mathfrak{O}$, but there is something that comes pretty close (hence the confusion for most people).
If you define the ordinal type $mathfrak{O}$ as a coinductive type with successor as coinductive operator, an equivalence relation, and a total order relation, you can define a pair of forward and backward operators once you have added the subtraction operator and the cardinal type $mathfrak{C}$, with $mathfrak{O} times mathfrak{C}$ as domain and $mathfrak{O}$ as codomain.
The forward operator is equivalent to performing multiple successor operations, while the backward operator is equivalent to performing multiple predecessor operations. And with these, you have operators that let you “add” and “subtract” temperature deltas to and from temperatures, without letting you add two temperatures together.
This line of thinking has some interesting consequences. For example, it suggests a different definition for the arithmetic mean. Indeed, if one defines the arithmetic mean as a sum divided by a count, how does one gets the average of two temperatures since they cannot be summed?
In order to work around this problem, one has to define the subdivision operator on an extension of the cardinal type $mathfrak{C}$. We will call it the fractional type $mathfrak{F}$. From there, the arithmetic mean of two ordinal terms (like two temperatures) is defined as:
$a, b : mathfrak{C} mid a < b \ mean(a, b) = forward(a, (b - a) / 2)$
I should also mention that all this should apply to both sets or types, but I do all my work with types instead of sets, just because non-mathematicians are more comfortable with them, and because it lends itself nicely to the approach of building a hierarchy of type whereby an extended type is defined from a primitive type by simply adding an extra relation or operation.
I do not know how far this could go, but I am putting my ideas together on this notebook.
arithmetic
arithmetic
edited Jan 10 at 19:09
Namaste
1
asked Dec 31 '18 at 1:47
ismaelismael
281216
edited Jan 10 at 19:09
Namaste
1
asked Dec 31 '18 at 1:47
ismaelismael
281216
edited Jan 10 at 19:09
Namaste
1
edited Jan 10 at 19:09
Namaste
1
edited Jan 10 at 19:09
Namaste
1
1
asked Dec 31 '18 at 1:47
ismaelismael
281216
asked Dec 31 '18 at 1:47
ismaelismael
281216
asked Dec 31 '18 at 1:47
ismaelismael
281216
281216
1
$begingroup$
This is a composition of properties (conmutative and associative)
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 1:53
3
$begingroup$
Why would it have a name?
$endgroup$
– fleablood
Dec 31 '18 at 1:59
1
$begingroup$
Even if you only have interval variables, you have a category error. Take "time" for instance. You can take the difference between two clock times, which gives you a duration. A duration is not a time. You can add a duration to a time and get another time. Adding two times is nonsensical. A time doesn't even have a well-defined unit of measure.
$endgroup$
– obscurans
Dec 31 '18 at 3:46
2
$begingroup$
In the context of affine geometry, the difference of two points in an affine space is a vector in the underlying vector space. In the context of electricity voltage is determined by a difference in electrical potential. In both these cases, however, you can't take take the difference of two elements and then take the difference of that with a third element. That is, $(a-b)-c$ makes no sense in these contexts.
$endgroup$
– Somos
Dec 31 '18 at 6:09
2
$begingroup$
You can, however, define $a-(b-c)$ which is equal to $c-(b-a)$, which is the equivalent form of $a+(b-c)=b+(a-c)$. The two operations that are defined are 1. subtraction of two points, forming a vector, and 2. adding a vector to a point, getting another point. Vectors have a zero so they can be scaled by $-1$.
$endgroup$
– obscurans
Dec 31 '18 at 7:10
|
show 8 more comments
1
$begingroup$
This is a composition of properties (conmutative and associative)
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 1:53
3
$begingroup$
Why would it have a name?
$endgroup$
– fleablood
Dec 31 '18 at 1:59
1
$begingroup$
Even if you only have interval variables, you have a category error. Take "time" for instance. You can take the difference between two clock times, which gives you a duration. A duration is not a time. You can add a duration to a time and get another time. Adding two times is nonsensical. A time doesn't even have a well-defined unit of measure.
$endgroup$
– obscurans
Dec 31 '18 at 3:46
2
$begingroup$
In the context of affine geometry, the difference of two points in an affine space is a vector in the underlying vector space. In the context of electricity voltage is determined by a difference in electrical potential. In both these cases, however, you can't take take the difference of two elements and then take the difference of that with a third element. That is, $(a-b)-c$ makes no sense in these contexts.
$endgroup$
– Somos
Dec 31 '18 at 6:09
2
$begingroup$
You can, however, define $a-(b-c)$ which is equal to $c-(b-a)$, which is the equivalent form of $a+(b-c)=b+(a-c)$. The two operations that are defined are 1. subtraction of two points, forming a vector, and 2. adding a vector to a point, getting another point. Vectors have a zero so they can be scaled by $-1$.
