Exponent Combination Laws/Power of Power/Proof 2

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

From Polynomial is Continuous:

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

From Power Function on Strictly Positive Base is Continuous:

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


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

From Closure of Rational Interval is Closed Real Interval:

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


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.

$\blacksquare$