Power is Well-Defined/Rational

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Theorem

Let $x \in \R_{> 0}$ be a (strictly) positive real number.

Let $q$ be a rational number.


Then $x^q$ is well-defined.


Proof

Let $x \in \R_{>0}$ be fixed.

Let $q \in \Q \setminus \set 0$.

Let $\dfrac r s$ and $\dfrac t u$ be two representations of $q$.

That is, $r, s, t, u$ are non-zero integers.

We now show that:

$\dfrac r s = \dfrac t u \implies x^{r / s} = x^{t / u}$


So:

\(\ds \dfrac r s = \dfrac t u\) \(\implies\) \(\ds r u = t s\)
\(\ds \) \(\implies\) \(\ds x^{r u} = x^{s t}\) Power is Well-Defined: Integer
\(\ds \) \(\implies\) \(\ds \paren {x^r}^u = \paren {x^t}^s\) Product of Indices of Real Number: Integers
\(\ds \) \(\implies\) \(\ds \sqrt [u] {\paren {x^r}^u} = \sqrt [u] {\paren {x^t}^s}\) Existence and Uniqueness of Positive Root of Positive Real Number
\(\ds \) \(\implies\) \(\ds x^r = \sqrt [u] {\paren {x^t}^s}\) Definition of Root
\(\ds \) \(\implies\) \(\ds \sqrt [s] {\paren {x^r} } = \sqrt [s] {\sqrt [u] {\paren {x^t}^s} }\) Existence and Uniqueness of Positive Root of Positive Real Number
\(\ds \) \(\implies\) \(\ds \sqrt [s] {\paren {x^r} } = \sqrt [u] {\sqrt [s] {\paren {x^t}^s} }\) Root is Commutative
\(\ds \) \(\implies\) \(\ds \sqrt [s] {\paren {x^r} } = \sqrt [u] {\paren {x^t} }\) Definition of Root
\(\ds \) \(\implies\) \(\ds x^{r / s} = x^{t / u}\) Definition of Rational Power

Hence the result.

$\blacksquare$