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Relation and Functions Revision Notes class 12th Cbse

Relation and Functions

In previous class , we initiated the study of relation and functions , where we studied about domain, co domain and range along with different types of specific real valued function and their graphs. all these concepts are the basics of relations and functions. In this chapter, we will continue our study with different types of relation and functions, composition of functions , invertible functions and binary operation.


Concept I Relations 

Ordered Pair 

A pair of elements listed in a specific order separated by comma and enclosing within the parentheses, is called an ordered pair e.g (a,b) is an ordered pair with a as first element and b as the second element.

Cartesian product 

the set of all ordered pairs (a,b) such that a `\in` A and b `\in` B is called the Cartesian product or cross product of sets A and B ; and it is denoted by A `\times` B . Similarly , the set of all ordered pair (b,a) such that b `\in` B and a `\in` A is called the cartesion product or cross product of sets B and A ; and it is denoted by B `\times` A . 

thus A `\times` B = { (a,b) : a `\in`  A, b `\in` B } and  B `\times` A = { (b,a) : b `\in`  B, a `\in` A }  

e.g if A= { 1,2,3} and B= {4,5} then A `\times` B is {(1,4) (1,5) (2,4) (2,5)(2,5)(3,4)(3,5)(3,5)} 

and B `\times` A is {(4,1)(4,2)(4,3)(5,1)(5,2)(5,2)(5,3)}

Relation 

In the mathematics, the concept of the term 'relation' has been drawn from the meaning of relation in English language , according to which two object or quantities are related , if there is a recognisable connection or link between two object or quantities  

Relation on set A and B

let A and B be two non-empty sets , then a relation R from set A and B is a subset of A `\times` B i.e. R `\subset` A `times` B .

The subset R is derived by describing a relationship between first and secound elements of ordered pairs in A`\times`B .the secound element is called image of first elements. 

 

Relation on a set

Let A be a non-empty set, then a relation from A to itself , i.e. a subset of A `\times` A , is called a relation on set A ( or a relation in set A ) .

 

example let A = {1,2,3,4} , then R = {(a,b) `\in` A`\times` A : a-b=3 } is a relation on set A . 

Domain , range and co domain of relations 

let us consider a relation R from set A to set B (i.e R `\subset` A `\times` B ) such that R= {(a,b) : a `\in` A and b `\in` B } then , the set of all first elements of the ordered pairs in R is called the domain of relation and the set of all element of ordered pair in R is called range of relation and set B element called co-domain 

 

Example 1 If R = {(x,y) : x+2y= 8} is a relation on a set of natural numbers (N) , then write the domain, range and co-domain of R {All India 2014}

solution 

Given , R = {(x,y): x+2y = 8 } on a set of natural numbers,=.

consider, x+y=8, which can be re written as y = `\frac {8-x}{2}`.

now substitute value of x  from natural numbers, such that y `\in`N. 

on putting x=2 , we get y= `\frac {8-2}{2}` =3

on putting x=4 , we get y= `\frac {8-4}{2}` =2

on putting x=6 , we get y= `\frac {8-6}{2}` =1

thus, R = {(2,3)(4,2)(6,1)}  {there is no value of x , for which y `\in` N}

domain of R = {2,4,5}, co-domain of R=N and range of R ={3,2,1}.

Type of Relations 

Empty or Void Relation

Relation R in set A is called an empty relation , if no element of A is related to any element of A , i.e R=  `\phi \subset A \times A` 

Universal Relation

Relation R in a set A is called an universal relation, if each element of A is related to every element of A  i.e R= A `\times` A.

Identity Relation 

Relation R in a set A is called an identity relation, if each element of A is related to itself only and it is denoted by I , i.e I = R = {(a,a): a `\in` A}

Reflexive Relation 

Relation R in set A is called reflexive relation if (a,a) `\in` R, every a `\in` A, i.e aRa , for all a `\in` A.

