Question -
Answer -
(i) (A ∪ B) x C = (A x C) = (A x C) ∪ (B x C)
Let (x, y) be an arbitrary element of (A ∪ B) × C
(x, y) ∈ (A ∪ B) C
Since, (x, y) are elements of Cartesian product of (A ∪ B) × C
x ∈ (A ∪ B) and y ∈ C
(x ∈ A or x ∈ B) and y ∈ C
(x ∈ A and y ∈ C) or (x ∈ Band y ∈ C)
(x, y) ∈ A × C or (x, y) ∈ B × C
(x, y) ∈ (A × C) ∪ (B × C) … (1)
Let (x, y) be an arbitrary element of (A × C) ∪ (B × C).
(x, y) ∈ (A × C) ∪ (B × C)
(x, y) ∈ (A × C) or (x, y) ∈ (B × C)
(x ∈ A and y ∈ C) or (x ∈ B and y ∈ C)
(x ∈ A or x ∈ B) and y ∈ C
x ∈ (A ∪ B) and y ∈ C
(x, y) ∈ (A ∪ B) × C … (2)
From 1 and 2, we get: (A ∪ B) × C = (A × C) ∪ (B × C)
(ii) (A ∩ B) x C = (A x C) ∩ (B x C)
Let (x, y) be an arbitrary element of (A ∩ B) × C.
(x, y) ∈ (A ∩ B) × C
Since, (x, y) are elements of Cartesian product of (A ∩ B) × C
x ∈ (A ∩ B) and y ∈ C
(x ∈ A and x ∈ B) and y ∈ C
(x ∈ A and y ∈ C) and (x ∈ Band y ∈ C)
(x, y) ∈ A × C and (x, y) ∈ B × C
(x, y) ∈ (A × C) ∩ (B × C) … (1)
Let (x, y) be an arbitrary element of (A × C) ∩ (B × C).
(x, y) ∈ (A × C) ∩ (B × C)
(x, y) ∈ (A × C) and (x, y) ∈ (B × C)
(x ∈A and y ∈ C) and (x ∈ Band y ∈ C)
(x ∈A and x ∈ B) and y ∈ C
x ∈ (A ∩ B) and y ∈ C
(x, y) ∈ (A ∩ B) × C … (2)
From 1 and 2, we get: (A ∩ B) × C = (A × C) ∩ (B × C)