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: 50

Answer

$A\bigtriangleup A = (A - A) \cup (A - A) \\ but(A-A=\varnothing \,\,\,by\,\,def.\,\,of\,\,set\,\,dif\! ference\,\,)\\ A\bigtriangleup A = (A - A) \cup (A - A) =\varnothing \cup \varnothing =\varnothing \\ \therefore A\bigtriangleup A =\varnothing $

Work Step by Step

$A\bigtriangleup A = (A - A) \cup (A - A) \\ but(A-A=\varnothing \,\,\,by\,\,def.\,\,of\,\,set\,\,dif\! ference\,\,)\\ A\bigtriangleup A = (A - A) \cup (A - A) =\varnothing \cup \varnothing =\varnothing \\ \therefore A\bigtriangleup A =\varnothing $
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