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?
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1$\begingroup$ Perhaps an absolute value is intended on the left side? $\endgroup$ – Ian Sep 24 '15 at 15:10
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$\begingroup$ Thank you. I did intend the absolute value on the left. $\endgroup$ – eeg710 Sep 24 '15 at 15:15
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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.
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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.
answered Sep 24 '15 at 15:16
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$\begingroup$ 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
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$\begingroup$ @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
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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
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$\begingroup$ 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
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$\begingroup$ 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
