Arithmetic in \(\symbb{N}\)

An Introduction to Real Analysis

2025-10-11

Katznelson and Katznelson (2024, sec 1.1.1) take the properties of addition and multiplication in \(\symbb{N}\) as axiomatic. These properties can in fact be derived from a more rudimentary set of assumptions called the Peano axioms. After stating these axioms and developing some understanding thereof, the derivation of all the properties taken by Katznelson and Katznelson (2024, sec 1.1.1) as axiomatic is carried out thence.

Peano systems

Let \(\symbb{N}\) be a non-empty set, \(1 \in \symbb{N}\) a distinguished element of \(\symbb{N}\), and \(\sigma \colon \symbb{N}\to \symbb{N}\) a function on \(\symbb{N}\). The Peano axioms are as follows:

  1. For \(a \in \symbb{N}\), \(1 \neq \sigma(a)\).
  2. For \(a, b \in \symbb{N}\), \(\sigma(a) = \sigma(b)\) implies that \(a = b\).
  3. If \(M \subseteq \symbb{N}\) is such that \(1 \in M\), and \(a \in M\) implies that \(\sigma(a) \in M\), then \(M = \symbb{N}\).

The function \(\sigma\) is called the successor function, and if \(a \in \symbb{N}\), then \(\sigma(a)\) is called the successor of \(a\).

The first Peano axiom states that \(1\) is not the successor of any natural number, or more formally, \(1 \notin \symup{img}(\sigma)\). The second Peano axiom states that the successor function is injective. The third Peano axiom, called the principle of induction or a similar name, captures the idea that \(\symbb{N}\) contains only the natural numbers.

A triplet \((\symbb{N}, \sigma, 1)\) satisfying the three Peano axioms is called a Peano system.

The following examples of a non-empty set, distinguished element and function do not satisfy the Peano axioms.

Example 1 Let \(N \coloneq \{ a \}\) and \(s \colon N \to N\) be the function such that \(a \mapsto a\). The triplet \((N, s, a)\) satisfies the second and third Peano axioms but not the first. \(\blacksquare\)

Example 2 Let \(N \coloneq \{ a, b \}\) and \(s \colon N \to N\) be the function such that \(x \mapsto b\) for all \(x \in N\). The triplet \((N, s, a)\) satisfies the first and third Peano axioms but not the second. \(\blacksquare\)

Example 3 Suppose \((\symbb{N}, \sigma, 1)\) is a Peano system. Introduce a new symbol \(\vartheta \notin \symbb{N}\) and consider the set \(\symbb{N}\uplus \{ \vartheta \}\) with distinguished element \(1\). Consider the function \(\tau \colon \symbb{N}\uplus \{ \vartheta \} \to \symbb{N}\uplus \{ \vartheta \}\) such that \(\tau\vert_{\symbb{N}} = \sigma\) and \(\tau(\vartheta) = \vartheta\). Then \((\symbb{N}\uplus \{ \vartheta \}, \tau, 1)\) satisfies the first and second Peano axioms, but not the third, as evidenced by taking \(M = \symbb{N}\subseteq \symbb{N}\uplus \{ \vartheta \}\).1 \(\blacksquare\)

Before developing addition and multiplication in \(\symbb{N}\) from a Peano system \((\symbb{N}, \sigma, 1)\), the following propositions provide some evidence that a Peano system behaves in accordance with intuition.

Proposition 1 The successor function \(\sigma\) is a bijection when viewed as a function \(\sigma \colon \symbb{N}\to \symbb{N}\setminus \{ 1 \}\).

