Axioms of algebraic structure - ring
If in the definition of ring $(R,+,times$) we insist that it has unit element $1$. Then we can show that addition $(+)$ is commutative operation. However, most of the proof which I've seen in MSE use the following trick: $$x+x+y+y=(1+1)x+(1+1)y=(1+1)(x+y)=x+y+x+y$$ which lead to $x+y=y+x$.
But what if we use another approach, namely: Suppose $a=x+y$ and $b=y+x$. Then consider the expression: $$a-b=a+(-b)=(x+y)+(-(y+x))=(x+y)+((-y)+(-x))$$ and then using associativity of $+$ we get that: $a-b=0$ which leads to desired result.
Note that the crucial moment in my proof is the property $-b=(-1)cdot b$ which can be proven easily without any reliance on abelian of addition.
I suppose that this approach is also correct.
But what do you think about it?
abstract-algebra proof-verification ring-theory
edited Dec 9 '18 at 22:48
José Carlos Santos
150k22121221
asked Dec 9 '18 at 22:29
ZFR
4,95631338
add a comment |
If in the definition of ring $(R,+,times$) we insist that it has unit element $1$. Then we can show that addition $(+)$ is commutative operation. However, most of the proof which I've seen in MSE use the following trick: $$x+x+y+y=(1+1)x+(1+1)y=(1+1)(x+y)=x+y+x+y$$ which lead to $x+y=y+x$.
But what if we use another approach, namely: Suppose $a=x+y$ and $b=y+x$. Then consider the expression: $$a-b=a+(-b)=(x+y)+(-(y+x))=(x+y)+((-y)+(-x))$$ and then using associativity of $+$ we get that: $a-b=0$ which leads to desired result.
Note that the crucial moment in my proof is the property $-b=(-1)cdot b$ which can be proven easily without any reliance on abelian of addition.
I suppose that this approach is also correct.
But what do you think about it?
abstract-algebra proof-verification ring-theory
edited Dec 9 '18 at 22:48
José Carlos Santos
150k22121221
asked Dec 9 '18 at 22:29
ZFR
4,95631338
1
It looks good, but you should be more explicit about the usage of the property, since ordinarily if you don't assume addition is commutative and don't use that property, then $-(y+x)=(-x)+(-y)$, so that step threw me for a bit till I got down to where you explained you're using $-b = (-1)cdot b$ there.
– jgon
Dec 9 '18 at 22:35
add a comment |
If in the definition of ring $(R,+,times$) we insist that it has unit element $1$. Then we can show that addition $(+)$ is commutative operation. However, most of the proof which I've seen in MSE use the following trick: $$x+x+y+y=(1+1)x+(1+1)y=(1+1)(x+y)=x+y+x+y$$ which lead to $x+y=y+x$.
But what if we use another approach, namely: Suppose $a=x+y$ and $b=y+x$. Then consider the expression: $$a-b=a+(-b)=(x+y)+(-(y+x))=(x+y)+((-y)+(-x))$$ and then using associativity of $+$ we get that: $a-b=0$ which leads to desired result.
Note that the crucial moment in my proof is the property $-b=(-1)cdot b$ which can be proven easily without any reliance on abelian of addition.
I suppose that this approach is also correct.
But what do you think about it?
abstract-algebra proof-verification ring-theory
edited Dec 9 '18 at 22:48
José Carlos Santos
150k22121221
asked Dec 9 '18 at 22:29
ZFR
4,95631338
If in the definition of ring $(R,+,times$) we insist that it has unit element $1$. Then we can show that addition $(+)$ is commutative operation. However, most of the proof which I've seen in MSE use the following trick: $$x+x+y+y=(1+1)x+(1+1)y=(1+1)(x+y)=x+y+x+y$$ which lead to $x+y=y+x$.
But what if we use another approach, namely: Suppose $a=x+y$ and $b=y+x$. Then consider the expression: $$a-b=a+(-b)=(x+y)+(-(y+x))=(x+y)+((-y)+(-x))$$ and then using associativity of $+$ we get that: $a-b=0$ which leads to desired result.
Note that the crucial moment in my proof is the property $-b=(-1)cdot b$ which can be proven easily without any reliance on abelian of addition.
I suppose that this approach is also correct.
But what do you think about it?
abstract-algebra proof-verification ring-theory
abstract-algebra proof-verification ring-theory
edited Dec 9 '18 at 22:48
José Carlos Santos
150k22121221
asked Dec 9 '18 at 22:29
ZFR
4,95631338
edited Dec 9 '18 at 22:48
José Carlos Santos
150k22121221
asked Dec 9 '18 at 22:29
ZFR
4,95631338
edited Dec 9 '18 at 22:48
José Carlos Santos
150k22121221
edited Dec 9 '18 at 22:48
José Carlos Santos
150k22121221
edited Dec 9 '18 at 22:48
José Carlos Santos
150k22121221
150k22121221
asked Dec 9 '18 at 22:29
ZFR
4,95631338
asked Dec 9 '18 at 22:29
ZFR
4,95631338
asked Dec 9 '18 at 22:29
ZFR
4,95631338
4,95631338
1
It looks good, but you should be more explicit about the usage of the property, since ordinarily if you don't assume addition is commutative and don't use that property, then $-(y+x)=(-x)+(-y)$, so that step threw me for a bit till I got down to where you explained you're using $-b = (-1)cdot b$ there.
