Q is a field – Jesse Schneider

In the Section 1.3 notes for Katznelson and Katznelson (2024) the rational numbers \(\symbb{Q}\) were constructed from the integers \(\symbb{Z}\). The goal of these notes is to develop arithmetic on \(\symbb{Q}\) that is a natural extension of the arithmetic on \(\symbb{Z}\) that was developed in the Section 1.2.1 notes for Katznelson and Katznelson (2024).

As the title suggests, the takeaway from these notes is that \((\symbb{Q}, +, \cdot)\) is a field. This means that addition is commutative, associative, has an identity element and additive-inverse elements; that multiplication is commutative, associative, has an identity element and multiplicative-inverse elements for any non-zero element; and that multiplication distributes over addition.¹ The system of integers \((\symbb{Z}, +, \cdot)\) does not have multiplicative-inverse elements in general, but it has all the other properties of a field. Because of this it might seem that the rational numbers are a strict upgrade from the integers, in the sense that \(\symbb{Q}\) has all the properties of \(\symbb{Z}\) and more. This is not true; in the Section 1.3.4 notes for Katznelson and Katznelson (2024) the order on \(\symbb{Q}\) will be developed, and it will be shown that \(\symbb{Q}\) loses some of the distinctive properties of \(\symbb{Z}\).

Addition in \(\symbb{Q}\)

In order to determine the correct definition of addition of rational numbers it is helpful to return to their informal development. The rational numbers were motivated in the Section 1.3 notes for Katznelson and Katznelson (2024) as a model for fractional quantities. For example, the concept of half an apple can be modeled by the rational number \([(1, 2)] \in \symbb{Q}\), and two-thirds of an apple can be modeled by \([(2, 3)] \in \symbb{Q}\). Imagine that one apple has been cut into halves and another apple has been cut into thirds, so \([(1, 2)]\) and \([(2, 3)]\) model one piece from the first apple and two pieces from the second apple, respectively.

Suppose these fractional quantities \([(1, 2)]\) and \([(2, 3)]\) of apple were at hand and one were interested in determining the total quantity of apple available. Mathematically this would be represented by the sum \[\begin{equation}\label{eq:add-1} [(1, 2)] + [(2, 3)]. \end{equation}\] Before determining how to evaluate this sum it will be helpful to evaluate a simpler sum. Suppose instead of half an apple and two-thirds of an apple there were half an apple and three-halves of an apple. In order to determine how much apple were available in this case it would be necessary to evaluate the sum \[\begin{equation}\label{eq:add-2} [(1, 2)] + [(3, 2)]. \end{equation}\] Because all the apples were originally cut into the same number of pieces the addition can take place as though it were integer addition, using units of half-apples instead of whole apples. That is, \(\eqref{eq:add-2}\) should be evaluated as \[\begin{align} [(1, 2)] + [(3, 2)] &\overset{!}{=}[(1 + 3, 2)] \label{eq:add-3} \\[0.75em] &= [(4, 2)] \notag \\[0.75em] &\overset{!}{=}2 \notag. \end{align}\] Importantly for the purpose of formalizing addition in \(\symbb{Q}\), the addition on the left-hand side of \(\eqref{eq:add-3}\) is addition in \(\symbb{Q}\) whereas the addition on the right-hand side is addition in \(\symbb{Z}\). This is a promising sign for the formalization of addition in \(\symbb{Q}\) because it suggests that addition in \(\symbb{Q}\) can be defined in terms of addition in \(\symbb{Z}\). This is similar to how addition in \(\symbb{Z}\) was defined in terms of addition in \(\symbb{N}\) in the Section 1.2.1 notes for Katznelson and Katznelson (2024).

Returning to the addition of half an apple and two-thirds of an apple in \(\eqref{eq:add-1}\), the addition in \(\eqref{eq:add-3}\) suggests that if each piece of apple in \(\eqref{eq:add-1}\) were the same size—that is, if both apples had initially been cut into the same number of pieces, and hence had common units—then the pieces of apple could be added as though they were whole apples by simply changing units from whole apples to their common fraction. For example, if both apples had initially been cut into sixths then half an apple should be equivalent to three-sixths of an apple, and two-thirds of an apple should be equivalent to four-sixths of an apple, so their sum should be \([(3, 6)] + [(4, 6)] \overset{!}{=}[(3 + 4, 6)] = [(7, 6)]\) apples. This conversion from half to three-sixths and from two-thirds to four-sixths can be modeled by the multiplications \([(3 \cdot 1, 3 \cdot 2)] \overset{!}{=}[(3, 6)]\) and \([(2 \cdot 2, 2 \cdot 3)] \overset{!}{=}[(4, 6)]\), where the the multiplications occur in \(\symbb{Z}\). These multiplications represent cutting the half apple into three equal pieces and cutting the two-thirds of an apple into two equal pieces each, so that the erstwhile half and two-thirds have been converted to common units of sixths of an apple.

To justify these conversions to common units it is necessary to confirm that \([(1, 2)] \overset{!}{=}[(3 \cdot 1, 3 \cdot 2)]\) and \([(2, 3)] \overset{!}{=}[(2 \cdot 2, 2 \cdot 3)]\), so that the unit conversions do not change the summands as rational numbers. Lemma 1 confirms that rational numbers are invariant under changes of units.

