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: 7

Answer

$(A\cap B)^{c}=A^{c}\cup B^{c} \\ to\,\,prove\,\,this\,\,we\,\,must\,\,prove\,\,\\ 1-(A\cap B)^{c}\subseteq A^{c}\cup B^{c} \\ 2-A^{c}\cup B^{c}\subseteq (A\cap B)^{c} \\ proof\,\,of\,\,1:\\ x\in (A\cap B)^{c}\Rightarrow x\notin(A\cap B)\\ \Rightarrow x\notin A \,or\,x\notin B \\ \Rightarrow x\in A^{c} \,\,or\,\,x\in B^{c} \\ \Rightarrow x\in A^{c}\cup B^{c}\\ \because x\in (A\cap B)^{c}\Rightarrow x\in A^{c}\cup B^{c}\\ \therefore (A\cap B)^{c}\subseteq A^{c}\cup B^{c}\\ $ $proof\,\,of\,\,2:\\ x\in A^{c}\cup B^{c} \Rightarrow x\in A^{c}\,or\,x\in B^{c}\\ \Rightarrow x\notin A \,or\,x\notin B \\ \Rightarrow x\notin (A\cap B)\\ \Rightarrow x\in (A\cap B)^{c} \\ \because x\in A^{c}\cup B^{c}\Rightarrow x\in (A\cap B)^{c}\\ \therefore A^{c}\cup B^{c}\subseteq (A\cap B)^{c}\\ from\,1\,,2\,\\ (A\cap B)^{c}=A^{c}\cup B^{c} \\ $

Work Step by Step

$(A\cap B)^{c}=A^{c}\cup B^{c} \\ to\,\,prove\,\,this\,\,we\,\,must\,\,prove\,\,\\ 1-(A\cap B)^{c}\subseteq A^{c}\cup B^{c} \\ 2-A^{c}\cup B^{c}\subseteq (A\cap B)^{c} \\ proof\,\,of\,\,1:\\ x\in (A\cap B)^{c}\Rightarrow x\notin(A\cap B)\\ \Rightarrow x\notin A \,or\,x\notin B \\ \Rightarrow x\in A^{c} \,\,or\,\,x\in B^{c} \\ \Rightarrow x\in A^{c}\cup B^{c}\\ \because x\in (A\cap B)^{c}\Rightarrow x\in A^{c}\cup B^{c}\\ \therefore (A\cap B)^{c}\subseteq A^{c}\cup B^{c}\\ $ $proof\,\,of\,\,2:\\ x\in A^{c}\cup B^{c} \Rightarrow x\in A^{c}\,or\,x\in B^{c}\\ \Rightarrow x\notin A \,or\,x\notin B \\ \Rightarrow x\notin (A\cap B)\\ \Rightarrow x\in (A\cap B)^{c} \\ \because x\in A^{c}\cup B^{c}\Rightarrow x\in (A\cap B)^{c}\\ \therefore A^{c}\cup B^{c}\subseteq (A\cap B)^{c}\\ from\,1\,,2\,\\ (A\cap B)^{c}=A^{c}\cup B^{c} \\ $
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