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 372: 15

Answer

$For\,\,all\,\,sets\,\,A,\,B,\,\,and\,C\,\,\\ if\,\,A \cap C = B \cap C\,\,and\,A \cup C = B \cup C\,\,\\ then\,\,A = B.\\ this\,\,is\,\,true:\\ proof:\\ to\,\,prove\,\,this\,we\,must\,show\,that\,:\\ A\subseteq B\,\,\,and\,\,B\subseteq A \\ proof\,\,(A\subseteq B)\\ x\in A \Rightarrow x\in C \,or\, x\notin C \\ first\,\,case\,\,(x\in A\,\,and\,\,x\in C)\\ x\in A\,\,and\,\,x\in C\Rightarrow x\in A\cap C \\ \because A \cap C = B \cap C \\ \therefore x\in A\cap C\Rightarrow x\in B \cap C\Rightarrow x\in B \\ second\,\,case\,\,:(x\in A \,\,and\,\,x\notin C)\\ x\in A \,\,and\,\,x\notin C \Rightarrow x\in A\cup C \\ \because A \cup C = B \cup C \\ \therefore x\in B\cup C\Rightarrow x\in B \,\,(as\,\,x\notin C) \\ so\,\,in\,\,both\,\,cases\,\,:\\ x\in A \Rightarrow x\in B \\ \therefore A\subseteq B $ $proof\,\,(B\subseteq A)\\ x\in B \Rightarrow x\in C \,or\, x\notin C \\ first\,\,case\,\,(x\in B\,\,and\,\,x\in C)\\ x\in B\,\,and\,\,x\in C\Rightarrow x\in B\cap C \\ \because A \cap C = B \cap C \\ \therefore x\in B\cap C\Rightarrow x\in A \cap C\Rightarrow x\in A \\ second\,\,case\,\,:(x\in B \,\,and\,\,x\notin C)\\ x\in B \,\,and\,\,x\notin C \Rightarrow x\in B\cup C \\ \because A \cup C = B \cup C \\ \therefore x\in A\cup C\Rightarrow x\in A \,\,(as\,\,x\notin C) \\ so\,\,in\,\,both\,\,cases\,\,:\\ x\in B \Rightarrow x\in A \\ \therefore B\subseteq A \\ \because B\subseteq A\,\,and\,\,A\subseteq B \\ \therefore A=B $

Work Step by Step

$For\,\,all\,\,sets\,\,A,\,B,\,\,and\,C\,\,\\ if\,\,A \cap C = B \cap C\,\,and\,A \cup C = B \cup C\,\,\\ then\,\,A = B.\\ this\,\,is\,\,true:\\ proof:\\ to\,\,prove\,\,this\,we\,must\,show\,that\,:\\ A\subseteq B\,\,\,and\,\,B\subseteq A \\ proof\,\,(A\subseteq B)\\ x\in A \Rightarrow x\in C \,or\, x\notin C \\ first\,\,case\,\,(x\in A\,\,and\,\,x\in C)\\ x\in A\,\,and\,\,x\in C\Rightarrow x\in A\cap C \\ \because A \cap C = B \cap C \\ \therefore x\in A\cap C\Rightarrow x\in B \cap C\Rightarrow x\in B \\ second\,\,case\,\,:(x\in A \,\,and\,\,x\notin C)\\ x\in A \,\,and\,\,x\notin C \Rightarrow x\in A\cup C \\ \because A \cup C = B \cup C \\ \therefore x\in B\cup C\Rightarrow x\in B \,\,(as\,\,x\notin C) \\ so\,\,in\,\,both\,\,cases\,\,:\\ x\in A \Rightarrow x\in B \\ \therefore A\subseteq B $ $proof\,\,(B\subseteq A)\\ x\in B \Rightarrow x\in C \,or\, x\notin C \\ first\,\,case\,\,(x\in B\,\,and\,\,x\in C)\\ x\in B\,\,and\,\,x\in C\Rightarrow x\in B\cap C \\ \because A \cap C = B \cap C \\ \therefore x\in B\cap C\Rightarrow x\in A \cap C\Rightarrow x\in A \\ second\,\,case\,\,:(x\in B \,\,and\,\,x\notin C)\\ x\in B \,\,and\,\,x\notin C \Rightarrow x\in B\cup C \\ \because A \cup C = B \cup C \\ \therefore x\in A\cup C\Rightarrow x\in A \,\,(as\,\,x\notin C) \\ so\,\,in\,\,both\,\,cases\,\,:\\ x\in B \Rightarrow x\in A \\ \therefore B\subseteq A \\ \because B\subseteq A\,\,and\,\,A\subseteq B \\ \therefore A=B $
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