Irrationality of √2

Contradiction. Suppose $\sqrt2=p/q$ with $p,q\in\mathbb{N}^*$ coprime.

Square. $p^2=2q^2$, so $p^2$ is even, hence $p$ is even: $p=2k$.

Substitute. $4k^2=2q^2$, so $q^2=2k^2$, hence $q$ is even too.

Contradiction. $p$ and $q$ both even contradicts $\gcd(p,q)=1$.

Conclusion. $\sqrt2\notin\mathbb{Q}$.