Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question


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 $\displaystyle \sqrt{3}$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.