Axiomatic Construction of the Real Numbers

Now we shall return to first concepts and make precise some ideas which have been left somewhat indefinite. In the following sections we supply proofs of the basic theorems which we have used without proof. It will be seen along the way that the tools available to you from your previous background were totally inadequate for proving, say, that a function continuous on an interval has a maximum there.

To begin with, we set down explicitly the properties of the numbers which we use. There are several ways of doing this. One of them sets down the properties of the positive integers, and builds from them a system with the properties which we list below. Another, which we prefer for this course, takes the form of establishing a set of ground rules, or axioms: "The system of numbers with which we shall work is to have the properties ... ." The axioms fall in a natural fashion into three groups, which we call the field axioms, the order axioms, and the axiom of continuity. Throughout these axioms, the letters $a, b, c, \dots$ represent members of a set of objects which we agree to call numbers. In this set, we assume that we can add $a$ to $b$ to get a number $a + b$ , and that we can multiply $a$ by $b$ to get a number $a b$ . The first group of axioms, the field axioms, express properties of these operations from which all other properties can be derived.

Group I. Field Axioms

A1) $a + b = b + a$
A2) $(a + b) + c = a + (b + c)$
A3) There is exactly one number $0$ such that $a + 0 = a$ for every $a$ .
A4) For each $a$ , there is exactly one number $- a$ such that $a + (- a) = 0$ .
M1) $a b = b a$
M2) $(a b) c = a (b c)$
M3) There is exactly one number $1 \neq 0$ such that for every $a$ , $a \cdot 1 = a$ .
M4) If $a \neq 0$ , there is exactly one number $a^{- 1}$ such that $a a^{- 1} = 1$ .
D) $a (b + c) = a b + a c$ .

Axioms A1) – 4) give the properties of addition,
M1) – 4) the properties of multiplication.
Axiom D), the distributive law, connects the two operations.
A1) and M1) express the commutative laws for addition and multiplication, respectively,
A2) and M2) the associative laws.
The number $0$ of A3) is of course called "zero" and the $1$ of M3) is called "one", although it is to be regarded as different from the "one" used in counting.
The number $- a$ of A4) is called the negative of $a$ ,
the number $a^{- 1}$ of M4) the inverse of $a$ .
One cannot define a system in which the field axioms are satisfied and which contains an inverse for $0$ .

Observe that the field axioms are satisfied by a much larger class of systems than our previous notion of real numbers; for example, they are satisfied by the rational numbers.

We supply some sample proofs to illustrate our assertion that all properties of addition and multiplication follow from the field axioms.

Theorem 1

- (- a) = a \cdot

$a + (- a) = 0$ ; $(- a) + a = a + (- a) = 0$ , by A1); But $- (- a)$ is the only number $b$ such that $- a + b = 0$ , by A4). Therefore $a = - (- a)$ .

Theorem 2

a \cdot 0 = 0 \cdot

$a = a \cdot 1$ , by M3); but $1 = 1 + 0$ , by A3). By D),

a = a \cdot 1 = a \cdot (1 + 0) = a \cdot 1 + a \cdot 0 = a + a \cdot 0 \cdot

Therefore

(- a) + a = (- a) + (a + a \cdot 0) = ((- a) + a) + a \cdot 0,

by A2). But $(- a) + a = 0$ , by A2) and A4), so we have $0 = 0 + a \cdot 0 = a \cdot 0$ , by A3).

Theorem 3

(- 1) \cdot a = - a \cdot

$0 = 1 + (- 1)$ . Therefore $a \cdot 0 = a (1 + (- 1)) = a \cdot 1 + a (- 1)$ , or by the last theorem, $0 = a \cdot 1 + a (- 1) = a + (- 1) a$ , by M3) and M1). Since $- a$ is the only number $b$ such that $a + b = 0$ , we have $(- 1) a = - a$ .

Theorem 4

(- a) (- b) = a b \cdot

$b + (- b) = 0$ . Therefore $a (b + (- b)) = a \cdot 0 = 0$ , or by D), $a b + a (- b) = 0$ . Thus $a (- b) = - (a b)$ . Also $a + (- a) = 0$ , so $(a + (- a)) (- b) = a (- b) + (- a) (- b) = 0$ . Thus $a (- b) = - (- a) (- b)$ , or $- (a b) = - (- a) (- b)$ . Taking negatives once more, we see by the first theorem that $a b = (- a) (- b)$ .

Observe that throughout these proofs we have used what you may have seen as the principle that "equals may be substituted for equals." But it has not been a case of substitution; the number $0$ is the same whether it is written as $0, a + (- a), (- b) + b$ , or what have you. It is only the representation of the number which changes, and we simply use the principle that if a rule holds for a number, it does not matter how we represent that number in applying the rule.

Exercise 1.

Prove that $- (a + b) = (- a) + (- b)$ ; $- (b - a) = a - b$ , where $a - b = a + (- b)$ .

The second group of axioms deals with the ordering of numbers. We give them the form of axioms about positive numbers, and then define $a < b$ to mean " $b - a$ is positive."