Proof. By definition \(\sigma \colon \symbb{N}\to \symbb{N}\) is injective, which implies that \(\sigma \colon \symbb{N}\to \symbb{N}\setminus \{ 1 \}\) is injective. Hence it suffices to show that \(\sigma \colon \symbb{N}\to \symbb{N}\setminus \{ 1 \}\) is surjective. Apply the third Peano axiom, i.e., the principle of induction, with \(M = \{ 1 \} \uplus \symup{img}(\sigma) \subseteq \symbb{N}\). (By the first Peano axiom, \(1 \notin \symup{img}(\sigma)\).) Then \(1 \in \{ 1 \} \uplus \symup{img}(\sigma)\), and \(a \in \{ 1 \} \uplus \symup{img}(\sigma)\) implies that \(a \in \symbb{N}\), so \(\sigma(a) \in \symup{img}(\sigma) \subseteq \{ 1 \} \uplus \symup{img}(\sigma)\). Hence by the third Peano axiom \(\{ 1 \} \uplus \symup{img}(\sigma) = \symbb{N}\). Therefore \(\symup{img}(\sigma) = \symbb{N}\setminus \{ 1 \}\), so \(\sigma \colon \symbb{N}\to \symbb{N}\setminus \{ 1 \}\) is surjective and therefore bijective. \(\blacksquare\)

Proposition 2 The successor function has no fixed points.

Proof. To show that for all \(a \in \symbb{N}\), \(a \neq \sigma(a)\), apply the third Peano axiom to the set \(M \coloneq \{ a \in \symbb{N}\colon a \neq \sigma(a) \} \subseteq \symbb{N}\). By the first Peano axiom, \(1 \neq \sigma(a)\) for all \(a \in \symbb{N}\), so taking \(a = 1\) implies that \(1 \in M\). If \(a \in M\), then \(a \neq \sigma(a)\). By the contrapositive of the second Peano axiom, \(\sigma(a) \neq \sigma(\sigma(a))\), so \(\sigma(a) \in M\) and therefore \(M = \symbb{N}\). \(\blacksquare\)

Similar to the proofs of the preceding propositions, the proofs of properties of addition and multiplication in \(\symbb{N}\) in what follows rely on the third Peano axiom. An apparent difficulty is that addition and multiplication involve two natural numbers, while the principle of induction seemingly applies when a statement or a set depends only on one natural number. For example, even if a definition of addition has been given, how would one prove that for all \(a, b \in \symbb{N}\), \(a + b = b + a\)? In general, this difficulty is surmounted by fixing \(a\) and applying the principle of induction with respect to \(b\). If this succeeds and the argument does not depend on the choice of \(a\), then the result has been proven for all \(a, b \in \symbb{N}\).

Addition in \(\symbb{N}\)

Before proving properties of addition in \(\symbb{N}\), it is necessary to define the binary operation \(+ \colon \symbb{N}\times \symbb{N}\to \symbb{N}\). The definition will be recursive, and the recursion applies the idea of fixing \(a\) and inducting on \(b\). The successor function is intended to formalize the idea of adding \(1\) to a natural number, i.e., for \(a \in \symbb{N}\), \(a + 1 \overset{!}{=}\sigma(a)\).2 To find the correct form for the recursion, work informally with summations as they ought to behave. Supposing that \(a + b\) has been defined, \[\begin{align*} a + \sigma(b) &\overset{!}{=}a + (b + 1)\\[0.75em] &\overset{!}{=}(a + b) + 1\\[0.75em] &\overset{!}{=}\sigma(a + b). \end{align*}\] To prove that such a binary operation \(+\) exists and is unique, the proof will first be carried out for a restricted form of addition with \(a\) fixed. For all \(a \in \symbb{N}\), \(f_a \colon \symbb{N}\to \symbb{N}\) will be defined such that \(f_a(b) \overset{!}{=}a + b\).

Lemma 1 For all \(a \in \symbb{N}\), there exists a unique function \(f_a \colon \symbb{N}\to \symbb{N}\) such that \[\begin{align} f_a(1) &= \sigma(a) \label{eq:1} \\[0.75em] f_a(\sigma(b)) &= \sigma(f_a(b)) \qquad \text{for all } b \in \symbb{N}. \label{eq:2} \end{align}\]

Proof. To prove existence, define the set \[ M \coloneq \{ a \in \symbb{N}\colon \exists \text{ at least one function } f_a \colon \symbb{N}\to \symbb{N}\text{ satisfying } \eqref{eq:1} \text{ and } \eqref{eq:2} \} \subseteq \symbb{N} \] and apply the principle of induction. Define \(f_1 \colon \symbb{N}\to \symbb{N}\) by \(f_1(b) = \sigma(b)\) for all \(b \in \symbb{N}\).3 Then \(f_1\) satisfies \(\eqref{eq:1}\) because \(f_1(1) = \sigma(1)\), and \(f_1\) satisfies \(\eqref{eq:2}\) because for all \(b \in \symbb{N}\), \(f_1(\sigma(b)) = \sigma(\sigma(b)) = \sigma(f_1(b))\). Since \(f_1\) satisfies \(\eqref{eq:1}\) and \(\eqref{eq:2}\), \(1 \in M\).

