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+^{1}=1+

Now suppose ~~
~~ and that the statement of the
lemma holds for ~~-1~~. Let ~~
~~~~
~~

by the induction hypothesis. Thus

and as ~~
~~~~-1)~~
~~-1)~~
^{2}

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 _{
} to converge to _{
} converges to _{
} will converge to some ^{=
}

Define (_{
)
} by induction as follows.
We let _{0=}_{1=0}. Now, supposing _{
} is defined
with both 0_{
}
~~<~~_{
+
}
^{}<_{
+1
} be _{
+
}.

Such a _{
+
}
^{
}
as _{
0
},
so by the Archimedean Property there is some integer
_{
+
}
^{
}.
If _{0}
_{0-1}
_{
0
} the number _{
+
}
^{
}
so we take the least such _{0}
_{0-1}

This construction of the values _{
} is
performed for all

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
~~<~~_{
}, and so
_{
+1
}
^{<
}.
By construction _{
+1=}_{
+
} where
~~<~~_{
+
}
_{
+
+1
}

contradicting our choice of

Therefore the limit _{
)
} satisfies
^{=
}

This was nice, but quite hard work. As a reward, we get
a definition of ^{
}
~~
~~
^{
}
~~th
root of ~~^{
}

Proposition.

If 0<~~
~~ then
0<^{1
<
1
}

**Proof.**

If not, ^{1
1
}
~~th power gives
(~~^{1
)}
^{=
(
1
)
=
}.