= 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 Sol: 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) [Distributive Law] =øU (A ∩ ∩ B) [i.e. A A1 = ø] ∩ ∩ = 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 ∩ Consider LHS, A U B = { x / { x / x A or x B } ε ε = { x / x ¢ A or x ¢ B } = { x / x ¢ (A B) } ∩
= 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 Sol: 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) [Distributive Law] =øU (A ∩ B) [i.e. A ∩ A1 = ø] = 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 Consider LHS, A U B = { x / { x / x ε A or x ε B } = { x / x ¢ A or x ¢ B } = { x / x ¢ (A ∩ B) }