Relations and Functions

Class XII

Relations and Functions

All functions are relations but all relations are not functions.

Relations

A relation R in a set A is a subset of AxA.

Empty relation φ and  universal relation AxA are extreme relations.

Empty Relation

A relation R in a set A is said to be empty, if no element of A is related to any element of A.

R = φ ⊂AxA

Universal Relation

A relation R in a set A is said to be universal relation, if each element of A is related to  every element of A.

R = AxA

Method to Represent a Relation

There are two methods to represent a Relation.

1. Roster Method 2. Set Builder Method.

For Example

Relation R in the set

R = {1,2,3,4}

is a relation represent roster method.

R = {(a,b):b = a +1}

is a relation represent set builder method and read as R is a relation as a related to b or aRb if and only if b = a+1.

Reflexive, Symmetric and Transitive and Equivalence are the types of Relation.

A relation R in set A is said to be reflexive, Symmetric and Transitive if and only if 

1. Reflexive, if (a,a) ε R for every aεA.
2. Symmetric, if  (a,b)εR implies that (b,a)εR for all elements of A.
3. Transitive, if (a,b)εR and (b,c)εR implies (a,c)εR for all elements of A.
4. Any relations which satisfy all of these conditions are said to be Equivalence Relations or other words the relation which is reflexive, symmetric and transitive is said to be Equivalence Relation.

Exercise 1.1

NCERT Text Books

Question No. 1 Determine whether each of the following relations are reflexive, symmetric and transitive.

(i) Relation R in the set A = {1,2,3, ................. 13,14}

defined as 

R = {(x,y):3x - y = 0}

Solution:

Roster form 

R = { (1,3), (2,6),(3,9), (4,12)}
(a,a) is not the element of  R for every elements of  A.
Hence, it is not reflexive.

(a,b)εR does not imply that (b,a)εR for all elements of A.
Hence, it is not symmetric.

 (a,b)εR and (b,c)εR does not imply (a,c)εR for all elements of A.
Hence, it is not transitive.

Therefore, this is none any one of these three.

(ii) Relation R in the set N of natural numbers defined as 
R = {(x,y):y = x + 5 and x<4}

Solution:
Roster form
R = { (1,6),(2,7), (3,8)}

(a,a) is not the element of  R for every elements of  A.
Hence, it is not reflexive.

(a,b)εR does not imply that (b,a)εR for all elements of A.
Hence, it is not symmetric.

 (a,b)εR and (b,c)εR does not imply (a,c)εR for all elements of A.
Hence, it is not transitive.

Therefore, this is none any one of these three.

(iii) A = {1,2,3,4,5,6}
R = {(x,y):y is divisible by x}

Solution:
Roster form
R = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (1,2), (1,3), (1,4), (1,6), (2,4), (2,6), (3,6)}

From roster form it is clear that;
(a,a) is the element of R for every element of A.
Hence, it is reflexive.

(a,b)εR does not imply that (b,a)εR for all elements of A.
Hence, it is not symmetric.

 (a,b)εR and (b,c)εR does not imply (a,c)εR for all elements of A.
Hence, it is not transitive.

Similarly we can do remain problems of this section.