If we append 0 and the set of additive inverses to the natural numbers, we
obtain the set of integers,

Z ={... − 2, −1, 0, 1, 2, ...} = {n ∈ N}∪{0}∪{−n ∈ N} . We denote the set of
nonnegative integers

by Z_{+} = {0, 1, 2, . . .} = N∪ {0} = {n ∈ Z : n ≥ 0} .

The discreteness of the natural numbers is inherited by the integers:

• If m, n ∈ Z, then m > n implies m ≥ n + 1.

An integer n is even if there is an integer m such that n = 2m. An integer that
is not even is odd.

• If n is even if and only if n + 1 and n − 1 are odd.

• If n is even then n^{2} is even, and if n is odd then n^{2} is odd.

Using the properties of N and the order and field properties, we may show that
the set of integers

is closed under addition and multiplication.

• For all, n, m ∈ Z, we have n + m ∈ Z, and n · m ∈ Z.

Although addition and multiplication are well defined on the restricted domain
Z, it is still not

a field because n^{-1} Z (unless n = 1). To obtain a field, we must include all
of the rational

numbers.

The set of rational numbers, which we denote by Q, consists of all elements of R
which may be

written as fractions:

Q ≡ {x ∈ R : x = m/n for m, n ∈ Z, and n ≠ 0}.

• Q is an ordered field and is a subset of any other ordered subfield of R.

· Left as an exercise.

Observe that the fractional representation of any x ∈ Q is not unique since
whenever

mp = nr. However, for any rational number, we may uniquely choose m and n such
that n > 0

and n ≤ p for any p ∈ N and r ∈ Z such that rn = pm (i.e. m and n have no common
divisor

greater than 1).

Since Q is an ordered subfield of R, all of the properties of R that depend only
on the field and

order axioms are inherited by Q. However, we will show later that certain
sequences in Q that

‘virtually’ converge do not have a limit in Q. The completeness axiom (C)
ensures that R fills

these gaps and thus distinguishes R from Q.

**Archimedean Axiom**

• (Axiom of Archimedes) For any x ∈ R, there is an n ∈ Z such that n > x.

· If x ≤ 0, then 1 > x. So suppose x > 0 and define A = {n ∈ N : n ≤ x} . To
prove the result it

is sufficient to show that N\A ≠
. Since A is bounded above by x, the
completeness axiom

implies that sup A exists. Then by the definition of sup A, it follows that sup
A − 1 is not an

upper bound for A and therefore there is an integer n > sup A − 1. But then n +
1 > sup A,

which implies that n +1
A. But then A is not an inductive set and hence N\A
≠
.

The Archimedean axiom can be used to show that Q is a ‘dense’ subset of R.

• For any x ∈ R, there is an n ∈ Z such that n ≤ x < n + 1.

· Let A = {n ∈ N : n > |x|} . Then since the Archimedean axiom implies A is
nonempty, the

well-ordering principle implies we may choose n = minA. Therefore n − 1 ≤ |x| <
n. For if

n > 1, we have n − 1 ∈ N and therefore n − 1 ≤ |x| , and if n = 1, then n − 1 =
0 ≤ |x| < n.

If |x| = x, then the theorem is proved. If |x| = −x, then 1 − n ≥ x > −n. If x <
1 − n, we

are done. Otherwise, x = 1− n in which case 2 − n > x ≥ 1 − n.

• If x < y, there is a q ∈ Q such that x < q < y.

· Choose N > (y − x)^{-1} . Then
Now choose n so that n ≤
< n + 1. Then

which implies

• For any x ∈ R and any ε > 0, there is a q ∈ Q such that |q − x| < ε.

· By the previous result there is a q such that x < q < x + ε. Therefore, 0 < q
− x < ε which

implies |q − x| < ε.

• If x ∈ R, then x = sup{q ∈ Q : q < x} .

