Problem with inequality: $ \left| \sqrt{2}-\frac{p}{q} \right| > \frac{1}{3q^2}$

Question

Prove that for for all $p,q\in \mathbb{Z}$, $q>0$ we have: $$ \left| \sqrt{2}-\frac{p}{q} \right| > \frac{1}{3q^2}. $$

To be honest, I do not know where to start - any help would be appreciated.

Answer 1

You can assume that $p>0$ and $q>1$, and $\sqrt 2 + p/q ≤ 3$, otherwise this is easy: if $\sqrt 2 + p/q > 3$ then $\sqrt 2-p/q < 2\sqrt 2 - 3<0$, so $$\left|\sqrt2 - \frac{p}{q}\right| > 3-2\sqrt 2 > 1/12 ≥ 1/(3q^2)$$

The highest power of $2$ dividing $2q^2$ is odd, while the highest power of $2$ dividing $p^2$ is even. Then, $p^2$ and $2q^2$ must be distinct integers, thus $|2 q^2 - p^2| \geq 1$. Then

$$\left|\sqrt2 - \frac{p}{q}\right| = \frac{|2p^2-q^2|}{q^2(\sqrt{2}+p/q)} \ge \frac{1}{q^2(\sqrt2 + p / q)} \ge \frac{1}{3q^2},$$

as desired.