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1800-102-2727In our day to day life, we very frequently come across several patterns that characterize relations such as mother and daughter,mother and son,student and teacher. Much of mathematics is also about finding a relation between quantities that change.
We come across many relations such as number p is greater than number q, line a is perpendicular to line b, set X is a proper subset of set Y. In all these scenarios, we observe that a relation involves pairs of objects arranged in a certain order. In this article, you will learn how to link the pairs of objects from any two sets and then to introduce relations between the two objects in the pair.
Table of Contents
A relation R from a non-empty set P to a non-empty set Q is a subset of the cartesian product PQ. The subset is obtained by describing a relationship between the first element and the second element of the ordered pairs in PQ. The second element is called the image of the first element.
A relation from A to B is expressed as R:AB and it is read as R is a relation such that it is from A to B. Also, Every subset of AB is a relation, i.e., RAB.
If (a,b)R then we say that a is related to b and we write, a R b.
If (a,b)R then we say that a is not related to b and we write, a b
Example: Let A={p, q, r} and B={Qimat, Priyansh, Ritu, Amit, Ritesh}
The cartesian product of A and B will contain 15 ordered pairs which can be listed as AB = {(p, Qimat), (p,Priyansh), (p, Ritu), ..... , (r, Ritesh)}.
Now we can get a subset of AB by introducing a relation R between the first element a & the second element b of each ordered pair (a,b) as
R= { (a,b): a is the first letter of the name b, aA, bB}.
R={(p,Priyansh), ( q,Qimat), (r,Ritu), ( r,Ritesh)}
A diagrammatic representation of this relation R (known as an arrow diagram) is shown in the figure given below.
Fig: Arrow diagram of a Relation
In the example given above, we have written the relation R by the following two methods:
In the first method, we have defined the relationship between the elements of set P and elements of set Q. This is called the Set-builder form.
In the second method, we have listed all the ordered pairs that belong to R. This is called the Roster form.
Example:Let A={1,3,7} and B={2,6,14} and R is a relation from A to B such that y=2x where xA,yB.
In set-builder form, we write R={(x,y):y=2x, xA,yB}
In roster form, we write R= {(1,2),(3,6),(7,14)}.
Domain, Codomain and Range
For a relation R: AB,
Domain is the set of all the first elements of the ordered pairs in the relation R, so here the domain is set A.
Range is the set of all the second elements of the ordered pairs in the relation R.
The second set i.e. set B is called the codomain of R and is the super-set of the range set.
Note that Range Codomain.
Example: Let R: PQ be a relation as shown in the figure.
R={(9,3), (9,-3), (4,2), (4,-2), (25,5), (25,-5)}
The domain of R is set P= {9,4,25}
The range of R is {5,3,2,-2,-3,-5}
The codomain of R is Q={5,3,2,1,-2,-3,-5}
Clearly, Range = {5,3,2,-2,-3,-5}Q
Number of Relations from set A to B
Number of relations from set A to B= Number of subsets of AB
If n(A)=p, n(B)=q and R:AB, then
Number of subsets of AB i.e. n(AB)=p.q
Hence, Number of relations from A to B=2n(AB)=2pq
Example: Let A={1,2,3,4} and B={4,5,6,7,8,9}
n(A)=4, n(B)=6
n(AB)=46=24
Number of relations from A to B=2n(AB)=224
Related Concept video:
Relations & Types of Relations - Relations and Functions Class 12 Maths | Target JEE 2023 Exam Prep
1.Void Relation
A relation R on set A is known as a void or empty relation, if no element of set A is related to any element of set A, i.e., R=AA.
Example: R is a relation on A={1,2,3} such that R={(a,b): a+b=12}
Clearly, we cannot find two numbers from set A such that their sum is 12. Hence, none of the elements in A satisfies the condition in R, therefore it is a void relation.
2.Universal Relation
A relation is called a universal relation, if each element of set A is related to every element of set A, i.e., R=AA.
