ProofWiki:Sandbox

Theorem
Let $a \in \R_+$ 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:
 * $\left({a^x}\right)^y = a^{xy}$

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

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

Suppose WLOG that $x < y$.

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

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

Let $M = \max \left\{ { \left\vert{ x }\right\vert, \left\vert{ y }\right\vert } \right\}$

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

From Polynomial is Continuous:
 * $\exists \delta' \in \R_{>0} : \left\vert{ a^{x} - a^{x'} }\right\vert < \delta' \implies \left\vert{ \left({ a^{x} }\right)^{y'} - \left({ a^{x'} }\right)^{y'} }\right\vert < \dfrac{\epsilon}{4}$

From Exponential with Arbitrary Base is Continuous:
 * $\exists \delta_1 \in \R_{>0} : \left\vert{ xx' - yy' }\right\vert < \delta_1 \implies \left\vert{ a^{xx'} - a^{xy'} }\right\vert < \dfrac{\epsilon}{4}$
 * $\exists \delta_2 \in \R_{>0} : \left\vert{ xy' - x'y' }\right\vert < \delta_2 \implies \left\vert{ a^{xy'} - a^{x'r'} }\right\vert < \dfrac{\epsilon}{4}$
 * $\exists \delta_3 \in \R_{>0} : \left\vert{ x' - x }\right\vert < \delta_3 \implies \left\vert{ a^{x'} - a^{x} }\right\vert < \delta'$
 * $\exists \delta_4 \in \R_{>0} : \left\vert{ y' - y }\right\vert < \delta_4 \implies \left\vert{ \left({ a^{x} }\right)^{y'} - \left({ a^{x} }\right)^{y} }\right\vert < \dfrac{\epsilon}{4}$

Let $\delta = \max \left\{ { \dfrac{\delta_1}{ \left\vert{ x }\right\vert }, \dfrac{\delta_2}{M}, \delta_3, \delta_4 } \right\}$.

From Closure of Rational Interval is Closed Real Interval:
 * $\exists r, s \in I_{\Q} : \left\vert{ x - r }\right\vert < \delta \land \left\vert{ y - s }\right\vert < \delta$

Thus:

Hence the result, by Real Plus Epsilon.