|A U B U C| = A + B + C |A B| |B C| |A C| + |A B C| – ∩ – ∩ – ∩ ∩ ∩ = 64 + 94 + 58 26 22 28 + 14 – – – = 154 i. |A U B U C| = U |A U B U C| – = 260 154 = 106 ii.|B1| = |B| - |B C| - |B A| + |A B C| – ∩ ∩ ∩ ∩ = 94 22 26 + 14 – – = 60 4. For any two sets A and B, prove the following (4m) jun 2013 A (A B) = A B – – ∩ Solution: A (A B) = A (A B1) [i.e A B = A B1] – – – ∩ – ∩ = A (A B1)1 [i.e A B = A B1] ∩ ∩ – ∩ = A (A1 U B) [Demorgan’s Law] ∩ = (A A1) U (A B) ∩ ∩ = ø U (A B) ∩ = A ∩B 5. State and prove De Morgan’s law of set theory. (6m) jun 2013 – Solution: The complement of union of 2 sets is equal to the intersection of complement of the sets. A U B = A B ∩ Consider RHS, A B = { x / x A and x B } ε ε = { x / x ¢ A and x ¢ B } = { x / x ¢ (A U B) } = A U B The union of compliment of 2 sets is equal to the compliment of intersection of the • 2 sets. A U B = A B ∩
|A U B U C| = A + B + C – |A ∩ B| – |B ∩ C| – |A ∩ C| + |A ∩ B ∩ C| = 64 + 94 + 58 – 26 – 22 – 28 + 14 = 154 i. |A U B U C| = U – |A U B U C| = 260 – 154 = 106 ii.|B1| = |B| - |B ∩ C| - |B ∩ A| + |A ∩ B ∩ C| = 94 – 22 – 26 + 14 = 60 4. For any two sets A and B, prove the following (4m) jun 2013 A – (A – B) = A ∩ B Solution: A – (A – B) = A – (A ∩ B1) [i.e A – B = A ∩ B1] = A ∩ (A ∩ B1)1 [i.e A – B = A ∩ B1] = A ∩ (A1 U B) [Demorgan’s Law] = (A ∩ A1) U (A ∩ B) = ø U (A ∩ B) = A ∩B 5. State and prove De – Morgan’s law of set theory. (6m) jun 2013 Solution: The complement of union of 2 sets is equal to the intersection of complement of the sets. A U B = A ∩ B Consider RHS, A B = { x / x ε A and x ε B } = { x / x ¢ A and x ¢ B } = { x / x ¢ (A U B) } = A U B • The union of compliment of 2 sets is equal to the compliment of intersection of the 2 sets. A U B = A ∩ B