Discrete Mathematics with Applications 4th Edition

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

Chapter 6 - Set Theory - Exercise Set 6.3 - Page 373: 47

Answer

$A\bigtriangleup B= (A - B) \cup (B - A) \\ =(B-A)\cup (A-B) \,\,\,\,\,\,\\(by\,\,commutative\,\,law\,\,for\,\,\cup)\\ but\,\,(B-A)\cup (A-B)=B\bigtriangleup A \\ so\,\,A\bigtriangleup B=B\bigtriangleup A $

Work Step by Step

$A\bigtriangleup B= (A - B) \cup (B - A) \\ =(B-A)\cup (A-B) \,\,\,\,\,\,\\(by\,\,commutative\,\,law\,\,for\,\,\cup)\\ but\,\,(B-A)\cup (A-B)=B\bigtriangleup A \\ so\,\,A\bigtriangleup B=B\bigtriangleup A $
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