Theorem: If N is not a perfect square, then \(\sqrt{N}\) is irrational.
Proof: Suppose \(\sqrt{N} = a/b\) for positive integers \(a,b\) with no common factor greater than 1. Then \(b/a = \sqrt{N}/N\), and so \(a/b = (bN)/a\). Since the first fraction is in lowest terms, the numerator and denominator of the second fraction must be a common integer multiple, say \(c\), of the numerator and denominator of the first. Hence \(a = cb\), and therefore, \(\sqrt{N} = c\), that is, \(N\) is a perfect square. QED
I learned this proof from a one paragraph insert in the American Mathematical Monthly (vol. 115, June-July 2008, p. 524) written by Geoffrey C. Berresford. I just love it.
Thanks, this is a proof of this fact I haven't seen.
(The way I've been thinking about this is that if \sqrt{N}=a/b, then \sqrt{Nb^2}=a and multiplying a perfect square with a nonperfect square yields a nonperfect square (which can be easily seen by looking at prime decomposition of both).)
