(idea) by tongpoo Thu Jul 31 2003 at 10:43:00
More proofs---
  1. A direct proof. Suppose the square root of a positive integer z is a rational x / y in lowest terms. Then xx = yyz which implies y divides xx. By a corollary of Euclid's First Theorem it follows that y divides x. The root x / y is therefore an integer.
  2. A proof by contradiction. Assume that x/y is a non-integer rational in lowest terms. Define z := (x/y)², and q := floor (x/y). Then: 
        x/y - 1 <   q    < x/y
      x - y <   yq   < x
         -y < yq - x < 0
          y > x - yq > 0

    x/y = x(x - yq) / y(x-yq)
    x/y = (xx -yqx) / y(x-yq)
    x/y = (yyz-yqx) / y(x-yq)
    x/y = (yz - qx) / (x-yq)
    An equivalent fraction with a denominator lower than y has been found. Contradiction: x/y is not in lowest terms.
  3. As a special case of the Rational Root Theorem (a.k.a. Rational Zeros Theorem).
Last update: June 22, 2005
I like it!