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Is there a proof that performing an operation on both sides of an equation preserves equality?
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so I was learning some abstract algebra and group theory, when they went over the proof of the cancellation law
$$
ab = acimplies a^-1(ab) = a^-1(ac)implies (a^-1a)b = (a^-1a)cimplies eb=ec implies b=c
$$
But the first step in which you add the additional term seems jarring to me, especially since I felt we were proving ever trivial thing from the ground up. Obviously I'm familiar with middle school pre-algebra, so I know that it is true that if we perform an operation on both sides it preserves equality, but I didn't know how we know this is the case always. Is it an axiom or is it proven?
Here is my attempt at a proof, let me know if I am going in the correct direction.
Assume via axiom that $x=x$ and if $a=b$ and $b=c$ then $a=c$, and prove that $a=bimplies ka=kb$
We know that $a=b$, we define $xmid x=a implies a=x$. Since $a=b$ and $a=x$, then $b=x$ which we can rewrite as $x=x$. Now we perform the operation on both sides $kx=kx$, which is true via our axiom. Then we re-substitute $x=a$ and $x=b$ to get $ka=kb$. Q.E.D
That was my original idea but I don't know if that's watertight. Thank you!
abstract-algebra group-theory proof-writing axioms
asked Aug 14 at 17:45
wjmccann
598117
add a comment |Â
up vote
12
down vote
favorite
so I was learning some abstract algebra and group theory, when they went over the proof of the cancellation law
$$
ab = acimplies a^-1(ab) = a^-1(ac)implies (a^-1a)b = (a^-1a)cimplies eb=ec implies b=c
$$
But the first step in which you add the additional term seems jarring to me, especially since I felt we were proving ever trivial thing from the ground up. Obviously I'm familiar with middle school pre-algebra, so I know that it is true that if we perform an operation on both sides it preserves equality, but I didn't know how we know this is the case always. Is it an axiom or is it proven?
Here is my attempt at a proof, let me know if I am going in the correct direction.
Assume via axiom that $x=x$ and if $a=b$ and $b=c$ then $a=c$, and prove that $a=bimplies ka=kb$
We know that $a=b$, we define $xmid x=a implies a=x$. Since $a=b$ and $a=x$, then $b=x$ which we can rewrite as $x=x$. Now we perform the operation on both sides $kx=kx$, which is true via our axiom. Then we re-substitute $x=a$ and $x=b$ to get $ka=kb$. Q.E.D
That was my original idea but I don't know if that's watertight. Thank you!
abstract-algebra group-theory proof-writing axioms
asked Aug 14 at 17:45
wjmccann
598117
8
One way of looking at it is to observe that multiplication in a group is a function. If $G$ is a group, and $a$ is a fixed element, then $f:xmapsto ax$ is a function $f$ from $G$ to itself. Because a function only has a single value at all the elements of the domain, we have the implication: $x=yimplies f(x)=f(y)$. This applies to all functions. Injective or not (injectivity gives you the converse implication also).
– Jyrki Lahtonen
Aug 14 at 18:43
Here's the answer I gave to a similar and much related question : math.stackexchange.com/questions/2808744/…
– Max
Aug 15 at 13:05
add a comment |Â
up vote
12
down vote
favorite
up vote
12
down vote
favorite
so I was learning some abstract algebra and group theory, when they went over the proof of the cancellation law
$$
ab = acimplies a^-1(ab) = a^-1(ac)implies (a^-1a)b = (a^-1a)cimplies eb=ec implies b=c
$$
But the first step in which you add the additional term seems jarring to me, especially since I felt we were proving ever trivial thing from the ground up. Obviously I'm familiar with middle school pre-algebra, so I know that it is true that if we perform an operation on both sides it preserves equality, but I didn't know how we know this is the case always. Is it an axiom or is it proven?
Here is my attempt at a proof, let me know if I am going in the correct direction.
Assume via axiom that $x=x$ and if $a=b$ and $b=c$ then $a=c$, and prove that $a=bimplies ka=kb$
We know that $a=b$, we define $xmid x=a implies a=x$. Since $a=b$ and $a=x$, then $b=x$ which we can rewrite as $x=x$. Now we perform the operation on both sides $kx=kx$, which is true via our axiom. Then we re-substitute $x=a$ and $x=b$ to get $ka=kb$. Q.E.D
That was my original idea but I don't know if that's watertight. Thank you!
abstract-algebra group-theory proof-writing axioms
asked Aug 14 at 17:45
wjmccann
598117
so I was learning some abstract algebra and group theory, when they went over the proof of the cancellation law
$$
ab = acimplies a^-1(ab) = a^-1(ac)implies (a^-1a)b = (a^-1a)cimplies eb=ec implies b=c
$$
But the first step in which you add the additional term seems jarring to me, especially since I felt we were proving ever trivial thing from the ground up. Obviously I'm familiar with middle school pre-algebra, so I know that it is true that if we perform an operation on both sides it preserves equality, but I didn't know how we know this is the case always. Is it an axiom or is it proven?
Here is my attempt at a proof, let me know if I am going in the correct direction.
