Arithmetic in \(\symbb{Z}\)
An Introduction to Real Analysis
2025-11-09
In the Section 1.2 notes for Katznelson and Katznelson (2024) the integers \(\symbb{Z}\) were constructed from the natural numbers \(\symbb{N}\). The goal of these notes is to develop arithmetic on \(\symbb{Z}\) that is a natural extension of the arithmetic on \(\symbb{N}\) that was developed in the Section 1.1.1 notes for Katznelson and Katznelson (2024).
Addition in \(\symbb{Z}\)
In order to determine the correct definition of addition of integers it is helpful to return to their informal development. As when defining the integers, informal work can provide guidance to rigorous definitions. For example, an informal calculation can use an example whose correct answer is obvious to help determine the correct formal definition.
Preceding their formal development integers were considered as differences of natural numbers. If \(a, b \in \symbb{N}\) and \(a < b\) then by the Section 1.1.2 notes for Katznelson and Katznelson (2024) there exists a unique \(k \in \symbb{N}\) such that \(b = a + k\). In the Section 1.1.4 notes for Katznelson and Katznelson (2024) the notation \(b - a \coloneq k\) was introduced for \(k\). Given this notation for \(k\) it was attempted to write “opposite \(k\)”, or negative \(k\), as \(-k \coloneq a - b\). These informal ideas of integers will provide guidance toward the definition of addition.
One constraint for the definition is that since addition in \(\symbb{Z}\) is meant to extend addition in \(\symbb{N}\), the informal work ought to lead to a definition of addition in \(\symbb{Z}\) in terms of addition in \(\symbb{N}\). Since \[\begin{align*} 0 &\overset{!}{=}k + (-k)\\[0.75em] &\overset{!}{=}(b - a) + (a - b)\\[0.75em] &\overset{!}{=}(b + a) - (a + b), \end{align*}\] with \(k \overset{!}{=}(a, b)\), \(-k \overset{!}{=}(b, a)\), this calculation strongly suggests that the ordered pairs \((a, b)\) and \((b, a)\) should be added according to the rule \[ (a, b) + (b, a) \overset{!}{=}(a + b, b + a). \] Since integers are properly equivalence classes of ordered pairs of natural numbers, the informal definition of addition using ordered pairs ought to be carried through to the equivalence classes. If \([(a, b)], [(c, d)] \in \symbb{Z}\) then \[\begin{equation} \label{eq:adddef} [(a, b)] + [(c, d)] \coloneq [(a + c, b + d)]. \end{equation}\]
In \(\eqref{eq:adddef}\) the addition of integers—equivalence classes of ordered pairs of natural numbers—is defined in terms of addition of the ordered pairs of natural numbers. This suggests that it is necessary to check that the result of the addition does not depend on the representatives of each equivalence class used when carrying out the addition.1 This is done in Proposition 1, but first it will be helpful to clarify in Lemma 1 that equivalence classes themselves can be referred to by any of their elements, or representatives.
Lemma 1 Let \(X\) be a non-empty set, \(x, y \in X\) and let \(\sim\) be an equivalence relation on \(X\). If \(x \sim y\) then \([x] = [y] \in X / {\sim}\).
Proof. By definition \([x] \coloneq \{ z \in X \colon z \sim x \}\) and \([y] \coloneq \{ z \in X \colon z \sim y \}\). If \(z \in [x]\) then \(z \sim x\), and by assumption \(x \sim y\), so by transitivity of the equivalence relation \(z \sim y\) whence \(z \in [y]\). Conversely if \(z \in [y]\) then \(z \sim y\). By symmetry of the equivalence relation \(x \sim y\) implies that \(y \sim x\), so again by transitivity \(z \sim x\) hence \(z \in [x]\). Since \([x] \subseteq [y]\) and \([y] \subseteq [x]\) it follows that \([x] = [y]\). \(\blacksquare\)
Proposition 1 Let \([(a, b)], [(c, d)], [(\tilde{a}, \tilde{b})], [(\tilde{c}, \tilde{d})] \in \symbb{Z}\). If \((a, b) \sim (\tilde{a}, \tilde{b})\) and \((c, d) \sim (\tilde{c}, \tilde{d})\) then \([(a, b)] + [(c, d)] = [(\tilde{a}, \tilde{b})] + [(\tilde{c}, \tilde{d})]\).