Proof. Because multiplication of integers is associative and commutative, \(m \cdot (o \cdot n) = (o \cdot n) \cdot m = o \cdot (n \cdot m) = o \cdot (m \cdot n) = (o \cdot m) \cdot n\). Furthermore, using Exercise 1.2.9 from Katznelson and Katznelson (2024) because \(n, o \neq 0\) it follows that \(o \cdot n \neq 0\) so \((o \cdot m, o \cdot n) \in \symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})\). By the definition of the equivalence relation \(\sim\) used in defining the rational numbers from the Section 1.3 notes from Katznelson and Katznelson (2024), it follows that \((m, n) \sim (o \cdot m, o \cdot n)\). By Lemma 1 from the Section 1.2.1 notes for Katznelson and Katznelson (2024), it follows that \([(m, n)] = [(o \cdot m, o \cdot n)]\). \(\blacksquare\)

By definition from the Section 1.3 notes for Katznelson and Katznelson (2024), the representative of a rational number \([(m, n)] \in \symbb{Q}\) is an element \((m, n) \in \symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})\). This allows for the possibility that the second integer in a representative of a rational number is negative, which unfortunately defies intuition. How should one interpret \([(3, -5)] \in \symbb{Q}\)? Three negative-fifths of an apple? Fortunately Lemma 2 shows that a rational number can always be represented by a pair of integers whose denominator is positive.

Proof. By definition \(n \in \symbb{Z}\setminus \{ 0 \}\), so by the trichotomy law for \(\symbb{Z}\) from the Section 1.2.2 notes for Katznelson and Katznelson (2024) it follows that either \(n > 0\) or \(n < 0\). If \(n > 0\) then taking \(o = m\) and \(p = n\) suffices. If \(n < 0\) then by Exercise 1.2.5 from Katznelson and Katznelson (2024) it follows that \(-n > 0\). Using Exercise 1.2.4 from Katznelson and Katznelson (2024), since \(m(-n) = m \cdot ((-1) \cdot n) = (m \cdot (-1)) \cdot n = ((-1) \cdot m) \cdot n = (-m)n\), it follows that \((m, n) \sim (-m, -n)\), so \([(m, n)] = [(-m, -n)]\) and therefore taking \(o = -m\) and \(p = -n\) suffices. \(\blacksquare\)

Because of Lemma 2 any representation of a rational number can be assumed without loss of generality to have a positive denominator.

Lemma 3 shows that a minus sign attached to the numerator in a representative of a rational number can be moved to the denominator, and vice-versa.

Proof. Using Exercise 1.2.4 and Lemma 1 from Exercise 1.2.5, both from Katznelson and Katznelson (2024), and the fact that multiplication of integers is associative and commutative, it follows that \[\begin{align*} (-m) \cdot (-n) &= ((-1) \cdot m) \cdot ((-1) \cdot n)\\[0.75em] &= \underbrace{((-1) \cdot (-1))}_{= 1} \cdot (m \cdot n)\\[0.75em] &= 1 \cdot (m \cdot n)\\[0.75em] &= m \cdot n, \end{align*}\] whence by definition \((-m, n) \sim (m, -n)\), so \([(-m, n)] = [(m, -n)]\). \(\blacksquare\)

The preceding is sufficient to formally define addition of rational numbers. If \([(m, n)], [(o, p)] \in \symbb{Q}\) then by Lemma 1 \([(m, n)] = [(pm, pn)]\) and \([(o, p)] = [(no, np)]\) so both rational numbers can be converted to common units. Then \[\begin{equation}\label{eq:adddef} [(m, n)] + [(o, p)] \coloneq [(pm + no, np)]. \end{equation}\] Similar to the definition of addition of integers in terms of natural-number addition in the Section 1.2.1 notes for Katznelson and Katznelson (2024), the addition of rational numbers is defined in terms of integer arithmetic. Again, similarly it is necessary to check that the result of the addition does not depend on the representative of each equivalence class used when carrying out the addition.

Proof. By the definition of addition in \(\symbb{Q}\), \([(m, n)] + [(o, p)] = [(pm + no, np)]\) and \([(\tilde{m}, \tilde{n})] + [(\tilde{o}, \tilde{p})] = [(\tilde{p} \tilde{m} + \tilde{n} \tilde{o}, \tilde{n} \tilde{p})]\). Using Lemma 1 from the Section 1.2.1 notes for Katznelson and Katznelson (2024) it suffices to show that \[\begin{equation}\label{eq:addwelldef-1} (pm + no, np) \sim (\tilde{p} \tilde{m} + \tilde{n} \tilde{o}, \tilde{n} \tilde{p}). \end{equation}\] By assumption \[\begin{align} m \tilde{n} &= \tilde{m} n, \text{ and} \label{eq:addwelldef-2} \\[0.75em] o \tilde{p} &= \tilde{o} p \label{eq:addwelldef-3}. \end{align}\] Multiplying \(\eqref{eq:addwelldef-2}\) on both sides by \(p \tilde{p}\) and using that multiplication in \(\symbb{Z}\) is associative and commutative gives \[\begin{align} m \tilde n p \tilde{p} &= \tilde{m} n p \tilde{p} \notag \\[0.75em] \implies pm \tilde{n} \tilde{p} &= \tilde{p} \tilde{m} np \label{eq:addwelldef-4} . \end{align}\] Similarly multiplying \(\eqref{eq:addwelldef-3}\) on both sides by \(n \tilde{n}\) and again using that multiplication in \(\symbb{Z}\) is associative and commutative gives \[\begin{align} o \tilde{p} n \tilde{n} &= \tilde{o} p n \tilde{n} \notag \\[0.75em] \implies no \tilde{n} \tilde{p} &= \tilde{n} \tilde{o} np \label{eq:addwelldef-5} . \end{align}\] Adding \(\eqref{eq:addwelldef-4}\) and \(\eqref{eq:addwelldef-5}\) together and using that multiplication in \(\symbb{Z}\) distributes over addition gives \[\begin{align*} \underbrace{pm \tilde{n} \tilde{p}}_{\eqref{eq:addwelldef-4}} + \underbrace{no \tilde{n} \tilde{p}}_{\eqref{eq:addwelldef-5}} &= \underbrace{\tilde{p} \tilde{m} np}_{\eqref{eq:addwelldef-4}} + \underbrace{\tilde{n} \tilde{o} np}_{\eqref{eq:addwelldef-5}}\\[0.75em] \implies (pm + no)\tilde{n} \tilde{p} &= (\tilde{p} \tilde{m} + \tilde{n} \tilde{o})np\\[0.75em] \implies (pm + no, np) &\sim (\tilde{p} \tilde{m} + \tilde{n} \tilde{o}, \tilde{n} \tilde{p}), \end{align*}\] which is exactly \(\eqref{eq:addwelldef-1}\). By the aforementioned lemma it follows that \([(pm + no, np)] = [(\tilde{p} \tilde{m} + \tilde{n} \tilde{o}, \tilde{n} \tilde{p})]\), whence \([(m, n)] + [(o, p)] = [(\tilde{m}, \tilde{n})] + [(\tilde{o}, \tilde{p})]\). \(\blacksquare\)