Assume via axiom that $x=x$ and if $a=b$ and $b=c$ then $a=c$, and prove that $a=bimplies ka=kb$
We know that $a=b$, we define $xmid x=a implies a=x$. Since $a=b$ and $a=x$, then $b=x$ which we can rewrite as $x=x$. Now we perform the operation on both sides $kx=kx$, which is true via our axiom. Then we re-substitute $x=a$ and $x=b$ to get $ka=kb$. Q.E.D
That was my original idea but I don't know if that's watertight. Thank you!
abstract-algebra group-theory proof-writing axioms
asked Aug 14 at 17:45
wjmccann
598117
asked Aug 14 at 17:45
wjmccann
598117
asked Aug 14 at 17:45
wjmccann
598117
asked Aug 14 at 17:45
wjmccann
598117
598117
8
One way of looking at it is to observe that multiplication in a group is a function. If $G$ is a group, and $a$ is a fixed element, then $f:xmapsto ax$ is a function $f$ from $G$ to itself. Because a function only has a single value at all the elements of the domain, we have the implication: $x=yimplies f(x)=f(y)$. This applies to all functions. Injective or not (injectivity gives you the converse implication also).
– Jyrki Lahtonen
Aug 14 at 18:43
Here's the answer I gave to a similar and much related question : math.stackexchange.com/questions/2808744/…
– Max
Aug 15 at 13:05
add a comment |Â
8
One way of looking at it is to observe that multiplication in a group is a function. If $G$ is a group, and $a$ is a fixed element, then $f:xmapsto ax$ is a function $f$ from $G$ to itself. Because a function only has a single value at all the elements of the domain, we have the implication: $x=yimplies f(x)=f(y)$. This applies to all functions. Injective or not (injectivity gives you the converse implication also).
– Jyrki Lahtonen
Aug 14 at 18:43
Here's the answer I gave to a similar and much related question : math.stackexchange.com/questions/2808744/…
– Max
Aug 15 at 13:05
8
8
One way of looking at it is to observe that multiplication in a group is a function. If $G$ is a group, and $a$ is a fixed element, then $f:xmapsto ax$ is a function $f$ from $G$ to itself. Because a function only has a single value at all the elements of the domain, we have the implication: $x=yimplies f(x)=f(y)$. This applies to all functions. Injective or not (injectivity gives you the converse implication also).
– Jyrki Lahtonen
Aug 14 at 18:43
One way of looking at it is to observe that multiplication in a group is a function. If $G$ is a group, and $a$ is a fixed element, then $f:xmapsto ax$ is a function $f$ from $G$ to itself. Because a function only has a single value at all the elements of the domain, we have the implication: $x=yimplies f(x)=f(y)$. This applies to all functions. Injective or not (injectivity gives you the converse implication also).
– Jyrki Lahtonen
Aug 14 at 18:43
Here's the answer I gave to a similar and much related question : math.stackexchange.com/questions/2808744/…
– Max
Aug 15 at 13:05
Here's the answer I gave to a similar and much related question : math.stackexchange.com/questions/2808744/…
– Max
Aug 15 at 13:05
add a comment |Â
4 Answers
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up vote
15
down vote
accepted
In first-order logic, we have the formal substitution principle:
Let $phi$ be a propositional formula with a free variable $v$, and let $Gamma$ be a context. Also, let $x, y$ be two terms representing values. Then:
beginalign* Gamma & vdash x = y \
Gamma & vdash phi[v := x] \
hline
Gamma & vdash phi[v := y]. endalign*
Informally, what this says is: if you can prove in some context that $x=y$, and you can also prove some statement is true for $x$, then you can conclude the same statement is true for $y$. (The notation $phi[v := x]$ just means the result of substituting $x$ in for $v$ in the proposition formula $phi$.)
Now, if we are working in a group, let us apply this to the formula $phi := (a^-1 (ab) = a^-1 v)$. Then in the context of the proof, we are assuming $ab = ac$. Also, $phi[v := ab]$ results in the proposition $a^-1(ab) = a^-1(ab)$, which is true by the first-order axiom (or in some formulations, the formal proof rule) of reflexivity of equality: $t = t$ for any term $t$. Therefore, the substitution principle allows us to conclude that $phi[v := ac]$ is true, which results in $a^-1 (ab) = a^-1 (ac)$.
To give another application which is implicitly used in the proof, let us see how to use the substitution principle to prove the transitivity of equality: if we have $x=y$ and $y=z$ then $x=z$. For this proof, we will use $phi := (x = v)$. Then we are assuming $y=z$. We also have that $phi[v := y]$ is true since it reduces to the assumption $x = y$. Therefore, we can conclude $phi[v := z]$ which is just $x = z$.
answered Aug 14 at 18:07
Daniel Schepler
7,0481514
1
No disrespect meant to the answer, but this explanation is always offered, and I never understand how it helps. The key of the problem seems to be just that the group operation is a function when one side is fixed. When people bring up "the substitution principle" I am always reminded of students who defined some $f$ and then concluded that $a=bimplies f(a)=f(b)$ by the substitution principle, when in fact their $f$ wasn't a function at all. In other words, I guess they didn't recognize their $f$ didn't satisfy the description given above. Very easy to misunderstand, comparatively.
– rschwieb
Aug 14 at 20:23
2
@rschwieb Your comment along the lines of "it's because group multiplication is assumed to be a function" would make a great separate answer.
– Daniel Schepler
Aug 14 at 21:06
Reluctant to, since the last sentence of the other solution and Jyrki's comment already say as much :/
– rschwieb
Aug 15 at 2:31
@rschwieb I'm quite sure the error in those proofs can be pointed out without referring to the concept of a function.
