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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

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  • $\begingroup$ Think about $|a-\frac mn|(a+\frac mn)$. $\endgroup$ – Hw Chu Sep 16 at 13:43
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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 $\left|\sqrt{2}-\frac{m}{n}\right|\leq \frac{1}{(2\sqrt{2}-\varepsilon)n^2}$.

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