Proof: [We take the negation and suppose it to be true.] √2 is irrational. Proof: [We take the negation and suppose it to be true.]

Suppose √2 is rational. √2 = m/n Then there are integers m and n with no common factors such that √2 = m/n Definition of a rational number (3.2, 141) [by dividing m and n by any common factor would do it].

Squaring both sides of the equation gives 2 = m2/n2 Or, equivalently, m2 = 2n2 Note that this implies that m2 is even (why?) It follows that m is even (why? Prop. 3.6.4, 176) [We file this away for future reference.]

We deduce that m2 = (2k)2 = 4k2 = 2n2. n2 = 2k2 m = 2k for some integer k. Substituting m = 2k into m2 = 2n2, we see that m2 = (2k)2 = 4k2 = 2n2. Dividing both sides of 4k2 = 2n2 by 2 yields n2 = 2k2

Consequently, n2 is even, and so n is even. But we also know that m is even. [We filed this away.] Hence, both m and n have a common factor of 2. But this is a contradiction. Our supposition was that m and n had no common factors. Therefore the supposition is false and so the theorem is true.