So I have to prove this proposition using axioms for integers for my math class. Is it enough to prove that if $x \cdot x = x$ and $x \neq 0$, then $x = 1$?
So far I have:
$x \cdot x = x$ by given proposition
$x = x \cdot 1$ by Multiplication Identity
$x \cdot x = x \cdot 1$ by Substitution
$x = 1$ by Cancellation
Have you learned that $\mathbb{R}$ is a field? If so:
Claim:
Any subring of a field $F$ is an integral domain ($ab= 0 \implies a = 0 \lor b= 0)$.
Proof:
Let $R$ be a subring of a field $F$. Suppose $a , b \in R$ and that $ab = 0$.
Then since $F$ is a field and assuming $a \neq 0$, $a$ has a multiplicative inverse $a^{-1}$ in $F$.
We get $0_R = 0_F = a^{-1}(ab) =_F (a^{-1}a)b =_F 1*b =_R b $ where the last equality is true in $R$ due to $R$ being a subring. $\blacksquare$
Next check that $\mathbb{Z}$ is a subring of the field $\mathbb{R}$.
Thus let $x \in \mathbb{Z}$ be with $x^2 = x$.
We get $x^2 - x = 0$ and so $x(x-1) = 0$. By the claim either $x = 0$ or $x =1$.