Show that for any rational a/b ∈ (0, 1), we have |a/b - 1/√2| > 1/(4b^{2}).

**Solution**

If |a/b - 1/√2| ≤ 1/(4b^{2}), then |a^{2}/b^{2} - 1/2| = |a/b - 1/√2| |a/b + 1/√2| < |a/b - 1/√2| (1 + 1) < 1/2b^{2}. So |2a^{2} - b^{2}| < 1. Hence 2 a^{2} = b^{2}. But that is impossible since √2 is irrational. Note that the restriction to the interval (0, 1) is unnecessary.

© John Scholes

jscholes@kalva.demon.co.uk

11 Mar 2002