Proving a statement about covered intervals

[littlemathquark]

Homework Statement

The rational number p/q in the inerval (0,1) covered by the closed interval [p/q-1/4q^2, p/q+1/4q^2]. How do I prove that none of these intervals cover the number (sqrt{2}) /2?

Relevant Equations

The rational number p/q in the inerval (0,1) covered by the closed interval [p/q-1/4q^2, p/q+1/4q^2]. How do I prove that none of these intervals cover the number (sqrt{2}) /2?

I need any idea. I haven't made any progress in solving the problem.

 

[PeroK]

The idea is that if $0 < \frac p q < \frac {\sqrt 2}{2}$, then $\frac p q + \frac 1 {4q^2} < \frac {\sqrt 2}{2}$. And, if $1 > \frac p q > \frac {\sqrt 2}{2}$, then $\frac p q - \frac 1 {4q^2} > \frac {\sqrt 2}{2}$.

Can you show that?

 

[littlemathquark]

I used Dirichlet approximation theorem but I can't find any progression.

 

[PeroK]

Looks tricky.

 

[nuuskur]

1. $\alpha := \sqrt{2}/2$ is a root of $f(x) = 2x^2-1$.
2. For $x\in (\alpha - 1/5, \alpha +1/5)$ we have $(\frac{1}{2} <) f'(x) < \frac{40}{11} =: C < 4$.
3. Conclude by Lagrange MVT that
   <br /> |x-\alpha| &lt; \frac{1}{5} \Rightarrow |f(x)-f(\alpha)| = |2x^2-1| &lt; C|x-\alpha|<br />

Claim. We have
<br /> \left\lvert \frac{m}{n}-\alpha \right\rvert &gt; \frac{1}{4n^2}<br />
for all $m,n\in\mathbb N$. Then it follows that either
<br /> \alpha &gt; \frac{m}{n} + \frac{1}{4n^2}\quad\mbox{or}\quad \frac{m}{n}-\frac{1}{4n^2}&gt; \alpha.<br />

Proof of claim. The nontrivial case is $|m/n - \alpha| < 1/5$. Then
<br /> \left\lvert f\left(\frac{m}{n}\right)\right\rvert = \left\lvert 2\frac{m^2}{n^2}-1 \right\rvert &lt; C\left\lvert \frac{m}{n}-\alpha \right\rvert \Rightarrow n^2\left\lvert 2\frac{m^2}{n^2}-1 \right\rvert &lt; Cn^2\left\lvert \frac{m}{n}-\alpha \right\rvert.<br />
Because $f$ has no rational roots, we conclude $n^2\left\lvert 2\frac{m^2}{n^2}-1 \right\rvert \geqslant 1$ and therefore
<br /> Cn^2\left\lvert \frac{m}{n}-\alpha \right\rvert &gt; 1 \Rightarrow \left\lvert \frac{m}{n}-\alpha \right\rvert &gt; \frac{1}{Cn^2} &gt; \frac{1}{4n^2}.<br />

This is an adaptation of a proof of Liouville's theorem. The trick is finding a small enough interval around $\alpha$ such that you get a suitable upper bound for the derivative.