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? The following is an argument that roots exist. You fill in
the details of the argument. Basically, roots exist because of completeness of the real line.
Here is a lemma.

Lemma 2.11.1Suppose n ∈ ℕ and a > 0. Then if x^{n}−a≠0, there exists δ > 0 such thatwhenever

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,

yn = (y− x+ x)n

n∑−1 ( )
= n (y − x)n−kxk + xn
k=0 k

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)

,

|yn − xn| ≤ C |x − y| < Cδ ≤ |xn − 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 2.11.2Let a > 0 and let n > 1. Then there exists a unique x > 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∑ ( n )
zn = (x + z − x)n = k xn−k(z − x)k
k=0
n∑−1( n ) k n∑−1( n ) k
= xn + k xn−k(z − x) = a + k xn−k (z − x) > 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.