– jgon
Dec 9 '18 at 22:35
add a comment |
1
It looks good, but you should be more explicit about the usage of the property, since ordinarily if you don't assume addition is commutative and don't use that property, then $-(y+x)=(-x)+(-y)$, so that step threw me for a bit till I got down to where you explained you're using $-b = (-1)cdot b$ there.
– jgon
Dec 9 '18 at 22:35
1
1
It looks good, but you should be more explicit about the usage of the property, since ordinarily if you don't assume addition is commutative and don't use that property, then $-(y+x)=(-x)+(-y)$, so that step threw me for a bit till I got down to where you explained you're using $-b = (-1)cdot b$ there.
– jgon
Dec 9 '18 at 22:35
It looks good, but you should be more explicit about the usage of the property, since ordinarily if you don't assume addition is commutative and don't use that property, then $-(y+x)=(-x)+(-y)$, so that step threw me for a bit till I got down to where you explained you're using $-b = (-1)cdot b$ there.
– jgon
Dec 9 '18 at 22:35
add a comment |
2 Answers
2
active
oldest
votes
Yes it is correct and much a more natural approach to think of . Good job. smn
answered Dec 9 '18 at 23:03
StuartMN
1,406410
add a comment |
One step in your proof actually assumes commutativity of addition, namely
$$-(y+x)=-y-x$$
Without establishing commutativity beforehand, we can only say that
$$-(y+x)=-x-y$$
To see why this is the case, consider how one would prove the last equality:
$$(y+x) + ((-x) + (-y)) = y + (x + (-x)) + (-y) = y + 0 + (-y) = 0$$
where we relied heavily on associativity.
answered Dec 9 '18 at 23:08
lisyarus
10.5k21433
But we have the property $(-1)cdot b=-b$ which I said earlier can be proven without commutativity of addition. Hence $-(y+x)=(-1)(y+x)=(-1)y+(-1)x=(-y)+(-x)$. So everything is OK
– ZFR
Dec 9 '18 at 23:17
@ZFR My apologies, I didn't realize how you are using this property here. I think this deserves mentioning in the question, - since we are working with bare axioms, such small derivations are easily missed. That being said, your proof looks OK indeed.
– lisyarus
Dec 9 '18 at 23:26
No worries! Remark such yours are very useful! Thank you )
– ZFR
Dec 9 '18 at 23:34
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes it is correct and much a more natural approach to think of . Good job. smn
answered Dec 9 '18 at 23:03
StuartMN
1,406410
add a comment |
Yes it is correct and much a more natural approach to think of . Good job. smn
answered Dec 9 '18 at 23:03
StuartMN
1,406410
add a comment |
Yes it is correct and much a more natural approach to think of . Good job. smn
answered Dec 9 '18 at 23:03
StuartMN
1,406410
Yes it is correct and much a more natural approach to think of . Good job. smn
answered Dec 9 '18 at 23:03
StuartMN
1,406410
answered Dec 9 '18 at 23:03
StuartMN
1,406410
answered Dec 9 '18 at 23:03
StuartMN
1,406410
answered Dec 9 '18 at 23:03
StuartMN
1,406410
1,406410
add a comment |
add a comment |
One step in your proof actually assumes commutativity of addition, namely
$$-(y+x)=-y-x$$
Without establishing commutativity beforehand, we can only say that
$$-(y+x)=-x-y$$
To see why this is the case, consider how one would prove the last equality:
$$(y+x) + ((-x) + (-y)) = y + (x + (-x)) + (-y) = y + 0 + (-y) = 0$$
where we relied heavily on associativity.
answered Dec 9 '18 at 23:08
lisyarus
10.5k21433
But we have the property $(-1)cdot b=-b$ which I said earlier can be proven without commutativity of addition. Hence $-(y+x)=(-1)(y+x)=(-1)y+(-1)x=(-y)+(-x)$. So everything is OK
– ZFR
Dec 9 '18 at 23:17
@ZFR My apologies, I didn't realize how you are using this property here. I think this deserves mentioning in the question, - since we are working with bare axioms, such small derivations are easily missed. That being said, your proof looks OK indeed.
