Proof

Show that, for all integers n, if 5n is odd, then n is odd.
What type of proof technique did you use?

I used a contrapositive proof.

Contrapositive: If n is even then 5n is even.

If n is even then there exists a k \in \mathbb{Z} such that n = 2k
Using this assumption, 5n = 2 * 5k, and any number multiplied by two is an even number, so 5n is even.
This means the contrapositive is true, so the original claim is true as well.

Using the contrapositive: For all integers n, if n is even, then 5n is even.
So if n is even, then there exists k as an element of all real numbers such that n = 2k.
Thus 5n = 5(2k), and any number multiplied by an even integer (2k) is even.
So 5n is even which means this statement is true and if the contrapositive is true, then the original claim must be true.

Assume 5n is odd, n\in \mathbb{Z}, and that n is even.

If n is even then by definition n = 2k such that \exists k\in\mathbb{Z}. So we have 5(2k)=10k pulling out a 2 we end up with 2(5k) which is even by definition so there’s no way 5n can be odd if n is even. Therefore the original statement stands.

proof by contradiction

I’m using a contrapositive proof.
If n is even, then 5n is even.
An even number means that there exists a k∈Z, so n = 2k. Then 5n = 5 * 2k which equals 2 * 5k. This is the same form as n = 2k, so 5n must also be even.