Proof. By the definition of addition in \(\symbb{Z}\), \([(a, b)] + [(c, d)] = [(a + c, b + d)]\) and \([(\tilde{a}, \tilde{b})] + [(\tilde{c}, \tilde{d})] = [(\tilde{a} + \tilde{c}, \tilde{b} + \tilde{d})]\). Using Lemma 1 it suffices to show that \((a + c, b + d) \sim (\tilde{a} + \tilde{c}, \tilde{b} + \tilde{d})\).
By assumption \(a + \tilde{b} = \tilde{a} + b\) and \(c + \tilde{d} = \tilde{c} + d\). Adding these two equations together and using that addition in \(\symbb{N}\) is associative and commutative shows that \[\begin{align*} \underbrace{(a + \tilde{b}) + (c + \tilde{d})}_{= (a + c) + (\tilde{b} + \tilde{d})} &= \underbrace{(\tilde{a} + b) + (\tilde{c} + d)}_{= (\tilde{a} + \tilde{c}) + (b + d)}\\[0.75em] \implies (a + c) + (\tilde{b} + \tilde{d}) &= (\tilde{a} + \tilde{c}) + (b + d)\\[0.75em] \implies (a + c, b + d) &\sim (\tilde{a} + \tilde{c}, \tilde{b} + \tilde{d}), \end{align*}\] so by Lemma 1 \([(a + c, b + d)] = [(\tilde{a} + \tilde{c}, \tilde{b} + \tilde{d})]\), whence \([(a, b)] + [(c, d)] = [(\tilde{a}, \tilde{b})] + [(\tilde{c}, \tilde{d})]\). \(\blacksquare\)
Since addition in \(\symbb{N}\) is commutative and associative, addition in \(\symbb{Z}\) ought to be commutative and associative as well. This is demonstrated in Theorem 1 and Theorem 2 respectively. Proving these theoremata is a matter of writing integers as elements of \((\symbb{N}\times \symbb{N}) / {\sim}\) and using the relevant properties of addition in \(\symbb{N}\).
Theorem 1 (Commutativity of addition) For all \(m, n \in \symbb{Z}\), \(m + n = n + m\).
Proof. Let \(m = [(a, b)]\) and \(n = [(c, d)]\) where \(a, b, c, d \in \symbb{N}\). Because addition of natural numbers is commutative, \[\begin{align*} m + n &= [(a, b)] + [(c, d)]\\[0.75em] &= [(a + c, b + d)]\\[0.75em] &= [(c + a, d + b)]\\[0.75em] &= [(c, d)] + [(a, b)]\\[0.75em] &= n + m, \end{align*}\] hence addition of integers is commutative. \(\blacksquare\)
Theorem 2 (Associativity of addition) For all \(m, n, o \in \symbb{Z}\), \((m + n) + o = m + (n + o)\).
Proof. Let \(m = [(a, b)], n = [(c, d)]\) and \(o = [(e, f)]\) where \(a, b, c, d, e, f \in \symbb{N}\). Because addition of natural numbers is associative, \[\begin{align*} (m + n) + o &= ([(a, b)] + [(c, d)]) + [(e, f)]\\[0.75em] &= [(a + c, b + d)] + [(e, f)]\\[0.75em] &= [((a + c) + e, (b + d) + f)]\\[0.75em] &= [(a + (c + e), b + (d + f))]\\[0.75em] &= [(a, b)] + [(c + e, d + f)]\\[0.75em] &= [(a, b)] + ([(c, d)] + [(e, f)])\\[0.75em] &= m + (n + o), \end{align*}\] hence addition of integers is associative. \(\blacksquare\)
A new property of addition in \(\symbb{Z}\) compared to addition in \(\symbb{N}\) is the existence of an additive identity element, zero. Zero is the equivalence class \(0 \coloneq [(1, 1)]\).2
Theorem 3 (Identity element of addition) For any \(m \in \symbb{Z}\), \(m + 0 = m\).
Proof. Let \(m = [(a, b)]\). Then by definition \[\begin{align*} m + 0 &= [(a, b)] + [(1, 1)]\\[0.75em] &= [(a + 1, b + 1)]. \end{align*}\]
Since addition of natural numbers is associative and commutative, \[\begin{align*} (a + 1) + b &= a + (1 + b)\\[0.75em] &= a + (b + 1), \end{align*}\] so \((a + 1, b + 1) \sim (a, b)\). By Lemma 1 then \([(a + 1, b + 1)] = [(a, b)]\), so it follows that \(m + 0 = m\). \(\blacksquare\)
Another new property of addition in \(\symbb{Z}\) is the existence of additive inverses.3
Theorem 4 (Additive inverses) For any \(m \in \symbb{Z}\), there exists \(-m \in \symbb{Z}\) such that \(m + (-m) = 0\).
