Product of nonzero integers is nonzero
An Introduction to Real Analysis
2025-10-26
Show that if \(m \neq 0\) and \(n \neq 0\), then \(mn \neq 0\). Conclude that if \(mn = 0\), then \(m = 0\) or \(n = 0\).
Hint: Proposition 1.2.2 (b) and the conclusions of exercises ex1.2.6 and ex1.2.8 will be useful here.
By the trichotomy law for \(\symbb{Z}\) from the Section 1.2.2 notes for Katznelson and Katznelson (2024), there are four cases:
- \(m > 0\), \(n > 0\); by Proposition 1.2.1 and Exercise 1.2.1 from Katznelson and Katznelson (2024), \(0 = 0 \cdot n < mn\);
- \(m > 0\), \(n < 0\); by Exercise 1.2.8 from Katznelson and Katznelson (2024), \(mn < 0\);
- \(m < 0\), \(n > 0\); by Exercise 1.2.8 from Katznelson and Katznelson (2024), \(mn < 0\); and
- \(m < 0\), \(n < 0\); b Exercise 1.2.6 from Katznelson and Katznelson (2024), \(mn > 0\).
In each case, either \(mn > 0\) or \(mn < 0\) has been established, so by the trichotomy law \(mn \neq 0\) for all four cases.
Since the statement \((m \neq 0) \land (n \neq 0) \implies (mn \neq 0)\) has been shown to be true, its contrapositive \((mn = 0) \implies (m = 0) \lor (n = 0)\) is also true.
Remark 1. Let \(R\) be a ring and \(a \in R\). If there exists \(x \in R \setminus \{ 0 \}\) such that \(x \cdot a = 0 \in R\) then \(a\) is called a (right) zero divisor in \(R\). It was shown in Exercise 1.2.4 for Katznelson and Katznelson (2024) that for \(r, 0 \in R\), \(r \cdot 0 = 0\), so \(a = 0\) is a zero divisor. Can a ring \(R\) have nonzero zero divisors? If there were such \(a \neq 0\), then \(x \cdot a = 0\) even though \(x, a \neq 0 \in R\).
This exercise shows that the ring \(\symbb{Z}\) has no nonzero zero divisors, but this is not true of all rings. As an example, consider the ring \(M_2(\symbb{Z})\) of \(2 \times 2\) matrices with entries in \(\symbb{Z}\). Then \[\begin{align*} \underbrace{ \begin{pmatrix*}[r] 1 & 0 \\ 0 & 0 \end{pmatrix*} }_{\neq 0 \in M_2(\symbb{Z})} \cdot \underbrace{ \begin{pmatrix*}[r] 0 & 0 \\ 0 & 1 \end{pmatrix*} }_{\neq 0 \in M_2(\symbb{Z})} &= \underbrace{ \begin{pmatrix*}[r] 0 & 0 \\ 0 & 0 \end{pmatrix*} }_{= 0 \in M_2(\symbb{Z})}\\[0.75em] \underbrace{ \begin{pmatrix*}[r] 1 & 1 \\ 2 & 2 \end{pmatrix*} }_{\neq 0 \in M_2(\symbb{Z})} \cdot \underbrace{ \begin{pmatrix*}[r] 1 & 1 \\ -1 & -1 \end{pmatrix*} }_{\neq 0 \in M_2(\symbb{Z})} &= \underbrace{ \begin{pmatrix*}[r] 0 & 0 \\ 0 & 0 \end{pmatrix*} }_{= 0 \in M_2(\symbb{Z})}. \end{align*}\]
As another example, consider the direct product \(R \times R\) of any ring \(R\). Since the ring operations are applied componentwise, it follows that \[\begin{align*} \underbrace{(1, 0)}_{\neq 0 \in R \times R} \cdot \underbrace{(0, 1)}_{\neq 0 \in R \times R} &= (1 \cdot 0, 0 \cdot 1)\\[0.75em] &= \underbrace{(0, 0)}_{= 0 \in R \times R}, \end{align*}\] hence the direct product of any ring has nonzero zero divisors. \(\blacksquare\)