Prove the following For all integers n if n2 is odd then n i

Prove the following:

For all integers n, if n^2 is odd, then n is odd.

Solution

For all integers n, if n² is odd, then n is odd.

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.]