Relation and function_xii

Relation and Function
Class - XII

Learning Objectives
i) definition, domain and range of relation
ii) classify them as reflexive, symmetric and transitive
relation
iii) Find the equivalence class of an element
iv) definition, domain and range of function
v) classify functions as injective, surjective and bijective
function.
vi) define composition of functions
vii) define and find inverse of functions

Cartesian products of sets : Given two non-
empty sets A and B, the set of all ordered pairs
(x, y), where x ∈ A and y ∈ B is called Cartesian
product of A and B;
symbolically, we write
A × B = {(x, y) | x ∈ A and y ∈ B}
For Example
If A = {1, 2, 3} and B = {4, 5}, then
A × B = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)} and
B × A = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)}

Relations
A Relation R from a non-empty set A to a non empty
set B is a subset of the Cartesian product set A × B
The set of all first elements in a relation R, is called the
domain of the relation R, and the set of all second
elements called images, is called the range of R.
For example,
the set R = {(1, 2), (– 2, 3), ( 1 /2 , 3)} is a relation; the
domain of R = {1, –2, 1/ 2 } and the range of R = {2, 3}.
If n (A) = p, n (B) = q; then the n (A × B) = pq
The total number of possible relations from the set
A to set B = 2pq.

Note: A relation on set A means a relation from set
A to A i.e. it is a subset of AXA
Example1 : Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by
R = {(x, y) : y = x + 1 }
(i) Depict this relation using an arrow diagram.
(ii) Write down the domain, codomain and range of R.
Solution (i) By the definition of the relation,
R = {(1,2), (2,3), (3,4), (4,5), (5,6)}.
(ii) We can see that the domain ={1, 2, 3, 4, 5,}Similarly,
the range = {2,3,4,5,6} and the codomain = {1, 2, 3, 4, 5, 6}.

Types Of Relations
1. Reflexive relation
A relation R in A is said to be reflexive if xRx for all x ∈ A.
We may write (x, x) ∈ R
For Example: A relation R on A = {1,2,4} as R ={(x,y) : x divides y } is
Reflexive
As R = {(1,1),(1,2),(1,4),(2,2),(2,4),(4,4)} so (1,1)∈ R,(2,2)∈R and (3,3) ∈R
Note: A relation R on A is not reflexive if there exist x ∈ A
such that (x, x) does not ∈ R or x R x
For Example, The Relation R in the set A = {1, 2, 3, ..., 13, 14}
defined as R = {(x, y) : 3x – y = 0} is not Reflexive as
R = {(1,3),(2,6),(3,9),(4,12)} so there exist x = 1 ∈ A such that
(1,1) ∉ R

ACTIVITY 1
OBJECTIVE: To find the number of Reflexive
relations that can be defined on a finite Set A
with n elements
MATERIAL REQUIRED: Grid Paper, white paper,
glue, Scissors, etc.
METHOD: Cut a 2X2, 3X3, 4X4 grid from a grid
paper
Consider a set A with
One element say A ={x1}
Two elements say A ={x1, x2}
Three elements say A = {x1,x2, x3}

Similarly take a 4X4 grid and fill it accordingly
by taking A = {x1, x2, x3}
Write all possible Reflexive relations in each of
the above cases and count them.
Paste these grids on the activity sheet.

Conclusion:
If a finite set A has n elements then number of
reflexive relations that can be defined on set A
is _____________
Application: This activity can further be
extended to find no. of symmetric relations on
a finite Set A.

Types of Relations
2. Symmetric relation
A relation R in A is said to be symmetric if xRy ⇒ yRx
i.e (x, y) ∈ R ⇒ (y, x) ∈ R
For Example: A relation R on A = {1,2,3} as R ={(x , y) : x +y is even } is
Symmetric
As R = {(1,1), (1,3) (1,4) (2,2) (3,1),(3,3)} so (1,3) ∈ R implies (3, 1) ∈ R
Note: A relation R on A is not symmetric if there exist x, y ∈ A such
that (x, y) ∈ R but (y, x) ∉ R
For Example, The Relation R in the set A = {1, 2, 3, ..., 13, 14} defined
as R = {(x, y) : 3x – y = 0} is not symmetric as
R = {(1,3),(2,6),(3,9),(4,12)}
there exist x = 1, y = 3 ∈ A such that (1,3) ∈ R but (3,1) ∉ R

