# 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 $\text {(1)}: \quad$ $\displaystyle \implies \ \$ $\displaystyle 0$ $<$ $\displaystyle y_1 - \epsilon$

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

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

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.

$\blacksquare$