The natural order on \(\symbb{Z}\)
An Introduction to Real Analysis
2025-11-14
In the Section 1.1.2 notes for Katznelson and Katznelson (2024), the order \(<\) on \(\symbb{N}\) was defined such that for all \(a, b \in \symbb{N}\), \(a < b\) if and only if there exists \(k \in \symbb{N}\) such that \(b = a + k\). In those notes the order was shown to be total and to work well with the arithmetic in \(\symbb{N}\) that was established in the Section 1.1.1 notes for Katznelson and Katznelson (2024).
This order mostly extends naturally to \(\symbb{Z}\), so the same symbol \(<\) can be used for the order on \(\symbb{Z}\) without risk of confusion. For \(m, n \in \symbb{Z}\), Katznelson and Katznelson (2024, 6) define the notation \(m < n\) if and only if \(n - m \in \symbb{N}\). Like the order on \(\symbb{N}\), the order on \(\symbb{Z}\) is also total and also works well with the arithmetic in \(\symbb{Z}\) that was established in the Section 1.2.1 notes for Katznelson and Katznelson (2024).
The totality of the order on \(\symbb{Z}\) is proven in Theorem 1. As in the Section 1.1.2 notes for Katznelson and Katznelson (2024), to prove that the order \(<\) on \(\symbb{Z}\) is total requires proving that \(<\) as a binary relation on \(\symbb{Z}\times \symbb{Z}\) is irreflexive, transitive and total.
Proposition 1 For all \(m \in \symbb{Z}\), \(m \nless m\).
Proof. Suppose there exists \(m \in \symbb{Z}\) such that \(m < m\). By definition of the order on \(\symbb{Z}\) and the properties of arithmetic in \(\symbb{Z}\), this implies that \(m - m = m + (-m) = 0 \in \symbb{N}\). However, by Lemma 1 from the Section 1.1.2 notes for Katznelson and Katznelson (2024), there is no additive identity in the natural numbers. \(\blacksquare\)
Proposition 2 For all \(m, n, o \in \symbb{Z}\), \((m < n) \land (n < o) \implies (m < o)\).
Proof. By definition \(m < n\) implies that \(n - m \in \symbb{N}\), and \(n < o\) implies that \(o - n \in \symbb{N}\). Then \[\begin{align*} o - m &= o + (-m)\\[0.75em] &= (o + (\underbrace{n + (-n)}_{= 0}) + (-m)\\[0.75em] &= (o + ((-n) + n)) + (-m)\\[0.75em] &= ((o + (-n)) + n) + (-m)\\[0.75em] &= (o + (-n)) + (n + (-m))\\[0.75em] &= (\underbrace{o - n}_{\in \symbb{N}}) + (\underbrace{n - m}_{\in \symbb{N}})\\[0.75em] &\in \symbb{N} \end{align*}\] because each summand in the penultimate line is a natural number by assumption. Since \(o - m \in \symbb{N}\) it follows that \(m < o\). \(\blacksquare\)
Proposition 3 gives a trichotomy law for \(\symbb{Z}\) equivalent to the trichotomy law for \(\symbb{N}\).
Proposition 3 (Trichotomy law) For all \(m, n \in \symbb{Z}\), exactly one of \(m < n\), \(m = n\) or \(m > n\) is true.
Proof. The proof is streamlined by observing that it’s not possible for more than one of \(m < n\), \(m = n\) or \(m > n\) to be true. From the Section 1.2.1 notes for Katznelson and Katznelson (2024) there is a partition \(\symbb{Z}\coloneq \symbb{N}\uplus \{ 0 \} \uplus \{ -a \colon a \in \symbb{N}\}\). If \(m < n\) then \(n - m \in \symbb{N}\). Using Exercise 1.2.4 and Lemma 1 from Exercise 1.2.5 from Katznelson and Katznelson (2024), \[\begin{align*} m - n &= m + (-n)\\[0.75em] &= (-1) \cdot n + 1 \cdot m\\[0.75em] &= (-1) \cdot n + (\underbrace{(-1) \cdot (-1)}_{= 1}) \cdot m\\[0.75em] &= (-1) \cdot n + (-1) \cdot ((-1) \cdot m)\\[0.75em] &= (-1) \cdot (n + (-1) \cdot m)\\[0.75em] &= -(n - m), \end{align*}\] so \(m - n \in \{ -a \colon a \in \symbb{N}\}\), whence \(m - n \notin \symbb{N}\) so \(m \ngtr n\). Furthermore, \(m \neq n\) because otherwise \(n - m = n - n = 0 \in \symbb{N}\), which was ruled out in the proof of Proposition 1.
