If a.a=a prove R is commutative

[gottfried]

Homework Statement

If a.a=a for all elements of R, prove R is commutative.

The Attempt at a Solution

(a+b)²=a+b=a²+ba+ab+b²=a+ba+ab+b

Then -ba=ab

Any suggestions of how to show that -b=b?

 

[HallsofIvy]

Are we to assume, here, that R is a group? Then it is always true that ae= a (e is the identity) so if it is also true that aa= a, then we have aa= ae and, multiplying on the left by the inverse of a, a= e. So we have only the "trivial" group containing only the identity.

If R is not a group, what sort of algebraic entity is it?

 

[gottfried]

Sorry I should have been more clear but R is ring. So we are trying to show that . is a commutative operation and therefore making (R,+,.) a commutative ring.(At least that is my understanding of a commutative ring)

If I understand correctly I don't think we can assume that R has a multiplicative identity(1).

 

[micromass]

The structure in your OP is called a Boolean ring. You can prove that any element a in your Boolean ring satisfies a+a=0.

Prove this by considering (a+a)(a+a)

 

[gottfried]

(a+a)(a+a)=a+a
a²+a²+a²+a²=a+a
a+a+a+a=a+a
a+a=0
a=-a for all a in R

Thanks, that makes sense.