Contradiction proof that root 2 is irrational

Show that there are no integers $p,q$ with $\sqrt2=p/q$.

Assume $\sqrt2=p/q$ in lowest terms. Then $p^2=2q^2$, so $p^2$ and hence $p$ is even; write $p=2m$, giving $q^2=2m^2$, so $q$ is even too. Both being even contradicts lowest terms.