Proof. Let \(m = [(a, b)]\) and define \(-m \coloneq [(b, a)] \in \symbb{Z}\). Then \[\begin{align*} m + (-m) &= [(a, b)] + [(b, a)]\\[0.75em] &= [(a + b, b + a)]\\[0.75em] &= [(a + b, a + b)]. \end{align*}\] Since \((a + b) + 1 = 1 + (a + b)\) it follows that \((a + b, a + b) \sim (1, 1)\), so by Lemma 1 \([(a + b, a + b)] = [(1, 1)] = 0\) and therefore \(m + (-m) = 0\). \(\blacksquare\)
In the Section 1.1.2 notes for Katznelson and Katznelson (2024) the cancellation law for addition was proven. The integers satisfy an identical cancellation law.
Theorem 5 (Cancellation law of addition) For all \(m, n, o \in \symbb{Z}\), \(m + n = m + o\) implies that \(n = o\).
The cancellation law for integers can be proven in two different ways. The first proof works with integers in terms of their definition. The second is more sophisticated, using properties of addition in \(\symbb{Z}\) developed hitherto.
Proof (First). Let \(m = [(a, b)], n = [(c, d)]\) and \(o = [(e, f)]\). Using the definition of addition of integers and the associated equivalence relation, \[\begin{align*} m + n &= m + o\\[0.75em] \implies [(a, b)] + [(c, d)] &= [(a, b)] + [(e, f)]\\[0.75em] \implies [(a + c, b + d)] &= [(a + e, b + f)]\\[0.75em] \implies (a + c, b + d) &\sim (a + e, b + f)\\[0.75em] \implies (a + c) + (b + f) &= (a + e) + (b + d). \end{align*}\] Because addition of natural numbers is associative and commutative, \[\begin{align*} (a + c) + (b + f) &= (a + e) + (b + d)\\[0.75em] \implies (c + f) + (a + b) &= (e + d) + (a + b). \end{align*}\] By the cancellation law for natural numbers, this implies that \(c + f = e + d\), whence \((c, d) \sim (e, f)\). By Lemma 1 \([(c, d)] = [(e, f)]\) so \(n = o\). \(\blacksquare\)
Proof (Second). By Theorem 4 \(m\) has an additive inverse. By Theorem 2 addition of integers is associative. By Theorem 3 zero can be removed as a summand from a sum. Putting these results together, \[\begin{align*} m + n &= m + o\\[0.75em] \implies -m + (m + n) &= -m + (m + o)\\[0.75em] \implies \underbrace{((-m) + m)}_{= 0} + n &= \underbrace{((-m) + m)}_{= 0} + o\\[0.75em] \implies 0 + n &= 0 + o\\[0.75em] \implies n &= o, \end{align*}\] so \(m\) cancels from each side of the equation. \(\blacksquare\)
An algebraic justification for extending \(\symbb{Z}\) to \(\symbb{N}\)
The Section 1.2 notes for Katznelson and Katznelson (2024) gave an intuitive justification for extending the system of numbers from \(\symbb{N}\) to \(\symbb{Z}\). The natural numbers are useful for recording quantities but they are of limited use for recording changes in quantities. In particular they are ill-suited for recording decreases in quantities and totally unsuited for recording negative quantities. Because the integers contain a Peano system they have all the expressive power of the natural numbers while rectifying their shortcomings.
For an algebraic perspective on how the integers rectify shortcomings of the natural numbers, consider the problem of solving equations. Suppose \(a, b \in \symbb{N}\) are fixed and consider the equation \(a + x = b\) where \(x \in \symbb{N}\). The number \(x\), if it exists, is called the solution of the equation. Informally, if \(a\) and \(b\) record quantities of some object then consider \(x\) to be the change in the quantity of \(a\) that leads to a quantity of \(b\). As was shown in the Section 1.1.2 notes for Katznelson and Katznelson (2024) if \(a < b\) then \(x\) exists and is unique. This shows that the natural numbers can model an increase in quantity from \(a\) to \(b\). However, if \(a > b\) then by the trichotomy law there can be no \(x \in \symbb{N}\) such that \(a + x = b\). If there were, this would imply that \(a < b\) which contradicts the assumption that \(a > b\).
Theorem 6 shows that the integers completely resolve this shortcoming of the natural numbers.
Theorem 6 For all \(m, n \in \symbb{Z}\), there exists a unique \(x \in \symbb{Z}\) such that \(m + x = n\).
