Answers:

(i) 

LHS,

A ∪ B = {x: x ∈ A or x ∈ B}

A ∪ B = {2, 3, 5, 7, 9}

(A∪B)’ = Complement of (A∪B) with U.

(A∪B)’ = U – (A∪B)’

U – (A∪B)’ = {x ∈ U: x ∉ (A∪B)’}

U = {2, 3, 5, 7, 9}

(A∪B)’ = {2, 3, 5, 7, 9}

U – (A∪B)’ = ϕ

RHS,

A’ = Complement of A with U.

A’ = U – A

(U – A) = {x ∈ U: x ∉ A}

U = {2, 3, 5, 7, 9}

A = {3, 7}

A’ = U – A

A’ = {2, 5, 9}

B’ = Complement of B with  U.

B’ = U – B

(U – B) = {x ∈ U: x ∉ B}

U = {2, 3, 5, 7, 9}

B = {2, 5, 7, 9}

B’ = U – B = {3}

A’ ∩ B’ = {x: x ∈ A’ and x ∈ C’}.

A’ ∩ B’ = ϕ

∴ LHS = RHS

Thus, verified.

(ii) 

LHS,

(A ∩ B) = {x: x ∈ A and x ∈ B}.

(A ∩ B) = {7}

(A∩B)’ = Complement of (A ∩ B) with U.

(A∩B)’ = U – (A ∩ B)

U – (A ∩ B) = {x ∈ U: x ∉ (A ∩ B)’}

U = {2, 3, 5, 7, 9}

(A ∩ B) = {7}

U – (A ∩ B) = {2, 3, 5, 9}

(A ∩ B)’ = {2, 3, 5, 9}

RHS,

A’ = Complement of A with U.

A’ = U – A

(U – A) = {x ∈ U: x ∉ A}

U = {2, 3, 5, 7, 9}

A = {3, 7}

A’ = U – A

A’ = {2, 5, 9}

B’ = Complement of B with U.

B’ = U – B

(U – B) = {x ∈ U: x ∉ B}

U = {2, 3, 5, 7, 9}

B = {2, 5, 7, 9}

B’ = U – B = {3}

A’ ∪ B’ = {x: x ∈ A or x ∈ B}

A’ ∪ B’ = {2, 3, 5, 9}

∴ LHS = RHS

Thus, verified.