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:

\(\ds \dfrac r s\)

\(=\)

\(\ds \dfrac t u\)

\(\ds \leadsto \ \ \)

\(\ds r u\)

\(=\)

\(\ds t s\)

\(\ds \leadsto \ \ \)

\(\ds x^{r u}\)

\(=\)

\(\ds x^{s t}\)

Power is Well-Defined: Integer

\(\ds \leadsto \ \ \)

\(\ds \paren {x^r}^u\)

\(=\)

\(\ds \paren {x^t}^s\)

Product of Indices of Real Number: Integers

\(\ds \leadsto \ \ \)

\(\ds \sqrt [u] {\paren {x^r}^u}\)

\(=\)

\(\ds \sqrt [u] {\paren {x^t}^s}\)

Existence and Uniqueness of Positive Root of Positive Real Number

\(\ds \leadsto \ \ \)

\(\ds x^r\)

\(=\)

\(\ds \sqrt [u] {\paren {x^t}^s}\)

Definition of Root of Number

\(\ds \leadsto \ \ \)

\(\ds \sqrt [s] {\paren {x^r} }\)

\(=\)

\(\ds \sqrt [s] {\sqrt [u] {\paren {x^t}^s} }\)

Existence and Uniqueness of Positive Root of Positive Real Number

\(\ds \leadsto \ \ \)

\(\ds \sqrt [s] {\paren {x^r} }\)

\(=\)

\(\ds \sqrt [u] {\sqrt [s] {\paren {x^t}^s} }\)

Root is Commutative

\(\ds \leadsto \ \ \)

\(\ds \sqrt [s] {\paren {x^r} }\)

\(=\)

\(\ds \sqrt [u] {\paren {x^t} }\)

Definition of Root of Number

\(\ds \leadsto \ \ \)

\(\ds x^{r / s}\)

\(=\)

\(\ds x^{t / u}\)

Definition of Rational Power

Hence the result.

$\blacksquare$