An Introduction to Real Analysis
Section 1.3 notes
2025-12-06
The Section 1.2.2 notes for Katznelson and Katznelson (2024) described the shortcomings of the integers \(\symbb{Z}\). The fact that for any \(m \in \symbb{Z}\) there is no \(n \in \symbb{Z}\) such that \(m < n < m + 1\) implies that the integers cannot model fractional quantities. The integers can count quantities of apples possessed or owed, but they cannot model an apple cut into halves without changing the units of measurement from apples to halves of apples.
Algebraically, the integers are well-suited for equations of the form \(m + x = n\) but not for equations of the form \(mx = n\). Informally, the integers are self-contained with respect to addition because for any integers \(m, n \in \symbb{Z}\) there is always a unique integer \(x \in \symbb{Z}\) that can be added to \(m\) so that the result is \(n\). The integers are not, however, self-contained with respect to multiplication. Abstractly there is always some unique quantity \(x\) of apples that can be added to any quantity \(m\) of apples that results in any other quantity \(n\) of apples. There is not always a unique factor \(x\) that can be multiplied by any quantity \(m\) of apples that results in any other quantity \(n\) of apples.1
The rational numbers \(\symbb{Q}\) rectify these shortcomings of the integers. Just as the integers \(\symbb{Z}\) were constructed from the more rudimentary natural numbers \(\symbb{N}\), subsuming the latter and effectively containing within themselves the natural numbers, the rational numbers are constructed from the integers and effectively contain within themselves the integers.
The set-theoretic construction of \(\symbb{Q}\) from \(\symbb{Z}\) is carried out in these Section 1.3 notes for Katznelson and Katznelson (2024). Arithmetic in \(\symbb{Q}\) is developed in the Section 1.3.1 notes. The order on \(\symbb{Q}\) is developed in the Section 1.3.4 notes. In between, the Section 1.3.2 notes discuss algebraic operations in a field, and the Section 1.3.3 notes discuss ordered fields.
Half an apple can be thought of as one piece of two. Framing the fraction in this way suggests that the mathematical representation of half should involve the integers \(1\) and \(2\). Furthermore this pair of numbers ought to be ordered because one piece of two is different from two pieces of one; the former refers to half an apple while the latter refers to two apples. This reasoning suggests looking at ordered pairs of integers, in particular the set \(\symbb{Z}\times \symbb{Z}\), in order to define the rational numbers. It also suggests that once the rational numbers have been formally constructed, in order to justify the convention of writing \(\symbb{Z}\subseteq \symbb{Q}\) one ought to look for a subset \(\symbb{Z}\overset{!}{=}\{ (m, 1) \colon m \in \symbb{Z}\} \overset{!}{\subseteq}\symbb{Q}\).2
Similar to the construction of the integers from the natural numbers in the Section 1.2 notes for Katznelson and Katznelson (2024), however, the catch is that the concept of “half” is not uniquely defined by the ordered pair \((1, 2) \in \symbb{Z}\times \symbb{Z}\). If the apple were cut into quarters then two quarters of an apple is still half an apple; the fact that the apple had been cut into four pieces, not two, is immaterial. Because of this \((1, 2), (2, 4) \in \symbb{Z}\times \symbb{Z}\) both seem equally suited to represent the same rational number, and indeed infinitely more ordered pairs of integers, such as \((3, 6)\), \((4, 8)\) and so on are also equally suited to represent half.3 Therefore, as in the Section 1.2 notes for Katznelson and Katznelson (2024) it is natural to look for an equivalence relation \(\sim\) to put on \(\symbb{Z}\times \symbb{Z}\) and then define \(\symbb{Q}\) as the quotient \((\symbb{Z}\times \symbb{Z}) / {\sim}\).
In order to find the right equivalence relation \(\sim\) to put on \(\symbb{Z}\times \symbb{Z}\) it is helpful to get a clearer idea of how the rational numbers should work algebraically. Since two halves are a whole apple, informally \((2, 2) \overset{!}{=}(1, 1) \overset{!}{=}1 \overset{!}{\in}\symbb{Q}\). Furthermore since two pieces of one should be two apples, \(2 \overset{!}{=}(2, 1)\), and double half an apple should be one apple, so \((2, 1) \cdot (1, 2) \overset{!}{=}(2, 2) \overset{!}{=}1\). This suggests that \((1, m)\) should act as a multiplicative inverse to \((m, 1)\). Adopting the notation \(n^{-1}\) for the multiplicative inverse of \(n \in \symbb{Z}\), this suggests if \((m, n)\) were a rational number it could be written as \(m \cdot n^{-1}\) or \(mn^{-1}\). Thus \((2, 2)\) would be written as \(2 \cdot 2^{-1} \overset{!}{=}1\), so the notation is consistent with intuition.
