Answer
See the proof below.
Work Step by Step
We alternatively defined $|x|$ as
$|x|=\sqrt{x^{2}}$
Apply on $x=-a:$
$|-a|=\sqrt{(-1\cdot a)^{2}}=\sqrt{(-1)^{2}a^{2}}=\sqrt{a^{2}}=|a|$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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
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