# 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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