Since addition in \(\symbb{Z}\) is commutative and associative, addition in \(\symbb{Q}\) 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 rational numbers as elements of \((\symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})) / {\sim}\) and using the relevant properties of arithmetic in \(\symbb{Z}\).

Proof. Let \(q = [(m, n)]\) and \(r = [(o, p)]\) where \(m, o \in \symbb{Z}\) and \(n, p \in \symbb{Z}\setminus \{ 0 \}\). Because addition and multiplication of integers are commutative, \[\begin{align*} q + r &= [(m, n)] + [(o, p)]\\[0.75em] &= [(pm + no, np)]\\[0.75em] &= [(no + pm, pn)]\\[0.75em] &= [(o, p)] + [(m, n)]\\[0.75em] &= r + q, \end{align*}\] hence addition of rational numbers is commutative. \(\blacksquare\)

Proof. Let \(q = [(m, n)], r = [(o, p)]\) and \(s = [(\alpha, \beta)]\) where \(m, o, \alpha \in \symbb{Z}\) and \(n, p, \beta \in \symbb{Z}\setminus \{ 0 \}\). Using several of the properties of arithmetic in \(\symbb{Z}\), \[\begin{align*} (q + r) + s &= ([(m, n)] + [(o, p)]) + [(\alpha, \beta)]\\[0.75em] &= [(pm + no, np)] + [(\alpha, \beta)]\\[0.75em] &= [(\beta(pm + no) + (np)\alpha, (np)\beta)]\\[0.75em] &= [(\beta (pm) + \beta (no) + n(p \alpha), n(p\beta))]. \end{align*}\] Continuing, \[\begin{align*} \MoveEqLeft[0] [(\beta (pm) + \beta (no) + n(p \alpha), n(p\beta))]\\[0.75em] &= [((p\beta) m + n(\beta o + p\alpha), n(p\beta))]\\[0.75em] &= [(m, n)] + [(\beta o + p\alpha, p\beta)]\\[0.75em] &= [(m, n)] + ([(o, p)] + [(\alpha, \beta)])\\[0.75em] &= q + (r + s), \end{align*}\] hence addition of rational numbers is associative. \(\blacksquare\)

As with the integers, the rational numbers also have an additive identity element, which is also called zero. Zero is the equivalence class \([(0, 1)]\).

Proof. Let \(q = [(m, n)]\). Then by definition \[\begin{align*} q + [(0, 1)] &= [(m, n)] + [(0, 1)]\\[0.75em] &= [(1 \cdot m + n \cdot 0, n \cdot 1)]\\[0.75em] &= [(m + 0, n)]\\[0.75em] &= [(m, n)]\\[0.75em] &= q, \end{align*}\] so \([(0, 1)]\) is the additive identity element of the rational numbers. \(\blacksquare\)

Proof. Let \(q = [(m, n)]\) and define \(-q \coloneq [(-m, n)] \in \symbb{Q}\). Then \[\begin{align*} q + (-q) &= [(m, n)] + [(-m, n)]\\[0.75em] &= [(nm + n(-m), nn)]\\[0.75em] &= [(n(m + (-m), nn))]\\[0.75em] &= [(n \cdot 0, nn)]\\[0.75em] &= [(0, nn)]. \end{align*}\] Because \(0 \cdot 1 = 0 \cdot nn = 0\), it follows that \((0, nn) \sim (0, 1)\), whence \([(0, nn)] = [(0, 1)]\). \(\blacksquare\)

It is mentioned in the Section 1.2.1 notes for Katznelson and Katznelson (2024) that the integral additive-identity is unique, as are integral additive-inverses. The same is true for the rational additive-identity and rational additive-inverses, but rather than proving this directly it is more succinct to observe that \(\symbb{Q}\) is a field and then prove uniqueness of the additive identity and additive inverses for a general field. This is taken care of in the Section 1.3.2 notes for Katznelson and Katznelson (2024).

