Problems on Operation on Sets
Solved problems on operation
on sets are given below to get a fair idea how to find the union and
intersection of two or more sets.
We know, the union of sets is a set which contains all the elements in those sets and intersection of sets is a set which contains all the elements that are common in those sets.
to know more about the two basic operations on sets.
Solved problems on operation on sets:
1. If A = {1, 3, 5}, B = {3, 5, 6} and C = {1, 3, 7}
(i) Verify that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
(ii) Verify A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Solution:
(i) A ∪ (B ∩ C) = (A ∪
B) ∩ (A ∪ C)
L.H.S. = A ∪ (B ∩ C)
B ∩ C = {3}
A ∪
(B ∩ C) = {1, 3, 5} ∪ {3} = {1, 3, 5} ……………….. (1)
R.H.S. = (A ∪ B) ∩ (A ∪ C)
A ∪
B = {1, 3, 5, 6}
A ∪
C = {1, 3, 5, 7}
(A ∪
B) ∩ (A ∪ C) = {1, 3, 5, 6} ∩ {1, 3, 5, 7} = {1, 3, 5}
……………….. (2)
From (1) and (2), we conclude that;
A ∪
(B ∩ C) = A ∪ B ∩ (A ∪ C) [verified]
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪
(A ∩ C)
L.H.S. = A ∩ (B ∪ C)
B ∪
C = {1, 3, 5, 6, 7}
A ∩ (B ∪ C) = {1, 3, 5} ∩ {1, 3, 5, 6, 7} = {1, 3, 5}
……………….. (1)
R.H.S. = (A ∩ B) ∪ (A ∩ C)
A ∩ B = {3, 5}
A ∩ C = {1, 3}
(A ∩ B) ∪ (A ∩ C) = {3, 5} ∪ {1, 3} = {1, 3, 5}
……………….. (2)
From (1) and (2), we conclude that;
A ∩ (B ⋃ C) = (A ∩ B) ⋃
(A ∩ C) [verified]
More worked-out problems on operation
on sets to find the union and
intersection of three sets.
2. Let A = {a, b, d, e}, B = {b, c, e,
f} and C = {d, e, f, g}
(i) Verify A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
(ii) Verify A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
Solution:
(i) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
L.H.S. = A ∩ (B ∪ C)
B ∪
C = {b, c, d, e, f, g}
A ∩ (B ∪ C) = {b, d, e} ……………….. (1)
R.H.S. = (A ∩ B) ∪ (A ∩ C)
A ∩ B = {b, e}
A ∩ C = {d, e}
(A ∩ B) ∪ (A ∩ C) = {b, d, e} ……………….. (2)
From (1) and (2), we conclude that;
A ∩ (B ⋃ C) = (A ∩ B) ⋃
(A ∩ C) [verified]
(ii) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
L.H.S. = A ∪ (B ∩ C)
B ∩ C = {e, f}
A ∪
(B ∩ C) = {a, b, d, e, f} ……………….. (1)
R.H.S. = (A ∪ B) ∩ (A ∪ C)
A∪B
= {a, b, c, d, e, f}
A∪C
= {a, b, d, e, f, g}
(A ∪
B) ∩ (A ∪ C) = {a, b, d, e, f} ……………….. (2)
From (1) and (2), we conclude that;
A ∪
(B ∩ C) = A ∪ B ∩ (A ∪ C) [verified]