I will present the full development as I have solved it.
Lemma. Let $a\in\mathbb{R}^{+}, a>1$ be fixed and $r=\frac{m}{n}\in\mathbb{Q},\, m\in\mathbb{Z},\, n\in\mathbb{N}$. Define $S_{r}(a)=\{x\in\mathbb{R}\mid 0\leq x^{n}\leq a^{m}\}$. $S_{r}(a)$ has supremum.
Proof. Let $m>0$. Define $S_{r}(a)=\{x\in\mathbb{R}\mid 0\leq x^{n}\leq a^{m}\}$. Because $1<a$ then $1=1^{n}=1^{m}=a^{0}<a^{m}$, that is $1\in S_{r}(a)$, thus $S_{r}(a)\neq\emptyset$. On the other hand, because $a>1$, $a^{m}>1$ and $x\geq a^{m}\longleftrightarrow x^{n}\geq a^{mn}>a^{m}$. Then $a^{m}$ is upper bound for $S_{r}(a)$.
Now, $m<0$. Let $S_{r}(a)$ as it was defined before. Because $1<a$ then $1=a^{0}>a^{m}\geq x^{n}>0$, and therefore $1$ is upper bound for $S_{r}(a)$. Also, because $a>1$, $0<a^{m}<1$ and $a^{mn}<a^{m}$. Thus $a^{m}\in S_{r}(a)$ and it follows that $S_{r}(a)\neq\emptyset$.
Then, by supremum axiom, $S_{r}(a)$ has supremum for each $r\in\mathbb{Q}$ and $a>1$.
The reason for taking the set $S_{r}(a)$ needs to be exposed. Intuitively the $x\in S_{r}(a)$ are such that $x<a^{r}$. But, since this is not already defined, we took the set of numbers that are such that $x^{n}<a^{m}$. Therefore, the former lemma says $a^{r}=\sup S_{r}(a)$. Thus, we can make the following
Definition. Let $a\in\mathbb{R}^{+},\, a>1$ fixed and $r\in\mathbb{Q}$. Define $\exp_{a}(r):\mathbb{Q}\longrightarrow\mathbb{R}$ as the function given by
\begin{align*}
r=\frac{m}{n}\neq 0 &\mapsto a^{r}=\sup S_{r}(a)=\sup\{x\in\mathbb{R}\mid 0\leq x^{n}\leq a^{m}\}\\
0 &\mapsto 1
\end{align*}
We need to show that this is a good definition. With the former definition we are setting positive solution to the equation $x^{n}=a^{m}$. It could be possible for other positive numbers to be solutions. We must, then, prove that this equation has one and only one rational solution.
Lemma. Let $a\in\mathbb{R}^{+},\, a>1$ fixed and $r\in\mathbb{Q}$. $\rho=\sup S_{r}(a)$ is the unique positive solution of $x^{n}=a^{m}$.
Proof. Let $0<\epsilon<1$, then
\begin{align*}
(1+\epsilon)^{n} &=\sum_{k=0}^{n}\dbinom{n}{k}\epsilon^{k}\\
&=1+\sum_{k=1}^{n}\dbinom{n}{k}\epsilon^{k}\\
&<1+\sum_{k=1}^{n}\dbinom{n}{k}\epsilon=1+K_{n}\epsilon
\end{align*}
with $K_{n}$ constant for each $n$.
Thus, take $\frac{\rho}{2}<\rho_{2,\epsilon}<\rho<\rho_{1,\epsilon}<2\rho$ tal que $\rho_{1,\epsilon}=\rho(1+\epsilon)$ y $\rho_{2,\epsilon}=\frac{\rho}{1+\epsilon}$. Suppose that $\rho^{n}<a^{m}$, then exists $\epsilon$ such that $\rho^{n}(1+K_{n}\epsilon)<a^{m}$. Therefore
\begin{align*}
\rho_{1,\epsilon}^{n} &=\rho^{n}(1+\epsilon)^{n}\\
&<\rho^{n}(1+K_{n}\epsilon)\\
&<a^{m}
\end{align*}
then $\rho$, which is lesser than $\rho_{1,\epsilon}\in S_{r}(a)$, would not be even a upper bound for $S_{r}(a)$. But $\rho$ is the supremum of this set. Therefore $\rho^{n}\geq a^{m}$. Now, suppose that $\rho^{n}>a^{m}$, then exists $\epsilon$ such that $\frac{\rho^{n}}{1+K_{n}\epsilon}>a^{m}$ and it follows that
\begin{align*}
\rho_{2,\epsilon}^{n} &=\frac{\rho^{n}}{(1+\epsilon)^{n}}\\
&>\frac{\rho^{n}}{1+K_{n}\epsilon}\\
&>a^{m}
\end{align*}
thus $\rho$, which is greater than $\rho_{2,\epsilon}$, would not be the least upper bound of $S_{r}(a)$, that is the supremum, as there would be a lesser upper bound. But $\rho$ is the supremum, by hypothesis. Then, it remains not choice but to $\rho^{n}=a^{m}$ as we wanted to show.