– JiK
Aug 15 at 10:01
1
@JiK Given that "function" is a prerequisite for defining a group, I don't see the merit in skirting what functions are. If the answer to a question already follows by definition, that should be mentioned foremost.
– rschwieb
Aug 15 at 13:11
 |Â
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11
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We usually take it as an axiom of the relation of equality. In particular, we assume that on a set $E$ there is a relation $=$ which satisfies the following properties for $a,b,cin E:$
(1) $a=a$ for all $ain E$ (egoism),
(2) If $a=b,$ then $b=a$ (reciprocity),
(3) If $a=b$ and $b=c,$ then $a=c$ (continuity),
(4) For any function $f$ defined on $E,$ we have that if $a=b,$ then $f(a)=f(b)$ (conservation).
answered Aug 14 at 18:05
Allawonder
1,847516
3
I suppose that reflexivity, symmetry, transitivity are more commonly used for the first three properties
– Hagen von Eitzen
Aug 14 at 18:19
@HagenvonEitzen Yes, that is correct.
– Allawonder
Aug 14 at 18:46
3
+1 for the last line alone. Sorry that I missed it in my first reading, when typing my comment under the question
– Jyrki Lahtonen
Aug 14 at 18:48
@Allawonder Where do 1-3 even come into play? As far as I can see it simply follows from 4) since multiplication on the left by an element is by definition a function.
– rschwieb
Aug 14 at 20:17
1
@Allawonder OK: thanks for the clarifying comment :)
– rschwieb
Aug 15 at 13:20
 |Â
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If the sign "$=$" was some relation that we had just defined on strings like "$ab$", then it would be necessary to establish what you ask for. That is not the case, however. The sign "$=$" denotes equality and juxtaposition of letters denotes multiplication, which is a function. And for every function $f$ you have that $x=y$ implies $f(x)=f(y)$. (Or more similar to the situation here: $x=y$ implies $g(z, x)=g(z, y)$.)
answered Aug 15 at 9:22
Carsten S
6,78911334
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Your proof seems fine, but a bit more complicated than necessary.
We start with x = x by the reflexive property of equality. Then we substitute each x with ka yielding ka = ka, since the reflexive property of equality holds for all x, and closure holds for any operation by definition.
Then, since a = b, we can choose to only replace the rightmost "a" with "b" and that yields,
ka = kb.
Since we've assumed an arbitrary binary operation here (not a group) which we suppressed writing, for any binary operation F, we have that if a = b, then F(k, a) = F(k, b).
The result has many corollaries, all assuming that x = y, such as F(a, k) = F(b, k), and for a unary operation U, U(x) = U(y), and for any trinary operation T, T(k, a, j) = T(k, b, j). The most general of which seems as follows:
Theoerm: Suppose that x = y, and that a$_1$, ..., a$_n$ is a complete list L of variables and constants in a formula with an operation F. Suppose for some a$_k$ in L, a$_k$ = x. (Note k could equal 1, or k could equal n). Then
F(a$_1$, ..., a$_k$, ..., a$_n$) = F(a$_1$, ..., y, ..., a$_n$).
Proof: I'll leave this as an exercise. See above for a hint.
answered Aug 17 at 22:59
Doug Spoonwood
7,80312043
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
15
down vote
accepted
In first-order logic, we have the formal substitution principle:
Let $phi$ be a propositional formula with a free variable $v$, and let $Gamma$ be a context. Also, let $x, y$ be two terms representing values. Then:
beginalign* Gamma & vdash x = y \
Gamma & vdash phi[v := x] \
hline
Gamma & vdash phi[v := y]. endalign*
Informally, what this says is: if you can prove in some context that $x=y$, and you can also prove some statement is true for $x$, then you can conclude the same statement is true for $y$. (The notation $phi[v := x]$ just means the result of substituting $x$ in for $v$ in the proposition formula $phi$.)
Now, if we are working in a group, let us apply this to the formula $phi := (a^-1 (ab) = a^-1 v)$. Then in the context of the proof, we are assuming $ab = ac$. Also, $phi[v := ab]$ results in the proposition $a^-1(ab) = a^-1(ab)$, which is true by the first-order axiom (or in some formulations, the formal proof rule) of reflexivity of equality: $t = t$ for any term $t$. Therefore, the substitution principle allows us to conclude that $phi[v := ac]$ is true, which results in $a^-1 (ab) = a^-1 (ac)$.
To give another application which is implicitly used in the proof, let us see how to use the substitution principle to prove the transitivity of equality: if we have $x=y$ and $y=z$ then $x=z$. For this proof, we will use $phi := (x = v)$. Then we are assuming $y=z$. We also have that $phi[v := y]$ is true since it reduces to the assumption $x = y$. Therefore, we can conclude $phi[v := z]$ which is just $x = z$.
answered Aug 14 at 18:07
Daniel Schepler
7,0481514
1
No disrespect meant to the answer, but this explanation is always offered, and I never understand how it helps. The key of the problem seems to be just that the group operation is a function when one side is fixed. When people bring up "the substitution principle" I am always reminded of students who defined some $f$ and then concluded that $a=bimplies f(a)=f(b)$ by the substitution principle, when in fact their $f$ wasn't a function at all. In other words, I guess they didn't recognize their $f$ didn't satisfy the description given above. Very easy to misunderstand, comparatively.