If \(m > n\) then similar reasoning shows that \(m \nless n\) and \(m \neq n\). If \(m = n\) then \(m \nless n\) and \(m \ngtr n\) because otherwise \(0 \in \symbb{N}\).
Therefore, in the remainder of the proof it is not necessary to entertain the possibility that more than one of \(m < n\), \(m = n\) or \(m > n\) is true. From the Section 1.2.1 notes for Katznelson and Katznelson (2024), the equation \[\begin{equation}\label{eq:trichlaw-1} m + x = n \end{equation}\] has a unique solution \(x \in \symbb{Z}\). Because of this and the fact that \(\symbb{Z}\coloneq \symbb{N}\uplus \{ 0 \} \uplus \{ -a \colon a \in \symbb{N}\}\), there are three mutually exclusive possibilities.
- If \(x \in \symbb{N}\) then adding \(-m\) to both sides of \(\eqref{eq:trichlaw-1}\) shows that \[\begin{align*} -m + (m + x) &= -m + n\\[0.75em] \implies \underbrace{(-m + m)}_{= 0} + x &= \underbrace{n + (-m)}_{= n - m}\\[0.75em] \implies 0 + x &= n - m\\[0.75em] \implies x &= n - m, \end{align*}\] so \(n - m \in \symbb{N}\), whence by definition \(m < n\).
- If \(x = 0\) then according to \(\eqref{eq:trichlaw-1}\) \(m + x = m + 0 = m = n\).
- If \(x \in \{ -a \colon a \in \symbb{N}\}\) then there exists \(a \in \symbb{N}\) such that \(x = -a\). By Exercise 1.2.4 from Katznelson and Katznelson (2024) it follows that \(x = -a = (-1) \cdot a\). Using the same exercise along with Lemma 1 from Exercise 1.2.5 for Katznelson and Katznelson (2024) shows that \[\begin{align*} -x &= (-1) \cdot x\\[0.75em] &= (-1) \cdot ((-1) \cdot a)\\[0.75em] &= (\underbrace{(-1) \cdot (-1)}_{= 1}) \cdot a\\[0.75em] &= 1 \cdot a\\[0.75em] &= a, \end{align*}\] so \(-x \in \symbb{N}\). Returning to \(\eqref{eq:trichlaw-1}\) and adding \(-x\) to both sides, \[\begin{align*} (m + x) + (-x) &= n + (-x)\\[0.75em] \implies m + (\underbrace{x + (-x)}_{= 0}) &= n + (-x)\\[0.75em] \implies m + 0 &= (-x) + n\\[0.75em] \implies m &= (-x) + n. \end{align*}\] Adding \(-n\) to both sides shows that \[\begin{align*} m - n &= m + (-n)\\[0.75em] &= ((-x) + n) + (-n)\\[0.75em] &= (-x) + (\underbrace{n + (-n)}_{= 0})\\[0.75em] &= (-x) + 0\\[0.75em] &= -x, \end{align*}\] so \(m - n \in \symbb{N}\) and therefore \(m > n\). \(\blacksquare\)
Theorem 1 confirms that like the order on \(\symbb{N}\), the order on \(\symbb{Z}\) is total.
Theorem 1 For all \(m, n \in \symbb{Z}\), define the binary relation \(<\) on \(\symbb{Z}\) by \(m < n\) if and only if \(n - m \in \symbb{N}\). Then \(<\) is a total order on \(\symbb{Z}\).