Symmetric Relation

Relation R in a set A is called symmetric Relation, if (a,b) `\in` R  `\Rightarrow` (b,a) `\in` R for every a,b `\in` A

Transitive Relation 

Relation R in a Set A is called Transitive Relation, If (a,b) `\in` R and (b,c) `in` R `\Rightarrow` (a,c) `\in` R, For all a,b,c `\in` A

 Method to Solve Problems Bases on Type of Relation 

In these types of problems, a set and a relation defined on that set is given to us and we have to check or show that given relation is Reflexive or symmetric or Transitive. For this , firstly we denote the given set as A and given relation as R. then, 

For Reflexive,

 we have to show that for all a `\in` A (a,a) `\in` R . for this, we take arbitrary element of set A in form of a variable and then check whether (x,x) satisfy the given condition , then R is reflexive otherwise not. 

For symmetric,

We have to show that for a,b `\in` A if (a,b) `\in` R , then (b,a) `\in` R. for this, we take two arbitrary elements of set A in the form of two variables ( says x and y ) such that (x,y) `\in` R 

and check (y,x) satisfy the given condition of R or not . If they satisfy the given condition, then R is symmetric otherwise not. 

For Transitive,

We have to show that for a,b,c `\in` A , if (a,b) `\in` R and (b,c) `\in` R then (a,c) `\in` R . for this, we take three arbitrary elements of set A in the form of three variables 

such that (x,y) `\in` R and (y,z) `\in` R and then check whether (x,z) satisfy the given condition, then R is Transitive otherwise not.

Example 2 Check Whether the rrlation R defined in the set A = {1,2,3,4,5,6} as R= {(x,y): y is divisible by x} is Reflexive symmetric and Transitive. {NCERT}

Solution : Given R = {(x,y): y is divisible by x }

and A = {1,2,3,4,5,6}

Reflexive let x `\in` A be any arbitrary element. We know that, x is divisible by x.
[Every real number except is divisible by itself]
`\Rightarrow` (x,x) `\in` R since , x `\in` A was arbitrary element,  therefore  (x,x) `\in` R, For all x `\in` A . So, R is Reflexive.
Symmetric 
Clearly 2,4 `\in` A and 4 is divisible by 2, but 2 is not divisible by 4.
(2,4) `\in` R but `\notin` R 
So , R is not Symmetric.
Transitive 
Let x,y,z `\in` A such that (x,y) `\in` R and (y,z) `\in` R. Now as (x,y) `\in` R, therefore y is divisible by x .
`\frac {y}{x}` = k ...(1)
Where k is a natural number 
And as (y,z) `\in` R, therefore z is divisible by y 
`\frac {z}{y}` = h ...(2)
Where h is a natural number 
On multiple equ (1) and (2) we get 
`\frac {y}{x}` x `\frac {z}{y}` = kh `\Rightarrow` `\frac {z}{x}= kh`
Where kh is a natural number 
z is divisible by x.
Thus (x,z) `\in` R for all (x,y)(y,z) `\in` R,
 i.e (x,y) `\in` R (y,z) `\in` R,  `\Rightarrow` (x,z)`\in` R 
So R is Transitive.

Example 3 Show that the relation R is the set {1,2,3} given by R = {(1,1)(2,2)(3,3)(1,2)(2,3)} is reflexive but neither symmetric nor transitive. {NCERT}

sol let given set A = {1,2,3}
and R= {(1,1)(2,2)(3,3)(1,2)(2,3)}
Reflexive here, 1,2,3 `\in` A and (1,1)(2,2)(3,3) `\in` R i.e. for all a `\in` A, (a,a) `\in` R so, R is reflexive 
Symmetric here, (1,2) `\in` R but (2,1) `\notin` R where 1,2 `\in` A. so R is not symmetric 
Transitive here {1,2} `\in` R and (2,3) `\in` R but (1,3) `\notin` R where 1,2,3 `\in` A so R is not transitive. 

