An inequality involving sqrt(2)

Prove that for all a,b \in \mathbb{Z}^+, | \sqrt{2} - \frac{a}{b} | \geq \frac{1}{3b^2}.


Let p(x) = x^2 - 2.

First we will show that the inequality holds for b = 1. Note that |\sqrt{2} - \frac{1}{1}| = 0.414 + \epsilon > 1/3 and |\sqrt{2} - \frac{2}{1}| = 0.585 + \epsilon > 1/3. For a > 2, we have |\sqrt{2} - \frac{a}{1}| > 1 > 1/3. So the inequality holds for b = 1. Henceforth, we will assume that b \geq 2.

Suppose |\sqrt{2} - \frac{a}{b}| \geq \frac{3 - 2\sqrt{2}}{2}. Note that since b \geq 2, b^2 \geq 4 > 3.88 + \epsilon  = \frac{2}{9 - 6\sqrt{2}}; so \frac{3 - 2\sqrt{2}}{2} > \frac{1}{3b^2}. Hence |\sqrt{2} - \frac{a}{b}| \geq \frac{1}{3b^2}.

Now suppose |\sqrt{2} - \frac{a}{b}| \leq \frac{3 - 2\sqrt{2}}{2}. By the Mean Value Theorem from calculus (which we will assume to be valid), there exists an element \xi between \sqrt{2} and a/b such that p^\prime(\xi) = \frac{p(\sqrt{2}) - p(a/b)}{\sqrt{2} - a/b}, and hence |p^\prime(\xi)| = \frac{|p(\sqrt{2}) - p(a/b)|}{|\sqrt{2} - a/b|}. Now \xi \in [\sqrt{2} - \frac{3-2\sqrt{2}}{2}, \sqrt{2} + \frac{3-2\sqrt{2}}{2}] = [1.328+\epsilon, 1.5]. Since p^\prime(x) = 2x is strictly increasing, we have |p^\prime(\xi)| \leq 3. Since p(\sqrt{2}) = 0, we have |p(a/b)| \leq 3|\sqrt{2} - \frac{a}{b}|.

Note that p(a/b) \neq 0, since (for example) p(x) is irreducible over \mathbb{Q}. Now |p(a/b)| = |\frac{a^2}{b^2} - 2| = \frac{|a^2 - 2b^2|}{b^2}. Since p(a/b) \neq 0, a^2 - 2b^2 is a nonzero integer. In particular, we have |a^2 - 2b^2| \geq 1, so that |p(a/b)| \geq \frac{1}{b^2}.

Hence |\sqrt{2} - \frac{a}{b}| \geq \frac{1}{3b^2}.

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