Suppose not. [We take the negation of the given statement and
suppose it to be true.] Assume, to the contrary, that ∃ an integer n such
that n2 is odd and n is
even. [We must deduce the contradiction.] By definition of even, we have
n = 2k for some integer k.
So, by substitution we have
n . n = (2k) . (2k)
= 2 (2.k.k)
Now (2.k.k) is an integer because products of integers are
integer; and 2 and k are integers. Hence,
n . n = 2 . (some integer)
n2 = 2. (some integer)
and so by definition of n2 even, is even.
So the conclusion is since n is even, n2,
which is the product of n with itself, is also even. This contradicts the
supposition that n2 is odd. [Hence, the supposition is false and the
proposition is true.]
Apr 20th, 2015
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