Example 4 Check whether the Relation R defined in the set A ={1,2,3,...,13,14} as R = {(x,y): 3x-y=0} is reflexive, symmetric and transitive. {NCERT}

Sol. Given , R = {(x,y): 3x - y = 0}
and A = {1,2,3,4,5,6,7,8,9,10,11,12,13,14}
Reflexive 
here, 2 `\in` A 
if (2,2) `\in` R, then 3(2)-2 =6-2 = 4 `\neq` 0
(2,2) `\notin` R
so, R is not Reflexive. 
Symmetric 
Here, 1,3 `\in` A
Clearly, (1,3) `\in` R as 3(1)-3 = 0
but if (3,1) `\in` R , then 3(3)-1=9-1=8 `\neq` 0 
thus (1,3) `\in` R `\Rightarrow` (3,1) `\in` R
so, R is not Symmetric 
Transitive 
Here, 1,3,9 `\in` A
Clearly (1,3) `\in` R as 3(1) -3= 0 
and (3,9) `\in` R , as 3(3)-9=0
but if (1,9) `\in` R, then 3(1)-9 = -6 `\neq` 0 `\rightarrow` R
thus, (1,3) `\in` R and (3,9) `\in` R `\Rightarrow` (1,9) `\in` R 
so R is not transitive 

Example 5 Let A be the set of all lines in a plane and R be a relation in A defined by R ={(L1 , L2): L1 ⟂ L2 }. (i) Show that R is symmetric but neither reflexive nor transitive. (ii) Give an example of symmetric relation in our real life  {NCERT}

sol (i) Given, R = {(L1, L2): L1 ⟂ L2} and A=set of all lines in a plane 
Reflexive Let L `\in` A be any arbitrary element . 
then , (L,L) `\in` R `\Rightarrow` L ⟂ L
But any line cannot be perpendicular to itself, so it is not true. i.e (L,L) `\notin` R, for all L `\in` A
so, R is not reflexive 
Symmetric 
Let L1 , L2 `\in` A, such that (L1 , L2 ) `\in` R `\Rightarrow` L1 ⟂ L2.
`\Rightarrow` L2 ⟂ L1
`\Rightarrow` (L2 , L1 ) `\in` R
so, R is symmetric 
Transitive 
Let L1, L2 , L3 `\in` A, such that (L1 ⟂ L2 ) `\in` R `\Rightarrow` L1⟂ L2 and (L2 ⟂ L3 ) `\in` R `\Rightarrow` L2⟂ L3
but L1 is not perpendicular to L3 as shown in figure 


i.e (L1, L3) `\notin` R
thus, ( L1 , L2 ) `\in` R,  (L2 , L3) `\in` R  not implies (L1, L3) `\in` R
so, R is not transitive 
(ii) in our real life , relation between two friend is a symmetric relation. 

Example 6 Give an example of a relation, which is (i) symmetric but neither reflexive nor transitive. (ii) transitive but neither reflexive nor symmetric. (iii) reflexive and symmetric but not transitive. (iv) reflexive and transitive but not symmetric. (v) symmetric and transitive but not reflexive.  {NCERT}

Sol: (i) let A ={1,2,3} and defined a relation R on A as R={(1,2)(2,1)} Then , R is symmetric, as (1,2)   `\in` R `\implies` (2,1) `\in` R. R is not reflexive, as 1 `\in` A but (1,1) `\notin` R, R is not transitive as (1,2) `\in` R, (2,1) `\in` R but (1,1) `\notin` R.
(ii) Let A = {1,2,3} and defined a relation R on A as R={(1,2)(2,2)}. then , R is transitive, as (1,2)(2,2) `\in` R `\implies`(1,2) `\in` R ,R is not reflexive, as 1 `\in` A but (1,1) `\notin` R,R is not symmetric, as (1,2) `\in` R but (2,1) `\notin` R.
(iii) Let A = {1,2,3} and defined a relation R on A, as R={(1,1)(2,2)(3,3)(1,2)(2,1)(2,3)(3,2)} then , R is reflexive, as for each a `\in` A, (a,a) `\in` R.
R is symmetric, as (1,2) `\in` R `\Rightarrow` (2,1) `\in` R and (2,3) `\in`R `\Rightarrow` (3,2) `\in` R.
R is not transitive, as (1,2) `\in` R , (2,3) `\in` R but (1,3) `\notin` R 
(iv) Let A = {1,2,3} and defined a relation R on A as R = {(1,1)(2,2)(3,3)(1,2)} then, R is reflexive, as for each a `\in` A, (a,a) `\in` R.
R is not symmetric, as (1,2) `\in` R but (2,1) `\notin` R
(v) Let A ={1,2,3} and defined a relation R on A as R={(2,2),(3,3)}
then, R is symmetric as R does not have any element of the type (a,b), where a `\ne` b
R is transitive, as R does not have element of the type (a,b) and (b,c).
R is not reflexive, as 1 `\in` A but (1,1) `\notin` R

