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?

  • 1
    $\begingroup$ 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

Yes it is correct and much a more natural approach to think of . Good job. smn


One step in your proof actually assumes commutativity of addition, namely


Without establishing commutativity beforehand, we can only say that


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.

  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.