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I could not prove that "Every Boolean ring is commutative", but I found it on the Internet. I am giving the idea of the proof that I "learned".

Let $R$ be the Boolean ring.

We can easily prove that $x+x = 0 \; \forall \; x \in R$. Then, we again use the same idea that $(x+y)^2 = (x+y), \; \forall \; x,y \in R$ to get that $xy=-yx$ and use the fact that $xy= -xy$ to get the result.

We had to find the new property for the Boolean ring that $x+x =0 $ to solve that it is commutative. Even with this solution, I cannot understand what makes a Boolean ring commutative. Is there any intuitive way to understand that Boolean ring is commutative? I am not looking for just a computational answer (if possible, there should be some idea behind the computation). I am hoping for any sort of intuition or idea (mathematical or in plain English) that tells us why Boolean ring is commutative.

P.S.: A Boolean ring $R$ is a ring in which $a^2 = a, \; \forall \; a \in R $.

MUH
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  • OK, let's agree the question is "A heuristic explanation for why a boolean ring is commutative in plain English, not merely just a computation." Then I suppose it is not a duplicate of https://math.stackexchange.com/q/10274/29335 (or any of its 17 duplicates) – rschwieb Feb 19 '19 at 14:46
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    I think the intuition comes from the correspondence between Boolean algebras and Boolean rings. The commutativity of multiplication in the Boolean ring is equivalent to commutativity of logical conjunction (the AND operator), and the latter does seems pretty intuitive. – Badam Baplan Feb 19 '19 at 14:55
  • @BadamBaplan That's a good handhold! If one is willing to stop there and say "it basically amounts to conjunction being symmetric" then we're done. One could object to it just being an equivalent rephrasing, though. – rschwieb Feb 19 '19 at 16:02

3 Answers3

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$$x+y = (x+y)^2 = x^2 + xy + yx + y^2 = x + xy + yx + y$$

This shows $xy + yx = 0$. That means $xy + \underbrace{xy + yx}_{=0} = xy$. From your statement ($X + X = 0$) we can rearrange to see:

$$\underbrace{xy + xy}_{=0} + yx = xy \Rightarrow yx = xy$$

naslundx
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    It is correct, of course: however, the user indicates they already know this, and they have requested a "reason in plain English, not a computation." So... this does not really address the question. – rschwieb Feb 19 '19 at 14:49
  • @rschwieb I see, that clarification came after this answer. I shall try to update. – naslundx Feb 19 '19 at 15:43
  • Sorry: then it is not your fault. I missed the difference in the time stamps. But if that was the case, it would really have been preferable to link this to the dustbunny of duplicates of this question, rather than answering. – rschwieb Feb 19 '19 at 15:56
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$(a+b)^2 = a+b \implies ab+ba=0$.

Now multiply $ab+ba=0$ with $a$ from left side we have $ab+aba=0$ and multiply $ab+ba=0$ with $a$ from right side $aba+ba=0$.

Now equating $(-aba)$'s value we have $ab=ba$ for all $a,b$ in the ring.

Arup Mazumder
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First a small claim: If $R$ is Boolean, then $-a=a$ for all $a \in R$.

$\textbf{Proof of claim:}$ We know that

$$-a=(-a)^2=(-1)^2a^2=a$$

and the proof of the claim is done.

Next let $a,b \in R$, we aim to show $ab=ba$. Consider the sum

\begin{align} a+b &=(a+b)^2 && \text{as $R$ is Boolean}\\ &=a^2+ab+ba+b^2 && \text{factoring}\\ &=a+ab+ba+b && \text{as $R$ is Boolean} \end{align}

This forces

$$ab+ba=0$$

replace $ab$ with $-ab$ and we conclude that

$$ba=ab$$

as needed.

MyMathYourMath
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