– rschwieb
Aug 14 at 20:23
2
@rschwieb Your comment along the lines of "it's because group multiplication is assumed to be a function" would make a great separate answer.
– Daniel Schepler
Aug 14 at 21:06
Reluctant to, since the last sentence of the other solution and Jyrki's comment already say as much :/
– rschwieb
Aug 15 at 2:31
@rschwieb I'm quite sure the error in those proofs can be pointed out without referring to the concept of a function.
– JiK
Aug 15 at 10:01
1
@JiK Given that "function" is a prerequisite for defining a group, I don't see the merit in skirting what functions are. If the answer to a question already follows by definition, that should be mentioned foremost.
– rschwieb
Aug 15 at 13:11
 |Â
show 4 more comments
up vote
15
down vote
accepted
In first-order logic, we have the formal substitution principle:
Let $phi$ be a propositional formula with a free variable $v$, and let $Gamma$ be a context. Also, let $x, y$ be two terms representing values. Then:
beginalign* Gamma & vdash x = y \
Gamma & vdash phi[v := x] \
hline
Gamma & vdash phi[v := y]. endalign*
Informally, what this says is: if you can prove in some context that $x=y$, and you can also prove some statement is true for $x$, then you can conclude the same statement is true for $y$. (The notation $phi[v := x]$ just means the result of substituting $x$ in for $v$ in the proposition formula $phi$.)
Now, if we are working in a group, let us apply this to the formula $phi := (a^-1 (ab) = a^-1 v)$. Then in the context of the proof, we are assuming $ab = ac$. Also, $phi[v := ab]$ results in the proposition $a^-1(ab) = a^-1(ab)$, which is true by the first-order axiom (or in some formulations, the formal proof rule) of reflexivity of equality: $t = t$ for any term $t$. Therefore, the substitution principle allows us to conclude that $phi[v := ac]$ is true, which results in $a^-1 (ab) = a^-1 (ac)$.
To give another application which is implicitly used in the proof, let us see how to use the substitution principle to prove the transitivity of equality: if we have $x=y$ and $y=z$ then $x=z$. For this proof, we will use $phi := (x = v)$. Then we are assuming $y=z$. We also have that $phi[v := y]$ is true since it reduces to the assumption $x = y$. Therefore, we can conclude $phi[v := z]$ which is just $x = z$.
answered Aug 14 at 18:07
Daniel Schepler
7,0481514
1
No disrespect meant to the answer, but this explanation is always offered, and I never understand how it helps. The key of the problem seems to be just that the group operation is a function when one side is fixed. When people bring up "the substitution principle" I am always reminded of students who defined some $f$ and then concluded that $a=bimplies f(a)=f(b)$ by the substitution principle, when in fact their $f$ wasn't a function at all. In other words, I guess they didn't recognize their $f$ didn't satisfy the description given above. Very easy to misunderstand, comparatively.
– rschwieb
Aug 14 at 20:23
2
@rschwieb Your comment along the lines of "it's because group multiplication is assumed to be a function" would make a great separate answer.
– Daniel Schepler
Aug 14 at 21:06
Reluctant to, since the last sentence of the other solution and Jyrki's comment already say as much :/
– rschwieb
Aug 15 at 2:31
@rschwieb I'm quite sure the error in those proofs can be pointed out without referring to the concept of a function.
– JiK
Aug 15 at 10:01
1
@JiK Given that "function" is a prerequisite for defining a group, I don't see the merit in skirting what functions are. If the answer to a question already follows by definition, that should be mentioned foremost.
– rschwieb
Aug 15 at 13:11
 |Â
show 4 more comments
up vote
15
down vote
accepted
up vote
15
down vote
accepted
In first-order logic, we have the formal substitution principle:
Let $phi$ be a propositional formula with a free variable $v$, and let $Gamma$ be a context. Also, let $x, y$ be two terms representing values. Then:
beginalign* Gamma & vdash x = y \
Gamma & vdash phi[v := x] \
hline
Gamma & vdash phi[v := y]. endalign*
Informally, what this says is: if you can prove in some context that $x=y$, and you can also prove some statement is true for $x$, then you can conclude the same statement is true for $y$. (The notation $phi[v := x]$ just means the result of substituting $x$ in for $v$ in the proposition formula $phi$.)
Now, if we are working in a group, let us apply this to the formula $phi := (a^-1 (ab) = a^-1 v)$. Then in the context of the proof, we are assuming $ab = ac$. Also, $phi[v := ab]$ results in the proposition $a^-1(ab) = a^-1(ab)$, which is true by the first-order axiom (or in some formulations, the formal proof rule) of reflexivity of equality: $t = t$ for any term $t$. Therefore, the substitution principle allows us to conclude that $phi[v := ac]$ is true, which results in $a^-1 (ab) = a^-1 (ac)$.
