Discrete Math 6 Prove: For all integers n, if n2 is odd, then n is odd

Mathematics
Tutor: None Selected Time limit: 1 Day

Apr 20th, 2015

Proof:

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)

or  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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