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Chapter 3
The Real Numbers
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 3.3, Slide 1
Section 3.3
The Completeness Axiom
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 3.3, Slide 2
There is one additional axiom that distinguishes from . It is called the
completeness axiom. Before presenting this axiom, let’s look at why it’s needed.
When we graph the function f (x) = x2 – 2, it appears to cross the horizontal axis
at a point between 1 and 2. But does it really?
f (x)
f (x) = x2 – 2
3
How can we be sure that there is a “number” x on
the axis such that x2 – 2 = 0?
2
It turns out that if the x-axis consists only of rational
numbers, then no such number exists.
1
That is, there is no rational number whose square is 2.
–2
1
–1
2
x
In fact, we can easily prove the more general result
that p is irrational (not rational) for any prime
number p.
Recall that an integer p > 1 is prime iff its only
divisors are 1 and p.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 3.3, Slide 3
Theorem 3.3.1
Let p be a prime number. Then
Proof: We suppose that
If
p is not a rational number.
p is rational and obtain a contradiction.
p is rational, then we can write
p = m / n, where m and n are integers
with no common factors other than 1. Then pn2 = m2, so p must be a factor of m2.
Now the prime factored forms for m and m2 have exactly the same prime factors,
so p is a factor of m. That is, m = kp for some integer k. But then pn2 = k 2 p 2,
so that n 2 = k 2 p. Thus p is a factor of n2, and as above we conclude that
p is also a factor of n. Hence p is a factor of both m and n, contradicting the fact
that they have no common factors other than 1. 
In Section 2.4 we learned that there are uncountably many irrational numbers.
Thus, if we were to restrict our analysis to rational numbers, our “number line”
would have uncountably many “holes” in it. It is these holes in the number line
that the completeness axiom fills.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 3.3, Slide 4
In order to state the completeness axiom for
, we need some preliminary definitions.
Upper Bounds and Suprema
Definition 3.3.2
Let S be a subset of . If there exists a real number m such that m  s for all s  S,
then m is called an upper bound of S, and we say that S is bounded above.
If m  s for all s  S, then m is a lower bound of S and S is bounded below.
The set S is said to be bounded if it is bounded above and bounded below.
If an upper bound m of S is a member of S, then m is called the maximum
(or largest element) of S, and we write m = max S.
Similarly, if a lower bound of S is a member of S, then it is called the minimum
(or least element) of S, denoted by min S.
While a set may have many upper and lower bounds, if it has a maximum or a
minimum, then those values are unique (Exercise 6).
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 3.3, Slide 5
Example 3.3.3
(a) Let S = {2, 4, 6, 8}.
min
0

2
max

4

6
lower bounds

8
10
upper bounds
Then S is bounded above by 8, 9, 8.5, 2, and any other real number greater than
or equal to 8. Since 8  S, we have max S = 8.
Similarly, S has many lower bounds, including 2, which is the largest of the
lower bounds and the minimum of S.
It is easy to see that any finite set is bounded and always has a maximum and
a minimum.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 3.3, Slide 6
Example 3.3.3
(b) The interval [0, ) is not bounded above.
min
no maximum
–2
[
0
2
4
6
no upper bounds
8
lower bounds
It is bounded below by any nonpositive number, and of these lower bounds,
0 is the largest. Since 0  [0, ), 0 is the minimum of [0, ).
(c) The interval (0, 4] has a maximum of 4, and this is the smallest of the upper bounds.
max
no min
(
0
]
2
4
lower bounds
upper bounds
It is bounded below by any nonpositive number, and of these lower bounds, 0 is the largest.
Since 0  (0,1], the set has no minimum.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 3.3, Slide 7
Example 3.3.3
(d) The empty set  is bounded above by any m 
. Note that the condition
m  s for all s   is equivalent to the implication “if s  , then m  s.”
This implication is true since the antecedent is false.
Likewise,  is bounded below by any real m.
Definition 3.3.5
Let S be a nonempty subset of . If S is bounded above, then the least upper bound
of S is called its supremum and is denoted by sup S. Thus m = sup S iff
(a) m  s, for all s  S,
That is, m is an upper bound of S.
(b) if m < m, then there exists s  S such that s > m.
Nothing smaller than m is an upper bound of S.
If S is bounded below, then the greatest lower bound of S is called its infimum
and is denoted by inf S.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 3.3, Slide 8

