We have seen the completeness axiom for the reals,
the idea of a monotonic sequence, and we have
seen why every bounded monotonic sequence converges to a limit. These are
important ideas and essential for much mathematics, including a proper
understanding of why square roots (and also cube roots, fourth roots, and
solutions of other algebraic equations) exist and how and why we can
define and use numbers such as
Some of this material is more difficult than much of the rest of the course. You should at a minimum read and appreciate the main theorem here, in particular noting that it follows in an essential way from the Completeness Axiom for the reals.
We aim to prove that th roots exist of all positive
. First we need a lemma
related to Bernoulli's Inequality.
Lemma.
Suppose 0<
is a natural number. Then
(1+<1+2
Proof.
This is proved by induction on . For =1
we have (1+
Now suppose
and that the statement of the
lemma holds for -1. Let
by the induction hypothesis. Thus
and as
-1)
-1)
as required.
We now prove
Theorem on roots.
Let
and =
Proof.
The idea is to mimic the proof that any
real number has a rational sequence converging to it but instead of
devising our sequence
=
Define (
<<
Such a
as
.
If
so we take the least such
This construction of the values
We claim that =
>
<
First:
Subproof.
Also:
Subproof.
Assume <
Then choose
Then
as
This last inequality is by choice of
-1
This gives a contradiction, for
<<.
By construction
<
contradicting our choice of
Therefore the limit =
This was nice, but quite hard work. As a reward, we get
a definition of
th
root of
Proposition.
If 0<
then
0<
Proof.
If not,
th power gives
(
=.
=