– lisyarus
Dec 9 '18 at 23:26
No worries! Remark such yours are very useful! Thank you )
– ZFR
Dec 9 '18 at 23:34
add a comment |
One step in your proof actually assumes commutativity of addition, namely
$$-(y+x)=-y-x$$
Without establishing commutativity beforehand, we can only say that
$$-(y+x)=-x-y$$
To see why this is the case, consider how one would prove the last equality:
$$(y+x) + ((-x) + (-y)) = y + (x + (-x)) + (-y) = y + 0 + (-y) = 0$$
where we relied heavily on associativity.
answered Dec 9 '18 at 23:08
lisyarus
10.5k21433
But we have the property $(-1)cdot b=-b$ which I said earlier can be proven without commutativity of addition. Hence $-(y+x)=(-1)(y+x)=(-1)y+(-1)x=(-y)+(-x)$. So everything is OK
– ZFR
Dec 9 '18 at 23:17
@ZFR My apologies, I didn't realize how you are using this property here. I think this deserves mentioning in the question, - since we are working with bare axioms, such small derivations are easily missed. That being said, your proof looks OK indeed.
– lisyarus
Dec 9 '18 at 23:26
No worries! Remark such yours are very useful! Thank you )
– ZFR
Dec 9 '18 at 23:34
add a comment |
One step in your proof actually assumes commutativity of addition, namely
$$-(y+x)=-y-x$$
Without establishing commutativity beforehand, we can only say that
$$-(y+x)=-x-y$$
To see why this is the case, consider how one would prove the last equality:
$$(y+x) + ((-x) + (-y)) = y + (x + (-x)) + (-y) = y + 0 + (-y) = 0$$
where we relied heavily on associativity.
answered Dec 9 '18 at 23:08
lisyarus
10.5k21433
One step in your proof actually assumes commutativity of addition, namely
$$-(y+x)=-y-x$$
Without establishing commutativity beforehand, we can only say that
$$-(y+x)=-x-y$$
To see why this is the case, consider how one would prove the last equality:
$$(y+x) + ((-x) + (-y)) = y + (x + (-x)) + (-y) = y + 0 + (-y) = 0$$
where we relied heavily on associativity.
answered Dec 9 '18 at 23:08
lisyarus
10.5k21433
answered Dec 9 '18 at 23:08
lisyarus
10.5k21433
answered Dec 9 '18 at 23:08
lisyarus
10.5k21433
answered Dec 9 '18 at 23:08
lisyarus
10.5k21433
10.5k21433
But we have the property $(-1)cdot b=-b$ which I said earlier can be proven without commutativity of addition. Hence $-(y+x)=(-1)(y+x)=(-1)y+(-1)x=(-y)+(-x)$. So everything is OK
– ZFR
Dec 9 '18 at 23:17
@ZFR My apologies, I didn't realize how you are using this property here. I think this deserves mentioning in the question, - since we are working with bare axioms, such small derivations are easily missed. That being said, your proof looks OK indeed.
– lisyarus
Dec 9 '18 at 23:26
No worries! Remark such yours are very useful! Thank you )
– ZFR
Dec 9 '18 at 23:34
add a comment |
But we have the property $(-1)cdot b=-b$ which I said earlier can be proven without commutativity of addition. Hence $-(y+x)=(-1)(y+x)=(-1)y+(-1)x=(-y)+(-x)$. So everything is OK
– ZFR
Dec 9 '18 at 23:17
@ZFR My apologies, I didn't realize how you are using this property here. I think this deserves mentioning in the question, - since we are working with bare axioms, such small derivations are easily missed. That being said, your proof looks OK indeed.
– lisyarus
Dec 9 '18 at 23:26
No worries! Remark such yours are very useful! Thank you )
– ZFR
Dec 9 '18 at 23:34
But we have the property $(-1)cdot b=-b$ which I said earlier can be proven without commutativity of addition. Hence $-(y+x)=(-1)(y+x)=(-1)y+(-1)x=(-y)+(-x)$. So everything is OK
– ZFR
Dec 9 '18 at 23:17
But we have the property $(-1)cdot b=-b$ which I said earlier can be proven without commutativity of addition. Hence $-(y+x)=(-1)(y+x)=(-1)y+(-1)x=(-y)+(-x)$. So everything is OK
– ZFR
Dec 9 '18 at 23:17
@ZFR My apologies, I didn't realize how you are using this property here. I think this deserves mentioning in the question, - since we are working with bare axioms, such small derivations are easily missed. That being said, your proof looks OK indeed.
– lisyarus
Dec 9 '18 at 23:26
@ZFR My apologies, I didn't realize how you are using this property here. I think this deserves mentioning in the question, - since we are working with bare axioms, such small derivations are easily missed. That being said, your proof looks OK indeed.
– lisyarus
Dec 9 '18 at 23:26
No worries! Remark such yours are very useful! Thank you )
– ZFR
Dec 9 '18 at 23:34
No worries! Remark such yours are very useful! Thank you )
– ZFR
Dec 9 '18 at 23:34
add a comment |
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1
It looks good, but you should be more explicit about the usage of the property, since ordinarily if you don't assume addition is commutative and don't use that property, then $-(y+x)=(-x)+(-y)$, so that step threw me for a bit till I got down to where you explained you're using $-b = (-1)cdot b$ there.
– jgon
Dec 9 '18 at 22:35