and does it even exist? You can ask for it on your calculator and the
calculator will give you a number which multiplied by itself 5 times will yield a number
which is close to 7 but it isn’t exactly right. Why should there exist a number which
works exactly? Every one you find, appears to be some sort of approximation at best. If
you can’t produce one, why should you believe it is even there? Are you to accept it on
faith like religion? Indeed, you must accept something without proof, but the appropriate
thing to accept is the completeness axiom of the real line on which every significant topic
in calculus depends. Basically, roots exist because of completeness of the real
line as do integrals and all the major existence theorems in calculus. Here is a
lemma.

Lemma 1.10.1Suppose n ∈ ℕ and a > 0. Then if x^{n}−a≠0, there exists δ > 0 suchthat whenever

y ∈ (x− δ,x+ δ),

it follows y^{n}− a≠0 and has the same sign as x^{n}− a.

Proof: Using the binomial theorem,

n n
y = (y− x + x)

( )
n∑−1 n n−k k n
= k (y− x) x + x
k=0

Let

|y− x|

< 1. Then using the triangle inequality, it follows that for

|y− x|

< 1,

n−1( )
|yn − xn| ≤ |y− x|∑ n |x|k ≡ C |x − y|
k=0 k

where, as indicated, C = ∑_{k=0}^{n−1}

( n )
k

|x|

^{k}. Let δ be small enough that the right side
is less than

|xn − a|

. For example, you could let

( )
|xn −-a|
δ = min 2C ,1

Then if y ∈

(x − δ,x + δ)

,

n n n
|y − x | ≤ C |x− y| < C δ ≤ |x − a|

It follows that on the number line, y^{n} must be between a and x^{n}. Consequently, y^{n}−a≠0
and has the same sign as x^{n}− a. (Draw a picture.) ■

PICT

Theorem 1.10.2Let a > 0 and let n > 1. Then there exists a uniquex > 0 such that x^{n} = a.

Proof: Let S denote those numbers y ≥ 0 such that t^{n}−a < 0 for all t ∈

[0,y]

. One
such number is 0. If a ≥ 1, then a short proof by induction shows a^{n}> a and so, in this
case, S is bounded above by a. If a < 1, then another short argument shows

(1∕a)

^{n}> a
and so S is bounded above by 1∕a. By completeness, there exists x, the least upper
bound of S. Thus for all y ≤ x,y^{n}− a < 0 since if this is not so, then x was not a least
upper bound to S. If x^{n}− a > 0, then by the lemma, y^{n}− a > 0 on some interval

(x − δ,x + δ)

. Thus x fails to be a the least upper bound because an upper bound is
x − δ∕2. If x^{n}− a < 0, then by the lemma, y^{n}− a < 0 on some interval

(x − δ,x + δ)

and so x is not even an upper bound because S would then contain [0,x + δ).
Hence the only other possibility is that x^{n}− a = 0. That is, x is an n^{th} root of
a.

This has shown that a has a positive n^{th} root. Could it have two? Suppose x,z both
work. If z > x, then by the binomial theorem,

∑n ( )
zn = (x+ z − x )n = n xn−k (z − x)k
k=0 k
n−∑ 1( n ) n∑−1( n )
= xn + k xn−k(z − x)k = a+ k xn− k(z − x)k > a.
k=0 k=0

Turning the argument around, it is also not possible that z < x. Thus the n^{th} root is also
unique. ■

From now on, we will use this fact that n^{th} roots exist whenever it is convenient to do
so.