Proof.
Let \(n\) be an integer. Assume \(n\) is odd. So \(n = 2k+1\) for some integer \(k\text{.}\) Then
\begin{equation*}
7n = 7(2k+1) = 14k + 7 = 2(7k +3) + 1\text{.}
\end{equation*}
Since \(7k + 3\) is an integer, we see that \(7n\) is odd.
Direct proof. Let \(n\) be an integer. Assume \(n\) is odd. So \(n = 2k+1\) for some integer \(k\text{.}\) Then Since \(7k + 3\) is an integer, we see that \(7n\) is odd.Proof.
\begin{equation*}
7n = 7(2k+1) = 14k + 7 = 2(7k +3) + 1\text{.}
\end{equation*}
The converse is: for all integers \(n\text{,}\) if \(7n\) is odd, then \(n\) is odd. We will prove this by contrapositive. Let \(n\) be an integer. Assume \(n\) is not odd. Then \(n = 2k\) for some integer \(k\text{.}\) So \(7n = 14k = 2(7k)\) which is to say \(7n\) is even. Therefore \(7n\) is not odd.Proof.