· Let A = {q ∈ Q : q < x} . Since the Archimedean axiom implies an integer n >
−x, it follows

that −n < x, which implies and that A is nonempty. Since it is also bounded
above by

x, it follows that sup A exists. Suppose sup A < x. Then there is a rational q
such that

sup A < q < x, which implies q ∈ A, but q > sup A, contradicting the definition
of sup A.

Similarly if sup A > x, then there is rational q such that x < q < sup A, which
implies that q

is an upper bound to A which is less than sup A, again contradicting the
definition of A.

**Existence of Square Roots**

In any ordered field F, the axioms imply that for any y > 0, there is a unique x
> 0 such that

x = y^{2}. However, the field and order axioms alone do not imply that x > 0
implies a y ∈ F such

that x = y^{2}. In fact, this is not true of the rational numbers.

Example: If q^{2} = 2, then q
Q.

Suppose q^{2} = 2 for some q ∈ Q. Then, we may choose m, n ∈ Z with no common
divisor so

that
Then m^{2} = 2n^{2} implies that m^{2} is even and hence m is even. Therefore,

n must be odd (otherwise 2 is a common divisor). But if m is even, then m = 2k
for some

integer k,which implies 4k^{2} = 2n^{2}. But this implies that 2k^{2} = n^{2}, which by the
argument

above implies that n is even. A contradiction.

One important implication of the completeness axiom is that the equation
= x
has a solution

y > 0 for any x > 0 and n ∈ N. Although this follows from results to be
established later (i.e. the

intermediate value theorem and the continuity the exponential function), it will
be useful to have

the result now for the case where n = 2.

• If x ∈ R_{+}, then there is a unique y ∈ R_{+}, such that y^{2} = x, which we denote by
and call

the square root of x.

· First observe that since 0 < w < z implies 0^{2} < w^{2} < z^{2}, there is at most one
y that

satisfies y^{2} = x. If x = 0, then 0 · 0 = 0 implies
= 0. So suppose x > 0 and
let

A = {
z ∈ R_{++} : z^{2} < x }. Then for 0 < z < min {1, x} , we have z^{2} < x, and therefore A is

not empty, which implies that y ≡ sup A exists.

· To show that y^{2} = x, suppose first that y^{2} < x. Then wemay choose 0 < z <

But then (y + z)^{2} = y^{2} +z (2y + z) < y^{2} +z (2y +1) < x, which implies y < y+z ∈
A, which

contradicts y = supA.

· Next suppose y^{2} > x. Then we may choose 0 < z <
But then (y − z)^{2} =

y^{2} − z (2y − z) > y^{2} − z (2y + 1) > x, which implies that y − z is an upper
bound of A.

Therefore sup A ≤ y − z < y. A contradiction.

For any integer b > 1, called the base, let
≡ {0, ..., b − 1} . A b-expansion
in a expression of the

form
where
is a sequence of integers with
∈ Z_{+}
and
for j > 0.

A b-expansion is rational if there is an integer n such that
= 0 for j ≥ n. A
b-expansion that is

not rational is called an irrational b-expansion.

For b = 10, a b-expansion is called a decimal expansion. For b = 2, so that each
∈ {0, 1} ,

it is called a binary expansion, and for b = 3, so that each
∈ {0, 1, 2} , it
is called a ternary

expansion.

To assign a real number to each b-expansion let

and define

(We show in the Appendix that is bounded above by + 1).

• Any rational b-expansion is a rational number. Why?

• Not all irrational expansions are irrational numbers. However, a b-expansion
is a rational number

if and only if it eventuallycycles, e.g. .111010101 . . . is a rational number.
[Can you state this

precisely and prove it?]

• For any rational number, there is a base b for which it is a rational b-expansion. Why?

• For any base b, any positive real number has a unique irrational b-expansion.

· A proof is provided in the Appendix.

It follows from this theorem that any rational expansion is equal to some
irrational expansion. In

fact, one may show that

As defined, an expansion can represent only nonnegative numbers. To define
expansions for the

negative numbers we define
whenever x ≥ 0.