Next suppose \(a \in M\), so there is at least one function \(f_a \colon \symbb{N}\to \symbb{N}\) satisfying \(\eqref{eq:1}\) and \(\eqref{eq:2}\). Using \(f_a\), define \(f_{\sigma(a)} \colon \symbb{N}\to \symbb{N}\) by \(f_{\sigma(a)}(b) = \sigma(f_a(b))\) for all \(b \in \symbb{N}\).4 Then \(f_{\sigma(a)}\) satisfies \(\eqref{eq:1}\) because \(f_{\sigma(a)}(1) = \sigma(f_a(1)) = \sigma(\sigma(a))\), and \(f_{\sigma(a)}\) satisfies \(\eqref{eq:2}\) because \[\begin{align*} f_{\sigma(a)}(\sigma(b)) &= \sigma(f_a(\sigma(b)))\\[0.75em] &= \sigma(\sigma(f_a(b)))\\[0.75em] &= \sigma(f_{\sigma(a)}(b)). \end{align*}\] Since \(f_{\sigma(a)}\) satisfies \(\eqref{eq:1}\) and \(\eqref{eq:2}\), \(\sigma(a) \in M\), so \(M = \symbb{N}\).

To prove uniqueness, for chosen \(a \in \symbb{N}\) suppose there are functions \(f_a \colon \symbb{N}\to \symbb{N}\) and \(g_a \colon \symbb{N}\to \symbb{N}\) both satisfying \(\eqref{eq:1}\) and \(\eqref{eq:2}\), and define the set \[ N \coloneq \{ b \in \symbb{N}\colon f_a(b) = g_a(b) \} \subseteq \symbb{N} \] and apply the principle of induction. By \(\eqref{eq:1}\), \(f_a(1) = \sigma(a) = g_a(1)\), so \(1 \in N\). Next suppose \(b \in N\), so \(f_a(b) = g_a(b)\). Then by \(\eqref{eq:2}\) and this assumption, \(f_a(\sigma(b)) = \sigma(f_a(b)) = \sigma(g_a(b)) = g_a(\sigma(b))\), so \(\sigma(b) \in N\). Therefore \(N = \symbb{N}\), and the uniqueness of \(f_a\) has been proven for the chosen \(a \in \symbb{N}\). Since \(a \in \symbb{N}\) was arbitrary, this argument applies for any \(a \in \symbb{N}\). \(\blacksquare\)

Proposition 3 There exists a unique binary operation \(+ \colon \symbb{N}\times \symbb{N}\to \symbb{N}\) such that \[\begin{alignat}{2} a + 1 &= \sigma(a) \qquad &&\text{for all } a \in \symbb{N}\label{eq:3} \\[0.75em] a + \sigma(b) &= \sigma(a + b) \qquad &&\text{for all } a, b \in \symbb{N}. \label{eq:4} \end{alignat}\]

Proof. To prove existence, for any \(a \in \symbb{N}\), define \(a + b \coloneq f_a(b)\) for all \(b \in \symbb{N}\) where \(f_a\) comes from Lemma 1. By Lemma 1, this is well-defined. Using this definition of addition, \(a + 1 = f_a(1) = \sigma(a)\) by \(\eqref{eq:1}\), so \(\eqref{eq:3}\) is satisfied. Again using the definition of addition, \(a + \sigma(b) = f_a(\sigma(b)) = \sigma(f_a(b)) = \sigma(a + b)\) by \(\eqref{eq:2}\), so \(\eqref{eq:4}\) is satisfied.