On the other hand, suppose that exists $0<\rho_{1}\neq\rho$ such that $\rho_{1}^{n}=a^{m}$. The condition $\rho_{1}\neq\rho$ implies $\rho_{1}<\rho$ or $\rho<\rho_{1}$, and by exponential properties for integers $\rho_{1}<\rho\longleftrightarrow \rho_{1}^{n}<\rho^{n}$ and in a similar way for the other option. But both are equal to $a^{m}$, then $\rho_{1}=\rho$ and the positive root of the equation $x^{n}=a^{m}$ is only $\rho$.
This makes formal the intuitive notion of fractional exponents or roots.
Let us make a lemma for handle some expressions more easily
Lemma. Let $a\in\mathbb{R}^{+},\, a>1$ fixed and $0<r=\frac{m}{n}\in\mathbb{Q}$. Then $\exp_{a}(-r)=\frac{1}{\exp_{a}(r)}$
Proof. We are stating that $\sup S_{-r}(a)=\frac{1}{\sup S_{r}(a)}$ or, with the notation of the previous lemma
\begin{align*}
\rho_{-r}=\frac{1}{\rho_{r}}
\end{align*}
We now by the previous lemma that $\rho_{r}$ is the unique positive solution of $x^{n}=a^{m}$
\begin{align*}
\rho_{r}^{n} &=a^{m}\therefore\\
\frac{1}{\rho_{r}^{n}} &=\frac{1}{a^{m}}\\
&=a^{-m}
\end{align*}
and, also we know that $\rho_{-r}$ es the unique positive solution to $x^{n}=a^{-m}$ then
\begin{align*}
\rho_{-r}^{n} &=\frac{1}{\rho_{r}^{n}}\\
&=\left(\frac{1}{\rho_{r}}\right)^{n}
\end{align*}
thus, we conclude
\begin{align*}
\rho_{-r}=\frac{1}{\rho_{r}}
\end{align*}
which is the result we wanted to prove.
We need to show three additional results for this definition to be a good one: Function does not depend on the representation for a given rational, it is the same as the one for integers and is well defined for $0<a<1$.
Lemma. Let $a\in\mathbb{R}^{+},\, a>1$ fixed and $r=\frac{m}{n}=\frac{p}{q}\in\mathbb{Q}$ such that $p,q$ are relatively prime. Then
\begin{align*}
\sup S_{r}(a)=\sup\{x\in\mathbb{R}\mid 0\leq x^{n}\leq a^{m}\}=\sup\{x\in\mathbb{R}\mid 0\leq x^{q}\leq a^{p}\}=\sup T_{r}(a)
\end{align*}
Proof. From the definition of equality of rationals $mq=np$, then $p$ divides $mq$, but, since $p,q$ son relatively prime then $p$ divides $m$ that is $m=kp$, thus $kqp=np$ and since $q\neq 0$ then $n=kq$ or $q$ divides $n$ and the factor is the same for $m,n$. Notice that $k\in\mathbb{N}$, since $m, p$ must have the same sign.
Thus $x\in T_{r}(a)\longleftrightarrow x^{q}\leq a^{p}\longleftrightarrow x^{kq}\leq a^{kp}\longleftrightarrow x^{n}\leq x^{m}\longleftrightarrow x\in S_{r}(a)$, by the exponential properties for naturals. Then $T_{r}(a)=S_{r}(a)$ and finally $\sup S_{r}(a)=\sup T_{r}(a)$.
This lemma solves the first point since every representation of a given rational is connected, via multiples, with the irreductible representation (which is unique, unless order, by the fundamental theorem of arithmetic applied to numerator and denominator).