Let T  q 

:0  q  2 .
Does T have a supremum? It depends on the context.
If we think of T as a subset of the real numbers, then sup T =
But
2.
2 is not a rational number. So T does not have a supremum in
.
When considering subsets of , it has been true that each set bounded above has had a
least upper bound. This supremum may be a member of the set, as in the interval [0,1],
or it may be outside the set, as in the interval [0,1), but in both cases the supremum
exists as a real number. This fundamental difference between
and
is the basis
for our final axiom of the real numbers, the completeness axiom:
Every nonempty subset S of
that is bounded above has a least upper bound.
That is, sup S exists and is a real number.
It follows readily from this that every nonempty subset S of that is bounded below
has a greatest lower bound. So, inf S exists and is a real number.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 3.3, Slide 9
Theorem 3.3.7
, let C denote the set C = {x + y : x  A and y  B}.
Given nonempty subsets A and B of
If A and B have suprema, then C has a supremum and sup C = sup A + sup B.
Proof: Let sup A = a and sup B = b. If z  C, then z  x + y for some x  A and y  B.
Thus z  x + y  a + b, so a + b is an upper bound of C. By the completeness axiom,
C has a least upper bound, say sup C  c. We must show that c  a + b. Since c is the
least upper bound of C, we have c  a + b.
To see that a + b  c, choose any  > 0. Since a  sup A, a –  is not an upper bound
of A, and there must exist x  A such that x > a –  . Similarly, since b = sup B, there
exists y  B such that y > b –  . Combining these inequalities, we have x + y > a + b – 2 .
But, x + y  C and c  sup C, so c > a + b – 2 . That is, a + b < c + 2 for every  > 0.
Thus by Theorem 3.2.8, a + b  c. Finally, since c  a + b and c  a + b, we conclude
that c  a + b. 
B
A
x
  
a–
a
A+B
y
  
b–
b

a + b – 2
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
x+y


c
Section 3.3, Slide 10
Theorem 3.3.8
Suppose that D is a nonempty set and that f : D  and g : D  . If for every
x, y  D, f (x)  g( y), then f (D) is bounded above and g(D) is bounded below.
Furthermore, sup f (D)  inf g(D).
Proof: Given any z  D, we have f (x)  g(z), for all x  D. Thus f (D) is bounded
above by g(z). It follows that the least upper bound of f (D) is no larger than g(z).
That is, sup f (D)  g(z). Since this last inequality holds for all z  D, g(D) is
bounded below by sup f (D). Thus the greatest lower bound of g(D) is no smaller
than sup f (D). That is, inf g(D)  sup f (D). 
D
z

f
f (D)
g
sup f (D)

g(z)
inf g(D)


