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

By definition, $a^x$ is continuous.

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

From Rationals are Everywhere Dense in Reals:
 * $\forall x \in \R : \forall \delta \in \R_{>0} : \exists r \in \Q : \left\vert{ x - r }\right\vert < \delta$

From Exponential with Arbitrary Base is Continuous:
 * $\exists \delta \in \R_{>0} : \left\vert{ x - r }\right\vert < \delta \implies \left\vert{ a^{x} - a^{r} }\right\vert < \epsilon$

Also:

So let $\delta_1 \in \R_{>0}$ be such that:
 * $ \left\vert{ \left({ x + y }\right) - \left({ s + t }\right) }\right\vert < \delta_1 \implies \left\vert{ a^{x + y} - a^{r + s} }\right\vert < \dfrac{\epsilon}{2}$

Thus:

Further, from the lemma:
 * $\left\vert{ a^{x} - a^{r} }\right\vert < \epsilon \land \left\vert{ a^{y} - a^{s} }\right\vert < \epsilon \implies \left\vert{ a^{x}a^{y} - a^{r}a^{s} }\right\vert < \epsilon \left({ 2a^{M} + 1 }\right)$

So let $\delta_2 \in \R_{>0}$ be such that:

Now let $\delta = \min \left\{ { \delta_1, \delta_2 } \right\} \right\}$.

Pick $s, t \in \Q \cap I$ such that:
 * $\left\vert{ x - r }\right\vert < \delta \land \left\vert{ y - s }\right\vert < \delta$.

Then:

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

Hence the result.