To prove uniqueness, suppose there are binary operations \(+ \colon \symbb{N}\times \symbb{N}\to \symbb{N}\) and \(\mathbin{\tilde{+}}\colon \symbb{N}\times \symbb{N}\to \symbb{N}\), both of which satisfy \(\eqref{eq:3}\) and \(\eqref{eq:4}\). For chosen \(a \in \symbb{N}\), define the set \(M_a \coloneq \{ b \in \symbb{N}\colon a + b = a \mathbin{\tilde{+}}b \} \subseteq \symbb{N}\). By \(\eqref{eq:3}\), \(a + 1 = \sigma(a) = a \mathbin{\tilde{+}}1\), so \(1 \in M_a\). Next assume that \(b \in M_a\), so \(a + b = a \mathbin{\tilde{+}}b\). Then by \(\eqref{eq:4}\), \(a + \sigma(b) = \sigma(a + b) = \sigma(a \mathbin{\tilde{+}}b) = a \mathbin{\tilde{+}}\sigma(b)\), so \(\sigma(b) \in M_a\) and therefore \(M_a = \symbb{N}\). Since \(a \in \symbb{N}\) was arbitrary, this argument applies for any \(a \in \symbb{N}\) and therefore \(a + b = a \mathbin{\tilde{+}}b\) for all \(a, b \in \symbb{N}\). \(\blacksquare\)

Theorem 1 (Associativity of addition) For all \(a, b, c \in \symbb{N}\), \((a + b) + c = a + (b + c)\).

Proof. For chosen \(a, b \in \symbb{N}\), define the set \(M_{a, b} \coloneq \{ c \in \symbb{N}\colon (a + b) + c = a + (b + c) \} \subseteq \symbb{N}\). Since \(+\) is a binary operation, \(a + b \in \symbb{N}\), and by definition of \(+\), \((a + b) + 1 = \sigma(a + b) = a + \sigma(b) = a + (b + 1)\), so \(1 \in M_{a, b}\). Next assume that \(c \in M_{a, b}\), so \((a + b) + c = a + (b + c)\). Then \[\begin{align*} (a + b) + \sigma(c) &= \sigma((a + b) + c)\\[0.75em] &= \sigma(a + (b + c))\\[0.75em] &= a + \sigma(b + c)\\[0.75em] &= a + (b + \sigma(c)), \end{align*}\] so \(\sigma(c) \in M_{a, b}\) and therefore \(M_{a, b} = \symbb{N}\). Since \(a, b \in \symbb{N}\) were arbitrary, this argument applies for any \(a, b \in \symbb{N}\). \(\blacksquare\)

Theorem 2 (Commutativity of addition) For all \(a, b \in \symbb{N}\), \(a + b = b + a\).

Proof. For chosen \(a \in \symbb{N}\), define the set \(M_a \coloneq \{ b \in \symbb{N}\colon a + b = b + a \} \subseteq \symbb{N}\). By \(\eqref{eq:3}\) and the definition of addition in terms of the functions \(f\) as in the proofs of Proposition 3 and Lemma 1, \(a + 1 = \sigma(a) = f_1(a) = 1 + a\), so \(1 \in M_a\). Next assume that \(b \in M_a\), so \(a + b = b + a\). Then by \(\eqref{eq:4}\) and again using the proofs of Proposition 3 and Lemma 1, \[\begin{align*} a + \sigma(b) &= \sigma(a + b)\\[0.75em] &= \sigma(b + a)\\[0.75em] &= \sigma(f_b(a))\\[0.75em] &= f_{\sigma(b)}(a)\\[0.75em] &= \sigma(b) + a, \end{align*}\] so \(\sigma(b) \in M_{a}\), and therefore \(M_a = \symbb{N}\). Since \(a \in \symbb{N}\) was arbitrary, this argument applies for any \(a \in \symbb{N}\). \(\blacksquare\)

Multiplication in \(\symbb{N}\)

After having defined addition in \(\symbb{N}\) and developed its properties, the procedure for defining multiplication in \(\symbb{N}\) and developing its properties will seem repetitive. The main difference is that multiplication will be defined partly in terms of addition, so the definitions and proofs in the development of multiplication will make repeated use of addition and its properties.

