Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 4 - Elementary Number Theory and Methods of Proof - Exercise Set 4.4 - Page 190: 44

Answer

See below.

Work Step by Step

For any real numbers $x$ and $y$, we only need to consider three cases: Case1: both are positive, we have $LHS=|x|\cdot |y|=xy=|xy|=RHS$; Case2: both are negative, we have $LHS=|x|\cdot |y|=(-x)(-y)=xy=|xy|=RHS$; Case3: they have oppsit signs, we have $LHS=|x|\cdot |y|=-xy=|xy|=RHS$; thus in any case, the statement is true.
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