To give another application which is implicitly used in the proof, let us see how to use the substitution principle to prove the transitivity of equality: if we have $x=y$ and $y=z$ then $x=z$. For this proof, we will use $phi := (x = v)$. Then we are assuming $y=z$. We also have that $phi[v := y]$ is true since it reduces to the assumption $x = y$. Therefore, we can conclude $phi[v := z]$ which is just $x = z$.
answered Aug 14 at 18:07
Daniel Schepler
7,0481514
In first-order logic, we have the formal substitution principle:
Let $phi$ be a propositional formula with a free variable $v$, and let $Gamma$ be a context. Also, let $x, y$ be two terms representing values. Then:
beginalign* Gamma & vdash x = y \
Gamma & vdash phi[v := x] \
hline
Gamma & vdash phi[v := y]. endalign*
Informally, what this says is: if you can prove in some context that $x=y$, and you can also prove some statement is true for $x$, then you can conclude the same statement is true for $y$. (The notation $phi[v := x]$ just means the result of substituting $x$ in for $v$ in the proposition formula $phi$.)
Now, if we are working in a group, let us apply this to the formula $phi := (a^-1 (ab) = a^-1 v)$. Then in the context of the proof, we are assuming $ab = ac$. Also, $phi[v := ab]$ results in the proposition $a^-1(ab) = a^-1(ab)$, which is true by the first-order axiom (or in some formulations, the formal proof rule) of reflexivity of equality: $t = t$ for any term $t$. Therefore, the substitution principle allows us to conclude that $phi[v := ac]$ is true, which results in $a^-1 (ab) = a^-1 (ac)$.
To give another application which is implicitly used in the proof, let us see how to use the substitution principle to prove the transitivity of equality: if we have $x=y$ and $y=z$ then $x=z$. For this proof, we will use $phi := (x = v)$. Then we are assuming $y=z$. We also have that $phi[v := y]$ is true since it reduces to the assumption $x = y$. Therefore, we can conclude $phi[v := z]$ which is just $x = z$.
answered Aug 14 at 18:07
Daniel Schepler
7,0481514
answered Aug 14 at 18:07
Daniel Schepler
7,0481514
answered Aug 14 at 18:07
Daniel Schepler
7,0481514
answered Aug 14 at 18:07
Daniel Schepler
7,0481514
7,0481514
1
No disrespect meant to the answer, but this explanation is always offered, and I never understand how it helps. The key of the problem seems to be just that the group operation is a function when one side is fixed. When people bring up "the substitution principle" I am always reminded of students who defined some $f$ and then concluded that $a=bimplies f(a)=f(b)$ by the substitution principle, when in fact their $f$ wasn't a function at all. In other words, I guess they didn't recognize their $f$ didn't satisfy the description given above. Very easy to misunderstand, comparatively.
– rschwieb
Aug 14 at 20:23
2
@rschwieb Your comment along the lines of "it's because group multiplication is assumed to be a function" would make a great separate answer.
– Daniel Schepler
Aug 14 at 21:06
Reluctant to, since the last sentence of the other solution and Jyrki's comment already say as much :/
– rschwieb
Aug 15 at 2:31
@rschwieb I'm quite sure the error in those proofs can be pointed out without referring to the concept of a function.
– JiK
Aug 15 at 10:01
1
@JiK Given that "function" is a prerequisite for defining a group, I don't see the merit in skirting what functions are. If the answer to a question already follows by definition, that should be mentioned foremost.
– rschwieb
Aug 15 at 13:11
 |Â
show 4 more comments
1
No disrespect meant to the answer, but this explanation is always offered, and I never understand how it helps. The key of the problem seems to be just that the group operation is a function when one side is fixed. When people bring up "the substitution principle" I am always reminded of students who defined some $f$ and then concluded that $a=bimplies f(a)=f(b)$ by the substitution principle, when in fact their $f$ wasn't a function at all. In other words, I guess they didn't recognize their $f$ didn't satisfy the description given above. Very easy to misunderstand, comparatively.
– rschwieb
Aug 14 at 20:23
2
@rschwieb Your comment along the lines of "it's because group multiplication is assumed to be a function" would make a great separate answer.
– Daniel Schepler
Aug 14 at 21:06
Reluctant to, since the last sentence of the other solution and Jyrki's comment already say as much :/
– rschwieb
Aug 15 at 2:31
@rschwieb I'm quite sure the error in those proofs can be pointed out without referring to the concept of a function.
– JiK
Aug 15 at 10:01
1
@JiK Given that "function" is a prerequisite for defining a group, I don't see the merit in skirting what functions are. If the answer to a question already follows by definition, that should be mentioned foremost.
– rschwieb
Aug 15 at 13:11
1
1
No disrespect meant to the answer, but this explanation is always offered, and I never understand how it helps. The key of the problem seems to be just that the group operation is a function when one side is fixed. When people bring up "the substitution principle" I am always reminded of students who defined some $f$ and then concluded that $a=bimplies f(a)=f(b)$ by the substitution principle, when in fact their $f$ wasn't a function at all. In other words, I guess they didn't recognize their $f$ didn't satisfy the description given above. Very easy to misunderstand, comparatively.
– rschwieb
Aug 14 at 20:23
No disrespect meant to the answer, but this explanation is always offered, and I never understand how it helps. The key of the problem seems to be just that the group operation is a function when one side is fixed. When people bring up "the substitution principle" I am always reminded of students who defined some $f$ and then concluded that $a=bimplies f(a)=f(b)$ by the substitution principle, when in fact their $f$ wasn't a function at all. In other words, I guess they didn't recognize their $f$ didn't satisfy the description given above. Very easy to misunderstand, comparatively.
– rschwieb
Aug 14 at 20:23
2
2
@rschwieb Your comment along the lines of "it's because group multiplication is assumed to be a function" would make a great separate answer.
