1 Proof that if n is an integer and n2 is odd then n is odd

1. Proof that, if n is an integer and n2 is odd, then n is odd.

a) True

b) False

Solution

a)

True.

Let us take the negation of the given statement and suppose that it should be true.

Assuming , to contrary,

that an integer n such that n² is odd and n is even.

To prove the given we must deduce the contradiction.

By definition of even, we have

                                                        n = 2k for some integer k.

By substitution we have,

n . n = (2k) . (2k)

= 2 (2.k.k)

Now in this case (2.k.k) is an integer because products of integers are integer; and 2 and k are integers. Hence,

                                                        n . n = 2 . (some integer value)

or                                                        n² = 2. (some integer value)

and so by definition of n² even, it is even i.

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.