14 1. From the Real Numbers to the Complex Numbers
all three values are the same. By contrast, let xn = (-1)n. Then lim inf(xn) = -1
and lim sup(xn) = 1. Occasionally in the subsequent discussion we can replace
limit by lim sup and things still work.
Next we turn to the concept of continuity, which we also define in terms of
sequences.
Definition 3.8. Let f : R R be a function. Then f is continuous at a if
whenever {xn} is a sequence and limn→∞xn = a, then limn→∞f(xn) = f(a). Also,
f is continuous on a set S if it is continuous at each point of the set. When S is
R or when S is understood from the context to be the domain of f, we usually say
“f is continuous” rather than the longer phrase “f is continuous on S”.
I
Exercise 1.19. Prove that the sum and product of continuous functions are
continuous. If c is a constant and f is continuous, prove that cf is continuous.
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Exercise 1.20. Prove that f is continuous at a if and only if the following holds.
For each 0, there is a δ 0 such that |x - a| δ implies |f(x) - f(a)| .
We close this section by showing how the completeness axiom impacts the
existence of square roots. First we recall the standard fact that there is no rational
square root of 2, by giving a somewhat unusual proof. See Exercise 1.22 for a
compelling generalization. These proofs are based on inequalities. For example,
the order axioms yield the following: 0 a b implies 0
a2
ab
b2;
we use
such inequalities without comment below.
Proposition 3.4. There is no rational number whose square is 2.
Proof. Seeking a contradiction, we suppose that there are integers m,n such that
(m
n
)2
= 2. We may assume that m and n are positive. Of all such representations
we may assume that we have chosen the one for which n is the smallest possible
positive integer. The equality
m2
=
2n2
implies the inequality 2n m n. Now
we compute
(10)
m
n
=
m(m - n)
n(m - n)
=
m2
- mn
n(m - n)
=
2n2
- mn
n(m - n)
=
2n - m
m - n
.
Thus
2n-m
m-n
is also a square root of 2. Since 0 m - n n, formula (10) provides
a second way to write the fraction
m;
n
the second way has a positive denominator,
smaller than n. We have therefore contradicted our choice of n. Hence there is no
rational number whose square is 2.
Although there is no rational square root of 2, we certainly believe that a
positive real square root of 2 exists. For example, the length of the diagonal of the
unit square should be

2. We next prove, necessarily relying on the completeness
axiom, that each positive real number has a square root.
Theorem 3.1. If t R and t 0, then there is an x R with
x2
= t.
Proof. This proof is somewhat sophisticated and can be omitted on first reading.
If t = 0, then t has the square root 0. Hence we may assume that t 0. Let S
denote the set of real numbers x such that x2 t. This set is nonempty, because
0 S. We claim that M = max(1,t) is an upper bound for S. To check the claim,
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