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 6.2 - Page 365: 8

Answer

$for\,\,all\,\,sets\,\,A,B\,\,and\,\,C \\ (A\cap B)\cup (A\cap B^{c})=A \\ to\,\,prove\,\,this\,\,we\,must\,prove\,that\,\\ 1-(A\cap B)\cup (A\cap B^{c})\subseteq A \\ 2-A\subseteq (A\cap B)\cup (A\cap B^{c}) \\ proof\,of\,\,1:\\ x\in(A\cap B)\cup (A\cap B^{c}) \\ by\,\,def.\,\,of\,\,union\,\,of\,\,sets \\ x\in (A\cap B)\,or\,x\in (A\cap B^{c}) \\ and\,by\,\,def.\,\,of\,\,inter\! section \\ (x\in A\,and\,x\in B)\,or\,(x\in A\,and\,x\in B^{c})\\ in\,\,both\,cases\,x\in A \\ \because x\in(A\cap B)\cup (A\cap B^{c}) \Rightarrow x\in A\\ \therefore (A\cap B)\cup (A\cap B^{c})\subseteq A \\$ $proof\,of\,\,2:\\ x\in A \,\Rightarrow x\in B\,\,or\,x\notin B \\ first\,case: (x\in A\,and\,x\in B)\\ \therefore x\in A\cap B \\ by\,\,def.\,\,of\,\,union \\ x\in (A\cap B)\cup (A\cap B^{c})\\ second\,case\,:(x\in A\,and\,x\notin B)\\ by\,\,def.\,\,of\,\,complement \\ x\in A\,and\,x\in B^{c} \\ \therefore x\in A\cap B^{c} \\ by\,\,def.\,\,of\,\,union \\ x\in (A\cap B^{c})\cup (A\cap B)\\ \because x\in A\Rightarrow x\in (A\cap B^{c})\cup (A\cap B)\\ (in\,both\,cases)\\ \therefore A\subseteq(A\cap B)\cup (A\cap B^{c}) $ $from\,\,1\,\,and\,\,2 \\ (A\cap B)\cup (A\cap B^{c})=A $

Work Step by Step

$for\,\,all\,\,sets\,\,A,B\,\,and\,\,C \\ (A\cap B)\cup (A\cap B^{c})=A \\ to\,\,prove\,\,this\,\,we\,must\,prove\,that\,\\ 1-(A\cap B)\cup (A\cap B^{c})\subseteq A \\ 2-A\subseteq (A\cap B)\cup (A\cap B^{c}) \\ proof\,of\,\,1:\\ x\in(A\cap B)\cup (A\cap B^{c}) \\ by\,\,def.\,\,of\,\,union\,\,of\,\,sets \\ x\in (A\cap B)\,or\,x\in (A\cap B^{c}) \\ and\,by\,\,def.\,\,of\,\,inter\! section \\ (x\in A\,and\,x\in B)\,or\,(x\in A\,and\,x\in B^{c})\\ in\,\,both\,cases\,x\in A \\ \because x\in(A\cap B)\cup (A\cap B^{c}) \Rightarrow x\in A\\ \therefore (A\cap B)\cup (A\cap B^{c})\subseteq A \\$ $proof\,of\,\,2:\\ x\in A \,\Rightarrow x\in B\,\,or\,x\notin B \\ first\,case: (x\in A\,and\,x\in B)\\ \therefore x\in A\cap B \\ by\,\,def.\,\,of\,\,union \\ x\in (A\cap B)\cup (A\cap B^{c})\\ second\,case\,:(x\in A\,and\,x\notin B)\\ by\,\,def.\,\,of\,\,complement \\ x\in A\,and\,x\in B^{c} \\ \therefore x\in A\cap B^{c} \\ by\,\,def.\,\,of\,\,union \\ x\in (A\cap B^{c})\cup (A\cap B)\\ \because x\in A\Rightarrow x\in (A\cap B^{c})\cup (A\cap B)\\ (in\,both\,cases)\\ \therefore A\subseteq(A\cap B)\cup (A\cap B^{c}) $ $from\,\,1\,\,and\,\,2 \\ (A\cap B)\cup (A\cap B^{c})=A $
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