 Irrationality of the square root of two

If $\sqrt{2}=p/q$ in lowest terms, then $p^2=2q^2$ makes $p$ even. Substituting $p=2r$ then makes $q$ even, contradicting that the integers are coprime. More quantitatively, $|2q^2-p^2|\ge 1$ for every nonzero integer $q$, and factoring the difference of squares gives $|\sqrt{2}-p/q|\ge 1/(4q^2)$. To verify the constant, for $p/q\in [1,2]$ the other factor $\sqrt{2}+p/q$ is less than four; outside this interval the approximation error is at least $\sqrt{2}-1>1/4$.