Equivalence Relation

A relation R on a set A is said to be an equivalence relation on A , iff it is 
  • Reflexive i.e. aRa or (a,a) `\in` R , `\forall` a `\in` A.
  • symmetric i.e. aRb `\Rightarrow`  bRa or (a,b) `\in` R `\Rightarrow` (b,a) `\in` R , where a,b `\in` A
  • transitive i.e if aRb and bRc, then aRc or (a.b) `\in` R and (b,c) `\in` R `\Rightarrow` (a,c) `\in` R , where a,b,c `\in` A  

Example 7 Let T be the set of all triangle in a plane with R is a relation in T given by R={(T1,T2): T1 is congruent to T2 and T1 ,T2 `\in` T } show that R is an equivalence relation.  {NCERT}

Sol Given , T = set of all triangle in a plane and R={(T1,T2): T1 is congruent to T2 and T1 ,T2 `\in` T }
we know that, two triangle are said to be congruent, if they have same shape and same size.
Reflexive 
let T1 `\in` T be any arbitrrary element. we know that, every triangle is congruent to itself. So, (T1,T1) `\in` R. 
Now , as T1 was arbitrary element of T, therefore R is reflexive. 
Symmetric 
Let T1 , T2 `\in` T , such that (T1,T2) `\in` R `\Rightarrow` T1 is congruent to T2.
`\Rightarrow` T2 is congruent to T1. [if a triangle is congruent to another triangle, then they are congruent to each other]
`\Rightarrow` (T2,T1) `\in` R
R is symmetric.
Transitive
let T1 , T2 , T3 `\in` T , such that (T1,T2) `\in` R and (T2,T3) `in` R.
`\Rightarrow` T1 is congruent to T2 and T2 is Congruent to T3
`\Rightarrow` T1 is congruent to T3
`\Rightarrow` (T1,T3) `\in` R 
R is transitive.
thus, relation R is Reflexive, symmetric and Transitive. so , R is equivalence Relation.

Equivalence Classes

Consider an equivalence relation R on the set A and let a `\in` A. then , the set of all those element of A which are related to a, is called the equivalence class of R determined by a ; and it is denoted by [a]. thus, [a] = {b `\in` A : (a,b) `\in` R}

Example 8 Let R be the equivalence relation in the set A ={0,1,2,3,4,5} given by R = {(a,b) : 2 divides (a-b)} then, write equivalence class {0} {Delhi 2014C}

sol : Given A = {0,1,2,3,4,5} and R={(a,b):2 divides (a-b)} clearly, [0]={b `\in` A : (0,b) `in` R}={b `\in` A : 2 divides (0-b)}= {b `\in` A : 2 divides -b}. So . b can be 0,2,4.
hence, equivalence class [0]={0,2,4}

Example 9 {HOTS} let A = {1,2,3,..,9} and R be the relation in A `\times` defined by (a,b)R(c,d), if a+d = b+c for (a,b)(c,d) in A `\times` A . Prove that R is an equivalence relation and also obtain the equivalence class [(2,5)] {Delhi 2014 ; NCERT Exemplar}