Proof. By Proposition 1 the binary relation is irreflexive. By Proposition 2 the binary relation is transitive. By Proposition 3 the binary relation is total. \(\blacksquare\)
The Section 1.2 notes for Katznelson and Katznelson (2024) informally introduced the integers via the partition \(\symbb{Z}\coloneq \symbb{N}\uplus \{ 0 \} \uplus \{ -a \colon a \in \symbb{N}\}\). The Section 1.2.1 notes for Katznelson and Katznelson (2024) justified this partition. Proposition 4 relates this partition of \(\symbb{Z}\) to the order on \(\symbb{Z}\).
Proposition 4 Let \(m \in \symbb{Z}\). Then \(m \in \symbb{N}\) if and only if \(m > 0\), and \(m \in \{ -a \colon a \in \symbb{N}\}\) if and only if \(m < 0\).
Proof. In Exercise 1.2.5 for Katznelson and Katznelson (2024) it was shown that \(-0 = 0\). Using this result, if \(m \in \symbb{N}\) then \(m = m + (-0) = m - 0 \in \symbb{N}\), so \(m > 0\). Conversely, if \(m > 0\) then by definition of the order on \(\symbb{Z}\), \(m - 0 = m + (-0) = m \in \symbb{N}\).
If \(m \in \{ -a \colon a \in \symbb{N}\}\) then \(m = -a\) where \(a \in \symbb{N}\). By the work done above it follows that \(a > 0\), and using Exercise 1.2.5 for Katznelson and Katznelson (2024), \(-a = m < 0\). Conversely, if \(m < 0\) then by definition of the order on \(\symbb{Z}\), \(0 - m = 0 + (-m) = -m \in \symbb{N}\). Using the same exercises again it follows that \(-(-m) = m \in \{ -a \colon a \in \symbb{N}\}\). \(\blacksquare\)
In light of Proposition 4, in the context of \(\symbb{Z}\) the elements of \(\symbb{N}\) are called the positive integers, and the elements of \(\{ -a \colon a \in \symbb{N}\}\) are called the negative integers.
As mentioned, the order on \(\symbb{Z}\) works well with the arithmetic on \(\symbb{Z}\), but there is an important caveat: the direction of an inequality is reversed when multiplying both sides by a negative number. Before proving that the order on \(\symbb{Z}\) works well with the arithmetic on \(\symbb{Z}\) in Proposition 5, it is helpful to prove in Lemma 1 that multiplication distributes over differences.1
Lemma 1 For all \(a, b, c \in \symbb{Z}\), \(ac - bc = (a - b)c\).
Proof. Using Exercise 1.2.4 for Katznelson and Katznelson (2024), \[\begin{align*} ac - bc &= ac + (-bc)\\[0.75em] &= ac + (-1) \cdot (bc)\\[0.75em] &= ac + ((-1) \cdot b) \cdot c\\[0.75em] &= (a + (-1) \cdot b) \cdot c\\[0.75em] &= (a + (-b)) \cdot c\\[0.75em] &= (a - b)c, \end{align*}\] so a common factor can be pulled out out of a difference. \(\blacksquare\)
Proposition 5 For all \(m, n, o \in \symbb{Z}\) and \(p > 0\), \(m < n\) if and only if \(m + o < n + o\), and \(m < n\) if and only if \(pm < pn\).
