# Showing that $|\sqrt{3} - m/n| \geq 1/(5n^2)$

Can you help me to prove this inequality

\begin{aligned} |\sqrt{3} - m/n| \geq 1/(5n^2) \end{aligned}

where m and n are integers. Hint:$sqrt(3)$ is irrational.

-

Hint:

Show that if $\displaystyle f(x) = x^2 -3$, then $$\left|f\left(\dfrac{m}{n}\right)\right| \ge \dfrac{1}{n^2}$$

Ok, I guess Vinod isn't interested in following up on the hints, here is almost a full proof.

For any rational number $\displaystyle m/n$, we have that $\displaystyle |3 - \frac{m^2}{n^2}| = \frac{|m^2 - 3n^2|}{n^2}$

Now since $\displaystyle \sqrt{3}$ is irrational, $\displaystyle |m^2 - 3n^2| \ge 1$ (it is an integer).

Thus $\displaystyle |3 - \frac{m^2}{n^2} | \ge \frac{1}{n^2}$

Now if $\displaystyle |\frac{m}{n} + \sqrt{3}| \le 5$, then we have that $\displaystyle |\frac{m}{n} - \sqrt{3}| \ge \frac{1}{5n^2}$.

The case $\displaystyle |\frac{m}{n} + \sqrt{3}| \ge 5$, is easier, as then $\displaystyle \frac{m}{n}$ will be not be very close to $\displaystyle \sqrt{3}$.

-
Still no clue. Can you lead me from real analysis point of view. – Vinod Dec 27 '10 at 6:33
@Vinod: What have you tried using the hint? – Aryabhata Dec 27 '10 at 6:38
I have reached m^2 >= 3n^2 + 1 – Vinod Dec 27 '10 at 6:54
@Vinod: So you were able to prove the hint? Another hint: $x^2 -3 = (x+ \sqrt{3})(x-\sqrt{3})$. – Aryabhata Dec 27 '10 at 7:40