# Power is Well-Defined/Rational

## 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.

$\blacksquare$