PROBLEM & SOLUTION OF SETS
Example.A)
Let A = { letters of the word BADMINTON } and B = { letters of the word CRICKET }
Find 1) A ∩ B, 2) A U B, 3) B U A, 4) B ∩ A, 5) A – B, 6) B – A
Answer)
1) A ∩ B
As per the given condition, we have got A = { letters of the word BADMINTON } and B = { letters of the word CRICKET }
A = { letters of the word BADMINTON } = { B, A, D, M, I, N, T, O }
B = { letters of the word CRICKET } = { C, R, I, K, E, T }
As per the given instruction, we have to find out
A ∩ B = { B, A, D, M, I, N, T, O } ∩ { C, R, I, K, E, T } = { I, T }
2) A U B
Ans.) As per the given condition, we have got A = { letters of the word BADMINTON } and B = { letters of the word CRICKET }
A = { letters of the word BADMINTON }
= { B, A, D, M, I, N, T, O }
B = { letters of the word CRICKET } = { C, R, I, K, E, T }
As per the given instruction, we have to find out
A U B = { B, A, D, M, I, N, T, O } U { C, R, I, K, E, T } = { B, A, D, M, I, N, T, O, C, R, K, E }
3) B U A
Ans.) As per the given condition, we have got A = { letters of the word BADMINTON } and B = { letters of the word CRICKET }
A = { letters of the word BADMINTON }
= { B, A, D, M, I, N, T, O }
B = { letters of the word CRICKET }
= { C, R, I, K, E, T }
As per the given instruction, we have to find out
B U A = { C, R, I, K, E, T } U { B, A, D, M, I, N, T, O } = { C, R, I, K, E, T, B, A, D, M, N, O }
4) B ∩ A
Answer) As per the given condition, we have got A = { letters of the word BADMINTON } and B = { letters of the word CRICKET }
A = { letters of the word BADMINTON }
= { B, A, D, M, I, N, T, O }
B = { letters of the word CRICKET }
= { C, R, I, K, E, T }
As per the given instruction, we have to find out
B ∩ A = { C, R, I, K, E, T } ∩ { B, A, D, M, I, N, T, O } = { I, T }
5) A – B
Answer) As per the given condition, we have got A = { letters of the word BADMINTON } and B = { letters of the word CRICKET }
A = { letters of the word BADMINTON }
= { B, A, D, M, I, N, T, O }
B = { letters of the word CRICKET } = { C, R, I, K, E, T }
As per the given instruction, we have to find out A – B
A – B = { B, A, D, M, I, N, T, O } – { C, R, I, K, E, T }
= { B, A, D, M, N, O }
6) B – A
Answer) As per the given condition, we have got A = { letters of the word BADMINTON } and B = { letters of the word CRICKET }
A = { letters of the word BADMINTON } = { B, A, D, M, I, N, T, O }
B = { letters of the word CRICKET } = { C, R, I, K, E, T }
As per the given instruction, we have to find out A – B
B – A = { C, R, I, K, E, T } – { B, A, D, M, I, N, T, O }
= { C, R, K, E }
Example.B)
If U = { x : x ∈ W and 6 ≤ x ≤ 12 } , A = { 6, 8, 9 } , B = { 7, 8, 11 } and C = { 6 } find the following sets.
1) A’ , 2) B’ , 3) C’ , 4) A – B , 5) (B U C )’ , 6) ( A ∩ B )’ , 7) A – ( B U C ), 8) A – ( B ∩ C ).
