Question

Show that, if n is an integer and n^3 + 5 is odd, then n is even, using:

Answer 1

Answer:

Explanation:

Let  n be an integer

Given that $n^3+5$ is odd

To prove that n is even

a) Proof by contraposition

Let $n^3+5$ be non odd

Then this would be a multiple of 2 being even

$n^3+5 = 2m\\n^3=2m-5\\n^3=2(m-3)+1$

i.e. we get cube of n is odd since gives remainder 1 when divided by 2

It follows that n is odd.

Thus proved by contraposition

b) contradiction method:

If possible let $n^3+5$ is odd for n odd.

Then we get

since n is odd,

$n^3$ is odd being the product of three odd numbers

When we add 5, we get

$n^3+5$ is even being the sum of two odd numbers

A contradiction

Hence our assumption was wrong

if n is an integer and n^3 + 5 is odd, then n is even