$endgroup$
– obscurans
Dec 31 '18 at 7:10
1
1
$begingroup$
This is a composition of properties (conmutative and associative)
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 1:53
$begingroup$
This is a composition of properties (conmutative and associative)
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 1:53
3
3
$begingroup$
Why would it have a name?
$endgroup$
– fleablood
Dec 31 '18 at 1:59
$begingroup$
Why would it have a name?
$endgroup$
– fleablood
Dec 31 '18 at 1:59
1
1
$begingroup$
Even if you only have interval variables, you have a category error. Take "time" for instance. You can take the difference between two clock times, which gives you a duration. A duration is not a time. You can add a duration to a time and get another time. Adding two times is nonsensical. A time doesn't even have a well-defined unit of measure.
$endgroup$
– obscurans
Dec 31 '18 at 3:46
$begingroup$
Even if you only have interval variables, you have a category error. Take "time" for instance. You can take the difference between two clock times, which gives you a duration. A duration is not a time. You can add a duration to a time and get another time. Adding two times is nonsensical. A time doesn't even have a well-defined unit of measure.
$endgroup$
– obscurans
Dec 31 '18 at 3:46
2
2
$begingroup$
In the context of affine geometry, the difference of two points in an affine space is a vector in the underlying vector space. In the context of electricity voltage is determined by a difference in electrical potential. In both these cases, however, you can't take take the difference of two elements and then take the difference of that with a third element. That is, $(a-b)-c$ makes no sense in these contexts.
$endgroup$
– Somos
Dec 31 '18 at 6:09
$begingroup$
In the context of affine geometry, the difference of two points in an affine space is a vector in the underlying vector space. In the context of electricity voltage is determined by a difference in electrical potential. In both these cases, however, you can't take take the difference of two elements and then take the difference of that with a third element. That is, $(a-b)-c$ makes no sense in these contexts.
$endgroup$
– Somos
Dec 31 '18 at 6:09
2
2
$begingroup$
You can, however, define $a-(b-c)$ which is equal to $c-(b-a)$, which is the equivalent form of $a+(b-c)=b+(a-c)$. The two operations that are defined are 1. subtraction of two points, forming a vector, and 2. adding a vector to a point, getting another point. Vectors have a zero so they can be scaled by $-1$.
$endgroup$
– obscurans
Dec 31 '18 at 7:10
$begingroup$
You can, however, define $a-(b-c)$ which is equal to $c-(b-a)$, which is the equivalent form of $a+(b-c)=b+(a-c)$. The two operations that are defined are 1. subtraction of two points, forming a vector, and 2. adding a vector to a point, getting another point. Vectors have a zero so they can be scaled by $-1$.
$endgroup$
– obscurans
Dec 31 '18 at 7:10
|
show 8 more comments
2 Answers
2
active
oldest
votes
$begingroup$
By definition $$ (a-b)-c = (a-b)+(-c) = (a+(-b))+(-c)$$
Now we apply the associative property of addition to get
$$ ((a+(-b))+(-c) = a+((-b)+(-c))$$
Now we use the commutative and associative properties of addition to get $$ a+((-b)+(-c))= a+((-c)+(-b)) =(a+(-c))+(-b)=(a-c)-b$$
answered Dec 31 '18 at 1:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
2
$begingroup$
I should have clarified the fact that my working set or type does not have any addition, sorry.
$endgroup$
– ismael
Dec 31 '18 at 2:10
$begingroup$
@ismael really? Then what doesa - (0 - b)(where 0 = c - c) do?
$endgroup$
– John Dvorak
Jan 2 at 1:39
$begingroup$
@JohnDvorak I have updated the original post. This should help.
$endgroup$
– ismael
Jan 2 at 16:12
add a comment |
$begingroup$
You may assume $a,b,c in G$ where $left< G, + right>$ is an abelian group.
$(a-b) - c $
$= (a + (-b ) ) + ( - c ) $ ($(-b),(-c)$ is the inverse for $b,c$, repectively)
$= a + ( (-b) + (-c) )$ (associativity)
$= a + ( (-c) + (-b) )$ (commutativity)
$= ( a + (-c) ) + (-b)$ (associativity)
$= ( a-c ) - b$
answered Dec 31 '18 at 2:01
Ryu Dae SickRyu Dae Sick
705
$endgroup$
3
$begingroup$
I should have clarified the fact that my working set or type does not have any addition, sorry.
