**Theorem 5. **There exists a real number c for which c^{2} = 2.

Proof. Define a sequence
of closed intervals in R
recursively as follows. First,

let

and let

= the midpoint of
.

Next, let

(We cannot have because no rational number
has square 2.) In general, once

has been constructed, let

= the midpoint of

and then let

(Since no rational number has square 2, in fact the case
never arises.)

By the construction, for all n = 0, 1, 2, . . . . Then

for all n = 0, 1, 2, . . . . (Why?)

By the Nested Interval Property of R, there exists some c with

We are going to show that c^{2} = 2.

Just suppose c^{2} ≠ 2. Define

so that ε > 0. By the Archimedean Ordering Property of R, there exits a positive
integer n for

which

(Why?)

Fix such an n. Since both c^{2} and 2 are in
, then

This is impossible because |c^{2} - 2| = ε.

**Exercise 6.** Use the method of the preceding proof to establish that
exists.

**
Exercise 7.** Prove uniqueness of the number c whose existence is guaranteed by
Theorem 5.

The method of proof of Theorem 5 does not merely establish existence of a real number c

whose square is 2. It also provides a method for approximating that number as closely as we

wish. Indeed, look again at the construction of the intervals . The number c we want

belongs to each of these intervals (and is different from the endpoints of each). Initially, all we

know about c is that , that is,

1 < c < 2.

If we use the midpoint = 1.5 as an approximation to c, the the error in that approximation—

the size of the difference between c and the approximation —satisfies

(half the length of = [1, 2]). Next, since , then

[1, 1.5]. So now we know c ∈ (1, 1.5), that is,

1 < c < 1.5.

If we use the midpoint as a new approximation to c, then the error in this

approximation satisfies

(half the length of . Next, since , then

. So now we know that c ∈ (1.25, 1.5), that is,

1.25 < c < 1.5.

If we use the midpoint as a new approximation to c, then the error in this

approximation satisfies

With each successive bisection of the interval, the number c is trapped inside an interval of

half the length of the interval used in the previous step, and the error in the approximating

midpoint is halved. This method is known as the "Bisection Method".

less than 0.01.

The method of proof used for Theorem 5—constructing a nested sequence of closed intervals—

is applicable for the next theorem, too.

Recall the relevant definitions (which we originally gave in N). Let A R. An

of A in R is a number b ∈ R such that a ≤ b for every a ∈ A. The set A is

R when there exists some upper bound of A in R.

About N you proved that each nonempty subset A of N that is bounded above in N has a

greatest element. Such a greatest element g of A is an upper bound of A in N that is no greater

than any other upper bound of A in N; in other words, g is the least element of the set of all

upper bounds of A in N. So this greatest element g of A is least upper bound of A in N.

A subset of R that is bounded above in R need not have a greatest element. For example,

the open interval (0, 1) in R has no greatest element (why?). But it does have a least upper

bound in R, namely, the real number 1. The following theorem generalizes this example.

**Theorem 9** (Order Completeness of R). Each nonempty subset
of R that is bounded above has

a least upper bound.

Proof. See the proof of Theorem 3.3.13. In that proof, just change F to R and
change the term

"supremum" to "least upper bound".