Note: Both the void and the universal relation are collectively called a trivial relation.
Example: Let A be the set of all real numbers & R={(a, b):a-b0, a, bA}
We know that a-b is a real number, so a-b will also be a non-negative real number.
a-b0 will hold true for every value of a and b.
Hence, each element of set A is related to every element of set A.Therefore, R will be a universal relation on set A
3.Identity Relation
Let A be a set. Then the relation IA={(a,a):aA} on A is the identity relation on A.
Simply, a relation is called the identity relation if every element of A is related to itself only.
Example:R is a relation on A={1,2,3,4,5} such that R={(a,b): a=b;a,bA }
R={(1,1),(2,2),(3,3),(4,4),(5,5)}
Clearly, R is an identity relation on A as every element of A is related to itself only.
4.Inverse Relation
Let A and B be two sets and R be a relation from A to B, then the inverse of R is a relation from B to A and is denoted by R-1. It is defined as R-1={(b, a):(a, b)R}.
Basically, the positions of elements in the ordered pair get interchanged.
Note: Domain of R-1= Range of R
Range of R-1= Domain of R
Example:A={1,2,3,4,5} and B={1,4,5}
Let R be a relation from A to B such that R={(x,y)AB:x R y iff x<y }
R={(1, 4),(1, 5),(2, 4),(2, 5),(3, 4),(3, 5),(4, 5)}
Then R-1 ={(4, 1),(5, 1),(4, 2),(5, 2),(4, 3),(5, 3),(5, 4)}
R-1 is a relation from B to A.
Example: If A={1,2,3} and B={4,5,6}, then which of the following can be a relation from A to B?
(a) R1={(1,4), (1,5), (1,6)}
(b) R2={(1,5), (2,4), (3,6)}
(c) R2={(1,4), (1,5), (1,6),(3,6),(2,6),(3,4)}
(d) R2={(4,2), (2,6), (5,1), (2,4)}
Solution:
Given, A={1,2,3} and B={4,5,6}
Therefore, AB={(1,4), (1,5), (1,6),(2,4), (2,5), (2,6), (3,4), (3,5), (3,6)}
R1ABR1is a relation from A to B
R2ABR2is a relation from A to B
R3ABR3 is a relation from A to B
R4is not a subset of ABR4is not a relation from A to B
Hence, (a),(b),and (c) are correct options.
Example:Let R={(a, b):a, bℤ, a2+3b28} be a relation defined on the set of integers ℤ, then find the domain of R-1.
Solution:
R={(a, b):a, bℤ, a2+3b28}
We know that the domain of an inverse relation is equal to the range of an original relation.
Domain of R-1= Range of R(Values of b)
Case 1: If a=0
3b28b283
b={-1, 0, 1} [∵ bℤ]
Case 2: If a= 1
1+3b28
b273
b={-1, 0, 1} [∵ bℤ]
Case 3: If a= 2
4+3b28
b243
b={-1, 0, 1} [∵ bℤ]
Case 4: If a= 3
9+3b28
3b2 -1, which is not possible.
b
Similarly for a=4,5,...... and so on , b
Therefore,Range of R= Domain of R-1={-1, 0, 1}
Example:Let P = {2, 3, 4} and Q = {8, 9, 10, 11}.Let R be the relation ‘is factor of’ from P to Q. Write R in the roster form. Also, find Domain and Range of R.
Solution:
R consists of elements of the form (a, b) where a is a factor of b.
Relation R in the roster form is R = {(2, 8); (2, 10); (3, 9); (4, 8)}
Domain of R={2, 3, 4} and Range of R = {8, 10, 9}
1.Is Relation always a function?
A. Every function is a relation but every relation is not a function
2.Does every relation have an inverse?
A. Yes, every relation has an inverse relation.
3.Is Relation a set?
A. A Relation is a set of ordered pairs.
4.What is an arrow diagram?
A. An arrow diagram is a pictorial representation of a relation, showing the relationship among the various elements.