Question -
Answer -
(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
Firstly let us consider the LHS
(B ∩ C) = {x: x ∈ B and x ∈ C}
= {5, 6}
A ∪ (B ∩ C) = {x: x ∈ A or x ∈ (B ∩ C)}
= {1, 2, 4, 5, 6}
Now, RHS
(A ∪ B) = {x: x ∈ A or x ∈ B}
= {1, 2, 4, 5, 6}.
(A ∪ C) = {x: x ∈ A or x ∈ C}
= {1, 2, 4, 5, 6, 7}
(A ∪ B) ∩ (A ∪ C) = {x: x ∈ (A ∪ B) and x ∈ (A ∪ C)}
= {1, 2, 4, 5, 6}
∴ LHS = RHS
Hence Verified.
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Firstly let us consider the LHS
(B ∪ C) = {x: x ∈ B or x ∈ C}
= {2, 3, 4, 5, 6, 7}
(A ∩ (B ∪ C)) = {x: x ∈ A and x ∈ (B ∪ C)}
= {2, 4, 5}
Now, RHS
(A ∩ B) = {x: x ∈ A and x ∈ B}
= {2, 5}
(A ∩ C) = {x: x ∈ A and x ∈ C}
= {4, 5}
(A ∩ B) ∪ (A ∩ C) = {x: x ∈ (A∩B) and x ∈ (A∩C)}
= {2, 4, 5}
∴ LHS = RHS
Hence verified.
(iii) A ∩ (B – C) = (A ∩ B) – (A ∩ C)
B–C is defined as {x ∈ B: x ∉ C}
B = {2, 3, 5, 6}
C = {4, 5, 6, 7}
B–C = {2, 3}
Firstly let us consider the LHS
(A ∩ (B – C)) = {x: x ∈ A and x ∈ (B – C)}
= {2}
Now, RHS
(A ∩ B) = {x: x ∈ A and x ∈ B}
= {2, 5}
(A ∩ C) = {x: x ∈ A and x ∈ C}
= {4, 5}
(A ∩ B) – (A ∩ C) is defined as {x ∈ (A ∩ B): x ∉ (A ∩ C)}
= {2}
∴ LHS = RHS
Hence Verified.
(iv) A – (B ∪ C) = (A – B) ∩ (A – C)
Firstly let us consider the LHS
(B ∪ C) = {x: x ∈ B or x ∈ C}
= {2, 3, 4, 5, 6, 7}.
A – (B ∪ C) is defined as {x ∈ A: x ∉ (B ∪ C)}
A = {1, 2, 4, 5}
(B ∪ C) = {2, 3, 4, 5, 6, 7}
A – (B ∪ C) = {1}
Now, RHS
(A – B)
A – B is defined as {x ∈ A: x ∉ B}
A = {1, 2, 4, 5}
B = {2, 3, 5, 6}
A – B = {1, 4}
(A – C)
A – C is defined as {x ∈ A: x ∉ C}
A = {1, 2, 4, 5}
C = {4, 5, 6, 7}
A – C = {1, 2}
(A – B) ∩ (A – C) = {x: x ∈ (A – B) and x ∈ (A – C)}.
= {1}
∴ LHS = RHS
Hence verified.
(v) A – (B ∩ C) = (A – B) ∪ (A – C)
Firstly let us consider the LHS
(B ∩ C) = {x: x ∈ B and x ∈ C}
= {5, 6}
A – (B ∩ C) is defined as {x ∈ A: x ∉ (B ∩ C)}
A = {1, 2, 4, 5}
(B ∩ C) = {5, 6}
(A – (B ∩ C)) = {1, 2, 4}
Now, RHS
(A – B)
A – B is defined as {x ∈ A: x ∉ B}
A = {1, 2, 4, 5}
B = {2, 3, 5, 6}
A–B = {1, 4}
(A – C)
A – C is defined as {x ∈ A: x ∉ C}
A = {1, 2, 4, 5}
C = {4, 5, 6, 7}
A – C = {1, 2}
(A – B) ∪ (A – C) = {x: x ∈ (A – B) OR x ∈ (A – C)}.
= {1, 2, 4}
∴ LHS = RHS
Hence verified.
(vi) A ∩ (B △ C) = (A ∩ B) △ (A ∩ C)
A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}.
Firstly let us consider the LHS
A ∩ (B △ C)
B △ C = (B – C) ∪ (C – B) = {2, 3} ∪ {4, 7} = {2, 3, 4, 7}
A ∩ (B △ C) = {2, 4}
Now, RHS
A ∩ B = {2, 5}
A ∩ C = {4, 5}
(A ∩ B) △ (A ∩ C) = [(A ∩ B) – (A ∩ C)] ∪ [(A ∩ C) – (A ∩ B)]
= {2} ∪ {4}
= {2, 4}
∴ LHS = RHS
Hence, Verified.