Squared nonzero integers are positive
An Introduction to Real Analysis
2025-10-25
Show that if \(m \neq 0\), then \(m^2 > 0\). Conclude that \(1 > 0\).
By the trichotomy law for \(\symbb{Z}\) from the Section 1.2.2 notes for Katznelson and Katznelson (2024), \(m \neq 0\) implies that either \(m > 0\) or \(m < 0\).
If \(m > 0\) then \(m - 0 \in \symbb{N}\), and \(m - 0 = m + (-0) = m + 0 = m\), so \(m \in \symbb{N}\). It follows that \(m^2 - 0 = m^2 \in \symbb{N}\), so \(m^2 > 0\).
If \(m < 0\) then Exercise 1.2.6 from Katznelson and Katznelson (2024) shows that \(m^2 > 0\).
In order to conclude that \(1 > 0\), it’s first necessary to determine that \(1 \neq 0\). By contradiction if \(1 = 0\), then for any \(m \in \symbb{Z}\), multiplying both sides by \(m\) gives that \(1 \cdot m = 0 \cdot m\), so \(m = 0\). This implies that for all \(m \in \symbb{Z}\), \(m = 0\), so \(\symbb{Z}= \{ 0 \}\). This is a contradiction because it implies that \(\symbb{N}\subseteq \{ 0 \}\), which means either that \(\symbb{N}= \emptyset\) or \(\symbb{N}= \{ 0 \}\), and both possibilities are violations of the Peano axioms. Therefore, since \(1 \neq 0\) and \(1 = 1 \cdot 1 = 1^2\), it follows that \(1 > 0\).
Remark 1. A simpler way to deduce that \(1 > 0\) would be to note that \(1 \in \symbb{N}\subseteq \symbb{Z}\) and \(1 = 1 - 0\), so \(1 - 0 \in \symbb{N}\) and therefore \(1 > 0\). \(\blacksquare\)