Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} be the universal set and let A = {2, 3, 4, 5, 6} and B = {3, 5, 7, 8} be its subsets.

Find:

(i) A'

(ii) B'

(iii) A' ∩ B'

(iv) A' ∪ B'

Verify that:

(v) (A ∪ B)' = (A' ∩ B')

(vi) (A ∩ B)' = (A' ∪ B')

Given:

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {2, 3, 4, 5, 6}

B = {3, 5, 7, 8}

(i) A'

A' = Elements in U which are not in A.

A' = U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {2, 3, 4, 5, 6} = {1, 7, 8, 9, 10}

∴ A' = {1, 7, 8, 9, 10}

(ii) B'

B' = Elements in U which are not in B.

B' = U - B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 5, 7, 8} = {1, 2, 4, 6, 9, 10}

∴ B' = {1, 2, 4, 6, 9, 10}

(iii) A' ∩ B'

We have:

A' = {1, 7, 8, 9, 10}

B' = {1, 2, 4, 6, 9, 10}

A' ∩ B' = {1, 7, 8, 9, 10} ∩ {1, 2, 4, 6, 9, 10} = {1, 9, 10}

∴ A' ∩ B' = {1, 9, 10}

(iv) A' ∪ B'

We have:

A' = {1, 7, 8, 9, 10}

B' = {1, 2, 4, 6, 9, 10}

A' ∪ B' = {1, 7, 8, 9, 10} ∪ {1, 2, 4, 6, 9, 10} = {1, 2, 4, 6, 7, 8, 9, 10}

∴ A' ∪ B' = {1, 2, 4, 6, 7, 8, 9, 10}

Verify

(v) (A ∪ B)' = (A' ∩ B')

We have:

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {2, 3, 4, 5, 6}

B = {3, 5, 7, 8}

A' = {1, 7, 8, 9, 10}

B' = {1, 2, 4, 6, 9, 10}

First let us find A ∪ B:

A ∪ B = {2, 3, 4, 5, 6} ∪ {3, 5, 7, 8} = {2, 3, 4, 5, 6, 7, 8}

(A ∪ B)' = Elements in U which are not in (A ∪ B).

LHS = (A ∪ B)' = U - (A ∪ B)

LHS = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {2, 3, 4, 5, 6, 7, 8}

LHS = {1, 9, 10}

RHS = (A' ∩ B') = {1, 7, 8, 9, 10} ∩ {1, 2, 4, 6, 9, 10}

RHS = {1, 9, 10}

Since LHS = RHS,

∴ The statement (A ∪ B)' = (A' ∩ B') is verified.

(vi) (A ∩ B)' = (A' ∪ B')

Given:

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {2, 3, 4, 5, 6}

B = {3, 5, 7, 8}

A' = {1, 7, 8, 9, 10}

B' = {1, 2, 4, 6, 9, 10}

First let us find A ∩ B:

A ∩ B = {2, 3, 4, 5, 6} ∩ {3, 5, 7, 8} = {3, 5}

(A ∩ B)' = Elements in U which are not in (A ∩ B).

LHS = (A ∩ B)' = U - (A ∩ B)

LHS = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 5}

LHS = {1, 2, 4, 6, 7, 8, 9, 10}

RHS = (A' ∪ B') = {1, 7, 8, 9, 10} ∪ {1, 2, 4, 6, 9, 10}

RHS = {1, 2, 4, 6, 7, 8, 9, 10}

Since LHS = RHS,

∴ The statement (A ∩ B)' = (A' ∪ B') is verified.