$endgroup$
– ismael
Dec 31 '18 at 2:10
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057369%2fhow-is-a-b-c-c-b-a-called%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By definition $$ (a-b)-c = (a-b)+(-c) = (a+(-b))+(-c)$$
Now we apply the associative property of addition to get
$$ ((a+(-b))+(-c) = a+((-b)+(-c))$$
Now we use the commutative and associative properties of addition to get $$ a+((-b)+(-c))= a+((-c)+(-b)) =(a+(-c))+(-b)=(a-c)-b$$
answered Dec 31 '18 at 1:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
2
$begingroup$
I should have clarified the fact that my working set or type does not have any addition, sorry.
$endgroup$
– ismael
Dec 31 '18 at 2:10
$begingroup$
@ismael really? Then what doesa - (0 - b)(where 0 = c - c) do?
$endgroup$
– John Dvorak
Jan 2 at 1:39
$begingroup$
@JohnDvorak I have updated the original post. This should help.
$endgroup$
– ismael
Jan 2 at 16:12
add a comment |
$begingroup$
By definition $$ (a-b)-c = (a-b)+(-c) = (a+(-b))+(-c)$$
Now we apply the associative property of addition to get
$$ ((a+(-b))+(-c) = a+((-b)+(-c))$$
Now we use the commutative and associative properties of addition to get $$ a+((-b)+(-c))= a+((-c)+(-b)) =(a+(-c))+(-b)=(a-c)-b$$
answered Dec 31 '18 at 1:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
2
$begingroup$
I should have clarified the fact that my working set or type does not have any addition, sorry.
$endgroup$
– ismael
Dec 31 '18 at 2:10
$begingroup$
@ismael really? Then what doesa - (0 - b)(where 0 = c - c) do?
$endgroup$
– John Dvorak
Jan 2 at 1:39
$begingroup$
@JohnDvorak I have updated the original post. This should help.
$endgroup$
– ismael
Jan 2 at 16:12
add a comment |
$begingroup$
By definition $$ (a-b)-c = (a-b)+(-c) = (a+(-b))+(-c)$$
Now we apply the associative property of addition to get
$$ ((a+(-b))+(-c) = a+((-b)+(-c))$$
Now we use the commutative and associative properties of addition to get $$ a+((-b)+(-c))= a+((-c)+(-b)) =(a+(-c))+(-b)=(a-c)-b$$
answered Dec 31 '18 at 1:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
By definition $$ (a-b)-c = (a-b)+(-c) = (a+(-b))+(-c)$$
Now we apply the associative property of addition to get
$$ ((a+(-b))+(-c) = a+((-b)+(-c))$$
Now we use the commutative and associative properties of addition to get $$ a+((-b)+(-c))= a+((-c)+(-b)) =(a+(-c))+(-b)=(a-c)-b$$
answered Dec 31 '18 at 1:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 31 '18 at 1:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 31 '18 at 1:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 31 '18 at 1:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
2
$begingroup$
I should have clarified the fact that my working set or type does not have any addition, sorry.
$endgroup$
– ismael
Dec 31 '18 at 2:10
$begingroup$
@ismael really? Then what doesa - (0 - b)(where 0 = c - c) do?
$endgroup$
– John Dvorak
Jan 2 at 1:39
$begingroup$
@JohnDvorak I have updated the original post. This should help.
$endgroup$
– ismael
Jan 2 at 16:12
add a comment |
2
$begingroup$
I should have clarified the fact that my working set or type does not have any addition, sorry.
$endgroup$
– ismael
Dec 31 '18 at 2:10
$begingroup$
@ismael really? Then what doesa - (0 - b)(where 0 = c - c) do?
$endgroup$
– John Dvorak
Jan 2 at 1:39
$begingroup$
@JohnDvorak I have updated the original post. This should help.
$endgroup$
– ismael
Jan 2 at 16:12
2
2
$begingroup$
I should have clarified the fact that my working set or type does not have any addition, sorry.
$endgroup$
– ismael
Dec 31 '18 at 2:10
$begingroup$
I should have clarified the fact that my working set or type does not have any addition, sorry.
$endgroup$
– ismael
Dec 31 '18 at 2:10
$begingroup$
@ismael really? Then what does
a - (0 - b) (where 0 = c - c) do?$endgroup$
– John Dvorak
Jan 2 at 1:39
$begingroup$
@ismael really? Then what does
a - (0 - b) (where 0 = c - c) do?$endgroup$
– John Dvorak
Jan 2 at 1:39
$begingroup$
@JohnDvorak I have updated the original post. This should help.
$endgroup$
– ismael
Jan 2 at 16:12
$begingroup$
@JohnDvorak I have updated the original post. This should help.
$endgroup$
– ismael
Jan 2 at 16:12
add a comment |
$begingroup$
You may assume $a,b,c in G$ where $left< G, + right>$ is an abelian group.
