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

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.

• What do you know about $\bigl\lvert\bigl(\frac{h}{k}\bigr)^2 - 2\bigr\rvert$? Commented Oct 29, 2015 at 14:09
• HINT : step 1 place h=$1/2$ and k=1 which satisfies . 0.91>=0.25 now assume it to be true for k then prove it true for k+1 . do it my mathematical induction Note here i have taken root2 as 1.41. Commented Oct 29, 2015 at 14:10
• @DanielFischer I tried the following already: $$\left| \sqrt2-\frac{h}{k} \right| = \left| \frac{(\frac{h}{k})^2 -2}{\frac{h}{k}+\sqrt{2}} \right|$$ and then estimate but just couldnt get the correct result..
– NPHA
Commented Oct 29, 2015 at 14:16
• That looks good so far. Can you give a lower bound for the numerator? And you need to distinguish the two cases a) $\frac{h}{k}$ is close to $\sqrt{2}$ and b) $\frac{h}{k}$ is far away from $\sqrt{2}$. Commented Oct 29, 2015 at 14:24
• @DanielFischer I can easily see the case where $\frac{h}{k}$ is close to $\sqrt{2}$. Continuation from above: $$\geq \frac{1}{k^2}\left|\frac{1}{\frac{h}{k}+\sqrt{2}}\right| \geq \frac{1}{2\sqrt{2}k^2}.$$ But I really cant do anything about the part b).
– NPHA
Commented Oct 29, 2015 at 14:25

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.

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

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 $y <\sqrt{ \frac1{2z(c-2z)}}$.