Lemma. Let $a\in\mathbb{R}^{+},\, a>1$ fixed and $r=\frac{m}{1}\in\mathbb{Q}$. Then
\begin{align*}
\sup S_{m}(a)=\sup\{x\in\mathbb{R}\mid 0\leq x\leq a^{m}\}=a^{m}
\end{align*}
Proof. Obviously we work with $S_{m}(a)=[0,a^{m}]$. It is clear that $a^{m}$ is an upper bound for this set. Moreover, there is no other which is strictly lesser, since if there would be one then we would construct$\sup S_{m}(a)<\sup S_{m}(a)+\frac{a^{m}-\sup S_{m}(a)}{2}<a^{m}$ and then we would see that $\sup S_{m}(a)$ is not an upper bound for $S_{m}(a)$, what is against the fact that it is a supremum. It follows $\sup S_{m}(a)=a^{m}$.
Then the exponential as we have defined reduces to the exponential for integers at integer exponents.
Now, we extend for $0<a<1$. We know how to calculate $\exp_{a}$ when $a>1$. But if $0<a<1$ we also know that $\frac{1}{a}>1$. Thus, in this case, for every $r$
\begin{align*}
a^{r} &=\left(\frac{1}{a}\right)^{-r}
\end{align*}
and this definition is good since we are writing it in terms of something that is well defined, as we have proven. We, then, get.
Definition. Let $a\in\mathbb{R}^{+}$ fixed and $r\in\mathbb{Q}$. Define $\exp_{a}(r):\mathbb{Q}\longrightarrow\mathbb{R}$ as the function given by
\begin{align*}
r=\frac{m}{n} &\mapsto a^{r}=\sup S_{r}(a)=\sup\{x\in\mathbb{R}\mid 0\leq x^{n}\leq a^{m}\},\, a>1\\
0 &\mapsto 1\\
r=\frac{m}{n} &\mapsto a^{r}=\left(\frac{1}{a}\right)^{-r},\, 0<a<1
\end{align*}
This definition must follow the same properties that the one for integers, that is
Theorem. Let $a,b\in\mathbb{R}^{+}$ fixed y $r,s\in\mathbb{Q}$ then
- $\exp_{a}(r)\exp_{a}(s)=\exp_{a}(r+s)$
- $\exp_{\exp_{a}(r)}(s)=\exp_{a}(rs)=\exp_{\exp_{a}(s)}(r)$
- $\exp_{ab}(r)=\exp_{a}(r)\exp_{b}(r)$
- $a<b\longleftrightarrow \exp_{a}(r)<\exp_{b}(r)$ for every $r\in\mathbb{Q}^{+}$.
- $a<b\longleftrightarrow \exp_{a}(r)>\exp_{b}(r)$ for every $r\in\mathbb{Q}^{-}$.
- $r<s\longleftrightarrow \exp_{a}(r)<\exp_{a}(s)$ for every $a>1$.
- $r<s\longleftrightarrow \exp_{a}(r)>\exp_{a}(s)$ for every $0<a<1$.
- $\exp_{a}(r)>0$
Proof.
- Given that $\rho_{r}=\exp_{a}(r)$ is the unique positiv solution to $x^{n}=a^{m}$, then $\rho_{r}^{n}=a^{m}$ y $\rho_{s}^{q}=a^{p}$ where $r=\frac{m}{n}$, $s=\frac{p}{q}$, $m,p\in\mathbb{Z}$ and $n,q\in\mathbb{N}$. Thus $\rho_{r}^{nq}=a^{mq}$ and $\rho_{s}^{nq}=a^{np}$. Therefore
\begin{align*}
\left(\rho_{r}\rho_{s}\right)^{nq} &=\rho_{r}^{nq}\rho_{s}^{nq}\\
&=a^{mq}a^{np}\\
&=a^{mq+np}
\end{align*}
But $\rho_{r+s}=\exp_{a}(r+s)$ is the unique positive solution for $\rho_{r+s}^{nq}=a^{mq+np}$, then $\exp_{r}(a)\exp_{s}(a)=\exp_{a}(r+s)$, as we wanted to show. Notice that we do not split into cases.