The definition of the binary operation \(\cdot \colon \symbb{N}\times \symbb{N}\to \symbb{N}\) will be recursive, and the recursion applies the idea of fixing \(a\) and inducting on \(b\). The only natural base case is \(a \cdot 1 \overset{!}{=}a\). To find the correct form for the recursion, work informally with products as they ought to behave. Supposing \(a \cdot b\) has been defined, \[\begin{align*} a \cdot \sigma(b) &\overset{!}{=}a \cdot (b + 1)\\[0.75em] &\overset{!}{=}(a \cdot b) + (a \cdot 1)\\[0.75em] &\overset{!}{=}(a \cdot b) + a. \end{align*}\]

To prove that such a binary operation \(\cdot\) exists and is unique, the proof will first be carried out for a restricted form of multiplication with \(a\) fixed. For all \(a \in \symbb{N}\), \(g_a \colon \symbb{N}\to \symbb{N}\) will be defined such that \(g_a(b) \overset{!}{=}a \cdot b\).

Lemma 2 For all \(a \in \symbb{N}\), there exists a unique function \(g_a \colon \symbb{N}\to \symbb{N}\) such that \[\begin{align} g_a(1) &= a \label{eq:5} \\[0.75em] g_a(\sigma(b)) &= g_a(b) + a \qquad \text{for all } b \in \symbb{N}. \label{eq:6} \end{align}\]

Proof. To prove existence, define the set \[ M \coloneq \{ a \in \symbb{N}\colon \exists \text{ at least one function } g_a \colon \symbb{N}\to \symbb{N}\text{ satisfying } \eqref{eq:5} \text{ and } \eqref{eq:6} \} \subseteq \symbb{N} \] and apply the principle of induction. Define \(g_1 \colon \symbb{N}\to \symbb{N}\) by \(g_1(b) = b\) for all \(b \in \symbb{N}\).5 Then \(g_1\) satisfies \(\eqref{eq:5}\) because \(g_1(1) = 1\), and \(g_1\) satisfies \(\eqref{eq:6}\) because for all \(b \in \symbb{N}\), \(g_1(\sigma(b)) = \sigma(b) = b + 1 = g_1(b) + 1\). Since \(g_1\) satisfies \(\eqref{eq:5}\) and \(\eqref{eq:6}\), \(1 \in M\).

Next suppose \(a \in M\), so there is at least one function \(g_a \colon \symbb{N}\to \symbb{N}\) satisfying \(\eqref{eq:5}\) and \(\eqref{eq:6}\). Using \(g_a\), define \(g_{\sigma(a)} \colon \symbb{N}\to \symbb{N}\) by \(g_{\sigma(a)}(b) = g_a(b) + b\) for all \(b \in \symbb{N}\).6 Then \(g_{\sigma(a)}\) satisfies \(\eqref{eq:5}\) because \(g_{\sigma(a)}(1) = g_a(1) + 1 = a + 1 = \sigma(a)\), and \(g_{\sigma(a)}\) satisfies \(\eqref{eq:6}\) because \[\begin{align*} g_{\sigma(a)}(\sigma(b)) &= g_a(\sigma(b)) + \sigma(b)\\[0.75em] &= (g_a(b) + a) + \sigma(b)\\[0.75em] &= g_a(b) + (a + \sigma(b))\\[0.75em] &= g_a(b) + \sigma(a + b)\\[0.75em] &= g_a(b) + \sigma(b + a)\\[0.75em] &= g_a(b) + (b + \sigma(a))\\[0.75em] &= (g_a(b) + b) + \sigma(a)\\[0.75em] &= g_{\sigma(a)}(b) + \sigma(a). \end{align*}\] Since \(g_{\sigma(a)}\) satisfies \(\eqref{eq:5}\) and \(\eqref{eq:6}\), \(\sigma(a) \in M\), so \(M = \symbb{N}\).

