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

## Theorem

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

Let $\epsilon \in \openint 0 {\min \set {y_1, y_2, 1} }$.

Then:

$\size {x_1 - y_1} < \epsilon \land \size {x_2 - y_2} < \epsilon \implies \size {x_1 x_2 - y_1 y_2} < \epsilon \paren {y_1 + y_2 + 1}$

## Proof

First:

 $\ds \epsilon$ $<$ $\ds \min \set {y_1, y_2, 1}$ $\ds \leadsto \ \$ $\ds \epsilon$ $<$ $\ds y_1$ Definition of Min Operation $\ds \leadsto \ \$ $\ds \epsilon - \epsilon$ $<$ $\ds y_1 - \epsilon$ subtracting $\epsilon$ from both sides $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds 0$ $<$ $\ds y_1 - \epsilon$

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

 $\ds \size {x_1 - y_1} < \epsilon$ $\implies$ $\ds y_1 - \epsilon < x_1 < y_1 - \epsilon$ $\ds \size {x_2 - y_2} < \epsilon$ $\implies$ $\ds y_2 - \epsilon < x_2 < y_2 - \epsilon$

Hence:

 $\ds \paren {y_1 - \epsilon} \paren {y_2 - \epsilon}$ $<$ $\, \ds x_1 x_2 \,$ $\, \ds < \,$ $\ds \paren {y_1 + \epsilon} \paren {y_1 + \epsilon}$ Inequality of Product of Unequal Numbers: from $(1)$ $\ds \leadsto \ \$ $\ds y_1 y_2 - \epsilon \paren {y_1 + y_2} + \epsilon^2$ $<$ $\, \ds x_1 x_2 \,$ $\, \ds < \,$ $\ds y_1 y_2 + \epsilon \paren {y_1 + y_2} + \epsilon^2$ Distributive Property $\ds \leadsto \ \$ $\ds y_1 y_2 - \epsilon \paren {y_1 + y_2} - \epsilon^2$ $<$ $\, \ds x_1 x_2 \,$ $\, \ds < \,$ $\ds y_1 y_2 + \epsilon \paren {y_1 + y_2} + \epsilon^2$ Square of Non-Zero Real Number is Strictly Positive

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

$-\epsilon \paren {y_1 + y_2} - \epsilon^2 < x_1 x_2 - y_1 y_2 < \epsilon \paren {y_1 + y_2} + \epsilon^2$

If follows that:

 $\ds \size {x_1 x_2 - y_1 y_2}$ $<$ $\ds \epsilon \paren {y_1 + y_2} + \epsilon^2$ Negative of Absolute Value: Corollary 1 $\ds$ $<$ $\ds \epsilon \paren {y_1 + y_2} + \epsilon$ Since $\epsilon < \min \size {y_1, y_2, 1} < 1$, we may apply Real Number between Zero and One is Greater than Square $\ds$ $<$ $\ds \epsilon \paren {y_1 + y_2 + 1}$ Distributive Property

Hence the result.

$\blacksquare$