Give an indirect proof of the theorem; “If 𝑛 is an integer and n3 +13is odd, then n is even.”?

Solution:

Indirect proof can be done as ‘proof by contraposition’.

Proof by contraposition:

The contraposition of the statement is "If n is odd then $n^{3}+13$ is even,".

Hence, to proof the contraposition, we need to assume that n is odd.

By the definition of odd numbers, there is an integer k such that

$n=2 k+1$

On substituting $n=2 k+1$ into $n^{3}+13$ , we get

$n^{3}+13=(2 k+1)^{3}+13=\left(8 k^{3}+12 k^{2}+6 k+1\right)+13=8 k^{3}+12 k^{2}+6 k+14=2\left(4 k^{3}+6 k^{2}+3 k+7\right)$

Thus, we can find an integer $m=4 k^{3}+6 k^{2}+3 k+7$ such that

$n^{3}+13=2 m$

It shows, $n^{3}+13$ is even.

Since the contraposition is true then the original statement is also true.