4. The complex numbers 15
we note ﬁrst that x2 1 implies x 1, because x ≥ 1 implies x2 ≥ 1. Therefore
if t 1, then 1 is an upper bound for S. On the other hand, if t ≥ 1, then t ≤ t2.
Therefore
x2
t implies
x2
≤
t2
and hence x ≤ t. Therefore in this case t is an
upper bound for S. In either case, S is bounded above by M and is nonempty. By
the completeness axiom, S has a least upper bound α. We claim that
α2
= t.
To prove the claim, we use the trichotomy property. We will rule out the cases
α2
t and
α2
t. In each case we use the Archimedean property to ﬁnd a positive
integer n whose reciprocal is suﬃciently small. Then we can add or subtract
1
n
to
α and obtain a contradiction. Here are the details. If α2 t, then Proposition 3.2
guarantees that we can ﬁnd an integer n such that
2α
n
α2
 t.
We then have
(α 
1
n
)2
=
α2

2α
n
+
1
n2
α2

2α
n
t.
Thus α 
1
n
is an upper bound for S, but it is smaller than α. We obtain a
contradiction. Suppose next that α2 t. We can ﬁnd n ∈ N (Exercise 1.21) such
that
(11)
2αn + 1
n2
t 
α2.
This time we obtain
(α +
1
n
)2
=
α2
+
2α
n
+
1
n2
t.
Since α +
1
n
is bigger than α and yet it is also an upper bound for S, again we
obtain a contradiction. By trichotomy we must therefore have α2 = t.
The kind of argument used in the proof of Theorem 3.1 epitomizes proofs
in basic real analysis. In this setting one cannot prove an equality by algebraic
reasoning; one requires the completeness axiom and analytic reasoning.
I
Exercise 1.21. For t 
α2
0, prove that there is an n ∈ N such that (11)
holds.
I
Exercise 1.22. Mimic the proof of Proposition 3.4 to prove the following state
ment. If k is a positive integer, then the square root of k must be either an integer
or an irrational number. Suggestion: Multiply
m
n
by
mnq
mnq
for a suitable integer q.
4. The complex numbers
We are ﬁnally ready to introduce the complex numbers C. The equation x2 +
1 = 0 will have two solutions in C. Once we allow a solution to this equation,
we ﬁnd via the quadratic formula and Lemma 4.1 below that we can solve all
quadratic polynomial equations. With deeper work, we can solve any (nonconstant)
polynomial equation over C. We will prove this result, called the fundamental
theorem of algebra, in Chapter 8.
Our ﬁrst deﬁnition of C arises from algebraic reasoning. As usual, we write
R2
for the set of ordered pairs (x,y) of real numbers. To think geometrically, we