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De Morgan\'s low |
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Answer» Thanks bhai ?? The complement of the union of two sets is equal to the intersection of their complements and the complement of the intersection of two sets is equal to the union of their complements. These are called\xa0De Morgan’s laws.For any two finite sets A and B;(i)\xa0(A U B)\' = A\' ∩ B\' (which is a De Morgan\'s law of union).(ii)\xa0(A ∩ B)\' = A\' U B\' (which is a De Morgan\'s law of intersection).\xa0Proof of De Morgan’s law:\xa0(A U B)\' = A\' ∩ B\'\xa0Let P = (A U B)\' and Q = A\' ∩ B\'Let x be an arbitrary element of P then x ∈ P ⇒ x ∈ (A U B)\'⇒ x ∉ (A U B)⇒ x ∉ A and x ∉ B⇒ x ∈ A\' and x ∈ B\'⇒ x ∈ A\' ∩ B\'⇒ x ∈ QTherefore, P ⊂ Q …………….. (i)Again, let y be an arbitrary element of Q then y ∈ Q ⇒ y ∈ A\' ∩ B\'⇒ y ∈ A\' and y ∈ B\'⇒ y ∉ A and y ∉ B⇒ y ∉ (A U B)⇒ y ∈ (A U B)\'⇒ y ∈ PTherefore, Q ⊂ P …………….. (ii)Now combine (i) and (ii) we get; P = Q i.e. (A U B)\' = A\' ∩ B\'\xa0Proof of De Morgan’s law:\xa0(A ∩ B)\' = A\' U B\'Let M = (A ∩ B)\' and N = A\' U B\'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)\'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B⇒ x ∈ A\' or x ∈ B\'⇒ x ∈ A\' U B\'⇒ x ∈ NTherefore, M ⊂ N …………….. (i)Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A\' U B\'⇒ y ∈ A\' or y ∈ B\'⇒ y ∉ A or y ∉ B⇒ y ∉ (A ∩ B)⇒ y ∈ (A ∩ B)\'⇒ y ∈ MTherefore, N ⊂ M …………….. (ii)Now combine (i) and (ii) we get; M = N i.e. (A ∩ B)\' = A\' U B\' |
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