Additive inverses have opposite signs
An Introduction to Real Analysis
2025-10-25
Show that if \(m > 0\), then \(-m < 0\) and vice versa.
Below are two proofs. The first is too complex; the second is conceptually clearer.
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 \((-1) \cdot (-1) = 1\). \(\blacksquare\)
Proof (Exercise 1.2.5; first). By definition \(m > 0\) implies that \(m - 0 = m + (-0) = m + 0 = m \in \symbb{N}\).1 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 \[\begin{align*} 0 - m &= 0 + (-m)\\[0.75em] &= (-m) + 0\\[0.75em] & (-m) + (-0)\\[0.75em] &= (-m) - 0\\[0.75em] &\in \symbb{N}, \end{align*}\] so \(-m > 0\). \(\blacksquare\)
Proof (Exercise 1.2.5; second). If \(m > 0\), then using Proposition 1.2.2 from Katznelson and Katznelson (2024, 6), \[\begin{align*} m &> 0\\[0.75em] \implies m + (-m) &> 0 + (-m)\\[0.75em] \implies 0 &> -m, \end{align*}\] so \(-m < 0\). Similarly, if \(m < 0\), then \[\begin{align*} m &< 0\\[0.75em] \implies m + (-m) &< 0 + (-m)\\[0.75em] \implies 0 &< -m, \end{align*}\] so \(-m > 0\). \(\blacksquare\)
References
Footnotes
Using Proposition 1.2.1 and Exercise 1.2.4 from Katznelson and Katznelson (2024), \(-0 = (-1) \cdot 0 = 0\), so \(0\) is its own additive inverse.↩︎