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’