Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 6 - Review - Review Exercises: 5

Answer

$A\cup B^{\prime}=\{a,b,d\}$ $A\times B^{\prime}=\{(a,a), (a,d), (b,a), (b,d)\}$

Work Step by Step

If $S$ is the universal set and $A\subseteq S$, then $A^{\prime}$ is the complement of A (in $S$), the set of all elements of $S$ not in $A$. $A^{\prime}=\{x\in S|x\not\in A\}$ For an element to be in $A\cup B$, it must be in $A$ or in $B$. For an element to be in $A\cap B$, it must be in $A$ and in $B$. A$\times B=\{(a, b)|a\in$A and $b\in B\}$ ------- $B^{\prime}=\{a,d\}$ $A\cup B^{\prime}=\{a,b,d\}$ $A=\{a,b\}, B^{\prime}=\{a,d\},$ $A\times B^{\prime}=\{(a,a), (a,d), (b,a), (b,d)\}$
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