Proof. That \(m < n\) implies \(m + o < n + o\) is stated in Proposition 1.2.2 (a) in Katznelson and Katznelson (2024, 6) and proven in Exercise 1.2.1 in Katznelson and Katznelson (2024). Using this result for the converse, \[\begin{align*} m + o &< n + o\\[0.75em] \implies (m + o) + (-o) &< (n + o) + (-o)\\[0.75em] \implies m + (\underbrace{o + (-o)}_{= 0}) &< n + (\underbrace{o + (-o)}_{= 0})\\[0.75em] \implies \underbrace{m + 0}_{= m} &< \underbrace{n + 0}_{= n}\\[0.75em] \implies m &< n. \end{align*}\]
That \(m < n\) implies \(pm < pn\) is Proposition 1.2.2 (b) in Katznelson and Katznelson (2024, 6). Using the work above and Lemma 1, \[\begin{align*} pm &< pn\\[0.75em] \implies pm + (-pn) &< pn + (-pn)\\[0.75em] \implies \underbrace{pm - pn}_{= p(m - n)} &< 0\\[0.75em] \implies p(m - n) < 0. \end{align*}\] By assumption \(p > 0\), and by the trichotomy law (Proposition 3), exactly one of \(m - n < 0, m - n = 0\) or \(m - n > 0\) is true. By Proposition 1.2.1 from Katznelson and Katznelson (2024, 6), \(m - n = 0\) would imply that \(p(m - n) = 0\), which violates the trichotomy law. If \(m - n > 0\) then by Proposition 4 \(p, m - n \in \symbb{N}\), whence \(p(m - n) \in \symbb{N}\) so \(p(m - n) > 0\), which again violates the trichotomy law. Therefore \(m - n < 0\); by the work above \((m - n) + n < 0 + n\), and after simplifying both sides this implies that \(m < n\). \(\blacksquare\)
Exercise 1.2.8 from Katznelson and Katznelson (2024) can be used to see what happens to the inequality \(m < n\) when multiplying both sides by \(p < 0\). If \(m < n\) then by definition \(n - m \in \symbb{N}\), and by Proposition 4 \(n - m > 0\). If \(p < 0\) then using that exercise it follows that \(p(n - m) < 0\), and by Lemma 1 it follows that \(pn - pm < 0\). Adding \(pm\) to both sides and simplifying shows that \(pn < pm\), so the direction of the original inequality \(m < n\) has been reversed when multiplying both sides by \(p < 0\).
Proposition 6 justifies the claim that the order on \(\symbb{Z}\) is a natural extension of the order on \(\symbb{N}\), In particular, if an inequality holds between two natural numbers then the same inequality holds when viewing the natural numbers as integers.
Proposition 6 Let \(m, n \in \symbb{N}\). If \(m < n\) with respect to the order on \(\symbb{N}\) then \(m < n\) with respect to the order on \(\symbb{Z}\).
Proof. By definition, \(m < n\) with respect to the order on \(\symbb{N}\) implies that there exists \(k \in \symbb{N}\) such that \(n = m + k\). Therefore \[\begin{align*} (-m) + n &= (-m) + (m + k)\\[0.75em] \implies n + (-m) &= ((-m) + m) + k\\[0.75em] \implies n - m &= 0 + k\\[0.75em] &= k \end{align*}\] so \(n - m \in \symbb{N}\), whence \(m < n\) with respect to the order on \(\symbb{Z}\). \(\blacksquare\)
In the Section 1.1.3 notes for Katznelson and Katznelson (2024) it was proven that the natural numbers are discrete, meaning that there is no natural number between a natural number and its successor. Proposition 7 shows that the integers are also discrete.
Proposition 7 For any \(m \in \symbb{Z}\), there is no \(n \in \symbb{Z}\) such that \(m < n < m + 1\).