Proof. By Theorem 4 the inverse \(-m\) exists. Let \(x = n + (-m)\). Then because addition of integers is commutative and associative and zero is the additive identity, \[\begin{align*} m + x &= m + (n + (-m))\\[0.75em] &= m + ((-m) + n)\\[0.75em] &= \underbrace{(m + (-m))}_{= 0} + n\\[0.75em] &= \underbrace{0 + n}_{= n}\\[0.75em] &= n, \end{align*}\] so at least one solution to the equation exists.
Suppose there were another \(y \in \symbb{Z}\) such that \(y \neq x\) and \(m + y = n\). Using the formal definition of integers, let \(m = [(a, b)], n = [(c, d)]\) and \(y = [(e, f)]\). Since \(x = n + (-m)\) it follows that \(x = [(c, d)] + [(b, a)] = [(c + b, d + a)]\). If \((e, f) \sim (c + b, d + a)\) then by Lemma 1 it would follow that \([(e, f)] = [(c + b, d + a)]\), or equivalently \(y = x\), which is a contradiction. Therefore \((e, f) \nsim (c + b, d + a)\), or equivalently \[\begin{equation}\label{eq:uniqsol-1} e + (d + a) \neq (c + b) + f. \end{equation}\] Because \(y\) is assumed to be a solution to the equation, \[\begin{align*} m + y &= n\\[0.75em] \implies [(a, b)] + [(e, f)] &= [(c, d)]\\[0.75em] \implies [(a + e, b + f)] &= [(c, d)]\\[0.75em] \implies (a + e, b + f) &\sim (c, d)\\[0.75em] \implies \underbrace{(a + e) + d}_{= e + (d + a)} &= \underbrace{c + (b + f)}_{= (c + b) + f}\\[0.75em] \implies e + (d + a) &= (c + b) + f, \end{align*}\] which contradicts \(\eqref{eq:uniqsol-1}\), so it must be that \(y = x\) and therefore the solution to the equation is unique. \(\blacksquare\)
Justifying the informal definition of the integers
In the Section 1.2 notes for Katznelson and Katznelson (2024) the integers were informally defined as the set \[\begin{equation}\label{eq:intinformdef} \symbb{Z}\coloneq \symbb{N}\uplus \{ 0 \} \uplus \{ -a \colon a \in \symbb{N}\}. \end{equation}\] Using the construction and properties of integers developed hitherto, the informal definition \(\eqref{eq:intinformdef}\) can be justified. Consider an integer \(m = [(a, b)] \in \symbb{Z}\). By the trichotomy law from the Section 1.1.2 notes for Katznelson and Katznelson (2024), there are three possibilities:
- If \(a < b\) then there exists \(k \in \symbb{N}\) such that \(b = a + k\). Then \(1 + b = b + 1 = (a + k) + 1 = a + (k + 1)\), so \((a, b) \sim (1, k + 1) = (1, \sigma(k))\). By Lemma 1 it follows that \(m = [(a, b)] = [(1, \sigma(k))]\). In the Section 1.2 notes for Katznelson and Katznelson (2024) the subset \(\tilde{\symbb{N}} \subseteq \symbb{Z}\) is defined to be \(\tilde{\symbb{N}} \coloneq \{ [(1, \sigma(a))] \colon a \in \symbb{N}\}\) so it follows that \(m\) is essentially a natural number.
- If \(a = b\) then \(a + 1 = b + 1 = 1 + b\) so \((a, b) \sim (1, 1)\) and by Lemma 1 it follows that \(m = [(a, b)] = [(1, 1)] = 0\).
- If \(a > b\) then there exists \(l \in \symbb{N}\) such that \(a = b + l\). Then \(a + 1 = (b + l) + 1 = b + (l + 1) = (l + 1) + b\), so \((a, b) \sim (l + 1, 1) = (\sigma(l), 1)\). By Lemma 1 and the proof of Theorem 4 it follows that \(m = [(a, b)] = [(\sigma(l), 1)] = -n\) where \(n = [(1, \sigma(l))] \in \tilde{\symbb{N}}\).
Because the foregoing shows that integers fall into one of three disjoint sets—natural numbers, zero, or inverses of natural numbers—the informal definition as in \(\eqref{eq:intinformdef}\) is justified. When the order on \(\symbb{Z}\) is discussed in the Section 1.2.2 notes for Katznelson and Katznelson (2024), the natural numbers will be called the positive integers and their inverses will be called the negative integers.