- $\sigma_{r}=\exp_{\exp_{a}(s)}(r)$ is the unique positive solution to $x^{n}=\left(a^{s}\right)^{m}$ and $\sigma_{s}=\exp_{\exp_{a}(r)}(s)$ is the unique positive solution to $x^{q}=\left(a^{r}\right)^{p}$. From the former point
\begin{align*}
\sigma_{r}^{n} &=\left(a^{s}\right)^{m}\\
\sigma_{r}^{nq} &=\left(a^{s}\right)^{mq}\\
&=\left(\left(a^{s}\right)^{q}\right)^{m}\\
&=\left(a^{s}a^{s}\dots a^{s}\right)^{m}\\
&=\left(a^{s+s+\dots+s}\right)^{m}\\
&=\left(a^{qs}\right)^{m}\\
&=\left(a^{\frac{pq}{q}}\right)^{m}\\
&=\left(a^{p}\right)^{m}\\
&=a^{mp}
\end{align*}
analogously
\begin{align*}
\sigma_{s}^{q} &=\left(a^{r}\right)^{p}\\
\sigma_{s}^{nq} &=\left(a^{r}\right)^{np}\\
&=\left(\left(a^{r}\right)^{n}\right)^{p}\\
&=\left(a^{r}a^{r}\dots a^{r}\right)^{p}\\
&=\left(a^{r+r+\dots+r}\right)^{p}\\
&=\left(a^{nr}\right)^{p}\\
&=\left(a^{\frac{mn}{n}}\right)^{p}\\
&=\left(a^{m}\right)^{p}\\
&=a^{mp}
\end{align*}
Also $\rho_{rs}=\exp_{a}(rs)$ is the unique positive solution to $x^{nq}=a^{mp}$. Then $\exp_{\exp_{a}(s)}(r)=\exp_{a}(rs)=\exp_{\exp_{a}(r)}(s)$.
- $\rho_{r,a}=\exp_{a}(r)$ is the unique positive solution to $x^{n}=a^{m}$ and $\rho_{r,b}=\exp_{b}(r)$ is the unique positive solution to $x^{n}=b^{m}$, for each $a,b>0$. Then
\begin{align*}
\rho_{r,a}^{n}\rho_{r,b}^{n} &=a^{m}b^{m}\\
\left(\rho_{r,a}\rho_{r,b}\right)^{n} &=\left(ab\right)^{m}
\end{align*}
but $\rho_{r,ab}$ is the unique positive solution to $x^{n}=\left(ab\right)^{m}$. We conclude
\begin{align*}
\rho_{r,a}\rho_{r,b}&=\rho_{r,ab}
\end{align*}
that is $\exp_{ab}(r)=\exp_{a}(r)\exp_{b}(r)$.
- As before,
\begin{align*}
\rho_{r,a}^{n}=a^{m},\, \rho_{r,b}^{n}=b^{m}
\end{align*}
but $a<b$ if and only if $a^{n}<b^{n}$ for every $n\in\mathbb{Z}^{+}$ as we have already shown. Then, since $m>0$ by hypothesis,
\begin{align*}
\rho_{r,a}^{n}=a^{m} &<b^{m}=\rho_{r,b}^{n}\\
\rho_{r,a}^{n} &<\rho_{r,b}^{n}
\end{align*}
but suppose $\rho_{r,a}^{n} <\rho_{r,b}^{n}$ implies $\rho_{r,a}\geq\rho_{r,b}$. Then
\begin{align*}
\rho_{r,a}^{n} &<\rho_{r,b}^{n}\leq\rho_{r,a}^{n}
\end{align*}
therefore
\begin{align*}
\rho_{r,a}^{n} &<\rho_{r,a}^{n}
\end{align*}
which is impossible, because a real number can not be strictly lesser than itself. Then $\rho_{r,a}<\rho_{r,b}$, that is
\begin{align*}
\exp_{a}(r) &<\exp_{b}(r)
\end{align*}
Using the same idea
\begin{align*}
\exp_{a}(r) &<\exp_{b}(r)\\
\rho_{r,a} &<\rho_{r,b}\longleftrightarrow\\
\rho_{r,a}^{n} &<\rho_{r,b}^{n}\longleftrightarrow\\
a^{m}=\rho_{r,a}^{n} &<\rho_{r,b}^{n}=b^{m}\\
a^{m} &<b^{m}\longrightarrow\\
a &<b
\end{align*}
- Let $a<b$. Now we have $r<0$, then $\exp_{a}(r)=\frac{1}{\exp_{a}(-r)}$ and analogously for $b$. But $-r>0$, using the former point $a<b$ if and only if $\exp_{a}(-r)<\exp_{b}(-r)$ therefore
\begin{align*}
\exp_{b}(-r) &>\exp_{a}(-r)\\
\frac{1}{\exp_{a}(-r)} &>\frac{1}{\exp_{a}(-r)}\\
\exp_{a}(r) &>\exp_{b}(r)
\end{align*}
as we need to show.
