Product of negative integers is positive
An Introduction to Real Analysis
2025-10-25
Show that if \(m < 0\) and \(n < 0\), then \(mn > 0\).
By definition \(m < 0\) and \(n < 0\) imply that \(0 - m, 0 - n \in \symbb{N}\). Using Exercise 1.2.4 and Lemma 1 from Exercise 1.2.5 from Katznelson and Katznelson (2024), \[\begin{align*} (0 - m) \cdot (0 - n) &= \underbrace{(0 + (-m))}_{= -m} \cdot \underbrace{(0 + (-n))}_{= -n}\\[0.75em] &= (-m) \cdot (-n)\\[0.75em] &= ((-1) \cdot m) \cdot ((-1) \cdot n)\\[0.75em] &= \underbrace{((-1) \cdot (-1))}_{= 1} \cdot (m \cdot n)\\[0.75em] &= 1 \cdot (m \cdot n)\\[0.75em] &= m \cdot n \end{align*}\] so by closure of multiplication in \(\symbb{N}\), \(mn \in \symbb{N}\). Further, it was shown in Exercise 1.2.5 from Katznelson and Katznelson (2024) that \(0 = -0\), so \(mn = mn + 0 = mn + (-0) = mn - 0\), so \(mn - 0 \in \symbb{N}\) whence \(mn > 0\).