Final answer:
To prove that if n^3 + 5 is odd, then n is even, we can use a proof by contraposition or a proof by contradiction. In both cases, we assume the opposite of what we want to prove and show that it leads to a contradiction.
Step-by-step explanation:
Proof by contraposition:
Let's assume that n is odd. Then n can be expressed as n = 2k + 1, where k is an integer. Substituting this into the given equation, we get:
n^3 + 5 = (2k + 1)^3 + 5 = 8k^3 + 12k^2 + 6k + 6
If we simplify this expression, we can see that it is an even number, because 8k^3 + 12k^2 + 6k is always divisible by 2. Therefore, the assumption that n is odd leads to a contradiction.
Therefore, by proof by contraposition, if n^3 + 5 is odd, then n must be even.
Proof by contradiction:
Let's assume that n is odd and n^3 + 5 is odd. We know that the sum of two odd numbers is always even. So if n^3 is odd, then 5 must be odd as well. But since 5 is odd and odd + odd is even, we have a contradiction.
Therefore, by proof by contradiction, if n^3 + 5 is odd, then n must be even.