# Proving an inequality between the difference of $\sqrt2$ and any rational number. [duplicate]

Let $$a = \sqrt{2}$$

Prove that for every $$m,n\in N$$

$$|a - \frac{m}{n}| \gt \frac{1}{(2\sqrt2+1)n^2}$$

Hint: Consider $$|a - \frac{m}{n}|\geq 1$$ and $$|a - \frac{m}{n}|\leq 1$$ as separate cases and consider the minimum of $$|m^2 - 2n^2|$$

Any and all help appreciated

• Think about $|a-\frac mn|(a+\frac mn)$. – Hw Chu Sep 16 at 13:43

By Lagrange's theorem, $$\left|\alpha-\frac{m}{n}\right|\leq \frac{1}{2n^2}$$ implies that $$\frac{m}{n}$$ is a convergent of the continued fraction of $$\alpha$$.
If $$\alpha=\sqrt{2}$$ these convergents are $$\frac{1}{1},\frac{3}{2},\frac{7}{5},\frac{17}{12},\frac{41}{29},\ldots$$ or $$\frac{p_n}{q_n} = \frac{\frac{1}{2}\left[(1+\sqrt{2})^n+(1-\sqrt{2})^n\right]}{\frac{1}{2\sqrt{2}}\left[(1+\sqrt{2})^n-(1-\sqrt{2})^n\right]},$$ since both $$\{p_n\}_{n\geq 1}$$ and $$\{q_n\}_{n\geq 1}$$ have the minimal polynomial $$\lambda^2-2\lambda-1$$. In explicit terms
$$q_n^2\left|\sqrt{2}-\frac{p_n}{q_n}\right|=\frac{1}{4\sqrt{2}}\left|(1+\sqrt{2})^{2n}+(1-\sqrt{2})^{2n}+2(-1)^n-(1+\sqrt{2})^{2n}+(1-\sqrt{2})^{2n}\right|$$ equals $$\frac{1}{2\sqrt{2}}\left|(1-\sqrt{2})^{2n}+(-1)^n\right|>\frac{1}{2\sqrt{2}+1}.$$ It follows that no rational approximations of $$\sqrt{2}$$ are so good that $$\left|\sqrt{2}-\frac{m}{n}\right|\leq \frac{1}{(2\sqrt{2}+1)n^2}$$.
The best we can do is to find approximations such that $$\left|\sqrt{2}-\frac{m}{n}\right|\leq \frac{1}{(2\sqrt{2}-\varepsilon)n^2}$$.