To prove uniqueness, for chosen \(a \in \symbb{N}\) suppose there are functions \(g_a \colon \symbb{N}\to \symbb{N}\) and \(h_a \colon \symbb{N}\to \symbb{N}\) both satisfying \(\eqref{eq:5}\) and \(\eqref{eq:6}\), and define the set \[ N \coloneq \{ b \in \symbb{N}\colon g_a(b) = h_a(b) \} \subseteq \symbb{N} \] and apply the principle of induction. By \(\eqref{eq:5}\), \(g_a(1) = a = h_a(1)\), so \(1 \in N\). Next suppose \(b \in N\), so \(g_a(b) = h_a(b)\). Then by \(\eqref{eq:6}\) and this assumption, \(g_a(\sigma(b)) = g_a(b) + a = h_a(b) + a = h_a(\sigma(b))\), so \(\sigma(b) \in N\). Therefore \(N = \symbb{N}\), and the uniqueness of \(g_a\) has been proven for the chosen \(a \in \symbb{N}\). Since \(a \in \symbb{N}\) was arbitrary, this argument applies for any \(a \in \symbb{N}\). \(\blacksquare\)

Proposition 4 There exists a unique binary operation \(\cdot \colon \symbb{N}\times \symbb{N}\to \symbb{N}\) such that \[\begin{alignat}{2} a \cdot 1 &= a \qquad &&\text{for all } a \in \symbb{N}\label{eq:7} \\[0.75em] a \cdot \sigma(b) &= (a \cdot b) + a \qquad &&\text{for all } a, b \in \symbb{N}. \label{eq:8} \end{alignat}\]

Proof. To prove existence, for any \(a \in \symbb{N}\), define \(a \cdot b \coloneq g_a(b)\) for all \(b \in \symbb{N}\) where \(g_a\) comes from Lemma 2. By Lemma 2, this is well-defined. Using this definition of multiplication, \(a \cdot 1 = g_a(1) = a\), so \(\eqref{eq:7}\) is satisfied. Again using the definition of multiplication, \(a \cdot \sigma(b) = g_a(\sigma(b)) = g_a(b) + a = (a \cdot b) + a\) by \(\eqref{eq:6}\), so \(\eqref{eq:8}\) is satisfied.

To prove uniqueness, suppose there are binary operations \(\cdot \colon \symbb{N}\times \symbb{N}\to \symbb{N}\) and \(\mathbin{\tilde{\cdot}}\colon \symbb{N}\times \symbb{N}\to \symbb{N}\), both of which satisfy \(\eqref{eq:7}\) and \(\eqref{eq:8}\). For chosen \(a \in \symbb{N}\), define the set \(M_a \coloneq \{ b \in \symbb{N}\colon a \cdot b = a \mathbin{\tilde{\cdot}}b \} \subseteq \symbb{N}\). By \(\eqref{eq:7}\), \(a \cdot 1 = a = a \mathbin{\tilde{\cdot}}1\), so \(1 \in M_a\). Next assume that \(b \in M_a\), so \(a \cdot b = a \mathbin{\tilde{\cdot}}b\). Then by \(\eqref{eq:8}\), \(a \cdot \sigma(b) = (a \cdot b) + a = (a \mathbin{\tilde{\cdot}}b) + a = a \mathbin{\tilde{\cdot}}\sigma(b)\), so \(\sigma(b) \in M_a\) and therefore \(M_a = \symbb{N}\). Since \(a \in \symbb{N}\) was arbitrary, this argument applies for any \(a \in \symbb{N}\) and therefore \(a \cdot b = a \mathbin{\tilde{\cdot}}b\) for all \(a, b \in \symbb{N}\). \(\blacksquare\)

Before proving that multiplication is associative and commutative, it will be helpful to prove that multiplication distributes over addition.

Theorem 3 (Distributivity of multiplication over addition) For all \(a, b, c \in \symbb{N}\), \(a \cdot (b + c) = (a \cdot b) + (a \cdot c)\).

Proof. For chosen \(a, b \in \symbb{N}\), define the set \(M_{a, b} \coloneq \{ c \in \symbb{N}\colon a \cdot (b + c) = (a \cdot b) + (a \cdot c) \} \subseteq \symbb{N}\). By definition of addition and multiplication, \(a \cdot (b + 1) = a \cdot \sigma(b) = (a \cdot b) + a = (a \cdot b) + (a \cdot 1)\), so \(1 \in M_{a, b}\). Next assume that \(c \in M_{a, b}\), so \(a \cdot (b + c) = (a \cdot b) + (a \cdot c)\). Then \[\begin{align*} a \cdot (b + \sigma(c)) &= a \cdot \sigma(b + c)\\[0.75em] &= (a \cdot (b + c)) + a\\[0.75em] &= ((a \cdot b) + (a \cdot c)) + a\\[0.75em] &= (a \cdot b) + ((a \cdot c) + a)\\[0.75em] &= (a \cdot b) + (a \cdot \sigma(c)), \end{align*}\] so \(\sigma(c) \in M_{a, b}\) and therefore \(M_{a, b} = \symbb{N}\). Since \(a, b \in \symbb{N}\) were arbitrary, this argument applies for any \(a, b, \in \symbb{N}\). \(\blacksquare\)

