# Exponent Combination Laws/Power of Power/Proof 2

## Theorem

Let $a \in \R_{>0}$ be a (strictly) positive real number.

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

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

Then:

$\paren {a^x}^y = a^{x y}$

## Proof

We will show that:

$\forall \epsilon \in \R_{>0}: \size {a^{x y} - \paren {a^x}^y} < \epsilon$

Without loss of generality, suppose that $x < y$.

Consider $I := \closedint x y$.

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

Let $M = \max \set {\size x, \size y}$

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

$\exists \delta' \in \R_{>0}: \size {a^x - a^{x'} } < \delta' \leadsto \size {\paren {a^x}^{y'} - \paren {a^{x'} }^{y'} } < \dfrac \epsilon 4$
 $\displaystyle \exists \delta_1 \in \R_{>0} : \ \$ $\displaystyle \size {x x' - y y'} \ \$ $\, \displaystyle < \,$ $\displaystyle \delta_1$ $\leadsto$ $\, \displaystyle \size { a^{x x'} - a^{x y'} } \,$ $\, \displaystyle <\,$ $\displaystyle \dfrac \epsilon 4$ $\displaystyle \exists \delta_2 \in \R_{>0} : \ \$ $\displaystyle \size {x y' - x' y'} \ \$ $\, \displaystyle < \,$ $\displaystyle \delta_2$ $\leadsto$ $\, \displaystyle \size {a^{x y'} - a^{x' y'} } \,$ $\, \displaystyle <\,$ $\displaystyle \dfrac \epsilon 4$ $\displaystyle \exists \delta_3 \in \R_{>0} : \ \$ $\displaystyle \size {x' - x} \ \$ $\, \displaystyle < \,$ $\displaystyle \delta_3$ $\leadsto$ $\, \displaystyle \size {a^{x'} - a^x} \,$ $\, \displaystyle <\,$ $\displaystyle \delta'$ $\displaystyle$ $\leadsto$ $\, \displaystyle \size {\paren {a^x}^{y'} - \paren {a^{x'} }^{y'} } \,$ $\, \displaystyle <\,$ $\displaystyle \dfrac \epsilon 4$ $\displaystyle \exists \delta_4 \in \R_{>0} : \ \$ $\displaystyle \size {y' - y} \ \$ $\, \displaystyle < \,$ $\displaystyle \delta_4$ $\leadsto$ $\, \displaystyle \size {\paren {a^x}^{y'} - \paren {a^x}^{y} } \,$ $\, \displaystyle <\,$ $\displaystyle \dfrac \epsilon 4$

Further:

 $\displaystyle \size {y - y'}$ $<$ $\displaystyle \frac {\delta_1} {\size x}$ $\displaystyle \leadsto \ \$ $\displaystyle \size {x y - x y'}$ $=$ $\displaystyle \size x \size {y - y'}$ Absolute Value Function is Completely Multiplicative $\displaystyle$ $<$ $\displaystyle \size x \frac {\delta_1} {\size x}$ multiplying both sides by $\size x \ge 0$ $\displaystyle$ $=$ $\displaystyle \delta_1$

and:

 $\displaystyle \size {x - x'}$ $<$ $\displaystyle \frac {\delta_2} M$ $\displaystyle \leadsto \ \$ $\displaystyle \size {x y' - x'y'}$ $=$ $\displaystyle \size {y'} \size {x - x'}$ Absolute Value Function is Completely Multiplicative $\displaystyle$ $\le$ $\displaystyle M \size {x - x'}$ Real Number Ordering is Compatible with Multiplication $\displaystyle$ $<$ $\displaystyle M \frac {\delta_1} M$ multiplying both sides by $M \ge 0$ $\displaystyle$ $=$ $\displaystyle \delta_2$

Let $\delta = \max \set {\dfrac {\delta_1} {\size x}, \dfrac {\delta_2} M, \delta_3, \delta_4}$.

$\exists r, s \in I_\Q: \size {x - r} < \delta \land \size {y - s} < \delta$

Thus:

 $\displaystyle \size {a^{x y} - \paren {a^x}^y}$ $\le$ $\displaystyle \size {a^{x y} - a^{x s} } + \size {a^{x s} - a^{r s} } + \size {a^{r s} - \paren {a^r}^s} + \size {\paren {a^r}^s - \paren {a^x}^s} + \size {\paren {a^x}^s - \paren {a^x}^y}$ Triangle Inequality for Real Numbers $\displaystyle$ $=$ $\displaystyle \size {a^{x y} - a^{x s} } + \size {a^{x s} - a^{r s} } + \size {\paren {a^r}^s - \paren {a^x}^s} + \size {\paren {a^x}^s - \paren {a^x}^y}$ Product of Indices of Real Number: Rational Numbers $\displaystyle$ $<$ $\displaystyle \frac \epsilon 4 + \frac \epsilon 4 + \frac \epsilon 4 + \frac \epsilon 4$ Definition of $r$ and $s$ $\displaystyle$ $=$ $\displaystyle \epsilon$

Hence the result, by Real Plus Epsilon.

$\blacksquare$