For any three sets A, B and C, (A – B) ∩ (C – B) is equal to __________ Solution \[ (A – B) \cap (C – B) \] \[ = (A \cap B’) \cap (C \cap B’) \] \[ = A \cap C \cap B’ \] \[ = (A \cap C) – B \] […]

For any three sets A, B and C, (A – B) – (B – C) is equal to __________ Solution \[ (A – B) – (B – C) \] \[ = (A \cap B’) – (B \cap C’) \] \[ = (A \cap B’) \cap (B \cap C’)’ \] \[ = (A \cap B’) \cap

For any two sets A and B, [B’ ∪ (B’ – A)]’ is equal to……. Solution \[ [B’ \cup (B’ – A)]’ \] Using \[ X – Y = X \cap Y’ \] \[ = [B’ \cup (B’ \cap A’)]’ \] Using absorption law, \[ B’ \cup (B’ \cap A’) = B’ \] Therefore, \[

If A and B are two sets, then ((A’ ∪ B’) – A)’ is equal to………. Solution \[ ((A’ \cup B’) – A)’ \] Using \[ X – Y = X \cap Y’ \] we get \[ = ((A’ \cup B’) \cap A’)’ \] Since \[ A’ \cap (A’ \cup B’) = A’ \] therefore,

If A and B are two sets, then A ∩ (A ∪ B)’ is equal to ___________ Solution \[ A \cap (A \cup B)’ \] By De Morgan’s law, \[ = A \cap (A’ \cap B’) \] Using associative law, \[ = (A \cap A’) \cap B’ \] \[ = \Phi \cap B’ \] \[

If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 5}, B = {2, 4, 6, 7} and C = {2, 3, 4, 8}, Then (i) \((B \cup C)’\) = …. (ii) \((C – A)’\) = …. Solution \[ U = \{1,2,3,4,5,6,7,8,9,10\} \] \[ B = \{2,4,6,7\}

Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8} then the universal set of all the three sets A, B and C can be…. Solution \[ A = \{1,3,5\} \] \[ B = \{2,4,6\} \] \[ C = \{0,2,4,6,8\} \] The universal set

For all sets A and B, B – (A ∩ B) is equal to…… Solution \[ B – (A \cap B) \] \[ = B \cap (A \cap B)’ \] \[ = B \cap (A’ \cup B’) \] \[ = (B \cap A’) \cup (B \cap B’) \] \[ = B \cap A’ \] \[

For all sets A and B, A – (A ∩ B) is equal to ….. Solution \[ A – (A \cap B) \] \[ = A \cap (A \cap B)’ \] \[ = A \cap (A’ \cup B’) \] \[ = (A \cap A’) \cup (A \cap B’) \] \[ = \Phi \cup (A \cap

When A = Φ, then the number of elements in P(P(A)) is ____________ Given: \[ A = \Phi \] Since the empty set has no elements, \[ n(A)=0 \] Therefore, \[ n(P(A)) = 2^0 = 1 \] Now, \(P(A)\) has 1 element. So the number of elements in \(P(P(A))\) is: \[ n(P(P(A))) = 2^1 =