An Introduction to Real Analysis
Exercise 1.2.5
2025-10-25
Show that if \(m > 0\), then \(-m < 0\) and vice versa.
Below are two proofs. The first proof is more rudimentary because it only relies on properties of arithmetic in \(\symbb{Z}\) and the definition of the order on \(\symbb{Z}\). The second proof is conceptually clearer but more sophisticated because it relies on a result—Proposition 1.2.2 from Katznelson and Katznelson (2024, 6)—concerning the interaction between arithmetic and the order. In particular, the key to the second proof is that an inequality between integers is preserved by adding the same integer to both sides of the inequality.
Lemma 1 It holds that \((-1) \cdot (-1) = 1\).
Proof. Using Exercise 1.2.4 from Katznelson and Katznelson (2024), \((-1) \cdot (-1) = -(-1)\), so \((-1) \cdot (-1)\) is an additive inverse of \(-1\). Since \(1 + (-1) = 0\), it follows that \(1\) is also an additive inverse of \(-1\). By Exercise 1.2.3 from Katznelson and Katznelson (2024), additive inverses are unique, whence \((-1) \cdot (-1) = 1\). \(\blacksquare\)
Lemma 2 It holds that \(0 = -0\).
Proof (First). By definition of the additive identity element, \(0 + 0 = 0\), whence \(0\) is its own additive inverse, so \(0 = -0\). \(\blacksquare\)
Proof (Second). Using Proposition 1.2.1 and Exercise 1.2.4 from Katznelson and Katznelson (2024), \(-0 = (-1) \cdot 0 = 0\). \(\blacksquare\)
Proof (Exercise 1.2.5; first). Using Lemma 2 and the definition of the order on \(\symbb{Z}\), \(m > 0\) implies that \(m - 0 = m + (-0) = m + 0 = m \in \symbb{N}\). Using Lemma 1 and Exercise 1.2.4 from Katznelson and Katznelson (2024), \[\begin{align*} m &= ((-1) \cdot (-1)) \cdot m\\[0.75em] &= (-1) \cdot \underbrace{((-1) \cdot m)}_{= -m}\\[0.75em] &= 0 + ((-1) \cdot (-m))\\[0.75em] &= 0 - (-m), \end{align*}\] hence \(0 - (-m) \in \symbb{N}\), and therefore \(-m < 0\).
On the other hand, if \(m < 0\), then \(0 - m \in \symbb{N}\), and \[\begin{align*} 0 - m &= 0 + (-m)\\[0.75em] &= (-m) + 0\\[0.75em] &= (-m) + (-0)\\[0.75em] &= (-m) - 0, \end{align*}\] so \((-m) - 0 \in \symbb{N}\) and therefore \(-m > 0\). \(\blacksquare\)
Proof (Exercise 1.2.5; second). If \(m > 0\), then using Proposition 1.2.2 from Katznelson and Katznelson (2024, 6), \[\begin{align*} m &> 0\\[0.75em] \implies \underbrace{m + (-m)}_{= 0} &> 0 + (-m)\\[0.75em] \implies 0 &> -m, \end{align*}\] so \(-m < 0\). Similarly, if \(m < 0\), then \[\begin{align*} m &< 0\\[0.75em] \implies \underbrace{m + (-m)}_{= 0} &< 0 + (-m)\\[0.75em] \implies 0 &< -m, \end{align*}\] so \(-m > 0\). \(\blacksquare\)