Question

Prove indirectly, if n^2 is an odd number and n is an integer, then n is an odd number?

Answer 1

Proof by Contradiction - see below

Explanation:

We are told that `n^2` is an odd number and `n in ZZ`

`:. n^2 in ZZ`

Assume that `n^2` is odd and `n` is even.

So `n=2k` for some `k∈ZZ`

and

`n^2= nxxn=2kxx2k`

`=2⋅(2k^2)` which is an even integer

`:. n^2` is even, which contradicts our assumption.

Hence we must conclude that if `n^2` is odd `n` must also be odd.