Axioms of the Real Numbers

Branko Ćurgus

The primary focus of this page is the Completeness Axiom. I want to present how I see the Completeness Axiom emerges naturally. This presentation is inspired by the Completeness Axiom stated in Mathematical Analysis by Vladimir Zorich.

To make the presentation of the Completeness Axiom more complete, I present all sixteen axioms of the Real Numbers.

In the axioms below we use the standard logical operators: the conjunction $\wedge$, the disjunction $\vee$, the exclusive disjunction $\oplus$, the implication $\Rightarrow$, universal quantifier $\forall,$ existential quantifier $\exists.$

We also use the standard set notation: the set membership $\in$, the subset $\subseteq$, the equality $=$, the set difference $\setminus$ and the Cartesian product $\times.$ For singleton sets instead of writing $\{a\} = \{b\}$ we write $a = b.$

The notation $f:A\to B$ stands for a function $f$ which is defined on a set $A$ with the values in $B.$

Axiom AE below establishes the existence of the addition function defined on $\mathbb{B}\!\times\!\mathbb{R}$ with the values in $\mathbb{R}.$ It is common to denote the value of $+$ at a pair $(a, b) \in \mathbb{R}\!\times \!\mathbb{R}$ by $a+b.$

Axiom ME establishes the existence of the multiplication function defined on $\mathbb{R}\!\times\!\mathbb{R}$ with the values in $\mathbb{R}.$ It is common to denote the value of this function at a pair $(a,b) \in \mathbb{R}\!\times\!\mathbb{R}$ by $a\cdot b$ which is almost always abbreviated as $ab.$

The meaning of the abbreviations is as follows: AE--addition exists, AA--addition is associative, AC--addition is commutative, AZ--addition has zero, AO--addition has opposites, ME--multiplication exists, MA--multiplication is associative, MC--multiplication is commutative, MO--multiplication has one, DL--distributive law, OE--order exists, OT--order is transitive, OA--order respects addition, OM--order respects multiplication, CA--Completeness Axiom.

The meaning of the expression $a \leq b$ is $a\lt b \lor a=b$. The meaning of the expression $a \leq b \leq c$ is $a \leq b \land b\leq c$.

A trichotomy is a compound proposition of three propositions which is true if and only if exactly one of the three propositions is true.


Definition. The set $\mathbb{R}$ of Real Numbers is a nonempty set satisfies the following 16 axioms.

Axiom AE. There exists a function $+: \mathbb{R}\!\times\!\mathbb{R} \to \mathbb{R}.$
Axiom AA. $\forall a \in \mathbb{R} \ \ \forall\mkern 1mu b \in \mathbb{R} \ \ \forall c \in \mathbb{R}$ we have $a+(b+c) = (a+b)+c$
Axiom AC. $\forall a \in \mathbb{R} \ \ \forall\mkern 1mu b \in \mathbb{R}$ we have $a+b = b+a$
Axiom AZ. $\exists\, 0 \in \mathbb{R}$ such that $\forall a \in \mathbb{R}$ we have $0+a = a$
Axiom AO. $\forall a \in \mathbb{R} \ \ \exists (-a) \in \mathbb{R}$ such that $a + (-a) = 0$
Axiom ME. There exists a function $\cdot : \mathbb{R} \times \mathbb{R} \to \mathbb{R}.$
Axiom MA. $\forall a \in \mathbb{R} \ \ \forall\mkern 1mu b \in \mathbb{R} \ \ \forall c \in \mathbb{R}$ we have $a(bc) = (ab)c$
Axiom MC. $\forall a \in \mathbb{R} \ \ \forall\mkern 1mu b \in \mathbb{R}$ we have $ab = ba$
Axiom MO. $\exists\, 1 \in \mathbb{R}\!\setminus\!\{0\}$ such that $\forall a \in \mathbb{R}$ we have $1a = a$
Axiom MR. $\forall a \in \mathbb{R}\!\setminus\!\{0\} \ \ \exists \mkern 2pt \dfrac{{1}}{a} \in \mathbb{R}\!\setminus\!\{0\}$ such that $a \dfrac{1}{a} = 1$.
Axiom DL. $\forall a \in \mathbb{R} \ \ \forall\mkern 1mu b \in \mathbb{R} \ \ \forall c \in \mathbb{R}$ we have $a(b+c) = ab + ac$
Axiom OE. $\forall a \in \mathbb{R} \ \ \forall\mkern 1mu b \in \mathbb{R}$ the propositions $a\lt b$ or $a=b$ or $b \lt a$ form a trichotomy.
Axiom OT. $\forall a \in \mathbb{R} \ \ \forall\mkern 1mu b \in \mathbb{R} \ \forall c \in \mathbb{R}$ we have $a \lt b \land b \lt c \Rightarrow a\lt c$.
Axiom OA. $\forall a \in \mathbb{R} \ \ \forall\mkern 1mu b \in \mathbb{R} \ \forall c \in \mathbb{R}$ we have $a \lt b \Rightarrow a+c\lt b+c$.
Axiom OM. $\forall a \in \mathbb{R} \ \ \forall\mkern 1mu b \in \mathbb{R} \ \forall c \in \mathbb{R}$ we have $a \lt b \land 0 \lt c \Rightarrow ac\lt bc$.
Axiom CA. Let $A \neq \emptyset$, $B \neq \emptyset$, $A \subset \mathbb{R}$, and $B \subset \mathbb{R}$. If $\forall a \in A$ and $\forall\mkern 1mu b \in B$ we have $a \leq b$, then $\exists\mkern 1.5mu c \in \mathbb{R}$ such that $\forall a \in A$ and $\forall\mkern 1mu b \in B$ we have $a \leq c \leq b$. Briefly, \[ \forall a \in A \ \ \forall\mkern 1mu b \in B \quad a \leq b \quad \Rightarrow \quad \exists\mkern 1.5mu c \in \mathbb{R} \ \ \text{such that} \ \ \forall a \in A \ \ \forall\mkern 1mu b \in B \quad a \leq c \leq b \]

