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We show
forall x in Q: x * x != 2.Take arbitrary x in Q. We assume (1) x * x = 2 and show a contradiction.
From the construction of Q, we know x = a/b for some a in Z and b in Z> 0 such that (2) N(a) and N(b) are relatively prime. We have a *Z a/b *Z b = 2 and thus (from now on we operate in Z and drop the corresponding subscripts):
(3) a * a = 2 * b * b.From (3) we know N(2) | N(a*a) and thus also (4) N(2) | N(a) (a proposition that has to be proved extra). Therefore there exists some c in Z such that
(5) a = 2*c.From (3) and (5) we have 2*c*2*c = 2*b*b, i.e., 2*c*c = b*b, thus (6) N(2) | N(b*b) and therefore (7) N(2) | N(b). (4) and (7) contradict (2).