Proof. By Theorem 4 \(-q\) exists, so adding \(-q\) to both sides of \(q + r = q + s\) and using Theorem 2 and Theorem 3 gives \[\begin{align*} (-q) + (q + r) &= (-q) + (q + s)\\[0.75em] \implies \underbrace{((-q) + q)}_{= [(0, 1)]} + r &= \underbrace{((-q) + q)}_{= [(0, 1)]} + s\\[0.75em] \implies [(0, 1)] + r &= [(0, 1)] + s\\[0.75em] \implies r &= s, \end{align*}\] so \(q\) cancels from each side of the equation. \(\blacksquare\)

Multiplication in \(\symbb{Q}\)

Determining the correct definition for multiplication in \(\symbb{Q}\) requires a different approach from that used for addition. For addition the insight was that addition of rational numbers, interpreted heuristically as fractional quantities of apple, has a natural definition in terms of integer addition when the two fractional quantities have common units. However, as suggested in the Section 1.3 notes for Katznelson and Katznelson (2024), it does not make sense to multiply together two quantities both having units of apples because the result would have units of square apples. Therefore when attempting to define multiplication of rational numbers there is no use converting both factors to common units.

Instead one can proceed heuristically with simple calculations for which the correct answer is clear. Consider the rational number \([(m, n)] \in \symbb{Q}\). Consistent with the informal development of \(\symbb{Q}\) in the Section 1.3 notes for Katznelson and Katznelson (2024), this rational number can be informally interpreted as “\(m\) pieces of \(n\)”, which should be the same as \(m\) times one piece of \(n\). In the formal development of \(\symbb{Q}\) it was suggested that an integer \(m\) interpreted as rational number should take the form \([(m, 1)] \in \symbb{Q}\). Furthermore, one piece of \(n\) interpreted as a rational number should take the form \([(1, n)] \in \symbb{Q}\). This implies that \([(m, n)] \overset{!}{=}[(m, 1)] \cdot [(1, n)]\), and this desired equality would be be true if \([(m, 1)] \cdot [(1, n)] \overset{!}{=}[(m \cdot 1, 1 \cdot n)]\), where the two multiplications in the last expression are multiplications of integers.

For arbitrary rational numbers \([(m, n)], [(o, p)] \in \symbb{Q}\) the foregoing would be enforced by the definition \[\begin{equation}\label{eq:multdef} [(m, n)] \cdot [(o, p)] \coloneq [(mo, np)]. \end{equation}\] Similar to the definition of rational addition in \(\eqref{eq:adddef}\) it is necessary to check that the result of the multiplication does not depend on the representative of each equivalence class used when carrying out the multiplication.

Proof. By the definition of multiplication in \(\symbb{Q}\), \([(m, n)] \cdot [(o, p)] = [(mo, np)]\) and \([(\tilde{m}, \tilde{n})] \cdot [(\tilde{o}, \tilde{p})] = [(\tilde{m} \tilde{o}, \tilde{n} \tilde{p} )]\). Using Lemma 1 from the Section 1.2.1 notes for Katznelson and Katznelson (2024) it suffices to show that \((mo, np) \sim (\tilde{m} \tilde{o}, \tilde{n} \tilde{p})\).

By assumption \(m\tilde{n} = \tilde{m}n\) and \(o\tilde{p} = \tilde{o}p\). Multiplying these two equations together and using that multiplication in \(\symbb{Z}\) is associative and commutative shows that \[\begin{align*} (m\tilde{n})(o\tilde{p}) &= (\tilde{m}n)(\tilde{o}p)\\[0.75em] \implies (mo)(\tilde{n} \tilde{p}) &= (\tilde{m} \tilde{o})(np)\\[0.75em] \implies (mo, np) &\sim (\tilde{m} \tilde{o}, \tilde{n} \tilde{p}), \end{align*}\] so by the aforementioned lemma it follows that \([(mo, np)] = [(\tilde{m} \tilde{o}, \tilde{n} \tilde{p})]\), whence \([(m, n)] \cdot [(o, p)] = [(\tilde{m}, \tilde{n})] \cdot [(\tilde{o}, \tilde{p})]\). \(\blacksquare\)

Proof. Let \(q = [(m, n)]\) and \(r = [(o, p)]\) where \(m, o \in \symbb{Z}\) and \(n, p \in \symbb{Z}\setminus \{ 0 \}\). Because multiplication of integers is commutative, \[\begin{align*} q \cdot r &= [(m, n)] \cdot [(o, p)]\\[0.75em] &= [(mo, np)]\\[0.75em] &= [(om, pn)]\\[0.75em] &= [(o, p)] \cdot [(m, n)]\\[0.75em] &= r \cdot q, \end{align*}\] hence multiplication of rational numbers is commutative. \(\blacksquare\)

Proof. Let \(q = [(m, n)], r = [(o, p)]\) and \(s = [(\alpha, \beta)]\) where \(m, o, \alpha \in \symbb{Z}\) and \(n, p, \beta \in \symbb{Z}\setminus \{ 0 \}\). Because multiplication of integers is associative, \[\begin{align*} (q \cdot r) \cdot s &= ([(m, n)] \cdot [(o, p)]) \cdot [(\alpha, \beta)]\\[0.75em] &= [(mo, np)] \cdot [(\alpha, \beta)]\\[0.75em] &= [((mo)\alpha, (np)\beta)]\\[0.75em] &= [(m(o\alpha), n(p\beta))]. \end{align*}\] Continuing, \[\begin{align*} [(m(o\alpha), n(p\beta))] &= [(m, n)] \cdot [(o\alpha, p\beta)]\\[0.75em] &= [(m, n)] \cdot ([(o, p)] \cdot [(\alpha, \beta)])\\[0.75em] &= q \cdot (r \cdot s), \end{align*}\] hence multiplication of rational numbers is associative. \(\blacksquare\)