– Daniel Schepler
Aug 14 at 21:06
@rschwieb Your comment along the lines of "it's because group multiplication is assumed to be a function" would make a great separate answer.
– Daniel Schepler
Aug 14 at 21:06
Reluctant to, since the last sentence of the other solution and Jyrki's comment already say as much :/
– rschwieb
Aug 15 at 2:31
Reluctant to, since the last sentence of the other solution and Jyrki's comment already say as much :/
– rschwieb
Aug 15 at 2:31
@rschwieb I'm quite sure the error in those proofs can be pointed out without referring to the concept of a function.
– JiK
Aug 15 at 10:01
@rschwieb I'm quite sure the error in those proofs can be pointed out without referring to the concept of a function.
– JiK
Aug 15 at 10:01
1
1
@JiK Given that "function" is a prerequisite for defining a group, I don't see the merit in skirting what functions are. If the answer to a question already follows by definition, that should be mentioned foremost.
– rschwieb
Aug 15 at 13:11
@JiK Given that "function" is a prerequisite for defining a group, I don't see the merit in skirting what functions are. If the answer to a question already follows by definition, that should be mentioned foremost.
– rschwieb
Aug 15 at 13:11
 |Â
show 4 more comments
up vote
11
down vote
We usually take it as an axiom of the relation of equality. In particular, we assume that on a set $E$ there is a relation $=$ which satisfies the following properties for $a,b,cin E:$
(1) $a=a$ for all $ain E$ (egoism),
(2) If $a=b,$ then $b=a$ (reciprocity),
(3) If $a=b$ and $b=c,$ then $a=c$ (continuity),
(4) For any function $f$ defined on $E,$ we have that if $a=b,$ then $f(a)=f(b)$ (conservation).
answered Aug 14 at 18:05
Allawonder
1,847516
3
I suppose that reflexivity, symmetry, transitivity are more commonly used for the first three properties
– Hagen von Eitzen
Aug 14 at 18:19
@HagenvonEitzen Yes, that is correct.
– Allawonder
Aug 14 at 18:46
3
+1 for the last line alone. Sorry that I missed it in my first reading, when typing my comment under the question
– Jyrki Lahtonen
Aug 14 at 18:48
@Allawonder Where do 1-3 even come into play? As far as I can see it simply follows from 4) since multiplication on the left by an element is by definition a function.
– rschwieb
Aug 14 at 20:17
1
@Allawonder OK: thanks for the clarifying comment :)
– rschwieb
Aug 15 at 13:20
 |Â
show 2 more comments
up vote
11
down vote
We usually take it as an axiom of the relation of equality. In particular, we assume that on a set $E$ there is a relation $=$ which satisfies the following properties for $a,b,cin E:$
(1) $a=a$ for all $ain E$ (egoism),
(2) If $a=b,$ then $b=a$ (reciprocity),
(3) If $a=b$ and $b=c,$ then $a=c$ (continuity),
(4) For any function $f$ defined on $E,$ we have that if $a=b,$ then $f(a)=f(b)$ (conservation).
answered Aug 14 at 18:05
Allawonder
1,847516
3
I suppose that reflexivity, symmetry, transitivity are more commonly used for the first three properties
– Hagen von Eitzen
Aug 14 at 18:19
@HagenvonEitzen Yes, that is correct.
– Allawonder
Aug 14 at 18:46
3
+1 for the last line alone. Sorry that I missed it in my first reading, when typing my comment under the question
– Jyrki Lahtonen
Aug 14 at 18:48
@Allawonder Where do 1-3 even come into play? As far as I can see it simply follows from 4) since multiplication on the left by an element is by definition a function.
– rschwieb
Aug 14 at 20:17
1
@Allawonder OK: thanks for the clarifying comment :)
– rschwieb
Aug 15 at 13:20
 |Â
show 2 more comments
up vote
11
down vote
up vote
11
down vote
We usually take it as an axiom of the relation of equality. In particular, we assume that on a set $E$ there is a relation $=$ which satisfies the following properties for $a,b,cin E:$
(1) $a=a$ for all $ain E$ (egoism),
(2) If $a=b,$ then $b=a$ (reciprocity),
(3) If $a=b$ and $b=c,$ then $a=c$ (continuity),
(4) For any function $f$ defined on $E,$ we have that if $a=b,$ then $f(a)=f(b)$ (conservation).
answered Aug 14 at 18:05
Allawonder
1,847516
We usually take it as an axiom of the relation of equality. In particular, we assume that on a set $E$ there is a relation $=$ which satisfies the following properties for $a,b,cin E:$
(1) $a=a$ for all $ain E$ (egoism),
(2) If $a=b,$ then $b=a$ (reciprocity),
(3) If $a=b$ and $b=c,$ then $a=c$ (continuity),
(4) For any function $f$ defined on $E,$ we have that if $a=b,$ then $f(a)=f(b)$ (conservation).
answered Aug 14 at 18:05
Allawonder
1,847516
edited Aug 14 at 19:06
answered Aug 14 at 18:05
Allawonder
1,847516
answered Aug 14 at 18:05
Allawonder
1,847516
answered Aug 14 at 18:05
Allawonder
1,847516
1,847516
3
I suppose that reflexivity, symmetry, transitivity are more commonly used for the first three properties
– Hagen von Eitzen
Aug 14 at 18:19
@HagenvonEitzen Yes, that is correct.
