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**nmatthies1** Let $\displaystyle R$ be a ring in which $\displaystyle x^2 =x ~ \forall x \in R$. Show that $\displaystyle x+x = 0 ~ \forall x \in R$ and that $\displaystyle R$ is commutative.

Using closure under addition I can show that $\displaystyle x+x = 0$. However, I'm not too sure about this. Do I use the fact that $\displaystyle x + (-x) = 0$ so that $\displaystyle 4x=2x \iff 2x=0 \iff x+x=0$ ?

I can't think of a way to show that $\displaystyle \forall x, y \in R, ~ xy=yx$.