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Prove that:
(i) (A ∪ B) x C = (A x C) = (A x C) ∪ (B x C)

(ii) (A ∩ B) x C = (A x C) ∩ (B x C)

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(i) (A ∪ B) x C = (A x C) = (A x C) ∪ (B x C)

Suppose (x, y) be an arbitrary element of (A ∪ B) × C

(x, y) ∈ (A ∪ B) C

Here, (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)

Suppose (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 the equation 1 and 2, we get: (A ∪ B) × C = (A × C) ∪ (B × C)

(ii) (A ∩ B) x C = (A x C) ∩ (B x C)

Suppose (x, y) be an arbitrary element of (A ∩ B) × C.

(x, y) ∈ (A ∩ B) × C

Here, (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)

Suppose (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 the equation (1) and (2), we get: (A ∩ B) × C = (A × C) ∩ (B × C)

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