Question -
Answer -
(i) (A ∪ B)’ = A’ ∩ B’
Firstly let us consider the LHS
A ∪ B = {x: x ∈ A or x ∈ B}
= {2, 3, 5, 7, 9}
(A∪B)’ means Complement of (A∪B) with respect to universal set U.
So, (A∪B)’ = U – (A∪B)’
U – (A∪B)’ is defined as {x ∈ U: x ∉ (A∪B)’}
U = {2, 3, 5, 7, 9}
(A∪B)’ = {2, 3, 5, 7, 9}
U – (A∪B)’ = ϕ
Now, RHS
A’ means Complement of A with respect to universal set U.
So, A’ = U – A
(U – A) is defined as {x ∈ U: x ∉ A}
U = {2, 3, 5, 7, 9}
A = {3, 7}
A’ = U – A = {2, 5, 9}
B’ means Complement of B with respect to universal set U.
So, B’ = U – B
(U – B) is defined as {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’}.
= ϕ
∴ LHS = RHS
Hence verified.
(ii) (A ∩ B)’ = A’ ∪ B’
Firstly let us consider the LHS
(A ∩ B)’
(A ∩ B) = {x: x ∈ A and x ∈ B}.
= {7}
(A∩B)’ means Complement of (A ∩ B) with respect to universal set U.
So, (A∩B)’ = U – (A ∩ B)
U – (A ∩ B) is defined as {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}
Now, RHS
A’ means Complement of A with respect to universal set U.
So, A’ = U – A
(U – A) is defined as {x ∈ U: x ∉ A}
U = {2, 3, 5, 7, 9}
A = {3, 7}
A’ = U – A = {2, 5, 9}
B’ means Complement of B with respect to universal set U.
So, B’ = U – B
(U – B) is defined as {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}
= {2, 3, 5, 9}
∴ LHS = RHS
Hence verified.