(last updated: 4:38:12 PM, October 04, 2020)
This is information that is not in Kosmala, but is provided as I feel that it is important.
Theorem: (square roots)
Let \(a\in\mathbb R\), \(a>0\). Then \(\sqrt a\in\mathbb R\).
To prove this, define \(S_a=\{x\in\mathbb R\mid x>0,x^2<a\}\). Then \(\sup S_a\in\mathbb R\) by completeness. Let \(b=\sup S_a\) and show that \(b^2=a\). An argument nearly identical to the argument shown for the existence of \(\sqrt 2\) works.
So now we have all non-negative square roots as follows: \[\sqrt a =\sup\{x\in\mathbb R\mid x \geq 0,x^2<a\}.\] Then we define \(-\sqrt a\) as the additive inverse of \(\sqrt a\).
Theorem: (natural number roots)
Let \(a\in\mathbb R\), \(a>0\), and \(n\in\mathbb N\). Then \(\sqrt[n] a\in\mathbb R\).
See the proof in this ArXiv paper. The proof is roughly as follows. Assume that \(\sqrt a\) exists (\(n=2\)), and then use induction to show \(\sqrt[2^n]a\) for all \(n\in\mathbb N\). Now reason that \(2^n=n+m\) for some \(m\in\mathbb N\) (since \(2^n>n\)). Let \(x=\sup\{y\in\mathbb R \mid y\geq0, y^n<a\}\) (make sure this is non-empty and has an upper bound). We know \(x\in\mathbb R\) by completeness. Then use a trichotomy argument to show that \(x^n=a\). This is a tricky argument.
Definition (rational exponents)
Define \(a^{\frac nm}\overset{def}{=}b\) such that \(b^m=a^n\).
It is not hard to prove that \(a^{\frac nm}=(a^n)^{1/m}=(a^{1/m})^n\).
Theorem: (comparison with a rational exponent)
Let \(a,b\in\mathbb R\) with \(a,b>0\) and let \(p\in\mathbb Q\) with \(p>0\). Then \(a<b\) if and only if \(a^p<b^p\).
(Note that if \(p<0\), then the inequality gets reversed: \(a<b\) if and only if \(a^p>b^p\).)
This follows from the previous results in a straightforward way.
Theorem: (comparison of different rational exponents)
Let \(a\in\mathbb R\) with \(a>1\) and let \(p,q\in\mathbb Q\). Then \(a^p<a^q\) if and only if \(p<q\).
Let \(a\in\mathbb R\) with \(0<a<1\) and let \(p,q\in\mathbb Q\). Then \(a^q<a^p\) if and only if \(p<q\).
(Note the reversed role of \(p\) and \(q\) in the last statement.)
This last result is a little tricky, but combining many of the previous results into a single argument works.
Definition (irrational exponents) Let \(a\in\mathbb R,a\geq1\) and \(b\in\mathbb R,b>0\). Define \(S=\{a^x \mid x\in\mathbb Q, 0<x<b\}\). We know that \(S\) is nonempty and bounded from above so we know that its supremum exists and is a real number. Define \(a^b=\sup S\). Similarly, for \(0<a<1\), define \(S=\{a^x \mid x\in\mathbb Q, 0<x<b\}\) and then define \(a^b=\inf S\).
Irrational exponentiation can be defined in various ways, and they are all equivalent in the sense that they give the same numerical values and satisfy all the same properties. The way we have done so here simply relies on the basic field, order, and completeness axioms. This is a bit of a mroe complicated way to construct general exponentiation. It is somewhat cleaner and simpler to develop differentiation and integration first and from that to define logarithms. Irrational exponents can then be defined in terms of logarithms.
It isn’t too hard to show that we can take odd roots of negative numbers, and that negative numbers can be raised to rational roots sometimes. Also, negative roots are defined as the inverse of the appropriate positive root.
Now you might want to go through deriving or convincing yourself that all of our familiar properties of exponents follow such as \(x^{a+b}=x^ax^b\). I’ll write more details here as I find time to do so. I do plan to eventually type up proofs for everything here. For now you can just assume that you are allowed to use these results whenever you find them to be useful.
Now that we have real exponentiation defined, we can develop every part of calculus that you are accustomed to.
\[ \diamond \S \diamond \]