Here is Definition 5.6.6 in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
(i) If $m, n \in \mathbb{N}$ and $x \geq 0$, we define $x^{m/n} \colon= \left( x^{1/n} \right)^m$. (ii) If $m, n \in \mathbb{N}$ and $x > 0$, we define $x^{-m/n} \colon= \left( x^{1/n} \right)^{-m}$.
And, here is Theorem 5.6.5 (Continuous Inverse Theorem):
Let $I \subset \mathbb{R}$ be an interval and let $f \colon I \to \mathbb{R}$ be strictly monotone and continuous on $I$. Then the function $g$ inverse to $f$ is strictly monotone and continuous on $J \colon= f(I)$.
However, the above theorem (perhaps also on the basis of the proof given by Bartle & Sherbert) could be restated more elaborately as follows:
Let $I \subset \mathbb{R}$ be an interval, and let $f \colon I \to \mathbb{R}$ be a strictly increasing (respectively strictly decreasing) continuous function. Then the range $f(I)$ of function $f$ is also an interval $J \subset \mathbb{R}$; $f$ is a bijective mapping of the interval $I$ onto the interval $J$; and the inverse function $g \colon J \to I$ is also a strictly increasing (respectively strictly decreasing) continuous function.
Am I right? I mean is there any problem with the result I've just stated?
Now suppose $n$ is a given natural number. Then as the function $f \colon [0, +\infty) \to \mathbb{R}$ defined by $f(x) \colon= x^n$ is a strictly increasing continuous function, and the range of $f$ equals $[0, +\infty)$ also, so the inverse $g$ of $f$ exists and is a function $g \colon [0, +\infty) \to [0, +\infty)$ that is also strictly increasing and continuous (on all of $[0, +\infty)$). This function is called the $n$-th root function. We write $$ g(x) = x^{1/n} \ \mbox{ for all } x \in [0, +\infty). $$ Thus, for each $x \in [0, +\infty)$, we have $$ g\big( f(x) \big) = x \qquad \mbox{ and } \qquad f\big( g(x) \big) = x, $$ which is the same as $$ \left( x^n \right)^{1/n} = x \qquad \mbox{ and } \qquad \left( x^{1/n} \right)^n = x. $$ The proof uses the fact that since $n$ and $q$ are natural numbers and since $x > 0$, therefore $$ \left( x^n \right)^q = x^{nq} = \left( x^q \right)^n. $$
Am I right?
So far so good!
On the basis of this wherewithal, how to prove the following?
MAIN RESULT
Suppose that $x$ is a positive real number, and suppose that $m, n, p, q$ are integers such that $n > 0$, $q > 0$, and such that $$ \frac{m}{n} = \frac{p}{q}. $$ Then $$ x^{m/n} = x^{p/q}, $$ that is, $$ \left( x^{1/n} \right)^m = \left( x^{1/q} \right)^p. $$
This result establishes that the notion of rational powers of positive real numbers is so to speak well-defined? Am I right?
I have managed to prove the following though:
Suppose that $x$ is a non-negative real number, and suppose that $n$ and $q$ are natural numbers. Then $$ \left( x^{1/n} \right)^{1/q} = \left( x^{1/q} \right)^{1/n}. $$
I think that with the help of the result just stated the MAIN RESULT could be more easily established if Definition 5.6.6 were to be stated as follows:
Definition 5.6.6 Modified
(i) If $m, n \in \mathbb{N}$ and $x \geq 0$, we define $$ x^{m/n} \colon= \left( x^m \right)^{1/n}$. (ii) If $m, n \in \mathbb{N}$ and $x > 0$, we define $x^{-m/n} \colon= \left( x^{-m} \right)^{1/n}$.
Am I right? If so, then (how) can we show rigorously that these two definitions for $x^{m/n}$ are equivalent (which indeed these are)?
Or else, is there another more direct way of coming up with a proof of our MAIN RESULT using Theorem 5.6.5 (or the more elaborate version thereof) and Definition 5.6.6 in Bartle & Sherbert?
PS:
Using @bangs idea, we can proceed as follows:
Proof of MAIN RESULT:
Let us put $$ N \colon= nq. $$
Now as the function $x \mapsto x^N$ is a bijective mapping of $[0, +\infty)$ onto $[0, +\infty)$ and as $0 \mapsto 0$, so this function also maps $(0, +\infty)$ bijectively onto $(0, +\infty)$.
Now using the familiar properties of integer exponents we find that $$ \begin{align} \left( x^{m/n} \right)^{N} &= \left[ \left( x^{1/n} \right)^m \right]^{nq} \qquad \mbox{ [ Definition 5.6.6 and the definition of $N$ ] }\\ &= \left[ \left\{ \left( x^{1/n} \right)^m \right\}^{n} \right]^q \\ &= \left[ \left( x^{1/n} \right)^{mn} \right]^q \\ &= \left[ \left\{ \left( x^{1/n} \right)^{n} \right\}^{m} \right]^q \\ &= \left[ \left( x^{1/n} \right)^n \right]^{mq} \\ &= x^{mq}, \tag{1} \end{align} $$ and also $$ \begin{align} \left( x^{p/q} \right)^{N} &= \left[ \left( x^{1/q} \right)^p \right]^{nq} \qquad \mbox{ [ Definition 5.6.6 and the definition of $N$ ] }\\ &= \left[ \left\{ \left( x^{1/q} \right)^p \right\}^{q} \right]^n \\ &= \left[ \left( x^{1/q} \right)^{pq} \right]^n \\ &= \left[ \left\{ \left( x^{1/q} \right)^{q} \right\}^{p} \right]^n \\ &= \left[ \left( x^{1/q} \right)^q \right]^{np} \\ &= x^{np}, \tag{2} \end{align} $$ But as $$ \frac{m}{n} = \frac{p}{q}, $$ so we must have $$ mq = np. $$ Therefore from (1) and (2) we obtain $$ \left( x^{m/n} \right)^{N} = x^{mq} = x^{np} = \left( x^{p/q} \right)^{N}. $$ from which it follows that $$ x^{m/n} = x^{p/q}, $$ because both $x^{m/n}$ and $x^{p/q}$ are in $(0, +\infty)$ and because the function $t \mapsto t^N$ maps $[0, +\infty)$ bijectively onto $(0, +\infty)$.
Is this proof satisfactory enough? If not, then where are problems in it as far as accuracy, rigor, or clarity go?