approximating $1/{\sqrt{2}}$ by rationals If $0<\ \frac{p}{q}<1$ is a rational number, then prove that $$\left|{\frac{1}{\sqrt{2}}}-{\frac{p}{q}}\right|> {\frac{1}{4q^2}}.$$ I found a proof using Liouville's inequality working with $f(x) = 2x^2 - 1$, but am wondering if there is a direct elementary proof.
 A: A short elementary proof as follows: Let $t = \frac{1}{\sqrt{2}} - \frac{p}{q}$. If $\lvert{t\rvert} > 1/4$ the result is trivial. Otherwise, note that
$$\lvert{t(\sqrt{2} - t)} \rvert = \left\lvert\frac{1}{2} - \frac{p^2}{q^2}\right\rvert \ge \frac{1}{2q^2}.$$
As we can easily bounded
$$\lvert\sqrt{2} - t\rvert < \lvert\sqrt{2} + 0.25\rvert < 2$$
We obtain the desired result.
A: (Inspired by Mr. Zhang!) Since $\sqrt{2}$ is irrational, for any rational $0< {\frac{p}{q}} < 1$, $|2p^2 - q^2|\geq 1$. Thus, $${\frac {|2p^2 - q^2|} {q^2}}\geq {\frac {1} {q^2}}.$$ We can write this as $$|2{\frac {p^2}{q^2}}-1|\geq {\frac {1}{q^2}}.$$ Dividing both sides by $2$,
$$|{\frac {p^2}{q^2}}-{\frac {1}{2}}|\geq {\frac {1}{2q^2}}.$$ We now factor the left hand side, and get $$|({\frac {p}{q}}-{\frac{1}{\sqrt{2}}})({\frac {p}{q}}+{\frac{1}{\sqrt{2}}})|\geq {\frac{1}{2q^2}}.$$ Now, we observe that $|{\frac {p}{q}}+{\frac{1}{\sqrt{2}}}|<2$. Thus, $$|{\frac {p}{q}}-{\frac{1}{\sqrt{2}}}|(2)> {\frac{1}{2q^2}},$$ which implies that $$|{\frac {p}{q}}-{\frac{1}{\sqrt{2}}}|> {\frac{1}{4q^2}}.$$