Relating addition in \(\symbb{Z}\) to addition in \(\symbb{N}\)
In order to demonstrate that addition in \(\symbb{Z}\) is a natural extension of addition in \(\symbb{N}\) it ought to be demonstrated that when adding two natural numbers \(a\) and \(b\), it makes no difference whether \(a\) and \(b\) are considered as elements of the original Peano sytem with its operation of addition, or as elements of the Peano system located within the integers with its operation of addition. More precisely, “it makes no difference” should mean that the result considered as an integer is identical.
From the Section 1.2 notes for Katznelson and Katznelson (2024) the isomorphism from the original Peano system to the Peano system within the integers maps the natural number \(a\) to the equivalence class \([(1, \sigma(a))]\). Similarly \(b\) is mapped to \([(1, \sigma(b))]\). Considered as integers, by definition \[\begin{equation}\label{eq:addintnat-1} [(1, \sigma(a))] + [(1, \sigma(b))] = [(\sigma(1), \sigma(a) + \sigma(b))]. \end{equation}\] Using the definition and properties of addition in \(\symbb{N}\) from the Section 1.1.1 notes for Katznelson and Katznelson (2024), \[\begin{align*} 1 + (\sigma(a) + \sigma(b)) &= ((a + 1) + \sigma(b)) + 1\\[0.75em] &= ((1 + a) + \sigma(b)) + 1\\[0.75em] &= (1 + \underbrace{(a + \sigma(b)}_{= \sigma(a + b)})) + 1\\[0.75em] &= (\sigma(a + b) + 1) + 1\\[0.75em] &= \sigma(a + b) + \underbrace{(1 + 1)}_{= \sigma(1)}\\[0.75em] &= \sigma(a + b) + \sigma(1)\\[0.75em] &= \sigma(1) + \sigma(a + b), \end{align*}\] so it follows that \((1,\sigma(a + b)) \sim (\sigma(1), \sigma(a) + \sigma(b))\). By Lemma 1 \[\begin{equation}\label{eq:addintnat-2} [(1, \sigma(a + b))] = [(\sigma(1), \sigma(a) + \sigma(b)], \end{equation}\] so the left-hand sides of \(\eqref{eq:addintnat-1}\) and \(\eqref{eq:addintnat-2}\) are equal. Since the isomorphism maps \(a + b\) to the left-hand side of \(\eqref{eq:addintnat-2}\), it makes no difference whether the addition of \(a\) and \(b\) is carried out as \(a + b\) and then mapped by the isomorphism to the corresponding integer \([(1, \sigma(a + b))]\), or the summands \(a\) and \(b\) are mapped to their corresponding integers \([(1, \sigma(a))]\) and \([(1, \sigma(b))]\) and then added. The resulting integer is the same either way.
Multiplication in \(\symbb{Z}\)
Similar to defining arithmetic in \(\symbb{N}\) in the Section 1.1.1 notes for Katznelson and Katznelson (2024), after defining addition in \(\symbb{Z}\), defining multiplication in \(\symbb{Z}\) will seem somewhat repetitive. Just like when defining addition it is helpful to work informally with integers as differences in order to find the right definition of multiplication. It is also again helpful to use an example whose correct is answer is obvious. Since \[\begin{align*} k &\overset{!}{=}k \cdot 1\\[0.75em] &\overset{!}{=}(b - a) \cdot (\sigma(1) - 1)\\[0.75em] &\overset{!}{=}b\sigma(1) - 1b - a\sigma(1) + 1a\\[0.75em] &\overset{!}{=}\underbrace{(b\sigma(1) + 1a)}_{= 2b + a} - \underbrace{(1b + a\sigma(1))}_{= b + 2a}, \end{align*}\] this suggests that \(k \overset{!}{=}(a, b)\) and \(1 \overset{!}{=}(1, \sigma(1))\) should be multiplied according to the rule \[ (a, b) \cdot (1, \sigma(1)) \overset{!}{=}(1b + a\sigma(1), b\sigma(1) + 1a). \] Since integers are properly equivalence classes of ordered pairs of natural numbers, the informal definition using ordered pairs ought to be carried through to the equivalence classes. If \([(a, b)], [(c, d)] \in \symbb{Z}\), then after rearranging terms \[\begin{equation}\label{eq:multdef} [(a, b)] \cdot [(c, d)] \coloneq [(ad + bc, ac + bd)]. \end{equation}\] As when defining addition of integers it is necessary to check that the calculation in \(\eqref{eq:multdef}\) does not depend on the representatives used for the equivalence classes when carrying out the multiplication.4
Proposition 2 Let \([(a, b)], [(c, d)], [(\tilde{a}, \tilde{b})], [(\tilde{c}, \tilde{d})] \in \symbb{Z}\). If \((a, b) \sim (\tilde{a}, \tilde{b})\) and \((c, d) \sim (\tilde{c}, \tilde{d})\) then \([(a, b)] \cdot [(c, d)] = [(\tilde{a}, \tilde{b})] \cdot [(\tilde{c}, \tilde{d})]\).
