SAUDI ARABIAN IMO Booklet 2023

Suppose the contrary that means for some rational number $\frac{p}{q}$, we have $$\left|\frac{\sqrt{2}}{2}-\frac{p}{q}\right|\le \frac{1}{4q^2}.$$ As $\frac{\sqrt{2}}{2}+\frac{p}{q}<2$, multiplying the last two inequalities gives $$\left|\frac{1}{2}-\frac{p^2}{q^2}\right|<\frac{1}{2q^2}.$$ However, as the LHS equals $\frac{|q^2-2p^2|}{2q^2}$, the above inequality reduces to $|q^2-2p^2|<1$, which implies that $q^2=2p^2$ and $\sqrt{2}=\frac{p}{q}$ is a rational number, which is a contradiction. □