Proof. Let \(q = [(m, n)], r = [(o, p)]\) and \(s = [(\alpha, \beta)]\) where \(m, o, \alpha \in \symbb{Z}\) and \(n, p, \beta \in \symbb{Z}\setminus \{ 0 \}\). Using the fact that multiplication distributes over addition for integers, \[\begin{align*} q \cdot (r + s) &= [(m, n)] \cdot ([(o, p)] + [(\alpha, \beta)])\\[0.75em] &= [(m, n)] \cdot [(\beta o + p\alpha, p\beta)]\\[0.75em] &= [(m(\beta o + p\alpha), n(p\beta))]\\[0.75em] &= [(m(\beta o) + m(p\alpha), n(p\beta))]. \end{align*}\] Next, using Lemma 1 and that multiplication of integers is commutative and associative, \[\begin{align*} [(m(\beta o) + m(p\alpha), n(p\beta))] &= [(n(\beta m o + pm\alpha), n(np\beta))]\\[0.75em] &= [((n\beta)(mo) + (np)(m\alpha), (np)(n\beta))]\\[0.75em] &= [(mo, np)] + [(m\alpha, n\beta)]\\[0.75em] &= [(m, n)] \cdot [(o, p)] + [(m, n)] \cdot [(\alpha, \beta)]\\[0.75em] &= q \cdot r + q \cdot s, \end{align*}\] hence multiplication distributes over addition for rational numbers. \(\blacksquare\)

As with the integers, the rational numbers also have a multiplicative identity element, which is also called one, or unity. Intuitively one is “one piece of one”, so it is the equivalence class \([(1, 1)]\).

Proof. Let \(q = [(m, n)]\). Then using that \(1\) is the multiplicative identity of the integers, \[\begin{align*} q \cdot [(1, 1)] &= [(m, n)] \cdot [(1, 1)]\\[0.75em] &= [(m \cdot 1, n \cdot 1)]\\[0.75em] &= [(m, n)]\\[0.75em] &= q, \end{align*}\] so \([(1, 1)]\) is the multiplicative identity element of the rational numbers. \(\blacksquare\)

The new property of the rational numbers compared to the integers is the existence of a multiplicative-inverse element for every number other than zero, the rational additive-identity.

Proof. Let \(q = [(m, n)]\). If \(m = 0\) then \(m \cdot 1 = 0 \cdot n\), so \((m, n) \sim (0, 1)\) whence \([(m, n)] = [(0, 1)]\), which is a contradiction. Therefore \(m \neq 0\) so define \(q^{-1} \coloneq [(n, m)] \in \symbb{Q}\). Because multiplication of integers is commutative, \[\begin{align*} q \cdot q^{-1} &= [(m, n)] \cdot [(n, m)]\\[0.75em] &= [(mn, nm)]\\[0.75em] &= [(mn, mn)]. \end{align*}\] Since \((mn) \cdot 1 = 1 \cdot (mn)\) it follows that \((mn, mn) \sim (1, 1)\), whence \([(mn, mn)] = [(1, 1)]\). \(\blacksquare\)

As with addition, the rational multiplicative-identity is unique, as are rational multiplicative-inverses. Again it is more succinct to observe that \(\symbb{Q}\) is a field and then prove uniqueness of the multiplicative identity and multiplicative inverses for a general field. This is also taken care of in the Section 1.3.2 notes for Katznelson and Katznelson (2024).

The existence of multiplicative-inverse elements for all non-zero rational numbers, proven in Theorem 10, is key to proving the cancellation law of multiplication for rational numbers in Theorem 11.

Proof. By Theorem 10 \(q^{-1}\) exists, so multiplying both sides of \(qr = qs\) by \(q^{-1}\) and using Theorem 7 and Theorem 9 gives \[\begin{align*} q^{-1}(qr) &= q^{-1}(qs)\\[0.75em] \implies \underbrace{(q^{-1}q)}_{= [(1, 1)]}r &= \underbrace{(q^{-1}q)}_{= [(1, 1)]}s\\[0.75em] \implies [(1, 1)] \cdot r &= [(1, 1)] \cdot s\\[0.75em] \implies r &= s, \end{align*}\] so \(q\) cancels from each side of the equation. \(\blacksquare\)

An algebraic justification for extending \(\symbb{Z}\) to \(\symbb{Q}\)

It was stated in the Section 1.3 notes for Katznelson and Katznelson (2024) that the integers are well-suited for equations of the form \(m + x = n\) but not for equations of the form \(mx = n\). Theorem 12 proves that the rational numbers are well-suited for both types of equations.