– Allawonder
Aug 14 at 18:46
3
+1 for the last line alone. Sorry that I missed it in my first reading, when typing my comment under the question
– Jyrki Lahtonen
Aug 14 at 18:48
@Allawonder Where do 1-3 even come into play? As far as I can see it simply follows from 4) since multiplication on the left by an element is by definition a function.
– rschwieb
Aug 14 at 20:17
1
@Allawonder OK: thanks for the clarifying comment :)
– rschwieb
Aug 15 at 13:20
 |Â
show 2 more comments
3
I suppose that reflexivity, symmetry, transitivity are more commonly used for the first three properties
– Hagen von Eitzen
Aug 14 at 18:19
@HagenvonEitzen Yes, that is correct.
– Allawonder
Aug 14 at 18:46
3
+1 for the last line alone. Sorry that I missed it in my first reading, when typing my comment under the question
– Jyrki Lahtonen
Aug 14 at 18:48
@Allawonder Where do 1-3 even come into play? As far as I can see it simply follows from 4) since multiplication on the left by an element is by definition a function.
– rschwieb
Aug 14 at 20:17
1
@Allawonder OK: thanks for the clarifying comment :)
– rschwieb
Aug 15 at 13:20
3
3
I suppose that reflexivity, symmetry, transitivity are more commonly used for the first three properties
– Hagen von Eitzen
Aug 14 at 18:19
I suppose that reflexivity, symmetry, transitivity are more commonly used for the first three properties
– Hagen von Eitzen
Aug 14 at 18:19
@HagenvonEitzen Yes, that is correct.
– Allawonder
Aug 14 at 18:46
@HagenvonEitzen Yes, that is correct.
– Allawonder
Aug 14 at 18:46
3
3
+1 for the last line alone. Sorry that I missed it in my first reading, when typing my comment under the question
– Jyrki Lahtonen
Aug 14 at 18:48
+1 for the last line alone. Sorry that I missed it in my first reading, when typing my comment under the question
– Jyrki Lahtonen
Aug 14 at 18:48
@Allawonder Where do 1-3 even come into play? As far as I can see it simply follows from 4) since multiplication on the left by an element is by definition a function.
– rschwieb
Aug 14 at 20:17
@Allawonder Where do 1-3 even come into play? As far as I can see it simply follows from 4) since multiplication on the left by an element is by definition a function.
– rschwieb
Aug 14 at 20:17
1
1
@Allawonder OK: thanks for the clarifying comment :)
– rschwieb
Aug 15 at 13:20
@Allawonder OK: thanks for the clarifying comment :)
– rschwieb
Aug 15 at 13:20
 |Â
show 2 more comments
up vote
3
down vote
If the sign "$=$" was some relation that we had just defined on strings like "$ab$", then it would be necessary to establish what you ask for. That is not the case, however. The sign "$=$" denotes equality and juxtaposition of letters denotes multiplication, which is a function. And for every function $f$ you have that $x=y$ implies $f(x)=f(y)$. (Or more similar to the situation here: $x=y$ implies $g(z, x)=g(z, y)$.)
answered Aug 15 at 9:22
Carsten S
6,78911334
add a comment |Â
up vote
3
down vote
If the sign "$=$" was some relation that we had just defined on strings like "$ab$", then it would be necessary to establish what you ask for. That is not the case, however. The sign "$=$" denotes equality and juxtaposition of letters denotes multiplication, which is a function. And for every function $f$ you have that $x=y$ implies $f(x)=f(y)$. (Or more similar to the situation here: $x=y$ implies $g(z, x)=g(z, y)$.)
answered Aug 15 at 9:22
Carsten S
6,78911334
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If the sign "$=$" was some relation that we had just defined on strings like "$ab$", then it would be necessary to establish what you ask for. That is not the case, however. The sign "$=$" denotes equality and juxtaposition of letters denotes multiplication, which is a function. And for every function $f$ you have that $x=y$ implies $f(x)=f(y)$. (Or more similar to the situation here: $x=y$ implies $g(z, x)=g(z, y)$.)
answered Aug 15 at 9:22
Carsten S
6,78911334
If the sign "$=$" was some relation that we had just defined on strings like "$ab$", then it would be necessary to establish what you ask for. That is not the case, however. The sign "$=$" denotes equality and juxtaposition of letters denotes multiplication, which is a function. And for every function $f$ you have that $x=y$ implies $f(x)=f(y)$. (Or more similar to the situation here: $x=y$ implies $g(z, x)=g(z, y)$.)
answered Aug 15 at 9:22
Carsten S
6,78911334
edited Aug 15 at 21:39
answered Aug 15 at 9:22
Carsten S
6,78911334
answered Aug 15 at 9:22
Carsten S
6,78911334
answered Aug 15 at 9:22
Carsten S
6,78911334
6,78911334
add a comment |Â
add a comment |Â
up vote
1
down vote
Your proof seems fine, but a bit more complicated than necessary.
We start with x = x by the reflexive property of equality. Then we substitute each x with ka yielding ka = ka, since the reflexive property of equality holds for all x, and closure holds for any operation by definition.
Then, since a = b, we can choose to only replace the rightmost "a" with "b" and that yields,
ka = kb.
Since we've assumed an arbitrary binary operation here (not a group) which we suppressed writing, for any binary operation F, we have that if a = b, then F(k, a) = F(k, b).