$(a-b) - c $
$= (a + (-b ) ) + ( - c ) $ ($(-b),(-c)$ is the inverse for $b,c$, repectively)
$= a + ( (-b) + (-c) )$ (associativity)
$= a + ( (-c) + (-b) )$ (commutativity)
$= ( a + (-c) ) + (-b)$ (associativity)
$= ( a-c ) - b$
answered Dec 31 '18 at 2:01
Ryu Dae SickRyu Dae Sick
705
$endgroup$
3
$begingroup$
I should have clarified the fact that my working set or type does not have any addition, sorry.
$endgroup$
– ismael
Dec 31 '18 at 2:10
add a comment |
$begingroup$
You may assume $a,b,c in G$ where $left< G, + right>$ is an abelian group.
$(a-b) - c $
$= (a + (-b ) ) + ( - c ) $ ($(-b),(-c)$ is the inverse for $b,c$, repectively)
$= a + ( (-b) + (-c) )$ (associativity)
$= a + ( (-c) + (-b) )$ (commutativity)
$= ( a + (-c) ) + (-b)$ (associativity)
$= ( a-c ) - b$
answered Dec 31 '18 at 2:01
Ryu Dae SickRyu Dae Sick
705
$endgroup$
3
$begingroup$
I should have clarified the fact that my working set or type does not have any addition, sorry.
$endgroup$
– ismael
Dec 31 '18 at 2:10
add a comment |
$begingroup$
You may assume $a,b,c in G$ where $left< G, + right>$ is an abelian group.
$(a-b) - c $
$= (a + (-b ) ) + ( - c ) $ ($(-b),(-c)$ is the inverse for $b,c$, repectively)
$= a + ( (-b) + (-c) )$ (associativity)
$= a + ( (-c) + (-b) )$ (commutativity)
$= ( a + (-c) ) + (-b)$ (associativity)
$= ( a-c ) - b$
answered Dec 31 '18 at 2:01
Ryu Dae SickRyu Dae Sick
705
$endgroup$
You may assume $a,b,c in G$ where $left< G, + right>$ is an abelian group.
$(a-b) - c $
$= (a + (-b ) ) + ( - c ) $ ($(-b),(-c)$ is the inverse for $b,c$, repectively)
$= a + ( (-b) + (-c) )$ (associativity)
$= a + ( (-c) + (-b) )$ (commutativity)
$= ( a + (-c) ) + (-b)$ (associativity)
$= ( a-c ) - b$
answered Dec 31 '18 at 2:01
Ryu Dae SickRyu Dae Sick
705
answered Dec 31 '18 at 2:01
Ryu Dae SickRyu Dae Sick
705
answered Dec 31 '18 at 2:01
Ryu Dae SickRyu Dae Sick
705
answered Dec 31 '18 at 2:01
Ryu Dae SickRyu Dae Sick
705
705
3
$begingroup$
I should have clarified the fact that my working set or type does not have any addition, sorry.
$endgroup$
– ismael
Dec 31 '18 at 2:10
add a comment |
3
$begingroup$
I should have clarified the fact that my working set or type does not have any addition, sorry.
$endgroup$
– ismael
Dec 31 '18 at 2:10
3
3
$begingroup$
I should have clarified the fact that my working set or type does not have any addition, sorry.
$endgroup$
– ismael
Dec 31 '18 at 2:10
$begingroup$
I should have clarified the fact that my working set or type does not have any addition, sorry.
$endgroup$
– ismael
Dec 31 '18 at 2:10
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057369%2fhow-is-a-b-c-c-b-a-called%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
This is a composition of properties (conmutative and associative)
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 1:53
3
$begingroup$
Why would it have a name?
$endgroup$
– fleablood
Dec 31 '18 at 1:59
1
$begingroup$
Even if you only have interval variables, you have a category error. Take "time" for instance. You can take the difference between two clock times, which gives you a duration. A duration is not a time. You can add a duration to a time and get another time. Adding two times is nonsensical. A time doesn't even have a well-defined unit of measure.
$endgroup$
– obscurans
Dec 31 '18 at 3:46
2
$begingroup$
In the context of affine geometry, the difference of two points in an affine space is a vector in the underlying vector space. In the context of electricity voltage is determined by a difference in electrical potential. In both these cases, however, you can't take take the difference of two elements and then take the difference of that with a third element. That is, $(a-b)-c$ makes no sense in these contexts.
$endgroup$
– Somos
Dec 31 '18 at 6:09
2
$begingroup$
You can, however, define $a-(b-c)$ which is equal to $c-(b-a)$, which is the equivalent form of $a+(b-c)=b+(a-c)$. The two operations that are defined are 1. subtraction of two points, forming a vector, and 2. adding a vector to a point, getting another point. Vectors have a zero so they can be scaled by $-1$.
$endgroup$
– obscurans
Dec 31 '18 at 7:10