Group II. Order Axioms

We take as basic for all these axioms the existence of a certain set $P$ of numbers, which we shall call "positive numbers." The axioms give properties that $P$ is to have.

P1) If $a$ and $b$ are in $P$ , so is $a + b$ .
P2) If $a$ and $b$ are in $P$ , so is $a b$ .
P3) If $a$ is any number $\neq 0$ , then either $a$ is in $P$ or $- a$ is in $P$ , but not both.
P4) $0$ is not in $P$ .

Definition 1.

Now we define $a < b$ to mean " $b - a$ is in $P$ ". $a \leq b$ is to mean "either $a < b$ or $a = b$ ". In place of $a < b$ we sometimes write $b > a$ and similarly $b \geq a$ for $a \leq b$ .

From the axioms of Groups I and II we can prove all the properties of $a < b$ :

Theorem 5

For any two numbers $a$ and $b$ , exactly one of the following holds: $a < b, b < a, a = b$ .

Consider the number $b - a$ . If $b - a$ is in $P$ , then $a < b$ , $- (b - a) = a - b$ is not in $P$ , and $a \neq b$ . Then $a < b$ is the case, and neither of $a = b, b < a$ is the case. Next suppose $b - a$ is not in $P$ . Then $a < b$ is not the case. But if $a \neq b$ , $- (b - a) = a - b$ is in $P$ , by P3); in other words, $b < a$ . The only remaining possibility is $a = b$ , in which case neither $b - a$ nor $a - b$ is in $P$ , since both are $0$ .

Theorem 6

If $a < b$ , then $a + c < b + c$ .

Let $a < b$ , i.e., $b - a$ is in $P$ . But $(b + c) - (a + c) = b + c - a - c = b - a$ is in $P$ . Therefore $a + c < b + c$ .

Theorem 7

If $a < b$ and $c > 0$ (i.e., $c$ is in $P$ ), then $c a < c b$ .

$b - a$ is in $P$ , and $c$ is in $P$ . Therefore $c (b - a) = c b - c a$ is in $P$ , by P2). Thus $c a < c b$ .

Theorem 8

If $a < b$ and $b < c$ , then $a < c$ .

If $a < b$ and $b < c$ , then both $b - a$ and $c - b$ are in $P$ . Then, by P1), so is $(c - b) + (b - a) = c - a$ , or $a < c$ .

Theorem 9

If $a \neq 0$ , then $a^{2}$ is in $P$ .

Either $a$ is in $P$ or $- a$ is in $P$ . If $a$ is in $P$ , then by P2), $a a = a^{2}$ is in $P$ . If $- a$ is in $P$ , then $(- a) (- a) = a \cdot a = a^{2}$ is in $P$ . This completes the proof.

Theorem 10

$1 > 0$ , i.e., $1$ is in $P$ .

$1 = 1^{2}$ .

Exercises

Prove the following:

Exercise 2.

If $a < b$ , then $- b < - a$ .

Exercise 3.

If $a < b$ and $c < 0$ , then $b c < a c$ .

Exercise 4.

If $a b > 0$ , then either both $a > 0$ and $b > 0$ or both $a < 0$ and $b < 0$ .

Exercise 5.

If $a > 0$ , then $a^{- 1} > 0$ ; if $a < 0$ , then $a^{- 1} < 0$ .

Exercise 6.

If $a < b$ and $c < d$ , then $a + c < b + d$ .

Exercise 7.

The integers are a system which satisfies all axioms of Groups I and II except M4). By considering pairs $(m, n)$ (or $\frac{m}{n}$ , if you wish) of integers with $n \neq 0$ , develop the operations and ordering of the rational numbers. Note that the rational number which we represent by $\frac{2}{3}$ can also be represented by $\frac{6}{9}, \frac{4}{6}, \frac{- 2}{- 3}$ , etc. Therefore $\frac{m}{n}$ will not itself be a rational number, but will be a way of representing a rational number. You should make it clear when two pairs \frac{m}{n}, \frac{m'}{n'} represent the same rational number.

Group III. Axiom of Continuity

All the axioms of Groups I and II are satisfied by the rational numbers. But within the rational numbers, we cannot prove the theorem which we have used in Chapter 12, namely that a continuous function $f (x)$ on $a \leq x \leq b$ , such that $f (a) < f (b)$ , takes on every value between $f (a)$ and $f (b)$ in the interval. For example, we consider the dilemma of the Greeks, who had no number to represent the length of the diagonal of a square of side $1$ . By the theorem of Pythagoras, the diagonal $d$ would satisfy $d^{2} = 1^{2} + 1^{2} = 2$ . But there is no rational number $d$ with this property; for if there were, we write $d = \frac{m}{n}$ as a fraction in lowest terms, where $m$ and $n$ are integers. Then

d^{2} = \frac{m^{2}}{n^{2}} = 2,

or $m^{2} = 2 n^{2}$ . Thus $m^{2}$ is divisible by $2$ . But if $m$ is odd, this is impossible, therefore $m$ is even, say $m = 2 p$ . Then $4 p^{2} = 2 n^{2}$ , or $n^{2} = 2 p^{2}$ , and so $n$ is also even. Therefore $\frac{m}{n}$ cannot have been in lowest terms, since we can cancel a $2$ . This is a contradiction, meaning that the original assumption that $d$ was rational is false.