least upper bound of f (D)
lower bound of g(D)
g(D)
upper bound of f (D)
greatest lower bound of g(D)
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 3.3, Slide 11
Theorem 3.3.9
The set
The Archimedean Property of
of natural numbers is unbounded above in
.
Proof:
If were bounded above, then by the completeness axiom it would have a least upper
bound, say sup = m. Since m is a least upper bound, m – 1 is not an upper bound of .
Thus there exists an n in such that n > m – 1. But then n + 1 > m, and since n + 1  ,
this contradicts m being an upper bound of . 
The Archimedean Property is widely used in analysis and there are several equivalent
forms (which are also sometimes referred to as the Archimedean Property).
We state them and establish their equivalence in the following theorem.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 3.3, Slide 12
Theorem 3.3.10
Each of the following is equivalent to the Archimedean property.
(a) For each z 
, there exists an n 
(b) For each x > 0 and for each y 
(c) For each x > 0, there exists an n 
such that n > z.
, there exists an n 
such that nx > y.
such that 0 < 1/n < x.
Proof:
We shall prove that Theorem 3.3.9  (a)  (b)  (c)  Theorem 3.3.9, thereby
establishing their equivalence.
Theorem 3.3.9  (a)
If (a) were not true, then for some z0  we would have n  z0 for all n 
But then z0 would be an upper bound of , contradicting Theorem 3.3.9.
Thus the Archimedean property implies (a).
.
(a)  (b)
Given x > 0 and y 
so that nx > y.
, let z = y /x. Then there exists n 
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
such that n > y/x,
Section 3.3, Slide 13
Theorem 3.3.10
Each of the following is equivalent to the Archimedean property.
(a) For each z 
, there exists an n 
(b) For each x > 0 and for each y 
(c) For each x > 0, there exists an n 
such that n > z.
, there exists an n 
such that nx > y.
such that 0 < 1/n < x.
Proof:
We shall prove that Theorem 3.3.9  (a)  (b)  (c)  Theorem 3.3.9, thereby
establishing their equivalence.
(b)  (c)
Given x > 0, take y = 1 in part (b). Then nx > 1, so that 1/n < x.
Since n  , n > 0 and also 1/n > 0.
(c)  Theorem 3.3.9
Suppose that were bounded above by some real number, say m.
That is, n  m for all n  . Let k = m + 1. Then n  k – 1 < k and 1/n > 1/k for all n.
This contradicts (c) with x = 1/k. Thus (c) implies the Archimedean property. 
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 3.3, Slide 14
In Theorem 3.3.1 we showed that
p is not rational when p is prime.
We are now in a position to prove there is a positive real number whose square is p,
thus illustrating that we actually have filled in the “holes” in .
Theorem 3.3.12
Let p be a prime number. Then there exists a positive real number x such that x2 = p.
Proof:
Let S = {r 
: r > 0 and r2 < p}. Since p > 1, 1  S and S is nonempty.
Furthermore, if r  S, then r2 < p < p2, so r < p. Thus S is bounded above by p,
and by the completeness axiom, sup S exists as a real number. Let x = sup S.
It is clear that x > 0, and we claim that x2 = p. To prove this, we shall show that
neither x2 < p nor x2 > p is consistent with our choice of x.
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Section 3.3, Slide 15
Theorem 3.3.12
Let p be a prime number. Then there exists a positive real number x such that x2 = p.
Proof:
Let S = {r 
: r > 0 and r2 < p} and x = sup S. Suppose that x2 < p.
2
x
1
x + n
2

p
We want to find something positive, say 1/n, that we can add on to x and still
have its square be less than p. This will contradict x as an upper bound of S.
We have (p – x2)/(2x + 1) > 0, so that Theorem 3.3.10(c) implies the existence of some
n
p  x2
1
<
.
such that
2x + 1
n
The text shows where this estimate comes from.
2
Then
1
2x 1
1
1

2
2
x
+

x
+
+

x
+
2
x
+




n
n n2
n
n

1
 x 2 +  2 x + 1 < x 2 + ( p  x 2 )  p.
n
It follows that x + 1/n  S, which contradicts our choice of x as an upper bound of S.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 3.3, Slide 16
Theorem 3.3.12
Let p be a prime number. Then there exists a positive real number x such that x2 = p.
Proof:
Let S = {r 
: r > 0 and r2 < p} and x = sup S. Now suppose that x2 > p.

p
1
x  m
2
2
x
We want to find something positive, say 1/m, that we can subtract from x and still
have its square be greater than p. This will contradict x as the least upper bound of S.
We have (x2 – p)/(2x) > 0, so there exists an m 
2
x2  p
1
<
.
such that
2x
m
1
2x 1
2x

+ 2 > x2 
> x 2  ( x 2  p)  p.
Then  x    x 2 
m
m m
m

This implies that x – 1/m > r, for all r  S, so x – 1/m is an upper bound of S.
Since x – 1/m < x, this contradicts our choice of x as the least upper bound of S.
Finally, since neither x2 < p nor x2 > p is a possibility, we conclude by the
trichotomy law that in fact x2 = p. 
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Section 3.3, Slide 17
Theorem 3.3.13
The density of
in
If x and y are real numbers with x < y, then there exists a rational number r such
that x < r < y.
Proof: We begin by supposing that x > 0. Using the Archimedean property 3.3.10(a),
there exists an n  such that n > 1/(y – x). That is, nx + 1 < ny. Since nx > 0, it is
not difficult to show (Exercise 9) that there exists m  such that m – 1  nx < m.
But then m  nx + 1 < ny, so that nx < m < ny. It follows that the rational number
r = m/n satisfies x < r < y.
Finally, if x  0, choose an integer k such that k > |x|. Then apply the argument
above to the positive numbers x + k and y + k. If q is a rational satisfying
x + k < q < y + k, then the rational r = q – k satisfies x < r < y. 
It follows easily from this that the irrational numbers are also dense in the real numbers
(Theorem 3.3.15). That is, if x and y are real numbers with x < y, then there exists an
irrational number w such that x < w < y.
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Section 3.3, Slide 18