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1. |
Prove that A ∩ (A∪B)’ = ϕ |
Answer» LHS = A ∩ (A∪B)’ Using De-Morgan’s law (A∪B)’ = (A’ ∩ B’) ⇒ LHS = A ∩ (A’ ∩ B’) ⇒ LHS = (A ∩ A’) ∩ (A ∩ B’) We know that A ∩ A’ = ϕ ⇒ LHS = ϕ ∩ (A ∩ B’) We know that intersection of null set with any set is null set only Let (A ∩ B’) be any set X hence ⇒ LHS = ϕ ∩ X ⇒ LHS = ϕ ⇒ LHS = RHS Hence proved |
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