We have seen: \[ P \Rightarrow Q \equiv \neg Q \Rightarrow \neg P \]
A possible way to build intuition for this is using a statement like:
“All humans are mammals”
which we can rephrase as
“If you are human, then you are a mammal”
If that’s the case, then the following statement rings true to our intuition:
“If you are not a mammal, then you are not human.”
The fact that \(P \Rightarrow Q\) and its contrapositive are logically equivalent means that showing \(\neg P \Rightarrow \neg Q\) is just as good as showing \(P \Rightarrow Q\). So we can prove \(P \Rightarrow Q\) as follows:
Definition: An integer \(n\) is even if \(n = 2a\) for some integer \(a \in \mathbb{Z}\).
Observation: by (confusing) convention, definitions are usually stated as a conditional but meant to imply a biconditional. That is, we interpret the above definition to mean: \[ n \text{ is even} \Leftrightarrow \exists a \in Z, n = 2a \] What is the distinction? We are saying that \(n\) is not even if it does not satisfy the given property, where the unidirectional implication would be inconclusive about that case.
Definition: An integer \(n\) is odd if \(n = 2a + 1\) for some integer \(a \in \mathbb{Z}\).
Definition: Suppose \(a\) and \(b\) are integers. We say that \(a\) divides \(b\), written \(a | b\), if \(b = ac\) for some \(c \in \mathbb{Z}\). In this case we also say that \(a\) is a divisor of \(b\), and that \(b\) is a multiple of \(a\).
Definition: Two integers have the same parity if they are both even or they are both odd. Otherwise they have opposite parity.