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 Real Polynomial Function 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:
| \(\ds \exists \delta_1 \in \R_{>0}: \, \) | \(\ds \size {x x' - y y'}\) | \(<\) | \(\ds \delta_1\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \size {a^{x x'} - a^{x y'} }\) | \(<\) | \(\ds \dfrac \epsilon 4\) | |||||||||||
| \(\ds \exists \delta_2 \in \R_{>0}: \, \) | \(\ds \size {x y' - x' y'}\) | \(<\) | \(\ds \delta_2\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \size {a^{x y'} - a^{x' y'} }\) | \(<\) | \(\ds \dfrac \epsilon 4\) | |||||||||||
| \(\ds \exists \delta_3 \in \R_{>0}: \, \) | \(\ds \size {x' - x}\) | \(<\) | \(\ds \delta_3\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \size {a^{x'} - a^x}\) | \(<\) | \(\ds \delta'\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \size {\paren {a^x}^{y'} - \paren {a^{x'} }^{y'} }\) | \(<\) | \(\ds \dfrac \epsilon 4\) | |||||||||||
| \(\ds \exists \delta_4 \in \R_{>0}: \, \) | \(\ds \size {y' - y}\) | \(<\) | \(\ds \delta_4\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \size {\paren {a^x}^{y'} - \paren {a^x}^{y} }\) | \(<\) | \(\ds \dfrac \epsilon 4\) |
Further:
| \(\ds \size {y - y'}\) | \(<\) | \(\ds \frac {\delta_1} {\size x}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \size {x y - x y'}\) | \(=\) | \(\ds \size x \size {y - y'}\) | Absolute Value of Product | ||||||||||
| \(\ds \) | \(<\) | \(\ds \size x \frac {\delta_1} {\size x}\) | multiplying both sides by $\size x \ge 0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \delta_1\) |
and:
| \(\ds \size {x - x'}\) | \(<\) | \(\ds \frac {\delta_2} M\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \size {x y' - x'y'}\) | \(=\) | \(\ds \size {y'} \size {x - x'}\) | Absolute Value of Product | ||||||||||
| \(\ds \) | \(\le\) | \(\ds M \size {x - x'}\) | Real Number Ordering is Compatible with Multiplication | |||||||||||
| \(\ds \) | \(<\) | \(\ds M \frac {\delta_1} M\) | multiplying both sides by $M \ge 0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \delta_2\) |
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:
| \(\ds \size {a^{x y} - \paren {a^x}^y}\) | \(\le\) | \(\ds \size {a^{x y} - a^{x s} } + \size {a^{x s} - a^{r s} } + \size {a^{r s} - \paren {a^r}^s} + \size {\paren {a^r}^s - \paren {a^x}^s} + \size {\paren {a^x}^s - \paren {a^x}^y}\) | Triangle Inequality for Real Numbers | |||||||||||
| \(\ds \) | \(=\) | \(\ds \size {a^{x y} - a^{x s} } + \size {a^{x s} - a^{r s} } + \size {\paren {a^r}^s - \paren {a^x}^s} + \size {\paren {a^x}^s - \paren {a^x}^y}\) | Product of Indices of Real Number: Rational Numbers | |||||||||||
| \(\ds \) | \(<\) | \(\ds \frac \epsilon 4 + \frac \epsilon 4 + \frac \epsilon 4 + \frac \epsilon 4\) | Definition of $r$ and $s$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \epsilon\) |
Hence the result, by Real Plus Epsilon.
$\blacksquare$