9th Putnam 1949

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Problem B1

Show that for any rational a/b ∈ (0, 1), we have |a/b - 1/√2| > 1/(4b2).

 

Solution

If |a/b - 1/√2| ≤ 1/(4b2), then |a2/b2 - 1/2| = |a/b - 1/√2| |a/b + 1/√2| < |a/b - 1/√2| (1 + 1) < 1/2b2. So |2a2 - b2| < 1. Hence 2 a2 = b2. But that is impossible since √2 is irrational. Note that the restriction to the interval (0, 1) is unnecessary.

 


 

9th Putnam 1949

© John Scholes
jscholes@kalva.demon.co.uk
11 Mar 2002