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: 28


See the proof below.

Work Step by Step

We alternatively defined $|x|$ as $|x|=\sqrt{x^{2}}$ Apply on $x=-a:$ $|-a|=\sqrt{(-1\cdot a)^{2}}=\sqrt{(-1)^{2}a^{2}}=\sqrt{a^{2}}=|a|$
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