Exponent Combination Laws/Product of Powers/Proof 2/Lemma

Theorem
Let $a, b \in \R_{> 0}$.

Let $r \in \Q$.

Let $a^r$ denote $a$ to the power of $r$. Let $I$ be some bounded interval.

Let $\alpha, \beta$ be the endpoints of $I$.

Let $a^r$ and $b^r$ be restricted to $I \cap \Q$.

Let $m = \min \left\{ { a^{\alpha}, a^{\beta}, b^{\alpha}, b^{\beta} } \right\}$, $M = \max \left\{ { a^{\alpha}, a^{\beta}, b^{\alpha}, b^{\beta} } \right\}$.

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

Then:
 * $\left\vert{a^x - a^r}\right\vert < \epsilon \land \left\vert{b^y - b^s}\right\vert < \epsilon \implies \left\vert{a^x b^y - a^r b^s}\right\vert < \epsilon \left({ 2 M + 1 }\right)$

Proof
Let $m' = \min \left\{ {m, 1} \right\}$.

Case 1: $\epsilon < m'$
Suppose $\epsilon < m'$.

We have that:

Hence:

Subtracting $a^r b^s$ from all sections of the inequality:
 * $- \epsilon \left({a^r + b^s}\right) - {\epsilon}^2 < a^x b^y - a^r b^s < \epsilon \left({ a^r + b^s}\right) + {\epsilon}^2$

It follows that:

Case 2: $\epsilon \ge m'$
Suppose $\epsilon \ge m'$.

Let $\delta \in \left({0 \,.\,.\, m'}\right)$.

Then:

Then as $0 < \delta < 1$ it follows that:

Hence the result.