Prove that $|\sqrt{2}-p/q|>1/(5q^2)$

Question

I need to prove that there is no rational number $\frac{p}{q}$ with:$\ |\sqrt{2}-\frac{p}{q}| \leq \frac{1}{5q^2}.$ I really don't know how to start because we didn't really learn how to solve equations with absolute in the lectures until now. We got a hint that we should somehow come to this: $\ |\sqrt{2}-\frac{p}{q}| \geq \frac{24}{50pq}$ and there would be a contradiction. As the title says, I only want to get some tips for solving it and not the solution. I want to know if it's just basic equation transforming or if there are other things to consider.

Answer 1

I reckon that $$\left|2-\frac{p^2}{q^2}\right|\ge\frac1{q^2}.$$ But $$\left|2-\frac{p^2}{q^2}\right| =\left|\sqrt2-\frac{p}{q}\right| \left|\sqrt2+\frac{p}{q}\right|$$ etc.

Answer 2

It's easy to see that

$$|\sqrt2-p/q|\le1/(5q^2)\implies5|2q^2-p^2|\le|\sqrt2+p/q|\implies5\le\sqrt2+|p/q|$$

since $|2q^2-p^2|$ is an integer not equal to $0$ (since $\sqrt2$ is irrational). But we also have

$$|\sqrt2-p/q|\le1/(5q^2)\implies|\sqrt2-p/q|\le1/5\implies|p/q|\le\sqrt2+1/5$$

Putting these together we find that

$$|\sqrt2-p/q|\le1/(5q^2)\implies5\le2\sqrt2+1/5$$

which is easily seen to be false (e.g., $2\sqrt2+1/5\approx2.82+0.2=3.02$) So we must have $|\sqrt2-p/q|\gt1/(5q^2)$.