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


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 $0 < y_2 - \epsilon$.


From Negative of Absolute Value: Corollary 3:

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


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.

$\blacksquare$