- Let $r<s$. Then $mq<np$. Now $\rho_{r}^{n}=a^{m}$ and $\rho_{s}^{q}=a^{p}$, thus $\rho_{r}^{nq}=a^{mq}<a^{np}=\rho_{s}^{nq}$. For the same reason $\rho_{r}<\rho_{s}$, that is $\exp_{a}(r)<\exp_{a}(s)$. The return follows trivially following the steps backwards.
- We have $0<a<1$. Then $\exp_{a}(r)=\exp_{\frac{1}{a}}(-r)$ and if $r<s$ then $-s<-r$ Applying the former point we get, necessarily and sufficiently, that $\exp_{\frac{1}{a}}(-s)<\exp_{\frac{1}{a}}(-r)$, that is $\exp_{a}(r)>\exp_{a}(s)$.
- $\exp_{a}(r)$ is the supremum of a set of positive elements then $\exp_{a}(r)>0$.
Best Answer
I don't think the question has really been answered yet, and it's been bothering me for almost a week. I have a partial solution, which I present below (it's really too long for a comment.) I hope others can help fill in a critical gap. To prove $b^xb^y=b^{x+y}$, we prove the two inequalities $b^xb^y\leq b^{x+y}$ and $b^{x+y}\leq b^xb^y$. Theoretical Economist provided an outline for a proof of the first inequality above which I believe is correct. (I can provide all the details if anyone is interested.)
The difficulty is the second inequality. I have a solution for the case where $x+y$ is irrational which I present below. The case where $x$ and $y$ are irrational, but $x+y$ is rational is proving, at least for me, to be quite difficult and I don't yet have a solution. Note that this can occur, for instance, if, say $x=\sqrt{2}$ and $y=2-\sqrt{2}$. As far as I'm able to determine, none of the proof outlines presented above work in that case. (This was noted by a commenter above.) So here is my solution for the proof of $b^{x+y}\leq b^xb^y$ when $x+y$ is irrational:
I will prove it by assuming the contrary and deriving a contradiction. So assume that: $$b^xb^y<b^{x+y}.$$ This means that: $$\sup B(x)\cdot\sup B(y) < \sup B(x+y)$$ This means there must exist a $b^t\in B(x+y)$ which is strictly greater than $b^xb^y$, for otherwise $b^xb^y$ would be an upper bound for $B(x+y)$ which would contradict the fact that $\sup B(x+y)$ is a least upper bound. So we have, then, that $$ b^xb^y<b^t\leq b^{x+y}.$$ Now, since $x+y$ is irrational, and $t$ is rational, we must have a strict inequality, $t<x+y$, so that: $$ b^xb^y<b^t<b^{x+y}.$$ Since $t<x+y$, then $t-x<y$ and since $\mathbb{Q}$ is dense in $\mathbb{R}$, we can find a rational $p$ such that: $$ t-x < p < y $$ Now, let $q=t-p$. Then it follows from $t-x<p$ that $q<x$. So we have found rationals $p$ and $q$ such that $t=p+q$, $p<y$ and $q<x$. Now, $$b^t=b^{p+q}=b^pb^q<b^xb^y,$$ but we've already asserted that $b^xb^y<b^t$ so we've arrived at at contradiction, proving that $b^{x+y}\leq b^xb^y$. Without the assumption that $x+y$ is irrational, this proof would not have worked.
I've also tried an approach similar to those suggested above, but they involve finding rationals $r$ and $s$ which satisfy inequalities like: $$b^{r+s}\leq b^t$$ for all rationals $t\leq x+y$, where $r\leq x$ and $s\leq y$. But such rationals $r$ and $s$ do not exist when $t=x+y$, which can occur if $x+y$ is rational, and $x$ and $y$ are irrational. So again, I don't think a complete solution to this problem has yet been offered. I would love to be corrected on this point, however.
I have been thinking about how to prove the inequality when $x+y$ is rational, but all my thoughts involve concepts beyond Chapter 1 in Rudin where this problem appears. One could invoke continuity or something like that, I suppose, having proven the identity for irrational numbers, but that isn't available in Chapter 1. A later problem in Chapter 1 introduces the existence of logarithms, which I suppose could be used to show that the set $\{b^q | q\in\mathbb{Q}\}$ is dense in $\mathbb{R}$ and thereby find a $b^t$ which satisfies the strict inequality in my proof, but again, we'd be using concepts introduced later than the problem. So can anyone offer a complete proof which uses only concepts found in Chapter 1 in Rudin?