prove that 
is irrational!
Lemma: If a perfect square is even, its square root is
also even.
For contradiction, assume that 
is even
and 
is odd:
Then there exist positive integers k and m
such that
and 
Substitute from the second equation into the first.
This says 1 is an even number, so we have a contradiction.
QED
----------------------------------
For contradiction assume
where 
is reduced to lowest terms.
The left side is even and the right side is a
perfect square, so by the lemma, p is even.
Therefore p is twice some integer, r, so let's
substitute 2r for p
The right side is even and the left side is a
perfect square, so by the lemma, q is even.
This is a contradiction because we assumed
was in lowest terms, so p and q can't
both be even.
QED
Edwin
