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Exercise: Complement Properties · Q22

Q.If U={1,2,…,12}U = \{1, 2, \dots, 12\} and A={2,4,6,8,10,12}A = \{2, 4, 6, 8, 10, 12\}, find A′A' and verify that (A′)′=A(A')' = A.

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U={1,…,12}U=\{1,\dots,12\}, A={2,4,6,8,10,12}A=\{2,4,6,8,10,12\}. A′A' contains every element of UU not in AA, i.e. the odd numbers 1,3,5,7,9,111,3,5,7,9,11. Now taking the complement again: (A′)′=U−A′=U−{1,3,5,7,9,11}={2,4,6,8,10,12}(A')' = U - A' = U - \{1,3,5,7,9,11\} = \{2,4,6,8,10,12\}, which is exactly AA again — confirming the law (A′)′=A(A')'=A. [!ANSWER] A′={1,3,5,7,9,11}A'=\{1,3,5,7,9,11\} and (A′)′=A={2,4,6,8,10,12}(A')'=A=\{2,4,6,8,10,12\}.

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