This notation reveals another catch that has to be addressed before proceeding to the formal construction of \(\symbb{Q}\). Proposition 1.2.1 in Katznelson and Katznelson (2024, 6) established that \(m \cdot 0 = 0\) for all \(m \in \symbb{Z}\), and \((m, 1)\) and \((1, m)\) should be multiplicative inverses of each other in \(\symbb{Q}\). If an analogous proposition is to hold for the rational numbers then zero as a rational number cannot have a multiplicative inverse. If \(0 \in \symbb{Q}\) did have a multiplicative inverse \(0^{-1} \in \symbb{Q}\), then \(0 \cdot 0^{-1} = 1 \overset{!}{\in}\symbb{Q}\) so there is some rational number \(0^{-1}\) whose product with zero is equal to one, which would violate the analogous proposition for \(\symbb{Q}\). Heuristically zero as a rational number should be zero pieces of one, or \((0, 1)\), so its inverse if it did exist would be \((1, 0)\). Because such a rational number cannot exist, this implies that zero cannot appear as the second integer in an ordered pair of integers informally representing a rational number. Therefore the set of ordered pairs under consideration should be \(\symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})\) instead of \(\symbb{Z}\times \symbb{Z}\), and given the right equivalence relation \(\sim\) on \(\symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})\), the rational numbers \(\symbb{Q}\) should be defined as the quotient \((\symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})) / {\sim}\).
To formalize the informal work above, the equivalence relation \(\sim\) on \(\symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})\) must be constructed in terms of integers and their arithmetic. If \((1, 2)\) and \((2, 4)\) are to be considered equivalent, algebraically this suggests that \(1 \cdot 2^{-1} \overset{!}{=}2 \cdot 4^{-1}\). This can be rewritten to avoid the use of inverses: \[\begin{align*} 1 \cdot 2^{-1} &\overset{!}{=}2 \cdot 4^{-1}\\[0.75em] \implies 1 \cdot 4 &\overset{!}{=}2 \cdot 2. \end{align*}\] The second equation suggests that if two ordered pairs of integers \((m, n), (o, p) \in \symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})\) are considered to be the same rational number, this can be expressed via the equation \(m \cdot p = o \cdot n\). Theorem 1 demonstrates that this indeed defines an equivalence relation \(\sim\) on \(\symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})\).
Theorem 1 Let the binary relation \(\sim\) on \(\symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})\) be defined by \((m, n) \sim (o, p)\) if and only if \(m \cdot p = o \cdot n\). Then \(\sim\) is an equivalence relation.
Proof. Since \(m \cdot n = m \cdot n\), it follows that \((m, n) \sim (m, n)\), so \(\sim\) is reflexive.
Next assume that \((m, n) \sim (o, p)\), so \(m \cdot p = o \cdot n\). Then \(o \cdot n = m \cdot p\), so \((o, p) \sim (m, n)\) and therefore \(\sim\) is symmetric.
