Let’s recall the truth table for an implication.
Table3.1.1.Truth Table for Implication
\(p\)
\(q\)
\(p\implies q\)
T
T
T
T
F
F
F
T
T
F
F
T
Note that an implication \(p \implies q\) is guaranteed to be true whenever the hypothesis is false. So, for a direct proof, we only need to worry about showing the implication is true when the hypothesis is true. Thus, our strategy becomes
Assume \(p\) is true.
Use this assumption (and definitions, theorems, or previous results) to show that \(q\) must be true, too.
Then, it follows that the truth value of the implication \(p \implies q\) is true.
Example3.1.2.
Claim. For all \(a \in \mathbb{Z}\text{,}\) if \(a\) is even, then \(a^2\) is even. (Recall that an integer \(a\) is defined to be even provided there exists an integer \(k\) for which \(a = 2k\text{.}\) Otherwise, \(a\) is odd, and there exists an integer \(k\) for whic \(a = 2k + 1\text{.}\))
Proof.
Let \(a \in \mathbb{Z}\text{.}\) Assume \(a\) is even. This means that there exists an integer \(k\) for which \(a = 2k\text{.}\) It follows that \(a^2 = 2(2k^2)\text{.}\) Since \(2k^2 \in \mathbb{Z}\text{,}\) we see that \(a^2\) is even.
