# Show that if n is an integer and n^3+5 is odd, then n is even using  a proof by contraposition and a proof by contradiction

This type of exercise tends to be easier, as it tells you and advance what method you should use.

Remember the beauty of proving a theorem, is that you have an arsenal of methods to use and most of the time, your skill identifying what method to use is what will take you to a successful proof.

## a) Using a proof by contraposition

In a proof by contraposition, we use the fact that p->q is logically equivalent to its contrapositive ¬q -> ¬p.

In this case, the contrapositive is: if n is an integer and it is not even, then n3 +5 is not odd.

In other words: if n is an integer and n is odd, then n3 +5 is even.

If n is odd, then n=2k+1 for a certain integer k.

Substituting n=2k+1 in n3 +5 we get the following:

n3 +5 = (2k+1)3 + 5 = (8k3 + 12k2 + 6k + 1) + 5 = 8k3 + 12k2 + 6k + 6 = 2(4k3 + 6k2 + 3k + 3)

Therefore, n3 +5 is even because it is two times an integer number ∎

## b) Using a proof by contradiction

To use proof by contradiction to prove a conditional statement(p->q), we do the following:

• Assume the negation of q (the conclusion of the conditional statement).
• Use p and ¬q to arrive at a contradiction.

This type of proofs is valid because p->q ≡ (p^q)-> F.

p= “n3 +5 is odd”

¬q = “n is odd”

Because n is odd, n=2k+1

Substituting n=2k+1 in n3 +5 we get the following:

n3 +5 = (2k+1)3 + 5 = (8k3 + 12k2 + 6k + 1) + 5 = 8k3 + 12k2 + 6k + 6 = 2(4k3 + 6k2 + 3k + 3)

Therefore, n3 +5 is even because it is two times an integer number.

In other words, p is false.

Because p is true (premise) and ¬p is also true, we have a contradiction.

This completes the proof by contradiction ∎

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