1.4The Integers and Real Numbers

Updated 11 May 2026Chapter (PDF)

We assume the real numbers exist and satisfy trivial axioms. From the reals we produce the integers.

[1.4.1]Definition(Inductive Subset, Positive Integers)#

A subset $A \subseteq \bR$ is said to be inductive if $1 \in A$ and if, for every $x \in A$ , the number $x + 1 \in A$ . Let $\mathcal{A}$ be the collection of all inductive subsets of $\bR$ . Define the positive integers to be the intersection of all elements of $\mathcal{A}$ .

\bZ_+ = \bigcap_{A \in \mathcal{A}} A.

Note this implies $\bZ_+$ is the smallest inductive subset of $\bR$ . Note that, of course, $\bZ_+$ itself is inductive. Note that if $A$ is any inductive set of positive integers, then it is necessarily $\bZ_+$ .

We define the set of integers $\bZ$ to be the set of positive integers $\bZ_+$ , $0$ , and the set of negative integers $-\bZ_+$ . The set of integers is closed under addition, subtraction, and multiplication, but is not under division—indeed, the set of rationals $\bQ$ is said to be the set of all possible quotients of integers.

[1.4.2]Definition(Section)#

A section of the positive integers, denoted $S_n$ , is such that

S_{n + 1} = \{1, \dots, n\}

for each $n \in \bZ_+$ .

[1.4.3]Theorem(Well-Ordering Property)#

Every nonempty subset of $\mathbb{Z}_+$ has a smallest element.

Proof.

We first show that for each $n \in \mathbb{Z}_+$ , every nonempty subset of the finite segment $\{1, \dots, n\}$ has a smallest element. For the base case $n = 1$ , the only nonempty subset is $\{1\}$ itself, which has 1 as its smallest element. Now, assume the statement holds for $n$ . Let $C$ be a nonempty subset of $\{1, \dots, n + 1\}$ . If $C$ consists only of the element $n + 1$ , then $n + 1$ is clearly the smallest element. Otherwise, $C$ must contain elements from $\{1, \dots, n\}$ . In this case, the intersection $C \cap \{1, \dots, n\}$ is a nonempty subset of $\{1, \dots, n\}$ , so by our inductive hypothesis, it has a smallest element $k$ . Since $k \le n < n + 1$ , $k$ is also the smallest element of the entire set $C$ . Thus, by induction, every nonempty subset of any finite segment $\{1, \dots, n\}$ has a smallest element.

Now let $D$ be any nonempty subset of $\mathbb{Z}_+$ . To find its smallest element, choose any element $n \in D$ . We only need to look at the elements of $D$ that are less than or equal to $n$ . Let $A = D \cap \{1, \dots, n\}$ . This set $A$ is nonempty and is a subset of a finite segment, so it must have a smallest element $k$ . Because $k$ is the smallest in $A$ , and every element in $D$ that is not in $A$ is strictly greater than $n$ , it follows that $k$ is the smallest element of $D$ .

[1.4.4]Theorem(Strong Induction Principle)#

Let $A \subseteq \mathbb{Z}_+$ . Suppose that for each $n \in \mathbb{Z}_+$ , the statement $S_n \subseteq A$ implies $n \in A$ . Then $A = \mathbb{Z}_+$ .

Proof.

Suppose $A$ does not equal $\mathbb{Z}_+$ . This means there are some positive integers missing from $A$ . By the Well-Ordering Property, there must be a smallest positive integer $n$ that is not in $A$ . Since $n$ is the smallest element missing, every positive integer less than $n$ must already be in $A$ . This is exactly the condition $S_n \subseteq A$ . Our hypothesis states that if $S_n \subseteq A$ , then $n$ must be in $A$ . This creates a contradiction: $n$ cannot be both missing from $A$ and required to be in $A$ . Therefore, there can be no missing elements, and $A = \mathbb{Z}_+$ .