The result has many corollaries, all assuming that x = y, such as F(a, k) = F(b, k), and for a unary operation U, U(x) = U(y), and for any trinary operation T, T(k, a, j) = T(k, b, j). The most general of which seems as follows:
Theoerm: Suppose that x = y, and that a$_1$, ..., a$_n$ is a complete list L of variables and constants in a formula with an operation F. Suppose for some a$_k$ in L, a$_k$ = x. (Note k could equal 1, or k could equal n). Then
F(a$_1$, ..., a$_k$, ..., a$_n$) = F(a$_1$, ..., y, ..., a$_n$).
Proof: I'll leave this as an exercise. See above for a hint.
answered Aug 17 at 22:59
Doug Spoonwood
7,80312043
add a comment |Â
up vote
1
down vote
Your proof seems fine, but a bit more complicated than necessary.
We start with x = x by the reflexive property of equality. Then we substitute each x with ka yielding ka = ka, since the reflexive property of equality holds for all x, and closure holds for any operation by definition.
Then, since a = b, we can choose to only replace the rightmost "a" with "b" and that yields,
ka = kb.
Since we've assumed an arbitrary binary operation here (not a group) which we suppressed writing, for any binary operation F, we have that if a = b, then F(k, a) = F(k, b).
The result has many corollaries, all assuming that x = y, such as F(a, k) = F(b, k), and for a unary operation U, U(x) = U(y), and for any trinary operation T, T(k, a, j) = T(k, b, j). The most general of which seems as follows:
Theoerm: Suppose that x = y, and that a$_1$, ..., a$_n$ is a complete list L of variables and constants in a formula with an operation F. Suppose for some a$_k$ in L, a$_k$ = x. (Note k could equal 1, or k could equal n). Then
F(a$_1$, ..., a$_k$, ..., a$_n$) = F(a$_1$, ..., y, ..., a$_n$).
Proof: I'll leave this as an exercise. See above for a hint.
answered Aug 17 at 22:59
Doug Spoonwood
7,80312043
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your proof seems fine, but a bit more complicated than necessary.
We start with x = x by the reflexive property of equality. Then we substitute each x with ka yielding ka = ka, since the reflexive property of equality holds for all x, and closure holds for any operation by definition.
Then, since a = b, we can choose to only replace the rightmost "a" with "b" and that yields,
ka = kb.
Since we've assumed an arbitrary binary operation here (not a group) which we suppressed writing, for any binary operation F, we have that if a = b, then F(k, a) = F(k, b).
The result has many corollaries, all assuming that x = y, such as F(a, k) = F(b, k), and for a unary operation U, U(x) = U(y), and for any trinary operation T, T(k, a, j) = T(k, b, j). The most general of which seems as follows:
Theoerm: Suppose that x = y, and that a$_1$, ..., a$_n$ is a complete list L of variables and constants in a formula with an operation F. Suppose for some a$_k$ in L, a$_k$ = x. (Note k could equal 1, or k could equal n). Then
F(a$_1$, ..., a$_k$, ..., a$_n$) = F(a$_1$, ..., y, ..., a$_n$).
Proof: I'll leave this as an exercise. See above for a hint.
answered Aug 17 at 22:59
Doug Spoonwood
7,80312043
Your proof seems fine, but a bit more complicated than necessary.
We start with x = x by the reflexive property of equality. Then we substitute each x with ka yielding ka = ka, since the reflexive property of equality holds for all x, and closure holds for any operation by definition.
Then, since a = b, we can choose to only replace the rightmost "a" with "b" and that yields,
ka = kb.
Since we've assumed an arbitrary binary operation here (not a group) which we suppressed writing, for any binary operation F, we have that if a = b, then F(k, a) = F(k, b).
The result has many corollaries, all assuming that x = y, such as F(a, k) = F(b, k), and for a unary operation U, U(x) = U(y), and for any trinary operation T, T(k, a, j) = T(k, b, j). The most general of which seems as follows:
Theoerm: Suppose that x = y, and that a$_1$, ..., a$_n$ is a complete list L of variables and constants in a formula with an operation F. Suppose for some a$_k$ in L, a$_k$ = x. (Note k could equal 1, or k could equal n). Then
F(a$_1$, ..., a$_k$, ..., a$_n$) = F(a$_1$, ..., y, ..., a$_n$).
Proof: I'll leave this as an exercise. See above for a hint.
answered Aug 17 at 22:59
Doug Spoonwood
7,80312043
answered Aug 17 at 22:59
Doug Spoonwood
7,80312043
answered Aug 17 at 22:59
Doug Spoonwood
7,80312043
answered Aug 17 at 22:59
Doug Spoonwood
7,80312043
7,80312043
add a comment |Â
add a comment |Â
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8
One way of looking at it is to observe that multiplication in a group is a function. If $G$ is a group, and $a$ is a fixed element, then $f:xmapsto ax$ is a function $f$ from $G$ to itself. Because a function only has a single value at all the elements of the domain, we have the implication: $x=yimplies f(x)=f(y)$. This applies to all functions. Injective or not (injectivity gives you the converse implication also).
– Jyrki Lahtonen
Aug 14 at 18:43
Here's the answer I gave to a similar and much related question : math.stackexchange.com/questions/2808744/…
– Max
Aug 15 at 13:05