**How to Check the Given Statement With Suitable Sets :**

Here we are going to see how to check the given statement with suitable sets.

**Question 1 :**

By taking suitable sets A,B,C, verify the following results:

(i) A × (B ∩ C) = (A × B) ∩ (A × C).

**Solution :**

Let A = {1, 2, 3}, B = {1, 2} and C = {2, 3 4}

L.H.S

(B n C) = {2}

A × (B ∩ C) = {1, 2, 3} × {2}

A × (B ∩ C) = { (1, 2)(2, 2) (3, 2) } ------(1)

R.H.S

(A × B) = {1, 2, 3} × {1, 2}

(A × B) = {(1, 1) (1, 2) (2, 1) (2, 2)(3, 1) (3, 2)}

(A × C) = {1, 2, 3} × {2, 3 4}

(A × C)

= {(1, 2) (1, 3) (1, 4) (2, 2) (2, 3) (2, 4) (3, 2) (3, 3) (3, 4)}

(A × B) ∩ (A × C) = { (1, 2)(2, 2) (3, 2) } ------(2)

(1) = (2)

Hence proved.

(ii) A × (B ∪ C) = (A × B) ∪ (A × C)

**Solution :**

Let A = {1, 2, 3}, B = {1, 2} and C = {2, 3 4}

L.H.S

(B U C) = {1, 2, 3, 4}

A × (B ∪ C) = {1, 2, 3} × {1, 2, 3, 4}

A × (B ∪ C) = {(1, 1) (1, 2) (1, 3) (1, 4)(2, 1) (2, 2) (2, 3) (2, 4)(3, 1) (3, 2) (3, 3) (3, 4) ------(1)

R.H.S

(A × B) = {1, 2, 3} × {1, 2}

(A × B) = {(1, 1) (1, 2) (2, 1) (2, 2)(3, 1) (3, 2)}

(A × C) = {1, 2, 3} × {2, 3 4}

(A × C)

= {(1, 2) (1, 3) (1, 4) (2, 2) (2, 3) (2, 4) (3, 2) (3, 3) (3, 4)}

(A × B) U (A × C) = {(1, 1) (1, 2) (1, 3) (1, 4)(2, 1) (2, 2) (2, 3) (2, 4)(3, 1) (3, 2) (3, 3) (3, 4) ------(2)

(1) = (2)

Hence proved.

(iii) (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A)

**Solution :**

Let A = {1, 2, 3}, B = {1, 2} and C = {2, 3 4}

L.H.S

(A × B) = {1, 2, 3} × {1, 2}

= {(1, 1) (1, 2) (2, 1) (2, 2) (3, 1) (3, 2)}

(B × A) = {1, 2} × {1, 2, 3}

= {(1, 1) (1, 2) (1, 3) (2, 1) (2, 2) (2, 3)}

(A × B) ∩ (B × A) = { (1, 1) (1, 2)(2, 1) (2, 2) } -----(1)

R.H.S

(A ∩ B) = {1, 2, 3} ∩ {1, 2} = {1, 2}

(B ∩ A) = {1, 2}

(A ∩ B) × (B ∩ A) = { (1, 1) (1, 2)(2, 1) (2, 2) } -----(2)

(1) = (2)

Hence proved.

(iv) C − (B − A) = (C ∩ A) ∪ (C ∩ B').

**Solution :**

Let U = {1, 2, 3, 4, 5}

A = {1, 2, 3}, B = {1, 2} and C = {2, 3, 4}

L.H.S

B' = {3, 4, 5}

(B - A) = { }

C − (B − A) = {2, 3, 4} ---(1)

(C ∩ A) = {2, 3 4} ∩ {1, 2, 3} = {2, 3}

(C ∩ B') = {2, 3, 4}

(C ∩ A) ∪ (C ∩ B') = {2, 3, 4} ---(2)

Hence proved.

(v) (B − A) ∩ C = (B ∩ C) − A = B ∩ (C − A).

**Solution :**

A = {1, 2, 3}, B = {1, 2} and C = {2, 3, 4}

Part 1 :

(B − A) = { }

(B − A) ∩ C = { } ------(1)

Part 2 :

(B ∩ C) = {2}

(B ∩ C) − A = { } ------(2)

Part 3 :

(C − A) = {4}

B ∩ (C − A) = { } ------(3)

(1) = (2) = (3)

Hence proved.

(vi) (B − A) ∪ C = (B ∪ C) − (A − C)

**Solution :**

A = {1, 2, 3}, B = {1, 2} and C = {2, 3, 4}

L.H.S

(B − A) = {3}

(B − A) ∪ C = {2, 3, 4} ------(1)

R.H.S

(B ∪ C) = {1, 2, 3, 4}

(A-C) = {1}

(B ∪ C) − (A − C) = {2, 3, 4} ------(2)

(1) = (2)

Hence proved.

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