Question 1 :
By taking suitable sets A,B,C, verify the following results:
(i) A × (B ∩ C) = (A × B) ∩ (A × C).
Solution :
Let A = {1, 2, 3}, B = {1, 2} and C = {2, 3 4}
L.H.S
(B n C) = {2}
A × (B ∩ C) = {1, 2, 3} × {2}
A × (B ∩ C) = { (1, 2)(2, 2) (3, 2) } ------(1)
R.H.S
(A × B) = {1, 2, 3} × {1, 2}
(A × B) = {(1, 1) (1, 2) (2, 1) (2, 2)(3, 1) (3, 2)}
(A × C) = {1, 2, 3} × {2, 3 4}
(A × C)
= {(1, 2) (1, 3) (1, 4) (2, 2) (2, 3) (2, 4) (3, 2) (3, 3) (3, 4)}
(A × B) ∩ (A × C) = { (1, 2)(2, 2) (3, 2) } ------(2)
(1) = (2)
Hence proved.
(ii) A × (B ∪ C) = (A × B) ∪ (A × C)
Solution :
Let A = {1, 2, 3}, B = {1, 2} and C = {2, 3 4}
L.H.S
(B U C) = {1, 2, 3, 4}
A × (B ∪ C) = {1, 2, 3} × {1, 2, 3, 4}
A × (B ∪ C) = {(1, 1) (1, 2) (1, 3) (1, 4)(2, 1) (2, 2) (2, 3) (2, 4)(3, 1) (3, 2) (3, 3) (3, 4) ------(1)
R.H.S
(A × B) = {1, 2, 3} × {1, 2}
(A × B) = {(1, 1) (1, 2) (2, 1) (2, 2)(3, 1) (3, 2)}
(A × C) = {1, 2, 3} × {2, 3 4}
(A × C)
= {(1, 2) (1, 3) (1, 4) (2, 2) (2, 3) (2, 4) (3, 2) (3, 3) (3, 4)}
(A × B) U (A × C) = {(1, 1) (1, 2) (1, 3) (1, 4)(2, 1) (2, 2) (2, 3) (2, 4)(3, 1) (3, 2) (3, 3) (3, 4) ------(2)
(1) = (2)
Hence proved.
(iii) (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A)
Solution :
Let A = {1, 2, 3}, B = {1, 2} and C = {2, 3 4}
L.H.S
(A × B) = {1, 2, 3} × {1, 2}
= {(1, 1) (1, 2) (2, 1) (2, 2) (3, 1) (3, 2)}
(B × A) = {1, 2} × {1, 2, 3}
= {(1, 1) (1, 2) (1, 3) (2, 1) (2, 2) (2, 3)}
(A × B) ∩ (B × A) = { (1, 1) (1, 2)(2, 1) (2, 2) } -----(1)
R.H.S
(A ∩ B) = {1, 2, 3} ∩ {1, 2} = {1, 2}
(B ∩ A) = {1, 2}
(A ∩ B) × (B ∩ A) = { (1, 1) (1, 2)(2, 1) (2, 2) } -----(2)
(1) = (2)
Hence proved.
(iv) C − (B − A) = (C ∩ A) ∪ (C ∩ B').
Solution :
Let U = {1, 2, 3, 4, 5}
A = {1, 2, 3}, B = {1, 2} and C = {2, 3, 4}
L.H.S
B' = {3, 4, 5}
(B - A) = { }
C − (B − A) = {2, 3, 4} ---(1)
(C ∩ A) = {2, 3 4} ∩ {1, 2, 3} = {2, 3}
(C ∩ B') = {2, 3, 4}
(C ∩ A) ∪ (C ∩ B') = {2, 3, 4} ---(2)
Hence proved.
(v) (B − A) ∩ C = (B ∩ C) − A = B ∩ (C − A).
Solution :
A = {1, 2, 3}, B = {1, 2} and C = {2, 3, 4}
Part 1 :
(B − A) = { }
(B − A) ∩ C = { } ------(1)
Part 2 :
(B ∩ C) = {2}
(B ∩ C) − A = { } ------(2)
Part 3 :
(C − A) = {4}
B ∩ (C − A) = { } ------(3)
(1) = (2) = (3)
Hence proved.
(vi) (B − A) ∪ C = (B ∪ C) − (A − C)
Solution :
A = {1, 2, 3}, B = {1, 2} and C = {2, 3, 4}
L.H.S
(B − A) = {3}
(B − A) ∪ C = {2, 3, 4} ------(1)
R.H.S
(B ∪ C) = {1, 2, 3, 4}
(A-C) = {1}
(B ∪ C) − (A − C) = {2, 3, 4} ------(2)
(1) = (2)
Hence proved.
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