Problems on Complement of a Set
Solved problems on complement
of a set are given below to get a fair idea how to find the complement
of two or more sets.
We know, when U be the universal set and A is a subset of U. Then the complement of A is the set all elements of U which are not the elements of A.
to know more about the complement of a set.
Solved problems on complement of a set:
1. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}.
(i) Find A’
(ii) Find B’
Solution:
(i) A’ = U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4}
= {5, 6, 7, 8, 9}
(ii) B’
= U – B
=
{1, 2, 3, 4, 5, 6, 7, 8, 9} – {2,
4, 6, 8}
= {1, 3, 5, 7, 9}
More worked-out problems on complement of a set.
2. Let A = {3, 5, 7}, B = {2, 3, 4, 6}
and C = {2, 3, 4, 5, 6, 7, 8}
(i) Verify (A ∩ B)’ = A’ ∪ B’
(ii) Verify (A ∪ B)’ = A’ ∩ B’
Solution:
(i) (A ∩ B)’ = A’ ∪ B’
L.H.S. = (A ∩ B)’
A ∩ B = {3}
(A ∩ B)’ = {2, 4, 5, 6, 7, 8} ……………….. (1)
R.H.S. = A’ ∪ B’
A’ = {5, 7, 8}
B’ = {2, 4, 6}
A’∪B’
= {2, 4, 5, 6, 7, 8} ……………….. (2)
From (1) and (2), we conclude that;
(A ∩ B)’ = (A’ ∪ B’)
(ii) (A ∪ B)’ = A’ ∩ B’
L.H.S. = (A ∪ B)’
A∪B
= {2, 3, 4, 5, 6, 7}
(A ∪
B)’ = {8} ……………….. (1)
R.H.S. = A’ ∩ B’
A’ = {2, 4, 6, 8}
B’ = {5, 7, 8}
A’ ∩ B’ = {8} ……………….. (2)
From (1) and (2), we conclude that;
(A ∪ B)’ = A’ ∩ B’