Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Appendices - Section A.1 - Real Numbers and the Real Line - Exercises A.1 - Page AP-6: 30


See the proof below.

Work Step by Step

$a.$ Using the alternative definition for $|x|,\ \quad |x|=\sqrt{x^{2}}$ $|\displaystyle \frac{1}{b}|=\sqrt{(\frac{1}{b})^{2}}=\sqrt{\frac{1^{2}}{b^{2}}}=\frac{1}{\sqrt{b^{2}}}=\frac{1}{|b|}$ $b.$ $|\displaystyle \frac{a}{b}|=\sqrt{(\frac{a}{b})^{2}}=\sqrt{\frac{a^{2}}{b^{2}}}=\frac{\sqrt{a^{2}}}{\sqrt{b^{2}}}=\frac{|a|}{|b|}$
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