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 and . Therefore I consider it essential for this course that you have seen and appreciate at least some of this sort of material, and that's what this web page starts to do.
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 , for all . First we need a lemma related to Bernoulli's Inequality.
Suppose where is a natural number. Then .
This is proved by induction on . For we have for all , so this case is clear.
Now suppose and that the statement of the lemma holds for . Let satisfy . Then certainly so we have
by the induction hypothesis. Thus
and as we have and
We now prove
Theorem on roots.
Let and . Then there is some such that .
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 we try to arrange that converges to . Then will converge to some with .
Define by induction as follows. We let . Now, supposing is defined with both and , let be the greatest integer such that and let be .
Such a exists by the Archimedean Property and the least number principle: in the case when we clearly have as , so by the Archimedean Property there is some integer with . If is the least such , then our is . If it is easier: again, since the number has so we take the least such and again.
This construction of the values is performed for all , and the result is clearly a monotonic nondecreasing sequence (since we only ever added nonnegative numbers, never subtracted them) which is bounded above by (if ) or by (if ) and hence converges to some limit by the Monotone Convergence Theorem.
We claim that , and we shall prove this by showing that and are both impossible.
Then let . As converges to there is some with . Then
so . This is impossible as it would mean that ; but by construction for all .
Then choose such that
as , using the lemma. So
This last inequality is by choice of such that .
This gives a contradiction, for and , and so . By construction where is largest possible giving . But and so
contradicting our choice of .
Therefore the limit of satisfies , as required.
This was nice, but quite hard work. As a reward, we get a definition of for positive and integer . (Define to be the th root of .) This idea of rational powers has a couple of nice properties that will be useful later.
If and then .
If not, , so by a result on ordered fields, raising to the th power gives .