1. 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.]

2. 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].

3. 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 , 176)
[We file this away for future reference.]

4. 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

5. 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.