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Show that if, for the ring $(A, \bigtriangleup, \bot)$, $\bot^2x=x$ then $(A, \bigtriangleup, \bot)$ is a commutative ring.

Since no elements of $A$ is provided and no operation is given, I am not sure how to proceed with this problem.

I have tried applying the simetric element $x^{\bigtriangleup}$ to the expanded equation:

$$ \bot^2x=x \\ x \bot x \bot x = x $$

But it is not working out. How do I proceed?

Omari Celestine
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1 Answers1

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First of all you have that for all $x\in X$ if $x^\sim$ is the opposite of $x$ with respect to $\Delta$ than $x^\sim=x$.

If you consider $x,y\in X$ than

$(x\Delta y^\sim)=\perp^2 (x\Delta y^\sim)=(x\Delta y^\sim)\perp (x\Delta y^\sim)=$ $(\perp^2x)\Delta(x\perp y)^\sim\Delta (y\perp x)^\sim\Delta(\perp^2y^\sim)=$ $ (x\Delta y^\sim) \Delta(x\perp y)^\sim\Delta (y\perp x)^\sim $

so

$(x\Delta y^\sim)= (x\Delta y^\sim) \Delta(x\perp y)^\sim\Delta (y\perp x)^\sim $

Than

$ (x\perp y)^\sim\Delta (y\perp x)^\sim =0$

And so $x\perp y=(y\perp x)^\sim=y\perp x$

If you want we can do the calculus also on $(R,+,*)$ :

$-x=(-x)^2=x^2=x$ and so $-x=x$ for all $x\in X$

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

And so

$x-y= x-y+(-xy-yx)$

$-xy-yx=0$

$xy=-(yx)=yx$

Federico Fallucca
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