The first fifteen axioms of $\mathbb{R}$ yield all the algebraic properties of $\mathbb{R}$. In two propositions below, we list a few.

Proposition 1.

Let $a,b,c \in \mathbb{R}$. The following statements hold.

  1. If $a + c = b + c,$ then $a = b.$
  2. $a\cdot 0 = 0$.
  3. $-a = a$ if and only if $a=0.$
  4. $-(-a) = a$.
  5. $(-a)b = - (ab)$.
  6. $(-a)(-b) = ab$.
  7. If $a \neq 0$, then $\dfrac{1}{\dfrac{1}{a}} = a$.
  8. If $ac = bc$ and $c \neq 0$, then $a = b.$
  9. $ab = 0$ if and only if $a = 0$ or $b = 0$.
  10. If $a \neq 0$ and $b \neq 0$, then $\dfrac{1}{ab} = \dfrac{1}{a}\mkern 1pt \dfrac{1}{b}$.

Proof.


Proposition 2.

Let $a,b,c \in \mathbb{R}$. The following statements hold.

  1. If $a \lt b$ then $-b \lt -a.$
  2. If $a < b$ and $c < 0$, then $bc \lt a c$.
  3. If $a \gt 0$ and $b \neq 0$, then $b \gt 0$ if and only if $ab \gt 0$.
  4. If $a \neq 0$, then $0 \lt aa$.
  5. $0 \lt 1.$
  6. If $a \gt 0$, then $\dfrac{1}{a} \gt 0$.
  7. If $0 \lt a \lt b$, then $\displaystyle 0 \lt \frac{1}{b} \lt \frac{1}{a}$.
  8. Let $0 \leq a, b.$ Then $ a \lt b $ if and only if $ a a \lt b b$.
  9. If $1 \lt a$ and $bb \lt a,$ then $b \lt a.$
  10. If $a \neq b$, then $4 ab \lt (a+b)(a+b)$.
  11. If $\forall\mkern 1pt b \gt 0$ we have $a \lt b$, then $a \leq 0$.

Proof.


The only specific numbers mentioned in Axiom AZ and Axiom MO are $0$ and $1$. In Proposition 2(v) it was proved that $0 \lt 1.$ The preceding inequality and Axiom OA repeated recursively yield \begin{align*} 1 &\lt 1+1 \\ 1+1 &\lt (1+1)+1 \\ (1+1)+1 &\lt \bigl((1+1)+1\bigr)+1 \\ \bigl((1+1)+1\bigr)+1 &\lt \Bigl(\bigl((1+1)+1\bigr)+1\Bigr) + 1 \\ \Bigl(\bigl((1+1)+1\bigr)+1\Bigr) + 1 &\lt \biggl(\Bigl(\bigl((1+1)+1\bigr)+1\Bigr) + 1 \biggr) + 1\\ \biggl(\Bigl(\bigl((1+1)+1\bigr)+1\Bigr) + 1 \biggr) + 1 & \lt \Biggl(\biggl(\Bigl(\bigl((1+1)+1\bigr)+1\Bigr) + 1 \biggr) + 1\Biggr) + 1 \\ & \mkern 5pt \vdots \end{align*}

