If a is a square, then
ordp(a) is even for each prime p.
Using this observation it is not difficult to prove that the
root of 2 is not in Q.
This means that there are no
integers a,
b such that (a/b)2 = 2.
For such a
and b, we would have 2b2 =
a2. Now ord2(2b2)
is odd. But ord2(a2) is even.
This is a contradiction. Therefore, the assumption that
a and b with
(a/b)2 = 2 exist is false.