Power Function to Rational Power permits Unique Continuous Extension

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

Let $f : \Q \to \R$ be the real mapping 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 $x, y \in I_k'$:

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

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

From Exponential Tends to One as Power Tends to Zero/Rational Number:

where $M$ is defined as above.

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

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.