Proof. By the definition of multiplication in \(\symbb{Z}\), \([(a, b)] \cdot [(c, d)] = [(ad + bc, ac + bd)]\) and \([(\tilde{a}, \tilde{b})] \cdot [(\tilde{c}, \tilde{d})] = [(\tilde{a}\tilde{d} + \tilde{b}\tilde{c}, \tilde{a}\tilde{c} + \tilde{b}\tilde{d})]\). Using Lemma 1 it suffices to show that \((ad + bc, ac + bd) \sim (\tilde{a}\tilde{d} + \tilde{b}\tilde{c}, \tilde{a}\tilde{c} + \tilde{b}\tilde{d})\).
By assumption \(a + \tilde{b} = \tilde{a} + b\) and \(c + \tilde{d} = \tilde{c} + d\). These two equations lead to four further equations \[\begin{align*} (a + \tilde{b})c &= (\tilde{a} + b)c\\[0.75em] (\tilde{a} + b)d &= (a + \tilde{b})d\\[0.75em] (c + \tilde{d})\tilde{a} &= (\tilde{c} + d)\tilde{a}\\[0.75em] (\tilde{c} + d)\tilde{b} &= (c + \tilde{d})\tilde{b}. \end{align*}\] Because multiplication distributes over addition for natural numbers, and multiplication of natural numbers is commutative, both sides of each equation expand to \[\begin{align*} ac + \tilde{b}c &= \tilde{a}c + bc\\[0.75em] \tilde{a}d + bd &= ad + \tilde{b}d\\[0.75em] \tilde{a}c + \tilde{a}\tilde{d} &= \tilde{a}\tilde{c} + \tilde{a}d\\[0.75em] \tilde{b}\tilde{c} + \tilde{b}d &= \tilde{b}c + \tilde{b}\tilde{d}. \end{align*}\] The left-hand sides of the four expansions can be added together, as can the right-hand sides. Because addition of natural numbers is associative and commutative, each sum is invariant under permutations. After re-ordering each sum as necessary, the summands on each side with only one factor with a tilde will cancel by the cancellation law of addition for natural numbers. After inserting parentheses as necessary, which is permitted because addition of natural numbers is associative, what remains is \[\begin{align*} \underbrace{(ac + bd) + (\tilde{a}\tilde{d} + \tilde{b}\tilde{c})}_{= (\tilde{a}\tilde{d} + \tilde{b}\tilde{c}) + (ac + bd)} &= \underbrace{(ad + bc) + (\tilde{a}\tilde{c} + \tilde{b}\tilde{d})}_{\text{Move to left-hand side}}\\[0.75em] \implies (ad + bc) + (\tilde{a}\tilde{c} + \tilde{b}\tilde{d}) &= (\tilde{a}\tilde{d} + \tilde{b}\tilde{c}) + (ac + bd)\\[0.75em] \implies (ad + bc, ac + bd) &\sim (\tilde{a}\tilde{d} + \tilde{b}\tilde{c}, \tilde{a}\tilde{c} + \tilde{b}\tilde{d}), \end{align*}\] so by Lemma 1 \([(ad + bc, ac + bd)] = [(\tilde{a}\tilde{d} + \tilde{b}\tilde{c}, \tilde{a}\tilde{c} + \tilde{b}\tilde{d})]\), whence \([(a, b)] \cdot [(c, d)] = [(\tilde{a}, \tilde{b})] \cdot [(\tilde{c}, \tilde{d})]\). \(\blacksquare\)
Theorem 7 (Commutativity of multiplication) For all \(m, n \in \symbb{Z}\), \(m \cdot n = n \cdot m\).
Proof. Let \(m = [(a, b)]\) and \(n = [(c, d)]\) where \(a, b, c, d \in \symbb{N}\). Because addition and multiplication of natural numbers are both commutative, \[\begin{align*} m \cdot n &= [(a, b)] \cdot [(c, d)]\\[0.75em] &= [(ad + bc, ac + bd)]\\[0.75em] &= [(cb + da, ca + db)]\\[0.75em] &= [(c, d)] \cdot [(a, b)]\\[0.75em] &= n \cdot m, \end{align*}\] hence multiplication of integers is commutative. \(\blacksquare\)
Theorem 8 (Associativity of multiplication) For all \(m, n, o \in \symbb{Z}\), \((m \cdot n) \cdot o = m \cdot (n \cdot o)\).
