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Exercise 5.1 · Q18

Q.If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8} and universal set X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then verify: A ∩ (BΔC) = (A∩B) Δ (A∩C).

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B−C={3,4,5,6}−{4,5,6,7,8}={3}B-C = \{3,4,5,6\}-\{4,5,6,7,8\} = \{3\}. C−B={4,5,6,7,8}−{3,4,5,6}={7,8}C-B = \{4,5,6,7,8\}-\{3,4,5,6\} = \{7,8\}. So BΔC=(B−C)∪(C−B)={3}∪{7,8}={3,7,8}B\Delta C = (B-C)\cup(C-B) = \{3\}\cup\{7,8\} = \{3,7,8\}. Left side: A∩(BΔC)={1,2,3,4}∩{3,7,8}={3}A\cap(B\Delta C) = \{1,2,3,4\}\cap\{3,7,8\} = \{3\}. Right side: A∩B={3,4}A\cap B = \{3,4\} and A∩C={1,2,3,4}∩{4,5,6,7,8}={4}A\cap C = \{1,2,3,4\}\cap\{4,5,6,7,8\} = \{4\}; their symmetric difference is ${3,4}\Delta{4} = ({3,4}-{4} …

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