, 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. □