0
$\begingroup$

I have to show that $$\left|\sqrt 2 - \frac m n\right| \ge \frac 1 {(2\sqrt 2 + 1)n^2}$$ given $m,n$ integers with $m,n$ greater than or equal to 1. I have shown it for the case where $|\sqrt 2 - \frac m n | \ge 1|$, but I need to show it when $|\sqrt 2 - \frac mn < 1|$. I am trying to use the minimum of $$n^2\left|\frac{m^2}{n^2}-2\right|$$ but I am stuck. Does anybody have any advice?

$\endgroup$
2
  • 1
    $\begingroup$ Perhaps an absolute value is intended on the left side? $\endgroup$ – Ian Sep 24 '15 at 15:10
  • $\begingroup$ Thank you. I did intend the absolute value on the left. $\endgroup$ – eeg710 Sep 24 '15 at 15:15
3
$\begingroup$

We have $|m^2-2n^2|\ge 1$ (as $m^2-2n^2$ is never zero) so $\left|\frac{m^2}{n^2}-2\right|\ge \frac1{n^2}$. Now $|\sqrt{2}-\frac mn|\,|\sqrt{2}+\frac mn|\ge \frac1{n^2}$. We also have (as $\sqrt{2}-1\le \frac mn \le \sqrt{2}+1$) and by the triangle inequality that $|\sqrt{2}+\frac mn|\le \sqrt{2}+\frac mn\le 2\sqrt{2}+1$. Hence $$|\sqrt{2}-\frac mn| \ge \frac1{n^2} \frac1{|\sqrt{2}+\frac mn|}\ge \frac1{(2\sqrt{2}+1)n^2} $$ as required.

$\endgroup$
0
$\begingroup$

Hint(s): The best rational approximation for $\sqrt{2}$ are given by the convergents of its continued fraction: $$ \sqrt{2}=[1;2,2,2,\ldots]\tag{1}$$ If $\frac{p_n}{q_n}$ and $\frac{p_{n+1}}{q_{n+1}}$ are two consecutive convergents, $\sqrt{2}$ is between them and: $$ \left|\frac{p_n}{q_n}-\frac{p_{n+1}}{q_{n+1}}\right|=\frac{1}{q_{n}q_{n+1}}.\tag{2}$$ The claim then follows by studying the rate of growth of Pell numbers.

$\endgroup$
5
  • $\begingroup$ Thanks, but I'm trying to use a method using the minimum of the expression given second in the question. $\endgroup$ – eeg710 Sep 24 '15 at 15:18
  • $\begingroup$ @eeg710: How do you plan to find such a minimum? As I said, the best rational approximations for $\sqrt{2}$ are given by the convergents of the continued fraction. $\endgroup$ – Jack D'Aurizio Sep 24 '15 at 15:21
  • 1
    $\begingroup$ @eeg710 A minimum over what, exactly? If you take the infimum over all of $\mathbb{N} \times \mathbb{N}$ you get zero (the positive rationals are dense in the positive reals). Do you want to take the infimum over $m$ for each fixed $n$? You will find that this infimum oscillates, coming to minima at the values of $n$ corresponding to the convergents of the continued fractions. $\endgroup$ – Ian Sep 24 '15 at 15:27
  • $\begingroup$ Even with the constraint $|\sqrt 2 -\frac {m}{n}| <1$ I would get 0 for the minimum? $\endgroup$ – eeg710 Sep 24 '15 at 15:44
  • $\begingroup$ Again, which minimum? If you minimize over both $m$ and $n$, then yes, you will get zero, since otherwise you would have a positive real that could not be approximated by positive rationals. But if you hold either $m$ or $n$ fixed then the infimum over the other one will be some positive number. Since the RHS of your inequality involves $n$, it is natural to try to take the infimum over $m$. This turns out to be hard, which is part of why we developed continued fractions, but at least at first it seems natural. $\endgroup$ – Ian Sep 24 '15 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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