Inequality in \(\symbb{Z}\) preserved under addition
An Introduction to Real Analysis
2025-10-24
Prove Proposition 1.2.2 (a).
Proposition 1.2.2 in full is as follows:
For all \(n, m \in \symbb{Z}\), (a) if \(m < n\), then for all \(l \in \symbb{Z}\), \(m + l < n + l\), and (b) if \(m < n\) and \(0 < k\), then \(mk < nk\).
Lemma 1 For all \(m, n \in \symbb{Z}\), \((-m) + (-n) = -(m + n)\).
Proof. Using associativity and commutativity of addition in \(\symbb{Z}\), \[\begin{align*} (m + n) + ((-m) + (-n)) &= \underbrace{(m + (-m))}_{= 0} + \underbrace{(n + (-n))}_{= 0}\\[0.75em] &= 0 + 0\\[0.75em] &= 0, \end{align*}\] so \((-m) + (-n)\) is the additive inverse of \(m + n\). In other words, \((-m) + (-n) = -(m + n)\). \(\blacksquare\)
To be precise, Lemma 1 proves that \((-m) + (-n)\) is an additive inverse of \(m + n\). Exercise 1.2.3 for Katznelson and Katznelson (2024) proves that additive inverses in \(\symbb{Z}\) are unique, so \((-m) + (-n)\) is the inverse of \(m + n\).
Proof (Proposition 1.2.2 (a)). By definition, \(m < n\) implies that \(n - m \in \symbb{N}\). Since \[\begin{align} n - m &= n + (-m) \notag \\[0.75em] &= (n + (-m)) + 0 \notag \\[0.75em] &= (n + (-m)) + (l + (-l)) \notag \\[0.75em] &= (n + l) + ((-m) + (-l)) \label{eq:1} \\[0.75em] &= (n + l) + (-(m + l)) \label{eq:2}\\[0.75em] &= (n + l) - (m + l) \notag , \end{align}\] where Lemma 1 was applied to proceed from \(\eqref{eq:1}\) to \(\eqref{eq:2}\), it follows that \((n + l) - (m + l) \in \symbb{N}\), whence \(m + l < n + l\). \(\blacksquare\)
Remark 1. The proof of Proposition 1.2.2 (b) in Katznelson and Katznelson (2024, 6) uses that \(kn - km = k(n - m)\), where \(k, n - m \in \symbb{N}\), to conclude that \(km < kn\). To justify this, first note that by distributivity and Proposition 1.2.1 in Katznelson and Katznelson (2024, 6), \(k(-m) = -(km) = -km\), because \(km + k(-m) = k(m + (-m)) = k \cdot 0 = 0\). Then \[\begin{align*} k(n - m) &= k(n + (-m))\\[0.75em] &= kn + k(-m)\\[0.75em] &= kn + (-km)\\[0.75em] &= kn - km, \end{align*}\] so \(k(n - m) \in \symbb{N}\) implies that \(kn - km \in \symbb{N}\), whence \(km < kn\). \(\blacksquare\)