# Power Function to Rational Power permits Unique Continuous Extension

## Theorem

Let $a \in \R_{> 0}$.

Let $f: \Q \to \R$ be the real-valued function defined as:

$f \left({q}\right) = a^q$

where $a^q$ denotes $a$ to the power of $q$.

Then there exists a unique continuous extension of $f$ to $\R$.

## Proof

Consider $I_k := \left({-k \,.\,.\, k }\right)$ for arbitrary $k \in \N$.

Let $I_k' = I_k \cap \Q$.

Note that, for all $x, y \in I_k'$:

 $\displaystyle \left\vert{a^x - a^y}\right\vert$ $=$ $\displaystyle \left\vert{a^{x - y + y} - a^y}\right\vert$ $\displaystyle$ $=$ $\displaystyle \left\vert{a^{x - y} a^y - a^y}\right\vert$ Powers of Group Elements $\displaystyle$ $=$ $\displaystyle \left\vert{a^y}\right\vert \left\vert{a^{x - y} - 1}\right\vert$ Absolute Value Function is Completely Multiplicative $\displaystyle$ $=$ $\displaystyle a^y \left\vert{a^{x - y} - 1}\right\vert$ Power of Positive Real Number is Positive over Rationals $\displaystyle$ $\le$ $\displaystyle M \left\vert{a^{x - y} - 1}\right\vert$ Power Function is Monotone over Rationals

where $M = \max \left({a^{-k}, a^k}\right)$.

Fix $\epsilon \in \R_{> 0}$.

 $\displaystyle \exists \delta \in \R_{> 0}: \ \$ $\displaystyle 0$ $<$ $\, \displaystyle \left\vert{x - y}\right\vert \,$ $\, \displaystyle <\,$ $\displaystyle \delta$ $\displaystyle \leadsto \ \$ $\displaystyle$  $\, \displaystyle \left\vert{a^{x - y} - a^0}\right\vert \,$ $\, \displaystyle <\,$ $\displaystyle \dfrac \epsilon M$ $\displaystyle \leadsto \ \$ $\displaystyle$  $\, \displaystyle \left\vert{a^{x - y} - 1}\right\vert \,$ $\, \displaystyle <\,$ $\displaystyle \dfrac \epsilon M$ Definition of Power of Zero

where $M$ is defined as above.

Therefore, $\exists \delta \in \R_{> 0}$ such that:

 $\displaystyle \left\vert{x - y}\right\vert$ $<$ $\displaystyle \delta$ $\displaystyle \leadsto \ \$ $\displaystyle \left\vert{a^x - a^y}\right\vert$ $\le$ $\displaystyle M \left\vert{a^{x - y} - 1}\right\vert$ $\displaystyle$ $<$ $\displaystyle M \dfrac \epsilon M$ $\displaystyle$ $=$ $\displaystyle \epsilon$

That is, $f$ is uniformly continuous on $I_k'$.

From Rationals are Everywhere Dense in Reals, $I_k'$ is everywhere dense in $I_k$.

From Continuous Extension from Dense Subset, there exists a unique continuous extension of $f$ to $I_k$.

Call this function $f_k$.

Define:

$\displaystyle F = \bigcup \left\{ {f_k : k \in \N}\right\}$

Note that $\left\langle{I_k}\right\rangle$ is an exhausting sequence of sets in $\R$.

By the Union of Functions Theorem, $F$ defines a function $\R \to \R$

From the Pasting Lemma, $F$ is continuous on $\R$.

Hence the result.

$\blacksquare$