Theorem 4 (Associativity of multiplication) For all \(a, b, c \in \symbb{N}\), \((a \cdot b) \cdot c = a \cdot (b \cdot c)\).

Proof. For chosen \(a, b \in \symbb{N}\), define the set \(M_{a, b} \coloneq \{ c \in \symbb{N}\colon (a \cdot b) \cdot c = a \cdot (b \cdot c) \} \subseteq \symbb{N}\). Since \(\cdot\) is a binary operation, \(a \cdot b \in \symbb{N}\), and by definition of \(\cdot\), \((a \cdot b) \cdot 1 = a \cdot b = a \cdot (b \cdot 1)\), so \(1 \in M_{a, b}\). Next assume that \(c \in M_{a, b}\), so \((a \cdot b) \cdot c = a \cdot (b \cdot c)\). Then \[\begin{align*} (a \cdot b) \cdot \sigma(c) &= ((a \cdot b) \cdot c) + (a \cdot b)\\[0.75em] &= (a \cdot (b \cdot c)) + (a \cdot b)\\[0.75em] &= a \cdot ((b \cdot c) + b)\\[0.75em] &= a \cdot (b \cdot \sigma(c)), \end{align*}\] so \(\sigma(c) \in M_{a, b}\) and therefore \(M_{a, b} = \symbb{N}\). Since \(a, b \in \symbb{N}\) were arbitrary, this argument applies for any \(a, b, \in \symbb{N}\). \(\blacksquare\)

Theorem 5 (Commutativity of multiplication) For all \(a, b \in \symbb{N}\), \(a \cdot b = b \cdot a\).

Proof. For chosen \(a \in \symbb{N}\), define the set \(M_a \coloneq \{ b \in \symbb{N}\colon a \cdot b = b \cdot a \} \subseteq \symbb{N}\). By \(\eqref{eq:7}\) and the definition of multiplication in terms of the functions \(g\) as in the proofs of Proposition 4 and Lemma 2, \(a \cdot 1 = a = g_1(a) = 1 \cdot a\), so \(1 \in M_a\). Next assume that \(b \in M_a\), so \(a \cdot b = b \cdot a\). Then by \(\eqref{eq:8}\) and again using the proofs of Proposition 4 and Lemma 2, \[\begin{align*} a \cdot \sigma(b) &= (a \cdot b) + a\\[0.75em] &= (b \cdot a) + a\\[0.75em] &= g_b(a) + a\\[0.75em] &= g_{\sigma(b)}(a)\\[0.75em] &= \sigma(b) \cdot a, \end{align*}\] so \(\sigma(b) \in M_a\) and therefore \(M_a = \symbb{N}\). Since \(a \in \symbb{N}\) was arbitrary, this argument applies for any \(a \in \symbb{N}\). \(\blacksquare\)

Remarks

The ursystem

If the Peano axioms are more rudimentary than natural numbers, one might ask what is more rudimentary than the Peano axioms. The standard answer is the Zermelo–Fraenkel axioms with the axiom of choice, commonly abbreviated as ZFC. ZFC is a formalization of set theory that was developed in the early 20th century to rectify the earlier, less formalized set theory of the 19th century. In the 19th century a naïve form of set theory was used in which it was assumed that there exists a set of objects satisfying any property at all. Around the turn of the 20th century it was realized that this formulation of set theory is so permissive that it permits contradictions. For example, using this permissive form of set theory, let \(R \coloneq \{ X \colon X \notin X \}\) be the set of all sets \(X\) which do not contain themselves. The question arises whether \(R \in R\). If \(R \in R\), then by definition \(R \notin R\). On the other hand, if \(R \notin R\), then \(R \in R\). This is known as Russell’s paradox, and its discovery stimulated the search for a formalized set theory which led to ZFC.

