A
direct proof of the
existence of
irrational real numbers:
Let S = {x ∈ R : x2 < 2}. S is bounded above (be eg. 3/2) so by the completeness of the real numbers S has a supremum, call it x. It is trivial to check from the definition of the supremum that neither x2 < 2 nor x2 > 2 are possible. Therefore x2 = 2, and we call x the positive square root of 2.
Now suppose that x = p/q for some integers p,q. Without loss of generality p, q are coprime. Then p2 = 2q2, so p is even. Hence q2 = p2/2, so q is even. This contradicts the choice of p, q coprime. Hence the real number x is irrational.