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I have proven yet that the following characterization of exponential function is well defined.

Let $a\in\mathbb{R}^{+}\cup\{0\}$ 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\\ r=\frac{m}{n} &\mapsto a^{r}=\sup S_{-r}\left(\frac{1}{a}\right),\, 0<a<1 \end{align*}

The rules for natural and integer arguments are given as true (I proved that too). Therefore I want to prove, for example, that \begin{align*} \exp_{a}(r)\exp_{a}(s)=\exp_{a}(r+s) \end{align*} that is \begin{align*} \sup S_{r}(a)\sup S_{s}(a) &=\sup S_{r+s}(a),\, a>1\\ \sup S_{-r}\left(\frac{1}{a}\right)\sup S_{-s}\left(\frac{1}{a}\right) &=\sup S_{-(r+s)}\left(\frac{1}{a}\right),\, 0<a<1\\ \end{align*} The idea of the proof begins showing that product of suprema in lhs is a upper bound for $S_{r+s}(a)$. However, I can't see a way to achieve this. Can anyone give me a hint?


I think I have found part of the proof by myself. Here it is:

Suppose the first case ($a>1$). Then

\begin{align*} x\in S_{r}(a) &\longleftrightarrow x^{n}\leq a^{m}\\ &\longleftrightarrow x^{nq}\leq a^{mq}\\ y\in S_{s}(a) &\longleftrightarrow y^{q}\leq a^{p}\\ &\longleftrightarrow y^{nq}\leq a^{np} \end{align*}

Thus,

\begin{align*} x\in S_{r}(a),y\in S_{s}(a)\longrightarrow xy\in S_{r+s}(a) \end{align*} and in the same manner, since $x\leq\sup S_{r}(a),\, y\leq\sup S_{s}(a)$, then \begin{align*} \sup S_{r}(a)\sup S_{s}(a)\in S_{r+s}(a) \end{align*} therefore \begin{align*} \sup S_{r}(a)\sup S_{s}(a)\leq\sup S_{r+s}(a) \end{align*}

It remains to be proved that $\sup S_{r}(a)\sup S_{s}(a)$ is an upper bound for $S_{r+s}(a)$. How can I do that?

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This problem is part of a greater one which I'm including into a course notes on Introductory Analysis as a project. This will serve me as didactic material for the class where I am assistant. It shows another form to construct exponential functions from scratch, and in this part how to apply suprema, archimedean property and the supremum axiom to this task. I prefer to solve every exercise I give to solve since it could be used as reference for me or for other instructors that use my notes. –  elessartelkontar Feb 6 at 0:56
    
May be I can prove it in more simple cases –  elessartelkontar Feb 11 at 4:00

1 Answer 1

up vote 1 down vote accepted

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

  1. $\exp_{a}(r)\exp_{a}(s)=\exp_{a}(r+s)$
  2. $\exp_{\exp_{a}(r)}(s)=\exp_{a}(rs)=\exp_{\exp_{a}(s)}(r)$
  3. $\exp_{ab}(r)=\exp_{a}(r)\exp_{b}(r)$
  4. $a<b\longleftrightarrow \exp_{a}(r)<\exp_{b}(r)$ for every $r\in\mathbb{Q}^{+}$.
  5. $a<b\longleftrightarrow \exp_{a}(r)>\exp_{b}(r)$ for every $r\in\mathbb{Q}^{-}$.
  6. $r<s\longleftrightarrow \exp_{a}(r)<\exp_{a}(s)$ for every $a>1$.
  7. $r<s\longleftrightarrow \exp_{a}(r)>\exp_{a}(s)$ for every $0<a<1$.
  8. $\exp_{a}(r)>0$

Proof.

  1. 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.
  2. $\sigma_{r}=\exp_{\exp_{a}(s)}(r)$ 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 $\exp_{\exp_{a}(s)}(r)=\exp_{a}(rs)=\exp_{\exp_{a}(r)}(s)$.
  3. $\rho_{r,a}=\exp_{a}(r)$ 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 $\exp_{ab}(r)=\exp_{a}(r)\exp_{b}(r)$.
  4. 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*}
  5. 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.
  6. 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.
  7. 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 $\exp_{a}(r)>\exp_{a}(s)$.
  8. $\exp_{a}(r)$ is the supremum of a set of positive elements then $\exp_{a}(r)>0$.
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