JEE Mathematics/ Lakshmikanta Satapathy/ Set Theory part 2/ Theory of Union Intersection and Difference of two sets Complement of a Set, Formulae necessary for solving practical problems involving Cardinality of Sets discussed in details

1. Physics Helpline
L K Satapathy
Set Theory 2

2. Physics Helpline
L K Satapathy
Universal set : It is a set such that all the sets used in a particular context are its
subsets . It is generally denoted as U.
Set Theory 2
(1) Union of Sets : The union of two sets A and B is
the set consisting of the elements which are either in
A or in B (including the elements which are in both)
In set notation :  :A B x x A or x B   
The Venn diagram for (AB) is shown in the figure
AU B
Properties of Union of Sets :
(i) A  B = B  A [ Commutative Law ]
(ii) (A  B)  C = A  (B  C) [ Associative Law ]
(iii) A   = A [ Identity Law ,  is the identity of Union]
(iv) A  U = U [ Law of U]
(iv) A  A = A [ Idempotent Law ]

3. Physics Helpline
L K Satapathy
Set Theory 2
(2) Intersection of Sets : The intersection of two sets A
and B is the set consisting of the elements which
belong to both A and B
In set notation :  :A B x x A and x B   
The Venn diagram for (AB) is shown in the figure
AU B
Properties of Intersection of Sets :
(i) A  B = B  A [ Commutative Law ]
(ii) (A  B)  C = A  (B  C) [ Associative Law ]
(iii)   A =  U  A = A [ Law of  and U ]
(iv) A  A = A [ Idempotent Law ]
(v) A  (B  C) = (A  B)  (A  C) [ Distributive Law ]
If A  B =  , then A and B are disjoint sets
 A and B do not have any common element
[Shown in the Venn diagram]
AU B

4. Physics Helpline
L K Satapathy
Set Theory 2
In set notation :  :A B x x A and x B   
The Venn diagram for (A – B ) is shown in the figure
AU B
(3) Difference of Sets : The difference of two sets (A – B )
is the set of elements which belong to A but not to B.
It may be noted that , (A – B ) , (A  B) and (B – A ) are mutually disjoint sets.
(4) Complement of a Set : If set A is a subset of the
universal set U , then the complement of A is the set of
elements of U which do not belong to A
In set notation :  :A x x U and x A   
The Venn diagram for (A) is shown in the figure
A
U
A

5. Physics Helpline
L K Satapathy
Set Theory 2
Properties of complement of Sets :
( ) ( )i A A U ii A A     
( )A A  
Complement Law :
De Morgan’s Law :
( ) ( )i A B A B     ( ) ( )ii A B A B    
Law of double complementation :
U Law of empty set :
U  Law of universal set :
AU B AU B

6. Physics Helpline
L K Satapathy
Set Theory 2
For solving Practical problems involving the Cardinality (number of elements) of
Sets , we need the following Results :
( ) ( ) ( ) ( ) ( )i n A B n A n B n A B    
For finite sets A and B :
( ) ( ) ( ) ( )ii n A B n A n B  
For disjoint sets A and B :
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
iii n A B C n A n B n C n A B n B C
n A C n A B C
        
    
For three finite sets A , B and C :
[ Since A  B =  ]

7. Physics Helpline
L K Satapathy
Set Theory 2
Question : In a survey of 600 students in a school , 150 students were found to be
taking tea , 225 students taking coffee and 100 students were taking both tea and
coffee . Then the number of students taking neither tea nor coffee is
(a) 225 (b) 275 (c) 325 (d) 350
( ) 600n U 
Answer :
Total number of students of the school = 600
( ) 150n T Number of students taking tea = 150
( ) 225n C Number of students taking coffee = 225
( ) 100n T C  Number of students taking both tea and coffee = 100
( ) ( ) ( ) ( ) 150 225 100 275n T C n T n C n T C         
( ) ( ) ( ) 600 275 32 ]5 [n T C n U n T nC A s      
Correct option = (c)
 Number of students taking neither tea nor coffee

8. Physics Helpline
L K Satapathy
Set Theory 2
Alternate Method : Using Venn diagram
Total number of students = 600
Number of students taking both T&C =100
( ) 600 275 [3 ]25 Ansn T C d     
Correct option = (c)
T C
a bcd
600 . . (1)a b c d    
Number of students taking T = 150
Number of students taking C = 225
150 . . (2)a c  
225 . . (3)b c  
100 . . (4)c 
(3) 125b  
(2) 50a 
50 125 100 275a b c      
T C
50 125100325

9. Physics Helpline
L K Satapathy
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