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.
$\blacksquare$