sol Given a relation R in A `\times` A, where A={1,2,3,...,9}. defined as R = {((a,b)(c,d)) : a+d = b+c} or (a,b)R(c,d), if a+d= b+c
Reflexive
let (a,b) be any arbitrary element of A `\times` A.
i.e. (a,b) `\in` A `\times` A, where a,b `\in` A
now as a+b = b+a 
(a,b)R(a,b)
so, R is Reflexive.
Symmetric
let (a,b)(c,d) `\in` A `\times` A, such that (a,b)R(c,d), then a+d = b+c
`\Rightarrow` b+c = a+d `\Rightarrow` c+b= d+a `\Rightarrow` (c,d)R(a,b)
so, R is symmetric. 
Transitive
let (a,b),(c,d),(e,f) `\in` A `\times` A such that (a,b)R(c,d) and (c,d)R(e,f).
then, a+d= b+c  and c+f  = d+e
on adding LHS to LHS Terms and RHS to RHS Terms , we get 
a+d+c+f=b+c+d+e
`\Rightarrow` a+f= b+e `\Rightarrow` (a,b)R(e,f)
so, R is transitive. thus, R is Reflexive, symmetric and transitive. hence, R is an Equivalence Relation.
Now equivalence class containing an element x of A is given by {y:xRy} Here, (a,b)R(c,d)
`\Rightarrow` a+d= b+c.
So for (2,5) , we will find (c,d) such that 2+d=5+c.
clearly
(2,5)R(1,4) as 2+4 = 5+1
(2,5)R(2,5) as 2+5 = 5+2
(2,5)R(3,6) as 2+6 = 5+3
(2,5)R(4,7) as 2+7 = 5+4
(2,5)R(5,8) as 2+8 = 5+5
(2,5)R(6,9) as 2+9 = 5+6
where 1,2,3...,9 `\in` A. hence, equivalence class [(2,5)]={(1,4)(2,5)(3,6)(4,7)(5,8)(6,8)},

Example 10 Show that the relation R in the set A={1,2,3,4,5} given by R ={(a,b): |a-b| is divisible by 2} is an equivalence relation. write all the equivalence classes of R. { All India 2015C}

Sol.  We have a relation R in set A = {1,2,3,4,5} defined as R = {(a,b): |a-b| is divisible by 2}
clearly R={(1,1)(1,3)(1,5)(2,2)(2,4)(3,1)(3,3)(3,5)(4,5)(4,2)(5,1)(5,3)(5,5)}
Reflexive 
For any `\in` A, we have |a-a|=0, which is divisible by 2.
`\Rightarrow` (a,a) `\in` R, \forall` a `\in`A
thus, R is reflexive.
Symmetric
let a,b `\in` A , such that (a,b) `\in` R
`\Rightarrow` |a-b| is divisible by 2
`\Rightarrow` |a-b| = 2 `\lambda` for some `\lambda` `\in` N 
`\Rightarrow` |b-a| = 2 `\lambda` for some `\lambda` `\in` N 
`\Rightarrow` (b,a) `\in` R
thus, R is symmetric. 
Transitive
Let a,b,c `\in` A, such that (a,b) `\in` R and (b,c) `\in` R
`\Rightarrow` |a-b| is divisible by 2  and |b-c| is divisible by 2.
`\Rightarrow` |a-b|=2 `\lambda` and |b-c|= 2 `\mu`
`\Rightarrow` a-b= `\pm` 2 `\lambda` and  b-c= `\pm` 2 `\mu`
Now, |a-c| = | (a-b) + (b-c) | = |`\pm` 2 `\lambda` + `\pm` 2 `\mu`| = 2 |`\pm` `\lambda` + `\pm` `\mu` | = 2(some positive number)
`\Rightarrow` |a-c| is divisible by 2 `\Rightarrow` (a,c) `\in` R
thus, R is Transitive. Now R being Reflexive, Symmetric and transitive. Hence, R is an equivalence relation . Set of all those elements of A , which are Related to a, is called the equivalence class of R determined by a.
i.e. [a]={x `\in` A : (a,x) `\in` R}
Equivalence class 
[1]={1,3,5}, [2]={2,4},[3]={1,3,5}, [4]={2,4} and [5]={1,3,5}

Concept II Functions and their Types

Function ( Mapping ) as a rule 

Function (Mapping ) as a set of Ordered Pairs 

Domain, Co-domain and Range of a function

Relation Between Range and Co-domain 

Geometrical Method to Check whether a given Expression is a function or not

Types of Functions

One-One function ( Injective function )

Many-One Function

Method to check whether a function is one or many-one 

Onto(Surjective) and Into Function

Method to check whether the Function is Onto or Into

One-One and Onto function ( Bijective )



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