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

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Theorem

Let $x_1, x_2, y_1, y_2 \in \R_{>0}$ be strictly positive real numbers.

Let $\epsilon \in \left({0, \,.\,.\, \min \left \{{y_1, y_2, 1}\right \} }\right)$.


Then:

$\left\vert{x_1 - y_1}\right\vert < \epsilon \land \left\vert{x_2 - y_2}\right\vert < \epsilon \implies \left \vert{x_1 x_2 - y_1 y_2}\right \vert < \epsilon \left({y_1 + y_2 + 1}\right)$

Proof

First:

\(\displaystyle \epsilon\) \(<\) \(\displaystyle \min \left\{ {y_1, y_2, 1}\right \}\)
\(\displaystyle \implies \ \ \) \(\displaystyle \epsilon\) \(<\) \(\displaystyle y_1\) Definition of min
\(\displaystyle \implies \ \ \) \(\displaystyle \epsilon - \epsilon\) \(<\) \(\displaystyle y_1 - \epsilon\) Subtract $\epsilon$ from both sides
\((1):\quad\) \(\displaystyle \implies \ \ \) \(\displaystyle 0\) \(<\) \(\displaystyle y_1 - \epsilon\)


The same logic, mutatis mutandis, shows that $0 < y_2 - \epsilon$.


From Negative of Absolute Value: Corollary 3:

\(\displaystyle \left \vert{x_1 - y_1}\right \vert < \epsilon\) \(\implies\) \(\displaystyle y_1 - \epsilon < x_1 < y_1 - \epsilon\)
\(\displaystyle \left \vert{x_2 - y_2}\right \vert < \epsilon\) \(\implies\) \(\displaystyle y_2 - \epsilon < x_2 < y_2 - \epsilon\)


Hence:

\(\displaystyle \left({y_1 - \epsilon}\right) \left({y_2 - \epsilon}\right)\) \(<\) \(\, \displaystyle x_1 x_2 \, \) \(\, \displaystyle <\, \) \(\displaystyle \left({y_1 + \epsilon}\right) \left({y_1 + \epsilon}\right)\) Inequality of Product of Unequal Numbers: from $(1)$
\(\displaystyle \implies \ \ \) \(\displaystyle y_1 y_2 - \epsilon \left({y_1 + y_2}\right) + \epsilon^2\) \(<\) \(\, \displaystyle x_1 x_2 \, \) \(\, \displaystyle <\, \) \(\displaystyle y_1 y_2 + \epsilon \left({y_1 + y_2}\right) + \epsilon^2\) Multiplication of Numbers Distributes over Addition
\(\displaystyle \implies \ \ \) \(\displaystyle y_1 y_2 - \epsilon \left({y_1 + y_2}\right) - \epsilon^2\) \(<\) \(\, \displaystyle x_1 x_2 \, \) \(\, \displaystyle <\, \) \(\displaystyle y_1 y_2 + \epsilon \left({y_1 + y_2}\right) + \epsilon^2\) Square of Non-Zero Real Number is Strictly Positive


Subtracting $y_1 y_2$ from all sections of the inequality:

$- \epsilon \left({y_1 + y_2}\right) - \epsilon^2 < x_1 x_2 - y_1 y_2 < \epsilon \left({y_1 + y_2}\right) + \epsilon^2$


If follows that:

\(\displaystyle \left\vert{x_1 x_2 - y_1 y_2}\right\vert\) \(<\) \(\displaystyle \epsilon \left({y_1 + y_2}\right) + \epsilon^2\) Negative of Absolute Value: Corollary 1
\(\displaystyle \) \(<\) \(\displaystyle \epsilon \left({y_1 + y_2}\right) + \epsilon\) Since $\epsilon < \min \left\{ {y_1, y_2, 1} \right\} < 1$, we may apply Real Number between Zero and One is Greater than Square
\(\displaystyle \) \(<\) \(\displaystyle \epsilon \left({y_1 + y_2 + 1}\right)\) Multiplication of Numbers Distributes over Addition


Hence the result.

$\blacksquare$