If $a^2=a$ find the inverse of $1+a$.
Note that $a\in R$ where $R$ is a ring and inverse of $1+a$ exists in $R$
I tried $(1-a)(1+a)=1-a^2=1-a$
But I need $1$ in the RHS.How to get it?
2026-03-25 17:19:48.1774459188
If $a^2=a$ find the inverse of $1+a$
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Hint: I assume that the $2$ is invertible in the ring, $(1+a)(2-a)=2-a+2a-a^2=2$.