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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.
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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 β© B)' = A' βͺ B'
(ii) Verify (A βͺ B)' = A' β© B'
Solution:
(i) (A β© B)' = A' βͺ B'
L.H.S. = (A β© B)'
A β© B = {3}
(A β© B)' = {2, 4, 5, 6, 7, 8} β¦β¦β¦β¦β¦β¦.. (1)
R.H.S. = A' βͺ B'
Aβ = {5, 7, 8}
Bβ = {2, 4, 6}
AββͺBβ
= {2, 4, 5, 6, 7, 8} β¦β¦β¦β¦β¦β¦.. (2)
From (1) and (2), we conclude that;
(A β© B)' = (A' βͺ B')
(ii) (A βͺ B)' = A' β© B'
L.H.S. = (A βͺ B)'
AβͺB
= {2, 3, 4, 5, 6, 7}
(A βͺ
B)' = {8} β¦β¦β¦β¦β¦β¦.. (1)
R.H.S. = A' β© B'
A' = {2, 4, 6, 8}
B' = {5, 7, 8}
A' β© B' = {8} β¦β¦β¦β¦β¦β¦.. (2)
From (1) and (2), we conclude that;
(A βͺ B)' = A' β© B'
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