1) A’
Answer) As per the given condition,
U = { x : x ∈ W and 6 ≤ x ≤ 12 } ,
U = { 6, 7, 8, 9, 10, 11, 12 } and A = { 6, 8, 9 }
As per the rule of complement of set
A’ = U – A = { 6, 7, 8, 9, 10, 11, 12 } – { 6, 8, 9 }
= { 7, 10, 11, 12 }
2) B’
Answer) As per the given condition,
U = { x : x ∈ W and 6 ≤ x ≤ 12 },
U = { 6, 7, 8, 9, 10, 11, 12 } and B = { 7, 8, 11 }
As per the rule of complement of set
A’ = U – A = { 6, 7, 8, 9, 10, 11, 12 } – { 7, 8, 11 } = { 6, 9, 10, 12 }
3) C’
Answer) As per the given condition,
U = { x : x ∈ W and 6 ≤ x ≤ 12 },
U = { 6, 7, 8, 9, 10, 11, 12 } and C = { 6 }
As per the rule of complement of set
A’ = U – A = { 6, 7, 8, 9, 10, 11, 12 } – { 6 }
= { 7, 8, 9, 10, 11, 12 }
4) A – B
Answer) As per the given condition, A = { 6, 8, 9 } , B = { 7, 8, 11 },
we have to find out A – B
So, A – B = { 6, 8, 9 } – { 7, 8, 11 } = { 6, 9 }
5) (B U C )’
Answer) As per the given condition,
U = { x : x ∈ W and 6 ≤ x ≤ 12 },
U = { 6, 7, 8, 9, 10, 11, 12 }, B = { 7, 8, 11 }, C = { 6 }
First, we have to find , B U C
So, B U C = { 7, 8, 11 } U { 6 } = { 6, 7, 8, 11 } [ As per laws of Union Set ]
Now, (B U C )’ = U - (B U C )
= { 6, 7, 8, 9, 10, 11, 12 } - { 6, 7, 8, 11 } [ As per Laws of Complement Set ]
= { 9, 10, 12 }
So, (B U C )’ = { 9, 10, 12 }
6) ( A ∩ B )’
Answer) As per the given condition,
U = { x : x ∈ W and 6 ≤ x ≤ 12 },
U = { 6, 7, 8, 9, 10, 11, 12 }, B = { 7, 8, 11 }, C = { 6 }
First, we have to find, B U C
So, B ∩ C = { 7, 8, 11 } ∩ { 6 }
= φ [ Overlapping rules ]
Now, ( A ∩ B )’ = U - ( A ∩ B )
= { 6, 7, 8, 9, 10, 11, 12 } - φ
= { 6, 7, 8, 9, 10, 11, 12 } = U
So, ( A ∩ B )’ = U
7) A – ( B U C ),
Answer) As per the given condition,
U = { x : x ∈ W and 6 ≤ x ≤ 12 },
A = { 6, 8, 9 }, B = { 7, 8, 11 } and C = { 6 }
We have to find out = ( B U C )
( B U C ) = { 7, 8, 11 } U { 6 } = { 6, 7, 8, 11 }
Now, A – ( B U C ) = { 6, 8, 9 } - { 6, 7, 8, 11 }
= { 9 }
8) A – ( B ∩ C )
Answer ) As per the given condition,
U = { x : x ∈ W and 6 ≤ x ≤ 12 },
A = { 6, 8, 9 }, B = { 7, 8, 11 } and C = { 6 }
B ∩ C = { 7, 8, 11 } ∩ { 6 } = φ
Now, A – ( B ∩ C ) = { 6, 8, 9 } – φ
= { 6, 8, 9 }
= A
So, A – ( B ∩ C ) = A
Example. C)
Let U = { x : x ∈ N and 2 ≤ x ≤ 12 }, A = { x : x is an even number }, B = { x : x is a multiple of 3}, and C = { x : x is a multiple of 4 }, verify the following 1) ( A U B )’ = A’ ∩ B’ , 2) ( A ∩ B )’ = A’ U B’ , 3) A U ( B ∩ C ) = ( A U B ) ∩ ( A U C ), 4) A ∩ ( B U C ) = ( A ∩ B ) U ( A ∩ C )
1) ( A U B )’ = A’ ∩ B’
Answer) As per the given condition, we can find,
U = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 },
A = { 2, 4, 6, 8, 10, 12 },
B = { 3, 6, 9, 12 },
C = { 4, 8, 12 }
Here we consider ( A U B )’ = 1st Part and A’ ∩ B’ = 2nd Part
1st Part
So, ( A U B ) = { 2, 4, 6, 8, 10, 12 } U { 3, 6, 9, 12 }
= { 2, 3, 4, 6, 8, 9, 10, 12 }
( A U B )’ = U - ( A U B )
= { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } – { 2, 3, 4, 6, 8, 9, 10, 12 }
= { 5, 7, 11 }
2nd Part
So, A’ ∩ B’ = ( U – A ) ∩ ( U – B )
= { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } – { 2, 4, 6, 8, 10, 12 } ∩ ( U – B )
= { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } – { 2, 4, 6, 8, 10, 12 } ∩ { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } – { 3, 6, 9, 12 }
= { 3, 5, 7, 9, 11 } ∩ { 2, 4, 5, 7, 8, 10, 11 }
= { 5, 7, 11 }
So, we can observe that 1st Part = 2nd Part
( A U B )’ = A’ ∩ B’ (Proven)
2) ( A ∩ B )’ = A’ U B’
Answer) As per the given condition, we can find , U = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }, A = { 2, 4, 6, 8, 10, 12 }, B = { 3, 6, 9, 12 }, C = { 4, 8, 12 }
Here we consider ( A ∩ B )’ = 1st Part and A’ U B’ = 2nd Part
1st Part
( A ∩ B ) = ( A ∩ B ) = { 2, 4, 6, 8, 10, 12 } ∩ { 3, 6, 9, 12 }
= { 6, 12 }
( A ∩ B )’ = U - ( A ∩ B )
= { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } – { 6, 12 }
= { 2, 3, 4, 5, 7, 8, 9, 10, 11 }
2nd Part
A’ U B’ = ( U – A ) U ( U – B )
= { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } – { 2, 4, 6, 8, 10, 12 } U { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } – { 3, 6, 9, 12 }
= { 3, 5, 7, 9, 11 } U { 2, 4, 5, 7, 8, 10, 11 }
= { 2, 3, 4, 5, 7, 8, 9, 10, 11 }
So, we can observe that 1st Part = 2nd Part
( A ∩ B )’ = A’ U B’ (Proven)
3) A U ( B ∩ C ) = ( A U B ) ∩ ( A U C )
Answer) As per the given condition, we can find, U = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }, A = { 2, 4, 6, 8, 10, 12 }, B = { 3, 6, 9, 12 }, C = { 4, 8, 12 }
Here we consider A U ( B ∩ C ) = 1st Part
and ( A U B ) ∩ ( A U C ) = 2nd Part
1st Part
A U ( B ∩ C ) = A U { 3, 6, 9, 12 } ∩ { 4, 8, 12 }
= A U { 12 }
= { 2, 4, 6, 8, 10, 12 } U { 12 }
= { 2, 4, 6, 8, 10, 12 }
So, A U ( B ∩ C ) = { 2, 4, 6, 8, 10, 12 }
2nd Part
( A U B ) ∩ ( A U C )
= { 2, 4, 6, 8, 10, 12 } U { 3, 6, 9, 12 } ∩ { 2, 4, 6, 8, 10, 12 } U { 4, 8, 12 }
= { 2, 3, 4, 6, 8, 9, 10, 12 } ∩ { 2, 4, 6, 8, 10, 12 }
= { 2, 4, 6, 8, 10, 12 }
So, we can see that, 1st Part = 2nd Part
So, now we can say -
A U ( B ∩ C ) = ( A U B ) ∩ ( A U C ) (Proven)
4) A ∩ ( B U C ) = ( A ∩ B ) U ( A ∩ C )
Answer) As per the given condition, we can find , U = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }, A = { 2, 4, 6, 8, 10, 12 }, B = { 3, 6, 9, 12 }, C = { 4, 8, 12 }
Here we consider A ∩ ( B U C ) = 1st Part
and ( A ∩ B ) U ( A ∩ C ) = 2nd Part
1st Part
A ∩ ( B U C ) = A ∩ { 3, 6, 9, 12 } U { 4, 8, 12 }
= A ∩ { 3, 4, 6, 8, 9, 12 }
= { 2, 4, 6, 8, 10, 12 } ∩ { 3, 4, 6, 8, 9, 12 }
= { 4, 6, 8, 12 }
So, A ∩ ( B U C ) = { 4, 6, 8, 12 }
2nd Part
( A ∩ B ) U ( A ∩ C )
= { 2, 4, 6, 8, 10, 12 } ∩ { 3, 6, 9, 12 } U ( A ∩ C )
= { 6, 12 } U { 2, 4, 6, 8, 10, 12 } ∩ { 4, 8, 12 }
= { 6, 12 } U { 4, 8, 12 }
= { 4, 6, 8, 12 }
So, we can observe that, 1st Part = 2nd Part
So, A ∩ ( B U C ) = ( A ∩ B ) U ( A ∩ C ) (Proven)