Proof.
Let \(n\) be an integer. Assume \(n\) is even. Then \(n = 2k\) for some integer \(k\text{.}\) Thus \(8n = 16k = 2(8k)\text{.}\) Therefore \(8n\) is even.
Direct proof. Let \(n\) be an integer. Assume \(n\) is even. Then \(n = 2k\) for some integer \(k\text{.}\) Thus \(8n = 16k = 2(8k)\text{.}\) Therefore \(8n\) is even.Proof.
The converse is false. That is, there is an integer \(n\) such that \(8n\) is even but \(n\) is odd. For example, consider \(n = 3\text{.}\) Then \(8n = 24\) which is even but \(n = 3\) is odd.