Question -
Answer -
(i) A ∩ B = A ∩ C need not imply B = C.
Let us consider, A = {1, 2}
B = {2, 3}
C = {2, 4}
Then,
A ∩ B = {2}
A ∩ C = {2}
Hence, A ∩ B = A ∩ C, where, B is not equal to C
(ii) A ⊂ B ⇒ C – B ⊂ C – A
Given: A ⊂ B
To show: C–B ⊂ C–A
Let us consider x ∈ C– B
⇒ x ∈ C and x ∉ B [by definition C–B]
⇒ x ∈ C and x ∉ A
⇒ x ∈ C–A
Thus x ∈ C–B ⇒ x ∈ C–A. This is true for all x ∈ C–B.
∴ A ⊂ B ⇒ C – B ⊂ C – A
7. For any two sets, prove that:
(i) A ∪ (A ∩ B) = A
(ii) A ∩ (A ∪ B) = A
Solution:
(i) A ∪ (A ∩ B) = A
We know union is distributive over intersection
So, A ∪ (A ∩ B)
(A ∪ A) ∩ (A ∪ B) [Since, A ∪ A = A]
A ∩ (A ∪ B)
A
(ii) A ∩ (A ∪ B) = A
We know union is distributive over intersection
So, (A ∩ A) ∪ (A ∩ B)
A ∪ (A ∩ B) [Since, A ∩ A = A]
A