ZFC is a list of nine axioms that stipulate, among other things, that two sets \(A\) and \(B\) are equal if and only if they have the same elements, and for any logical statement \(P(a)\) depending on \(a \in A\), the set \(\{ a \in A \colon P(a) \text{ is true} \} \subseteq A\) is well defined. Informal statements of the nine axioms can be found in Katznelson and Katznelson (2024, app. A) and Schröder (2008, app. B).

Existence and uniqueness

The ZFC axioms imply the existence of the empty set \(\emptyset\) and an infinite set \(S\). The set \(S\) has the properties that \(\emptyset \in S\) and if \(A \in S\), where \(A\) is a set, then \(A \cup \{ A \} \in S\). Using these ingredients, a Peano system can be constructed.

To construct the Peano system, \(S\) is taken as the non-empty set and \(\emptyset \in S\) as the distinguished element. The successor function is given by \(f \colon S \to S\) where \(f(A) = A \cup \{ A \}\). The natural numbers will then be as follows: \[\begin{align*} \emptyset\\[0.75em] f(\emptyset) &= \emptyset \cup \{ \emptyset \} = \{ \emptyset \}\\[0.75em] f(\{ \emptyset \}) &= \{ \emptyset \} \cup \{ \{ \emptyset \} \} = \{ \emptyset, \{ \emptyset \} \}\\[0.75em] f(\{ \emptyset, \{ \emptyset \} \}) &= \{ \emptyset, \{ \emptyset \} \} \cup \{ \{ \emptyset, \{ \emptyset \} \} \} = \{ \emptyset, \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \} \}\\[0.75em] &\vdotswithin{=} . \end{align*}\]

Given the Peano system \((S, f, \emptyset)\), the question arises whether there exist other Peano systems that are meaningfully different. The Peano system \((S, f, \emptyset)\) is in fact unique up to isomorphism, which means that any other Peano system \((\symbb{N}, \sigma, 1)\) is not meaningfully different from \((S, f, \emptyset)\). However, the theory to prove this is most naturally developed when constructing the integers \(\symbb{Z}\) from the natural numbers \(\symbb{N}\). This is done in the Section 1.2.1 notes for Katznelson and Katznelson (2024).

References

Katznelson, Yitzhak, and Yonatan Katznelson. 2024. An Introduction to Real Analysis. Pure and Applied Undergraduate Texts 65. American Mathematical Society.
Schröder, Bernd S. W. 2008. Mathematical Analysis: A Concise Introduction. John Wiley & Sons.

Footnotes

  1. Note that \(1 \in \symbb{N}\) by assumption. If \(a \in \symbb{N}\), then \(\tau(a) = \sigma(a) \in \symbb{N}\) because \(\tau\vert_{\symbb{N}} = \sigma\), and yet \(\symbb{N}\neq \symbb{N}\uplus \{ \vartheta \}\).↩︎

  2. The symbol \(\overset{!}{=}\) is to be read “ought to be equal to”, or “should be equal to”; it is an expression of a desired equality when using intuition to guide rigor.↩︎

  3. This is a natural definition because \(f_1(b) \overset{!}{=}1 + b \overset{!}{=}b + 1 \overset{!}{=} \sigma(b)\).↩︎

  4. This is also a natural definition because \(f_{\sigma(a)}(b) \overset{!}{=}\sigma(a) + b \overset{!}{=}(a + 1) + b \overset{!}{=}(a + b) + 1 \overset{!}{=}\sigma(f_a(b))\).↩︎

  5. This is a natural definition because \(g_1(b) \overset{!}{=}1 \cdot b \overset{!}{=}b \cdot 1 \overset{!}{=}b\).↩︎

  6. This is also a natural definition because \(g_{\sigma(a)}(b) \overset{!}{=}\sigma(a) \cdot b \overset{!}{=}(a + 1) \cdot b \overset{!}{=}(a \cdot b) + (1 \cdot b) \overset{!}{=}g_a(b) + b\).↩︎