Vtyjkyuk

Here is Definition 5.6.6 in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:

(i) If $m, n in mathbbN$ and $x geq 0$, we define $x^m/n colon= left( x^1/n right)^m$. (ii) If $m, n in mathbbN$ 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 $I subset mathbbR$ be an interval and let $f colon I to mathbbR$ 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 $I subset mathbbR$ be an interval, and let $f colon I to mathbbR$ be a strictly increasing (respectively strictly decreasing) continuous function. Then the range $f(I)$ of function $f$ is also an interval $J subset mathbbR$; $f$ 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 $f colon [0, +infty) to mathbbR$ 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
$$ gbig( f(x) big) = x qquad mbox and qquad fbig( g(x) big) = x, $$
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 $$ fracmn = fracpq. $$
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 $m, n in mathbbN$ and $x geq 0$, we define
$$ x^m/n colon= left( x^m right)^1/n$. (ii) If $m, n in mathbbN$ 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
$$
beginalign
left( x^m/n right)^N &= left[ left( x^1/n right)^m right]^nq qquad mbox [ Definition 5.6.6 and the definition of $N$ ] \
&= left[ left left( x^1/n right)^m right^n right]^q \
&= left[ left( x^1/n right)^mn right]^q \
&= left[ left left( x^1/n right)^n right^m right]^q \
&= left[ left( x^1/n right)^n right]^mq \
&= x^mq, tag1
endalign
$$
and also
$$
beginalign
left( x^p/q right)^N &= left[ left( x^1/q right)^p right]^nq qquad mbox [ Definition 5.6.6 and the definition of $N$ ] \
&= left[ left left( x^1/q right)^p right^q right]^n \
&= left[ left( x^1/q right)^pq right]^n \
&= left[ left left( x^1/q right)^q right^p right]^n \
&= left[ left( x^1/q right)^q right]^np \
&= x^np, tag2
endalign
$$
But as $$ fracmn = fracpq, $$
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?