Proof. By Theorem 4 \(-r\) exists, and by Theorem 10 \(q^{-1}\) exists. Let \(x \coloneq q^{-1}(s + (-r))\). Then using Theorem 7, Theorem 10, Theorem 9, Theorem 2, Theorem 4 and Theorem 3 shows that \[\begin{align*} qx + r &= q(q^{-1}(s + (-r))) + r\\[0.75em] &= \underbrace{(qq^{-1})}_{= [(1, 1)]}(s + (-r)) + r\\[0.75em] &= [(1, 1)] \cdot (s + (-r)) + r\\[0.75em] &= s + \underbrace{((-r) + r)}_{= [(0, 1)]}\\[0.75em] &= s + [(0, 1)]\\[0.75em] &= s, \end{align*}\] so at least one solution to the equation exists.

Suppose there were another \(y \in \symbb{Q}\) such that \(y \neq x\) and \(qy + r = s\). Then using the same theoremata, \[\begin{align*} qy + r &= s\\[0.75em] \implies (qy + r) + (-r) &= s + (-r)\\[0.75em] \implies qy + (r + (-r)) &= s + (-r)\\[0.75em] \implies qy &= s + (-r)\\[0.75em] \implies q^{-1}(qy) &= q^{-1}(s + (-r)). \end{align*}\] Continuing, \[\begin{align*} \implies (q^{-1}q)y &= q^{-1}(s + (-r))\\[0.75em] \implies y &= \underbrace{q^{-1}(s + (-r))}_{= x}\\[0.75em] \implies y &= x, \end{align*}\] which is a contradiction. Therefore \(y = x\) so the solution to the equation is unique. \(\blacksquare\)

A subset of \(\symbb{Q}\) isomorphic to \(\symbb{Z}\)

The Section 1.2 notes for Katznelson and Katznelson (2024) established an isomorphism between \(\symbb{N}\) and a special subset of \(\symbb{Z}\), which justified the convention of writing \(\symbb{N}\subseteq \symbb{Z}\) even though \(\symbb{N}\) is not actually a subset of \(\symbb{Z}= (\symbb{N}\times \symbb{N}) / {\sim}\). Similarly, \(\symbb{Z}\) is not actually a subset of \(\symbb{Q}= (\symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})) / {\sim}\), but there is a subset of \(\symbb{Q}\) that is not meaningfully different from \(\symbb{Z}\) so it is conventional to write \(\symbb{Z}\subseteq \symbb{Q}\). The Section 1.3 notes for Katznelson and Katznelson (2024) explained that justifying this convention couldn’t be done until arithmetic in \(\symbb{Q}\) had been developed. Now that this has been done, an isomorphism can be established between \(\symbb{Z}\) and a special subset of \(\symbb{Q}\).

Because integers are, informally, whole quantities, the corresponding subset of the rational numbers should also represent whole quantities. The rational numbers that represent whole quantities are “\(m\) pieces of one” where \(m\) is an integer, since these are degenerate fractional quantities. Therefore \[ \symbb{Z}\overset{!}{=}\underbrace{\{ [(m, 1)] \colon m \in \symbb{Z}\}}_{\eqcolon \tilde{\symbb{Z}}} \subseteq \symbb{Q}. \] Since this cannot be a literal equality, what is needed instead is an isomorphism \(\iota \colon \symbb{Z}\to \tilde{\symbb{Z}}\), that is, a bijection whose properties show that it makes no difference whether integers are interpreted as elements of \(\symbb{Z}\) or elements of \(\tilde{\symbb{Z}} \subseteq \symbb{Q}\).

Definition 1 An isomorphism between \(\symbb{Z}\) and \(\tilde{\symbb{Z}}\) is a bijection \(\iota \colon \symbb{Z}\to \tilde{\symbb{Z}}\) such that for all \(m, n \in \symbb{Z}\), \[\begin{align} \iota(m + n) &= \iota(m) + \iota(n) \label{eq:isodef-1} \\[0.75em] \iota(m \cdot n) &= \iota(m) \cdot \iota(n) \label{eq:isodef-2}, \end{align}\] where the binary operations on the left-hand sides of \(\eqref{eq:isodef-1}\) and \(\eqref{eq:isodef-2}\) operate on integers, and those on the right-hand sides operate on rational numbers.

Theorem 13 establishes the desired isomorphism \(\iota \colon \symbb{Z}\to \tilde{\symbb{Z}}\). The significance of the theorem can be explained as follows. If \(m, n \in \symbb{Z}\) are interpreted as integers, and their integer sum or product is interpreted as its corresponding rational number in \(\tilde{\symbb{Z}} \subseteq \symbb{Q}\), this leads to exactly the same result as if \(m\) and \(n\) were first interpreted as their corresponding rational numbers in \(\tilde{\symbb{Z}} \subseteq \symbb{Q}\) and then added or multiplied together in \(\symbb{Q}\). Therefore there is no meaningful difference between \(\symbb{Z}\) and \(\tilde{\symbb{Z}}\).

Proof. It’s necessary to verify that \(\iota\) is a bijection and that it satisfies \(\eqref{eq:isodef-1}\) and \(\eqref{eq:isodef-2}\). That \(\iota\) is a surjection is clear because for any \([(m, 1)] \in \tilde{\symbb{Z}}\), \(m \in \symbb{Z}\) is such that \(\iota(m) = [(m, 1)]\). To see that \(\iota\) is an injection, note that \(\iota(m) = \iota(n)\) implies that \([(m, 1)] = [(n, 1)]\). By the definition of \(\symbb{Q}\) this implies that \((m, 1) \sim (n, 1)\), so \(m \cdot 1 = n \cdot 1\), so \(m = n\), hence \(\iota\) is an injection.

