Show that $\left| \sqrt2-\frac{h}{k} \right| \geq \frac{1}{4k^2},$ for any $k \in \mathbb{N}$ and $h \in \mathbb{Z}$.

Question

Show that $$\left| \sqrt2-\frac{h}{k} \right| \geq \frac{1}{4k^2},$$ for any $k \in \mathbb{N}$ and $h \in \mathbb{Z}$.

I tried many different ways to expand left side and estimate it but always got stuck at some point.

Answer 1

As in How find the value $\beta$ such $\left|\frac{p}{q}-\sqrt{2}\right|<\frac{\beta}{q^2}$ (and the above comments), $$ \left| \sqrt2-\frac{h}{k} \right| \, \left| \sqrt2+\frac{h}{k} \right| = \left| 2-\frac{h^2}{k^2} \right| = \frac{|2k^2 - h^2|}{k^2} \ge \frac{1}{k^2} \tag{1} $$ for all $k \in \mathbb{N}$ and $h \in \mathbb{Z}$, because $\sqrt 2$ is irrational.

Now assume that there exist some $k \in \mathbb{N}$ and $h \in \mathbb{Z}$ such that $$ \left| \sqrt2-\frac{h}{k} \right| < \frac {1}{4k^2} \quad . \tag 2 $$ Then the other factor can be estimated as $$ \left| \sqrt2+\frac{h}{k} \right| = \left| 2 \sqrt 2 - \bigl(\sqrt 2 - \frac{h}{k} \bigr) \right| \le 2 \sqrt 2 + \left| \sqrt2-\frac{h}{k} \right| \\ < 2 \sqrt 2 + \frac {1}{4k^2} \le 2 \sqrt 2 + \frac 14\approx 3.078 < 4 \quad . \tag 3 $$ Multiplying the inequalities $(2)$ and $(3)$ gives a contradiction to $(1)$. Therefore $(2)$ must be false for any $k \in \mathbb{N}$ and $h \in \mathbb{Z}$, which is what you wanted to prove.

Answer 2

I will show that $|\sqrt{n}-\frac{x}{y}| >\frac1{(\sqrt{n}+\sqrt{n+1})y^2} $.

For $n=2$, $\sqrt{2}+\sqrt{3} < 3.15 $, so $|\sqrt{2}-\frac{x}{y}| >\frac1{3.15 y^2} $.

I also show that if $|\sqrt{n}-\frac{x}{y}| <\frac1{(2\sqrt{n}+\epsilon)y^2} $. then $y <\sqrt{ \frac1{2\epsilon\sqrt{n}}} $.

In general, if $n$ is not a perfect square, then $|x^2-ny^2| \ge 1 $ for integers $x$ and $y$.

Therefore, if $z = \sqrt{n}$,

$\begin{align*} 1 &\le |x^2-ny^2|\\ &=|x-zy||x+zy|\\ \text{so}\\ |x-zy| &\ge \frac1{|x+zy|}\\ &= \frac1{x+zy}\\ \text{or}\\ |z-\frac{x}{y}| &\ge \frac1{y}\frac1{x+zy}\\ &\ge \frac1{y^2}\frac1{z+x/y}\\ \end{align*} $

If $|z-\frac{x}{y}| < \frac1{cy^2} $, then $\frac{x}{y} < z+\frac1{cy^2} $ and $\frac1{z+x/y} \le y^2|z-\frac{x}{y}| <\frac1{c} $ so that $c <z+x/y <z+z+\frac1{cy^2} =2z+\frac1{cy^2} $.

Therefore $c^2-2zc < \frac1{y^2} \le 1 $ or $c^2-2zc+z^2 < z^2+1 $ or $(c-z)^2 < n+1 $ or $c < z+\sqrt{n+1} =\sqrt{n}+\sqrt{n+1} $.

Therefore $|\sqrt{n}-\frac{x}{y}| >\frac1{(\sqrt{n}+\sqrt{n+1})y^2} $.

For $n=2$, $c < \sqrt{2}+\sqrt{3} < 3.15 $.

(added later)

Note that, since $c^2-2zc < \frac1{y^2} $, $c < z+\sqrt{z^2+\frac1{y^2}} =z+z\sqrt{1+\frac1{z^2y^2}} <z+z(1+\frac1{2z^2y^2}) =2z+\frac1{2zy^2}) $. Therefore, if $c > 2z$, $\frac1{2zy^2} > c-2z $ or $2zy^2 < \frac1{c-2z} $ or $y <\sqrt{ \frac1{2z(c-2z)}} $.