Proof. Let \(m = [(a, b)], n = [(c, d)]\) and \(o = [(e, f)]\). Because addition and multiplication of natural numbers are both associative and commutative, and multiplication distributes over addition for natural numbers, \[\begin{align*} \MoveEqLeft[0] (m \cdot n) \cdot o\\[0.75em] &= ([(a, b)] \cdot [(c, d)]) \cdot [(e, f)]\\[0.75em] &= [(ad + bc, ac + bd)] \cdot [(e, f)]\\[0.75em] &= [((ad + bc)f + (ac + bd)e, (ad + bc)e + (ac + bd)f)]\\[0.75em] &= [((adf + bcf) + (ace + bde), (ade + bce) + (acf + bdf))]\\[0.75em] &= [((ace + adf) + (bcf + bde), (acf + ade) + (bce + bdf))]\\[0.75em] &= [(a(ce + df) + b(cf + de), a(cf + de) + b(ce + df))]\\[0.75em] &= [(a, b)] \cdot [(cf + de, ce + df)]\\[0.75em] &= [(a, b)] \cdot ([(c, d)] \cdot [(e, f)])\\[0.75em] &= m \cdot (n \cdot o), \end{align*}\] hence multiplication of integers is associative. \(\blacksquare\)
Theorem 9 (Distributivity of multiplication over addition) For all \(m, n, o \in \symbb{Z}\), \(m \cdot (n + o) = m \cdot n + m \cdot o\).
Proof. Let \(m = [(a, b)], n = [(c, d)]\) and \(o = [(e, f)]\). Because addition of natural numbers is associative and commutative, and multiplication distributes over addition for natural numbers, \[\begin{align*} \MoveEqLeft[0] m \cdot (n + o)\\[0.75em] &= [(a, b)] \cdot ([(c, d)] + [(e, f)])\\[0.75em] &= [(a, b)] \cdot [(c + e, d + f)]\\[0.75em] &= [(a(d + f) + b(c + e), a(c + e) + b(d + f))]\\[0.75em] &= [((ad + af) + (bc + be), (ac + ae) + (bd + bf))]\\[0.75em] &= [((ad + bc) + (af + be), (ac + bd) + (ae + bf))]\\[0.75em] &= [(ad + bc, ac + bd)] + [(af + be, ae + bf)]\\[0.75em] &= [(a, b)] \cdot [(c, d)] + [(a, b)] + [(e, f)]\\[0.75em] &= m \cdot n + m \cdot o, \end{align*}\] hence multiplication distributes over addition for integers. \(\blacksquare\)
Theorem 10 (Identity element of multiplication) For any \(m \in \symbb{Z}\), \(m \cdot 1 = m\).
Proof. Let \(m = [(a, b)]\). Then by the definition of multiplication of integers and multiplication of natural numbers, \[\begin{align} m \cdot 1 &= [(a, b)] \cdot [(1, \sigma(1))] \notag \\[0.75em] &= [(a\sigma(1) + b, a + b\sigma(1))] \notag \\[0.75em] &= [((a + a) + b, a + (b + b))] \notag \\[0.75em] &= [(a + (a + b), (a + b) + b)] \notag \\[0.75em] &= [(a + (a + b), b + (a + b))] \notag \\[0.75em] &= [(a, b)] + [(a + b, a + b)] \label{eq:multid-1}. \end{align}\] Because \((a + b) + 1 = 1 + (a + b)\), it follows that \((a + b, a + b) \sim (1, 1)\). Hence by Lemma 1, \([(a + b, a + b)] = [(1, 1)] = 0\). Applying this to \(\eqref{eq:multid-1}\) and using Theorem 3, \[\begin{align*} m \cdot 1 &= \underbrace{[(a, b)]}_{= m} + \underbrace{[(a + b, a + b)]}_{= 0}\\[0.75em] &= m + 0\\[0.75em] &= m, \end{align*}\] so \(1\) is the multiplicative identity element. \(\blacksquare\)
Relating multiplication in \(\symbb{Z}\) to multiplication in \(\symbb{N}\)
In order to demonstrate that multiplication in \(\symbb{Z}\) is a natural extension of multiplication in \(\symbb{N}\) it ought to be demonstrated that when multiplying two natural numbers \(a\) and \(b\), it makes no difference whether \(a\) and \(b\) are considered as elements of the original Peano sytem with its operation of multiplication, or as elements of the Peano system located within the integers with its operation of multiplication. More precisely, “it makes no difference” should mean that the result considered as an integer is identical.
