Question

If A = {-1, 0, 1), B = {2, 3) and C = (-2, 3}, verify that,
A x (B-C) = (A x B) - (AXC).
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Answer 1

Note :

★ Set : A well defined collection of distinct objects is called a set .

★ Union of two sets : The union of two sets A and B is the set of all those elements which are either in A or in B or in both .

→ This set is denoted by A U B .

★ Intersection of two sets : The intersection of two sets A and B is the set of all those elements which are in common in both A and B .

→ This set is denoted by A ∩ B .

★ Difference of sets : The difference of two sets A and B in the order ( also called relative complement of B in A ) is the set of all those elements of A which are not the elements of B .

→ It is denoted by (A - B) .

★ Cartesian Product : If A and B are any two non empty sets then the set of all ordered pairs (a,b) such that a € A and b € B is called the cartesian product of sets a with set B and it is denoted by A×B .

→ A×B = { (a,b) : a € A and b € B } .

Solution :

• Given : A = { -1 , 0 , 1 } , B = { 2 , 3 } , C = { -2 , 3 }
• To verify : A×(B - C) = (A×B) - (A×C)

Verification :

→ B - C = { 2 , 3 } - { -2 , 3 }

→ B - C = { 2 }

Now ,

→ LHS = A×(B - C)

→ LHS = { -1 , 0 , 1 } × { 2 }

→ LHS = { (-1 , 2) , (0 , 2) , (1 , 2) }

Now ,

→ A×B = { -1 , 0 , 1 } × { 2 , 3 }

→ A×B = { (-1,2) , (0,2) , (1,2) , (-1 , 3) , (0 , 3) , (1 , 3) }

Also ,

→ A×C = { -1 , 0 , 1 } × { -2 , 3 }

→ A×C = { (-1 , -2) , (0 , -2) , (1 , -2) , (-1 , 3) , (0 , 3) , (1 , 3) }

Now ,

→ RHS = (A×B) - (A×C)

→ RHS = { (-1,2) , (0,2) , (1,2) , (-1 , 3) , (0 , 3) , (1 , 3) } – { (-1 , -2) , (0 , -2) , (1 , -2) , (-1 , 3) , (0 , 3) , (1 , 3) }

→ RHS = { (-1 , 2) , (0 , 2) , (1 , 2) }

Clearly ,

LHS = RHS

Hence verified .

Answer 2

Step-by-step explanation:

(i) To verify : A×(B∩C)=(A×B)∩(A×C)

We have B∩C={1,2,3,4}∩{5,6}=ϕ

∴ L.H.S = A×(B∩C)=A×ϕ=ϕ

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

A×C={(1,5),(1,6),(2,5),(2,6)}

∴R.H.S.=(A×B)∩(A×C)=ϕ

∴L.H.S=R.H.S

Hence A×(B∩C)=(A×B)∩(A×C)

(ii) To verify: A×C is a subset of B×D

A×C={(1,5),(1,6),(2,5),(2,6)}

B×D={(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),

(3,8),(4,5),(4,6),(4,7),(4,8)}

We can observe that all the elements of set A×C are the elements of set B×D

Therefore A×C is a subset of B×D

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