How find the value $\beta$ such $\left|\frac{p}{q}-\sqrt{2}\right|<\frac{\beta}{q^2}$

Question

Find all positive real number $\beta$,there are infinitely many relatively prime integers $(p,q)$ such that $$\left|\dfrac{p}{q}-\sqrt{2}\right|<\dfrac{\beta}{q^2}$$

maybe this problem background is Hurwitz's theorem: $$\left|\sqrt{2}-\dfrac{p}{q}\right|<\dfrac{1}{\sqrt{5}q^2}$$

so I guess my problem ? $\beta\ge\dfrac{1}{\sqrt{5}}$

and this problem is Germany National Olympiad 2013 last problem (1), see: http://www.mathematik-olympiaden.de/aufgaben/52/4/A52124b.pdf

Answer 1

Proof sketch. First write

$$\left|\frac{p}{q}-\sqrt{2}\right| = \frac{|p^2-2q^2|}{q^2\left|\frac{p}{q} + \sqrt{2}\right|}$$

The Pell equation $$p^2 - 2q^2 = 1$$

is known to have infinitely many solutions (should be proven) so

$$\left|\frac{p}{q}-\sqrt{2}\right| = \frac{|p^2-2q^2|}{q^2\left|\frac{p}{q} + \sqrt{2}\right|} = \frac{1}{q^2}\frac{1}{\left|\sqrt{2 + \frac{1}{q^2}} + \sqrt{2}\right|}$$

for infinitely many $p,q$ so all $\beta \geq \frac{1}{\sqrt{8}}$ are possible. Now try to show that it fails for $\beta < \frac{1}{\sqrt{8}}$. Let $p^2 - 2q^2 = k$ then $p/q = \sqrt{2 + k/q^2}$ and by inserting this into the inequality above show that it cannot hold for large enough $q$.

Answer 2

Hints:

• For all $p,q$ we have $$\left|\left(\frac pq-\sqrt2\right)\left(\frac pq+\sqrt2\right)\right|=\frac{|p^2-2q^2|}{q^2}\ge\frac1{q^2},$$ because $\sqrt2$ is irrational.
• The Pell equations $p^2-2q^2=\pm1$ have infinitely many solutions $(p_n,q_n)$. For example those determined by $$(\sqrt2-1)^n=p_n-q_n\sqrt2.$$
• When $p_n$ and $q_n$ are large, $p_n/q_n\approx\sqrt2$.

Answer 3

To supplement @Winther's good answer, regarding the constraint on viable constants $\beta$: this is an instance of Liouville's theorem on the not-good-approximability of non-rational algebraic numbers... using the mean value theorem from calculus, charmingly enough:

Let $f(x)=x^2-2$. For every rational $p/q$, by the mean value theorem, $$ f(p/q)-f(\sqrt{2}) \;=\; f'(y)\cdot ({p\over q}-\sqrt{2}) $$ for some $y$ between $p/q$ and $\sqrt{2}$. Since $f(\sqrt{2})=0$, and since $|f(p/q)|=|{p^2-2q^2\over q^2}|\ge{1\over q^2}$, $$ {1\over q^2}\;\le\; |f'(y)|\cdot |{p\over q}-\sqrt{2}| $$ As $p/q\to \sqrt{2}$, necessarily also $y\to \sqrt{2}$, and $f'(y)\to f'(\sqrt{2})=2\sqrt{2}$...