JEE Mathematics/ Lakshmikanta Satapathy/ Relations and Functions theory part 1/ Theory of Cartesian product Relation in a Set Types of Relations Equivalence Relation and Equivalence Class explained with examples

1. Physics Helpline
L K Satapathy
Relations & Functions Theory 1

2. Physics Helpline
L K Satapathy
Cartesian Product of Sets :
Cartesian Product of two non-empty Sets A and B is the set of all
ordered pairs (x , y) , such that x  A and y  B
In set notation we write A  B = { (x , y) : x  A , y  B }
Properties :
(i) Two ordered pairs are equal if and only if the two elements of one is equal to
the corresponding elements of the other.
 (x , y) = (a , b)  x = a and y = b [ (x , y)  (y , x) ]
(ii) n(AB) = n(A).n(B)
If A has (p) elements and B has (q) elements , then (AB) has (pq) elements
(iii) A  A  A = { (x , y , z) : x , y , z  A }
Here (x , y , z) is called an ordered triplet
Relations & Functions Theory 1

3. Physics Helpline
L K Satapathy
2 5 1
1 , ,
3 3 3 3
x
y
   
     
   
Example : Find the values of x and y , if
Answer :
Two ordered pairs are equal if and only if the corresponding elements are equal
2 5 1
1 , ,
3 3 3 3
x
y
   
      
   
5
1 3 5 2 [ ]
3 3
Anx x s
x
       
2 1 1 2
1
3 3 3 3
[ ]and y y Ans     
5 2 1
1
3 3 3 3
x
and y    
Relations & Functions Theory 1

4. Physics Helpline
L K Satapathy
Answer :
Example : The cartesian product A  A has 9 elements which includes the
ordered pairs (– 1 , 0) and (0 , 1) . Then find AA
It is given that
2
( ) ( )Let n A p n A A p p p     
2
( ) 9 9 3n A A p p     
Also ( 1, 0) 1 0A A A and A     
(0 , 1) 1A A A   
{ 1, 0 , 1}A  
{( 1 , 1),( 1,0),( 1,1),(0, 1),
(0,0),(0,1),(1, 1),(1,0),( ) ]1,1 } [Ans
A A       

Relations & Functions Theory 1
And

5. 10/11/2015
Physics Helpline
L K Satapathy
If (x , y)  R , then y is called the image of x in R
Relation : A relation R from a non-empty set A to a Non-empty set B
is a subset of the cartesian product A  B
Relations & Functions Theory 1
Domain of R : It is the set of the 1st elements of the ordered pairs in R
Range of R : It is the set of the 2nd elements of the ordered pairs in R
Co-domain of R : If R  (A  B) , then set B is the co-domain of R
2pq
If A has (p) elements and B has (q) elements , then (AB) has (pq) elements
 Number of Subsets of A  B =
 Total number of Relations from Set A to Set B = 2pq

6. 10/11/2015
Physics Helpline
L K Satapathy
(iii) Arrow diagram :
Relations & Functions Theory 1
Domain of R = { 1 , 2 , 3 , 4 , 5 }1
2
3
4
5
6
1
2
3
4
5
6
Range of R = { 2 , 3 , 4 , 5 , 6 }
Co-domain of R = { 1 , 2 , 3 , 4 , 5 , 6 }
Representation of a relation : Consider a set A = { 1 , 2 , 3 , 4 , 5 , 6 }
A relation on set A may be represented as follows :
(i) Set builder form : R = { (x , y) : y = x + 1 ; x , y  A }
(ii) Roster form : R = { (1 , 2) , (2 , 3) , (3 , 4) , (4 , 5) , (5 , 6) }

7. 10/11/2015
Physics Helpline
L K Satapathy
Example : Consider a relation R on N , in set builder form , given by
Relations & Functions Theory 1
R = { (x , y) : y = x + 5 ; x , y  N , x < 4 }
Represent this relation in Roster form. Write its domain and range.
Answer :
In Roster form R = { (1 , 6) , (2 , 7) , (3 , 8) }
Domain of R = { 1 , 2 , 3 }
Range of R = { 6 , 7 , 8 }
Let us draw the Arrow diagram for better understanding:
1
2
3
.
.
.
.
.
.
.
6
7
8
.
.
.
NN

8. Physics Helpline
L K Satapathy
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