Item 3.3.9.a.
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*}