For all \(m, n \in \symbb{Z}\), \[\begin{align*} \iota(m + n) &= [(m + n, 1)]\\[0.75em] &= [(1 \cdot m + 1 \cdot n, 1 \cdot 1)]\\[0.75em] &= [(m, 1)] + [(n, 1)]\\[0.75em] &= \iota(m) + \iota(n), \end{align*}\] so \(\iota\) satisfies \(\eqref{eq:isodef-1}\), and \[\begin{align*} \iota(m \cdot n) &= [(m \cdot n, 1)]\\[0.75em] &= [(m \cdot n, 1 \cdot 1)]\\[0.75em] &= [(m, 1)] \cdot [(n, 1)]\\[0.75em] &= \iota(m) \cdot \iota(n), \end{align*}\] so \(\iota\) satisfies \(\eqref{eq:isodef-2}\). \(\blacksquare\)

However, one aspect of the relationship between \(\symbb{Z}\) and \(\tilde{\symbb{Z}} \subseteq \symbb{Q}\) that is not addressed by Theorem 13 is the order on \(\symbb{Z}\), which was established in the Section 1.2.2 notes for Katznelson and Katznelson (2024). This will be rectified in the Section 1.3.4 notes for Katznelson and Katznelson (2024), which will define an order on \(\symbb{Q}\) such that for the isomorphism \(\iota \colon \symbb{Z}\to \tilde{\symbb{Z}}\) defined in Definition 1 and all \(m, n \in \symbb{Z}\), \(m < n\) if and only if \(\iota(m) < \iota(n)\). (Note that the former inequality is in \(\symbb{Z}\) whereas the latter inequality will be in \(\symbb{Q}\).) Once this result is proven then the full equivalence of \(\symbb{Z}\) and \(\tilde{\symbb{Z}} \subseteq \symbb{Q}\) will be established.

Notation

As proven in Proposition 1 and Proposition 2, arithmetic operations on rational numbers are insensitive to the choice of representatives used for the summands or factors. Because of this, the equivalence relation defining the rational numbers is often not made explicit in arithmetic in \(\symbb{Q}\). It is common to do arithmetic on elements of \(\symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})\) directly. Furthermore, instead of using the notation \((m, n) \in \symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})\) it is conventional to write \(m / n\) or \[ \frac{m}{n} \] instead. Also, common alternative notation for the multiplicative inverse of \(q \in \symbb{Q}\) is \(1 / q\) instead of \(q^{-1}\).

The “division” of one rational number \(q\) by another \(r\), where \(r \neq [(0, 1)]\), is written as \(q \cdot r^{-1}\). Alternative notations for division of \(q\) by \(r\) are \(q / r\) or \(q \cdot (1 / r)\).

Rational numbers in lowest terms

Every rational number has many representatives. This can be seen from Lemma 1, because if \(q = [(m, n)] \in \symbb{Q}\) and \(o \in \symbb{Z}\), then \([(o \cdot m, o \cdot n)] = [(m, n)]\). For example, \([(1, 2)] = [(2 \cdot 1, 2 \cdot 2)] = [(3 \cdot 1, 3 \cdot 2)] = \dots\), or in the less formal notation, \(1/2 = 2/4 = 3/6 = \dots\) . For a given calculation a specific choice of representative may be advantageous. The definition of rational addition in \(\eqref{eq:adddef}\) was in fact primarily motivated through the use of specific representatives for the summands so that rational addition could be reduced to integer addition.

Nonetheless, every rational number has one representation that deserves special attention: its reduced form. Even though Lemma 1 suggests multiplying numerator and denominator by a non-zero integer, it can also be applied in the reverse direction. If both numerator and denominator of a given rational number share a common factor then Lemma 1 implies that this common factor can be “cancelled” or removed from the numerator and denominator.

By Lemma 2, for any \(q = [(m, n)] \in \symbb{Q}\) it can be assumed that \(n > 0\). From the Section 1.2.2 notes for Katznelson and Katznelson (2024) it follows that \(n \in \symbb{N}\). By the well-ordering princple from the Section 1.1.3 notes for Katznelson and Katznelson (2024), therefore, the set \[\begin{equation}\label{eq:lowterms-1} \{ p \in \symbb{N}\colon \exists o \in \symbb{Z}\text{ such that } [(o, p)] = [(m, n)] \} \end{equation}\] has a least element. The pair \((\alpha, \beta)\), where \(\beta\) is the least element of the set defined in \(\eqref{eq:lowterms-1}\), are said to represent \(q\) in lowest terms, or in reduced form.

Proof. Let \(q = [(m, n)]\) where \(m \in \symbb{Z}\) and \(n \in \symbb{Z}\setminus \{ 0 \}\). Because of Lemma 2 and the well-ordering principle, the reduced form \((\alpha, \beta)\) exists for any \(q \in \symbb{Q}\). Suppose \((\alpha, \beta)\) were not unique, so there exists \((\gamma, \delta)\) such that \((\gamma, \delta)\) expresses \(q\) in lowest terms and \((\alpha \neq \gamma) \lor (\beta \neq \delta)\). By the trichotomy law for the natural numbers, if \(\beta \neq \delta\) then either \(\beta < \delta\) or \(\beta > \delta\). Either possibility contradicts the fact that both \(\beta\) and \(\delta\) are the least element of the set \[ \{ p \in \symbb{N}\colon \exists o \in \symbb{Z}\text{ such that } [(o, p)] = [(m, n)] \}. \] Therefore \(\beta = \delta\) so it must be that \(\alpha \neq \gamma\). By the trichotomy law for the integers, either \(\alpha < \gamma\) or \(\alpha > \gamma\).