Lastly assume \((m, n) \sim (o, p)\) and \((o, p) \sim (q, r)\), so \(m \cdot p = o \cdot n\) and \(o \cdot r = q \cdot p\). Then using standard properties of arithmetic in \(\symbb{Z}\), \[\begin{align*} (m \cdot p) \cdot r &= (o \cdot n) \cdot r\\[0.75em] \implies m \cdot (p \cdot r) &= o \cdot (n \cdot r)\\[0.75em] \implies m \cdot (r \cdot p) &= o \cdot (r \cdot n)\\[0.75em] \implies (m \cdot r) \cdot p &= \underbrace{(o \cdot r)}_{= q \cdot p} \cdot n. \end{align*}\] Continuing, \[\begin{align*} (m \cdot r) \cdot p &= (q \cdot p) \cdot n\\[0.75em] &= q \cdot (p \cdot n)\\[0.75em] &= q \cdot (n \cdot p)\\[0.75em] &= (q \cdot n) \cdot p. \end{align*}\] Using the cancellation law of multiplication for integers from the Section 1.2.1 notes for Katznelson and Katznelson (2024), because \(p \in \symbb{Z}\setminus \{ 0 \}\) it follows that \(m \cdot r = q \cdot n\). Therefore \((m, n) \sim (q, r)\), whence \(\sim\) is transitive. \(\blacksquare\)
Because \(\sim\) is an equivalence relation on \(\symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})\), this set can be partitioned into the equivalence classes of the equivalence relation. The rational numbers are defined to be this set of equivalence classes: \(\symbb{Q}\coloneq (\symbb{Z}\times (\symbb{Z}\setminus \{ 0 \} )) / {\sim}\). Therefore a rational number \([(m, n)] \in \symbb{Q}\) is the equivalence class containing \((m, n) \in \symbb{Z}\times (\symbb{Z}\setminus \{ 0 \})\), that is \[ [(m, n)] \coloneq \{ (o, p) \in \symbb{Z}\times (\symbb{Z}\setminus \{ 0 \}) \colon (o, p) \sim (m, n) \}. \]
In the representative \((m, n)\) of the rational number \([(m, n)]\), the first integer \(m\) is called the numerator and the second non-zero integer \(n\) is called the denominator.
The Section 1.2 notes for Katznelson and Katznelson (2024) constructed the integers \(\symbb{Z}\) from the natural numbers \(\symbb{N}\) and then devoted a considerable amount of time to developing the notion of an isomorphism between Peano systems and proving that the integers contain a Peano system. This did not rely on the development of arithmetic in \(\symbb{Z}\), which was done later in the Section 1.2.1 notes for Katznelson and Katznelson (2024), because a Peano system by definition does not require concepts of addition and multiplication.
In contrast, the essential characteristics of the integers cannot be captured without reference to their arithmetic operations. Therefore proving that, informally, the rational numbers contain the integers requires establishing that the arithmetic of a particular subset of the rational numbers works just like the arithmetic of the integers. Hence proving that there is a subset of \(\symbb{Q}\) isomorphic to \(\symbb{Z}\) cannot be done until arithmetic in \(\symbb{Q}\) is developed in the Section 1.3.1 notes for Katznelson and Katznelson (2024).
For the purposes of modeling phenomena another consideration is the units associated with the terms in the equations \(m + x = n\) and \(mx = n\). In the first equation \(m + x = n\) all three terms should have units of apples, because apples added to apples should result in apples. However, in the second equation \(mx = n\), \(x\) ought not have units of apples. If it did, then \(mx\) as apples times apples would have units of square apples, which is not consistent with the units of \(n\), which ought to be apples. Instead \(x\) ought to be a unitless quantity. For example, if \(m = 1\) then doubling the quantity of apples results in \(1_{\text{apples}} \cdot 2 = 2_{\text{apples}}\), not square apples, so the doubling factor needs to be unitless. This suggests that \(x\) needs to be drawn from a separate set of numbers than \(m\) and \(n\), at least with respect to their units. This foreshadows the development of vector spaces, which are an abelian group \(V\) and field \(F\) which are compatible in ways that can be made precise. The elements of \(V\) are called vectors and the elements of \(F\) are called scalars. In particular, if \(v \in V\) is a vector and \(f \in F\) is a scalar then \(fv \in V\) ought to be another vector. The development of vector spaces is given in the notes for Katznelson and Katznelson (2008).↩︎
Justifying this convention will be postponed until arithmetic in \(\symbb{Q}\) is developed in the Section 1.3.1 notes for Katznelson and Katznelson (2024). Establishing that there is a subset of \(\symbb{Q}\) that behaves just like \(\symbb{Z}\) necessarily involves proving that the arithmetic operations in \(\symbb{Z}\) are just like those of a special subset of \(\symbb{Q}\).↩︎
Here integers are written in their base-ten representations. It would be equivalent to write them using any other radix, for example in base-three as \((1, 2)\), \((2, 11)\), \((10, 20)\) and \((11, 22)\), respectively. In general the choice of radix will be clear from context.↩︎