Proof. By contradiction suppose there were such \(n\). Then adding \(-m\) to both sides of both inequalities shows that \[\begin{align*} \MoveEqLeft[0] m < n < m + 1\\[0.75em] \implies &(m < n) \land (n < m + 1)\\[0.75em] \implies &(\underbrace{m - m}_{= 0} < n - m) \land (n - m < (m + 1) - m)\\[0.75em] \implies &(0 < n - m) \land (n - m < (1 + (\underbrace{m - m}_{= 0})))\\[0.75em] \implies &(0 < n - m) \land (n - m < 1)\\[0.75em] \implies &0 < n - m < 1. \end{align*}\] By Proposition 4, \(0 < n - m\) implies that \(n - m \in \symbb{N}\), but the Section 1.1.3 notes for Katznelson and Katznelson (2024) showed that for all \(a \in \symbb{N}\), \(a \geq 1\). Since \(n - m \geq 1\) and \(n - m < 1\) is a contradiction of the trichotomy law for \(\symbb{N}\), it follows that there can be no such \(n\). \(\blacksquare\)
Shortcomings of \(\symbb{Z}\)
It is worth taking stock of what has been developed to this point. Conceptually, the integers extend the natural numbers through the addition of an additive identity element \(0\) along with additive inverses \(-a\) for every natural number \(-a\). The Section 1.2.1 notes for Katznelson and Katznelson (2024) validated this perspective and demonstrated that arithmetic in \(\symbb{Z}\) is an extension of arithmetic in \(\symbb{N}\). These Section 1.2.2 notes for Katznelson and Katznelson (2024) demonstrate that the order on \(\symbb{Z}\) is a natural extension of the order on \(\symbb{N}\).
An intuitive description of natural numbers as abstractions for counting objects was given in the Section 1.2.1 notes for Katznelson and Katznelson (2024). The integers were motivated as abstractions that generalized the natural numbers by including “opposite” numbers that can represent negative quantities, and decreases in quantities. However, the integers cannot model fractions of quantities, or parts of a whole. This is clear from Proposition 7 because there is no integer between \(0\), representing nothing, and \(1\), representing one unit of whatever is being counted. For example, if a villager picks an apple, cuts it in half and gives half to each of his two children, neither the natural numbers nor the integers can model the amount of apple possessed by each child.
This shortcoming of \(\symbb{Z}\) can also be seen from the algebraic perspective. The Section 1.2.1 notes for Katznelson and Katznelson (2024) showed that when \(m, n \in \symbb{Z}\), the equation \(m + x = n\) always has a unique solution \(x \in \symbb{Z}\). This is an improvement compared to the natural numbers, but other equations are in turn not solvable in \(\symbb{Z}\). For example, it is not possible to solve all equations \(mx = n\) in the integers. Considering the example of the villager’s children having half an apple each, suppose \(m = \sigma(1)\) and \(n = 1\), and suppose \(x\) is intended to represent the amount of apple possessed by each child. If the equation \(\sigma(1)x = 1\) did have a solution \(x \in \symbb{Z}\), then by Proposition 3, either \(x < 0\), \(x = 0\) or \(x > 0\). Because \(1, \sigma(1) \in \symbb{N}\) it follows from Proposition 4 that \(1, \sigma(1) > 0\).
If \(x < 0\) then Exercise 1.2.8 from Katznelson and Katznelson (2024) shows that \(\sigma(1)x < 0\), which is a contradiction. If \(x = 0\) then Proposition 1.2.1 from Katznelson and Katznelson (2024, 6) shows that \(\sigma(1)x = 0\), which is another contradiction. Therefore \(x > 0\), so by Proposition 4 \(x \in \symbb{N}\). Because \(\sigma(1) = 1 + 1\), it follows that \((1 + 1)x = x + x = 1\). If \(x = 1\) then \(\sigma(1) \cdot 1 = \sigma(1) = 1\), which is a contradiction. If \(x \neq 1\) then there exists \(x^- \in \symbb{N}\) such that \(\sigma(x^-) = x\). Using the definition of addition from the Section 1.1.1 notes for Katznelson and Katznelson (2024) it follows that \(x + x = \sigma(x^-) + \sigma(x^-) = \sigma(\sigma(x^-) + x^-) = 1\), which is another contradiction. Therefore there can be no \(x \in \symbb{Z}\) such that \(\sigma(1)x = 1\), so the amount of apple possessed by each child cannot be modeled with the integers.
The rational numbers rectify both the intuitive and algebraic shortcomings of the integers; they are constructed from the integers in the Section 1.3 notes for Katznelson and Katznelson (2024).
References
Footnotes
The reversal of an inequality when multiplying both sides by a negative number will be established after Proposition 5.↩︎