ProofWiki:Sandbox

Theorem
Let $a \in \R_{> 0}$ be a positive real number.

Let $x, y \in \R$ be real numbers.

Let $a^x$ be defined as $a$ to the power of $x$.

Then:
 * $a^x a^y = a^{x + y}$

Proof
This proof uses Definition 2 of $a^x$.

We will show that:
 * $\forall \epsilon \in \R_{>0} : \left\vert{ a^{x + y} - a^{x}a^{y} }\right\vert < \epsilon$

So fix $\epsilon \in \R_{>0}$.

Suppose WLOG that $x < y$

Consider $I := \left [{x - 1 \,.\,.\, y + 1} \right]$.

Let $I_{\Q} = I \cap \Q$.

From Exponential with Arbitrary Base is Continuous:
 * $\exists \delta_1 \in \R_{>0} : \left\vert{ \left({ x + y }\right) - \left({ x' + y' }\right) }\right\vert < \delta_1 \implies \left\vert{ a^{x + y} - a^{x}a^{y} }\right\vert < \epsilon$

From Triangle Inequality/Real Numbers:
 * $\left\vert{ x - x' }\right\vert < \dfrac{\delta_1}{2} \land \left\vert{ y - y' }\right\vert < \dfrac{\delta_1}{2} \implies \left\vert{ \left({ x + y }\right) - \left({ x' + y' }\right) }\right\vert < \delta_1$

From Exponential with Arbitrary Base is Continuous:
 * $\exists \delta_2 \in \R_{>0} : \left\vert{ x - x' }\right\vert < \delta_2 \land \left\vert{ y - y' }\right\vert < \delta_2 \implies \left\vert{ a^{x} - a^{x'} }\right\vert < \dfrac{1}{2}\dfrac{\epsilon}{2M + 1} \land \left\vert{ a^{y} - a^{y'} }\right\vert < \dfrac{1}{2}\dfrac{\epsilon}{2M + 1}$

From the lemma:
 * $\left\vert{ a^{x} - a^{x'} }\right\vert < \dfrac{1}{2}\dfrac{\epsilon}{2M + 1} \land \left\vert{ a^{y} - a^{y'} }\right\vert < \dfrac{1}{2}\dfrac{\epsilon}{2M + 1} \implies \left\vert{ a^{x}a^{x'} - a^{y}a^{y'}  }\right\vert < \epsilon$

where $M$ is defined as in the lemma.

So let $\delta = \min \left\{ { \dfrac{\delta_1}{2}, \delta_2 } \right\}$.

From Rationals are Everywhere Dense in Reals:
 * $\exists r, s \in \Q : \left\vert{ x - r }\right\vert < \delta \land \left\vert{ y - s }\right\vert < \delta \implies \left\vert{ a^{x+y} - a^{x}a^{y} }\right\vert$

Thus:

Hence the result, by Real Plus Epsilon.