By assumption \([(\alpha, \beta)] = [(\gamma, \delta)] = [(m, n)]\), which implies that \((\alpha, \beta) \sim (\gamma, \delta)\) so \(\alpha \cdot \delta = \gamma \cdot \beta\). If \(\alpha < \gamma\) then because \(\delta > 0\) it follows from the Section 1.2.2 notes for Katznelson and Katznelson (2024) that \(\alpha \cdot \delta < \gamma \cdot \delta\). Furthermore, because it has been deduced that \(\beta = \delta\) it follows that \(\alpha \cdot \delta < \gamma \cdot \beta\), which is a contradiction. Similarly, if \(\alpha > \gamma\) then it follows that \(\alpha \cdot \delta > \gamma \cdot \delta\), and then \(\alpha \cdot \delta > \gamma \cdot \beta\), which is again a contradiction. \(\blacksquare\)

As defined by Katznelson and Katznelson (2024, 6), \(m, n \in \symbb{Z}\) have a common factor \(f \in \symbb{Z}\setminus \{ 0 \}\) if there exist \(k_1, k_2 \in \symbb{Z}\) such that \(m = fk_1\) and \(n = fk_2\). If there exists no such \(f \in \symbb{Z}\setminus \{ -1, 0, 1 \}\) then \(m\) and \(n\) are said to have no non-trivial common factors, or to be relatively prime.

Proposition 4 shows that for a rational number in reduced form, its numerator and denominator are relatively prime.

Proof. If \(\alpha\) and \(\beta\) were not relatively prime then by the definition of relative primality there exists \(f \in \symbb{Z}\setminus \{ -1, 0, 1 \}\) and \(k_1, k_2 \in \symbb{Z}\) such that \(\alpha = fk_1\) and \(\beta = fk_2\). First note that \(k_2 \neq 0\) otherwise using Proposition 1.2.1 from Katznelson and Katznelson (2024, 6) it follows that \(\beta = fk_2 = 0\), which is a contradiction. By the trichotomy law for integers from the Section 1.2.2 notes from Katznelson and Katznelson (2024) it follows that either \(k_2 > 0\) or \(k_2 < 0\). By the same law it also follows that \(f > 0\) or \(f < 0\).

If \(f > 0\) then from the same notes it is not possible that \(0 < f < 0 + 1 = 1\), so \(f \geq 1\), and by assumption therefore \(f > 1\). In this case \(k_2 > 0\), because if \(k_2 < 0\) then by Exercise 1.2.8 from Katznelson and Katznelson (2024) it follows that \(\beta = fk_2 < 0\). Therefore \(\beta = fk_2 > k_2\), so using Lemma 1 \(q = [(\alpha, \beta)] = [(fk_1, fk_2)] = [(k_1, k_2)]\) which is a contradiction because \((\alpha, \beta)\) was assumed to be in lowest terms.

If \(f < 0\) then from the same notes again it is not possible that \(-1 < f < (-1) + 1 = 0\), so \(f \leq -1\), and by assumption therefore \(f < -1\). In this case \(k_2 < 0\), because if \(k_2 > 0\) then by the same exercise as before it follows that \(\beta = fk_2 < 0\). Therefore using Exercise 1.2.4 and Exercise 1.2.5 from Katznelson and Katznelson (2024) \(\beta = fk_2 > (-1)k_2 = -k_2 > 0\). Using Lemma 1 and Lemma 2 it follows that \(q = [(\alpha, \beta)] = [(fk_1, fk_2)] = [(k_1, k_2)] = [(-k_1, -k_2)]\), which is a contradiction because \((\alpha, \beta)\) was assumed to be in lowest terms. \(\blacksquare\)

One reason for the importance of reduced forms of rational numbers is that reduced forms can be used to deduce an important shortcoming of \(\symbb{Q}\). In particular, there is no \(q \in \symbb{Q}\) such that \(q^2 = 2\). This is discussed in the Section 1.3.4 notes for Katznelson and Katznelson (2024), and proven explicitly in Exercise 1.3.4 for Katznelson and Katznelson (2024).

Using \(\symbb{Z}\) and \(\symbb{Q}\) together

If \(m \in \symbb{Z}\) and \(q \in \symbb{Q}\) then arithmetic expressions such as \(m + q\) and \(m \cdot q\) are not well-defined. However, because of the isomorphism \(\iota \colon \symbb{Z}\to \tilde{\symbb{Z}}\) established in Theorem 13 this is an acceptable notational shorthand because \(m\) can be replaced with \(\iota(m) = [(m, 1)] = m / 1\) in order to evaluate the expressions. For example, using this shorthand the statement of Theorem 3 can be rewritten from \(q + [(0, 1)] = q\) to \(q + 0 = q\). Similarly, the statement of Theorem 9 can be rewritten from \(q \cdot [(1, 1)] = q\) to \(q \cdot 1 = q\).

References

Ironically these notes do not quite show that \(\symbb{Q}\) is a field. These notes show that \(\symbb{Q}\) is almost a field because they show that \(\symbb{Q}\) has all the properties of a field \(F\), as defined by Katznelson and Katznelson (2024, 9), except that the additive and multiplicative identity elements of \(\symbb{Q}\) are distinct. This last task is completed in the Section 1.3.2 notes for Katznelson and Katznelson (2024).↩︎