From the Section 1.2 notes for Katznelson and Katznelson (2024) the isomorphism from the original Peano system to the Peano system within the integers maps the natural number \(a\) to the equivalence class \([(1, \sigma(a))]\). Similarly \(b\) is mapped to \([(1, \sigma(b))]\). Considered as integers, by definition \[\begin{equation}\label{eq:multintnat-1} [(1, \sigma(a))] \cdot [(1, \sigma(b))] = [(\sigma(b) + \sigma(a), 1 + \sigma(a) \cdot \sigma(b))]. \end{equation}\] Using properties of arithmetic in \(\symbb{N}\), \[\begin{align*} \MoveEqLeft[0] 1 + (1 + \sigma(a) \cdot \sigma(b))\\[0.75em] &= 1 + (1 + \underbrace{(a + 1) \cdot (b + 1)}_{= (a + 1) \cdot b + (a + 1)})\\[0.75em] &= 1 + (1 + (\underbrace{(a + 1) \cdot b}_{= a \cdot b + b} + (a + 1)))\\[0.75em] &= 1 + (1 + ((a \cdot b + b) + (a + 1)))\\[0.75em] &= 1 + (((b + 1) + a \cdot b) + (a + 1))\\[0.75em] &= 1 + (((b + 1) + (a + 1)) + a \cdot b)\\[0.75em] &= ((\underbrace{(b + 1)}_{= \sigma(b)} + \underbrace{(a + 1)}_{= \sigma(a)}) + a \cdot b) + 1\\[0.75em] &= (\sigma(b) + \sigma(a)) + (a \cdot b + 1)\\[0.75em] &= (\sigma(b) + \sigma(a)) + \sigma(a \cdot b), \end{align*}\] so it follows that \((1, \sigma(a \cdot b)) \sim (\sigma(b) + \sigma(a), 1 + \sigma(a) \cdot \sigma(b)))\). By Lemma 1 \[\begin{equation}\label{eq:multintnat-2} [(1, \sigma(a \cdot b))] = [(\sigma(b) + \sigma(a), 1 + \sigma(a) \cdot \sigma(b))] \end{equation}\] so the left-hand sides of \(\eqref{eq:multintnat-1}\) and \(\eqref{eq:multintnat-2}\) are equal. Since the isomorphism maps \(a \cdot b\) to the left-hand side of \(\eqref{eq:multintnat-2}\), it makes no difference whether the multiplication of \(a\) and \(b\) is carried out as \(a \cdot b\) and then mapped by the isomorphism to the corresponding integer \([(1, \sigma(a \cdot b))]\), or the factors \(a\) and \(b\) are mapped to their corresponding integers \([(1, \sigma(a))]\) and \([(1, \sigma(b))]\) and then multiplied. The resulting integer is the same either way.
References
Footnotes
As an aside the binary operation of addition on integers \(+ \colon \symbb{Z}\times \symbb{Z}\to \symbb{Z}\) is not the same as the binary operation of addition on natural numbers \(+ \colon \symbb{N}\times \symbb{N}\to \symbb{N}\). However there are a couple of reasons why it is not worth distinguishing the two binary operations. First addition in \(\symbb{Z}\) is defined in terms of addition in \(\symbb{N}\). Second addition in \(\symbb{Z}\) extends addition in \(\symbb{N}\) so when adding natural numbers \(a, b \in \symbb{N}\subseteq \symbb{Z}\), the sum \(a + b\) does not depend on whether the addition is viewed as occuring in \(\symbb{N}\) or in \(\symbb{Z}\).↩︎
Exercise 1.2.2 for Katznelson and Katznelson (2024) shows that the additive identity is unique.↩︎
Exercise 1.2.3 for Katznelson and Katznelson (2024) shows that additive inverses are unique.↩︎
As in the case of addition, the binary operation of multiplication on integers \(\cdot \colon \symbb{Z}\times \symbb{Z}\to \symbb{Z}\) is not the same as the binary operation of multiplication on natural numbers \(\cdot \colon \symbb{N}\times \symbb{N}\to \symbb{N}\). For similar reasons it is not worth distinguishing the two binary operations. First multiplication in \(\symbb{Z}\) is defined in terms of addition and multiplication in \(\symbb{N}\). Second multiplication in \(\symbb{Z}\) extends multiplication in \(\symbb{N}\) so when adding natural numbers \(a, b \in \symbb{N}\subseteq \symbb{Z}\), the sum \(a + b\) does not depend on whether the multiplication is viewed as occuring in \(\symbb{N}\) or in \(\symbb{Z}\).↩︎