Motivated by the preceding inequalities we define:

Definition. We define the following eight real numbers \begin{alignat*}{4} 2 & = 1+1, \hspace{8mm} & 3 &= 2+1, \hspace{8mm} & 4 &= 3+1, \hspace{8mm} & 5 &=4+1, \\ 6 & = 5+1, & 7 &= 6+1, & 8 &= 7+1, & 9 &=8+1. \end{alignat*} The numbers $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ are called digits and we set \[ \mathbb{D} = \bigl\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\bigr\}. \]

The next proposition follows from the preceding definition, Proposition 2(v), (vii), and Axiom OA

Proposition 3.

The following inequalities hold: \[ 1 \lt 2\lt 3\lt 4\lt 5\lt 6\lt 7\lt 8\lt 9 \] and \[ \frac{1}{9} \lt \frac{1}{8} \lt \frac{1}{7} \lt \frac{1}{6} \lt \frac{1}{5} \lt \frac{1}{4} \lt \frac{1}{3} \lt \frac{1}{2} \lt 1. \]

Three Subsets of the Real Numbers

In the preceding section, we introduced ten real numbers $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$, the digits. We defined the positive digits by adding $1$ to the preceding one. We could have continued this process and obtained more special real numbers of this kind. Let us consider the family of all nonempty subsets $S$ of $\mathbb{R}$ which have the following property \[ 1 \in S \ \ \land \ \ \forall x \in \mathbb{R} \ \ x \in S \ \ \Rightarrow \ \ x+1 \in S. \]

Many subsets of $\nR$ have the preceding two properties. For example, one such set is the set of positive real numbers, that is the open infinite interval, \[ (0, +\infty). \] Another such set is the closed infinite interval \[ [1, +\infty), \] and also the union \[ \{ 1 \} \cup [2, +\infty). \] Denote by $\mathcal{N}$ the family of all subsets of $\nR$ with the two properties stated in the preceding paragraph. That is, \[ \mathcal{N} = \Bigl\{ S \subseteq \mathbb{R} \, : \, 1 \in S \ \ \land \ \ x \in S \ \Rightarrow \ x + 1 \in S \Bigr\}. \]

Now we are ready to define the set of natural numbers (or, positive integers). Intuitively, the set of natural numbers is the smallest set in $\mathcal{N}$.

Definition. We define $\mathbb{N}$ to be the intersection of the family $\mathcal{N}$: \[ \mathbb{N} := \bigcap \bigl\{ S \, : \, S \in \mathcal{N} \bigr\}. \] That is, $k \in \nN$ if and only if $k \in S$ for all $S \in \mathcal{N}$. The elements of the set $\nN$ are called natural numbers or positive integers.

Theorem 1.

The following are the basic properties of $\nN$.

  1. If $K \subseteq \nN$ has the following two properties \begin{gather*} 1 \in K, \\ \forall n \in K \quad n \in K \ \ \Rightarrow \ \ n+1 \in K, \end{gather*} then $K = \nN$.
  2. The function $\sigma: \{0\}\cup\nN \to \nN$ defined by ${\sigma}(x) = x+1$ is a bijection.
  3. $1, 2, 3, 4, 5, 6, 7, 8, 9 \in \nN$.

Proof.


The following five special features of the set $\mathbb{N}$ of natural numbers (positive integers) are worth emphasising in an ordered list. Each of the features will be discussed below.

In the next theorem we prove the Principle of Mathematical Induction.

Theorem 2. (The Principle of Mathematical Induction)

Let $P(n)$, with $n\in \nN$, be a family of statements such that

  1. $P(1)$ is true,
  2. For all $n \in \nN$ we have $P(n) \ \Rightarrow \ P(n+1)$.
Then the statement $P(n)$ is true for all $n \in \nN.$

Proof.


The same role that the Principle of Mathematical Induction plays in proving statements involving positive integers, the Principle of Recursive Definition plays in defining functions. We have defined several functions on $\mathbb{R}$. Please review the definitions of the functions $\operatorname{us}$, $\operatorname{sgn}$, $\operatorname{abs}$. In addition to these functions we have defined fu

Theorem 3. (The Principle of Recursive Definition)

Let $P(n)$, with $n\in \nN$, be a family of statements such that

  1. $P(1)$ is true,
  2. For all $n \in \nN$ we have $P(n) \ \Rightarrow \ P(n+1)$.
Then the statement $P(n)$ is true for all $n \in \nN.$

Proof.