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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?

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    $\begingroup$ 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. $\endgroup$ – jgon Dec 9 '18 at 22:35
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Yes it is correct and much a more natural approach to think of . Good job. smn

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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.

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  • $\begingroup$ 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 $\endgroup$ – ZFR Dec 9 '18 at 23:17
  • $\begingroup$ @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. $\endgroup$ – lisyarus Dec 9 '18 at 23:26
  • $\begingroup$ No worries! Remark such yours are very useful! Thank you ) $\endgroup$ – ZFR Dec 9 '18 at 23:34

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