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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.

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    $\begingroup$ Interesting that we can only get about $|\sqrt{2}-\frac ab|>\frac 1{5b^2}$ using Liouville's theorem $\endgroup$ – vrugtehagel Feb 14 '16 at 20:47
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    $\begingroup$ Sorry about the closure see-saw. I misread the question initially. Anyway, here is a related answer by Winther. $\endgroup$ – Jyrki Lahtonen Feb 15 '16 at 13:10
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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.

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    $\begingroup$ $|2q^2-p^2|\ge 1$ can also by explained by the irrationality of $\sqrt{2}$ (irrationality gives it nonzero, the difference between 2 integers is at least 1). $\endgroup$ – tong_nor Feb 14 '16 at 20:49
  • $\begingroup$ @Watson Not really sure why the inequality is trivial if $\sqrt{2}+p/q>3$. $\endgroup$ – egreg Feb 14 '16 at 20:52
  • $\begingroup$ @egreg : I hope this is clearer now. You're right, I should have written a little bit more about this part. $\endgroup$ – Watson Feb 14 '16 at 20:59

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