Problem with inequality: $ \left| \sqrt{2}-\frac{p}{q} \right| > \frac{1}{3q^2}$ 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.
 A: 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.
A: $\sqrt{2}$ is an algebraic number of order $2$ for it is irrational and satisfies  $f(x)=x^2-2=0$. For $x\in [\sqrt{2}-1,\sqrt{2}+1]$,  $\frac{1}{4} <f'(x)=2x<5$. Thus, by the mean value theorem
$$ |f(\sqrt{2})-f(x)|=|x^2-2|< 5|\sqrt{2}-x|$$
Claim: for any integers $p$, $q$ ($q\geq1$)
$$
\begin{align}
\big|\sqrt{2}-\frac{p}{q}\big|>\frac{1}{5q^2}\tag{1}\label{one}
\end{align}
$$
This clearly holds true if $|\sqrt{2}-\tfrac{p}{q}|> 1$. If $|\sqrt{2}-\tfrac{p}{q}|\leq 1$, then by $\eqref{one}$
$$
\big|f(p/q)|=\big|\tfrac{p^2}{q^2}-2|< 5|\sqrt{2}-\tfrac{p}{q}|
$$
and so
$$
q^2|f(p/q)|<5q^2|\sqrt{2}-\tfrac{p}{q}|
$$
Clearly $q^2|f(p/q)|=|p^2-2q^2|$ is a nonnegative integer. Since the equation $f(x)=x^2-2=0$ has no rational solution,  $q^2|f(p/q)|\geq1$. Putting things together,
$$
\big|\sqrt{2}-\tfrac{p}{q}\big|>\frac{1}{5q^2}
$$
for all $p,q\in\mathbb{Z}$ with $q\geq1$.

Comment: This solution I outlined above is an adaptation of standard technique used by Liouville to prove the following result:
Theorem: For any real algebraic number of degree $n>1$, there exists a positive integers $M$ such that
$$
\Big|z-\frac{p}{q}\Big|>\frac{1}{Mq^n}
$$
doe all integers $p$ and $q$, $q\geq1$.
