Prove the following proposition using contrapositive proof.
For \(x \in \mathbb{R}\), if \(x^2 + 5x < 0\), then \(x < 0\).
Try to prove the proposition from #1 using direct proof; if you reach a point where you’re convinced that this approach is harder, you may give up.
Proposition: Suppose \(a, b \in \mathbb{Z}\). If both \(ab\) and \(a+b\) are even, then both \(a\) and \(b\) are even.
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.