In terms of the intermediate value theorem, this can be interpreted as follows: $y = f (x) = x^{2}$ is continuous for $0 \leq x \leq 2$ , and $f (0) = 0 < 4 = f (2)$ . If the intermediate value theorem were to hold for rational numbers, there would be a rational number $x_{0}$ between $0$ and $2$ such that $x_{0}^{2} = 2$ . But there is no such rational number. In order to prove the intermediate value theorem we thus need a further property that a system satisfying only the field and order axioms need not possess. It is the fact that you had no experience with this property which made it impossible to prove the theorem at the time it was stated. Before we give the axiom, however, we must make a few definitions.

Definition 2.

Let $S$ be any set of numbers. A number $M$ is called an upper bound for $S$ if $x \leq M$ whenever $x$ is in $S$ . A number $N$ is a lower bound for $S$ if $N \leq x$ whenever $x$ is in $S$ . The set of all numbers has no upper or lower bounds, for instance, if $M$ were to be an upper bound, then $M + 1$ would be a number and $M + 1 > M$ . The set of all negative numbers has many upper bounds, for instance $41, 29 \frac{3}{8}, 0, π, \sqrt{2}$ , but no lower bound.

Definition 3.

A number $M$ is called a least upper bound for $S$ if:
a) $M$ is an upper bound for $S$ ; and
b) there is no upper bound M' for $S$ such that M' < M.

Greatest lower bound is similarly defined. A least upper bound for the set of negative numbers is $0$ ; this is also a least upper bound for the set of non-positive numbers. If $S$ contains no numbers, then any number is an upper bound for $S$ , so $S$ cannot have a least upper bound. It is evident that if $S$ has a least upper bound, it has only one.

Now we can state our final axiom:

C) If a non-empty set $S$ has an upper bound, it has a least upper bound.

It follows that if $𝑺$ has a lower bound, it has a greatest lower bound. For if $- S$ denotes the set of negatives of the elements of $S$ , and if $M$ is a lower bound for $S$ , then $- M$ is an upper bound for $- S$ . Therefore $- S$ has a least upper bound M', and -M' is a greatest lower bound for $S$ . As a first application, we prove that the integers have no upper bound.

Theorem 11

There is no upper bound for the set $1, 2, 3, \dots$ .

Suppose the theorem is false; then by C), the integers have a least upper bound $M$ . Then $M - 1$ cannot be an upper bound, so there is an integer $n > M - 1$ . But then $n + 1 > (M - 1) + 1 = M$ , and $n + 1$ is an integer. This contradicts the assumption that $M$ was an upper bound, and so the theorem is proved. (Note that it follows that if $ϵ$ is any positive number, there is a positive integer $n$ such that $\frac{1}{n} < ϵ$ ; for otherwise, $\frac{1}{ϵ}$ would be an upper bound for the positive integers.)

Now we know that there is in our system a number $d$ such that $d^{2} = 2$ . This will follow from the continuity of $x^{2}$ and the intermediate value theorem, but let us give an indication of how it could be shown directly. Let $S$ be the set of positive rational numbers $r$ such that $r^{2} < 2$ . Then $1$ is in $S$ , so that $S$ is non-empty. $2$ is an upper bound for $S$ , for if $r > 2$ , then $r^{2} > 2^{2} = 4 > 2$ , and $r$ is not in $S$ . So let $d$ be the least upper bound for $S$ . If $d^{2} > 2$ , let $d^{2} - 2 = ϵ > 0$ . Let $n$ be an integer such that $\frac{4}{n} < ϵ$ , or $\frac{1}{n} < \frac{ϵ}{4}$ . Then

{(d - \frac{1}{n})}^{2} = d^{2} - \frac{2 d}{n} + \frac{1}{n^{2}} \geq d^{2} - \frac{4}{n} + \frac{1}{n^{2}},

since $d \leq 2$ , or

{(d - \frac{1}{n})}^{2} \geq d^{2} - ϵ + \frac{1}{n^{2}} > d^{2} - ϵ = 2 \cdot

Thus $d - \frac{1}{n}$ is an upper bound for $S$ and is less than $d$ , which is impossible. Therefore $d^{2} \leq 2$ . If $d^{2} < 2$ , we find a rational number $r$ in $S$ such that $r > d$ . (If $2 - d^{2} = δ > 0$ , let $n$ be a positive integer such that $\frac{1}{n^{2}} < δ$ ; then let $q$ be a positive integer so large that there is a square of an integer, $m$ , between $(2 n^{2} - 1) q^{2}$ and $2 n^{2} q^{2}$ . Let $r = \frac{m}{q n} \cdot$ Then $r^{2} < 2$ , and $r > d \cdot$ ) The only remaining possibility is $d^{2} = 2$ .

The Real Number System

Group I. Field Axioms

Group II. Order Axioms

Exercises

Group III. Axiom of Continuity

Quick Links

Social Media