supremum is the only positive root of $z^n=a^m$

Question

I'm trying to define $a^x$ with $x$ rational. Let $r=\frac{m}{n}$ with $m,n \in \mathbb{Z}$, and let $a>1$ be a real number. We define $S_{r}(a)= \left \{ x \in \mathbb{R} | 0\leq x^n\leq a^m \right \} $ and $a^r=\sup S_{r}(a)$. I need to show that $z=a^r$ is the only positive root of the equation $z^n=a^m$ .

Answer 1

Let $E:=\{x\in\Bbb R:0\le x^n\le a^m\}$ be a subset of $\Bbb R$, and let $a^{m/n}:=\sup E$.

Basically we need to prove $$z=a^{m/n}\quad\text{if and only if}\quad z^n=a^m.$$

$(\Rightarrow)$ First we prove the direct. Suppose for sake of contradiction that we have $z=a^{m/n}$ and $z^n\ne a^m$. We have two cases, $z^n<a^m$ or $z^n>a^m$, and we show that both lead a contradiction.

First suppose $z^n<a^m$. You can use three formulae to prove that there exists a $0<\epsilon<1$ such that $(z^n+\epsilon)^m<a^m$, this means that $z^n+\epsilon$ lies in $E$; but this contradicts the fact that $z^n$ is a upper bound of $E$.

The formulae are (replace $x:=z^n$):

• First (Binomial formula): $$\sum_{k=0}^m\binom{m}{k}x^{m-k}\epsilon^k.$$ Then use the expansion to obtain $(x+\epsilon)^m\le x^m+x^{m-1}\epsilon+\dotsb<a^m$.
• Second: $$(x+\epsilon)^m\le x^m+\epsilon((x+1)^m-x^m).$$ Then you can find $\epsilon$ such that $(x+\epsilon)^m<x$, just take a $\epsilon$ such that $$0<\epsilon<\min\left\{\frac{x-x^m}{(x+1)^m-x^m},1\right\},$$ which surely exists. Therefore such $x$ cannot be the supremum of the set $E$.
• Third: $$(x+\epsilon)^m\le x^m+k\epsilon$$ for some $k\in\Bbb R$. Then you obtain $(x+\epsilon)^m\le x^m+k\epsilon<a^m$, as desired.

All formulae are proved by induction on $m$.

The other case, $z^n>a^m$, is similar; but we need to work with $(z^n-\epsilon)^m>a^m$.

From these two contradictions we see that $z^n=a^m$, as desired.

$(\Leftarrow)$ To prove the converse use the definition of $a^{m/n}$.

---

EXTRA. Now, I think your definition seems a little complicated. Maybe you can work with the following (it is similar to definition):

Suppose we have defined the exponentiation for rationals and we have proved the laws of exponentiation.

Defintion 1 (Exponentiating a real by a natural number). Let $x$ be a real number. To raise $x$ to the power $0$, we define $x^0:=1$. Now suppose recursively that $x^n$ has been defined for some natural number $n$, then we define $x^{n+1}:=x^n\times x$.

Definition 2 (Exponentiating a real by an integer). Let $x$ be a non-zero real number. Then for any negative integer $-n$, we define $x^{-n}:=1/x^n$.

Defintion 3. Let $x>0$ be a positive real, and let $n\ge1$ be a positive integer. We define $x^{1/n}$, also known as the $n^{th}$ root of $x$, by the formula $$x^{1/n}:=\sup\{y\in\Bbb R:y\ge0\quad\text{and}\quad y^n\le x\}.$$

Lemma 4 (Existence of $n^{th}$ roots). Let $x>0$ be a positive real, and let $n\ge1$ be a positive integer. Then the set $E:=\{y\in\Bbb R:y\ge0\quad\text{and}\quad y^n\le x\}$ is non-empty and is also bounded above. In particular, $x^{1/n}$ is a real number.

Definition 5. Let $x>0$ be a positive real number, and let $q$ be a rational number. To define $x^q$, we write $q=a/b$ for some integer $a$ and positive integer $b$, and define $$x^q:=\left(x^{1/b}\right)^a.$$

It looks like your definition, but I think it is more simple to work. Note that the Definition 3 only uses real numbers and natural numbers, and in Definition 5 we applied the laws of exponentiation of rational numbers. Thus, one can to prove the main properties of $x^{a/b}$; for instance, $x^{q+r}=x^qx^r$, $x^{-q}=1/x^q$, etc.