Types Of Relations
3. Transitive Relation
A relation R in A is said to be transitive if xRy, yRz ⇒ xRz
i.e. (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R
For Example: A relation R on A = {1,2,3} as R ={(x , y) : x + y is even } is
transitive as R = {(1,1), (1,3) (1,4) (2,2) (3,1),(3,3)}
as (1,3) ,(3,1)∈ R ⇒(1,1) ∈ R and (3,1), (1,3) ∈ R ⇒(3,3) ∈ R
Note: A relation R on A is not transitive if there exist x, y, z ∈ A such
that (x, y) ∈ R and (y, z) ∈ R but (x, z) ∉ R
For Example, The Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as
R = {(x, y) : 3x – y = 0} is not transitive as R = {(1,3),(2,6),(3,9),(4,12)}
So there exist x= 1,y = 3,z=9 ∈A such that (1,3) ∈ R (3,9) ∈ R but (1,9) ∉ R
Note: A relation R on A = {1,2,3} as R = {(1,2), (1,3) } is also transitive
As we are not able to find x, y, z ∈ A such that (x, y) ∈ R and
(y, z) ∈ R but (x, z) ∉ R

Equivalence Relation
A relation R in a set A is said to be an equivalence relation if R is
reflexive, symmetric and transitive.
Given an arbitrary equivalence relation R in an arbitrary set X,
R divides X into mutually disjoint subsets Ai called partitions or
subdivisions of X satisfying:
(i) all elements of Ai are related to each other, for all i.
(ii) no element of Ai is related to any element of Aj , i ≠ j.
(iii) ∪ Aj = X and Ai ∩ Aj = φ, i ≠ j.
The subsets Ai are called equivalence classes.
NOTE: Equivalence class of an element x is denoted by [x]
y ∈ [x] ⇔ (y, x) ∈ R and [x]= [y]

Example2 : Show that the relation R in the set Z of integers given by
R = {(a, b) : 2 divides a – b}is an equivalence relation.
Further find all equivalence classes of Z
Solution: R is reflexive, as 2 divides (a – a) = 0 for all a ∈ Z.
R is Symmetric
Further,Let (a,b)∈ R,then2 divides a – b therefore a- b = 2m ⇒b-a = 2(-m)
∴ 2 divides b – a. Hence, (b, a) ∈ R, which shows that R is symmetric.
R is Transitive
Let (a, b)∈R and (b, c)∈ R,then a–b and b– c are divisible by2.T.P (a,c) ∈ R
a-b= 2m, b-c = 2n . Now a-c = (a-b) + (b-c) = 2m +2n=2(m+n) which is
divisible by 2. Hence R is transitive
Further note that all even integers are related to zero, as (0, ± 2), (0, ± 4)
etc., lie in R
Similarly, all odd integers are related to one as (1, ± 3), (1, ± 5) etc., lie in R
Let E = [0] = {0, ± 2, ± 4------------} and O = [1] = { ± 1, ± 3------------}
Also E U O = Z

Example 3
Solution 2
Taking a = 10, b = 5, c = 3, we get (10, 5) and (5,3) ∈ R but (10, 3) ∉ R
So R is not transitive

Example . Do as H.W.
Example 4
Solution 4

Let N denote the set of all natural numbers and R be the relation
on N × N defined by
Show that R is an equivalence relation.
EXAMPLE 5 (HOTS)
          a, b R c, d ad b c bc a d .
Solution
Reflexive: (a, b) R (a, b) As ab(b + a) = ba(a + b)  Reflexive
Symmetric: Let (a, b) R (c, d) ad(b + c) = bc(a + d)
cb(d + a) = da(c + b) (c, d) R (a, b) R is symmetric
Transitive: Let (a, b) R (c, d) and (c, d) R (e, f)
ad(b + c) = bc(a + d) and cf(d + e) = de(c + f)
      
1 1 1 1 1 1 1 1
and
b c a d d e c f       
1 1 1 1 1 1 1 1
and
c d a b c d e f
       
1 1 1 1 1 1 1 1
a b e f a f b e
af(b+e) = be(a+f) (a, b)R(e, f) 
 Transitive Hence, R is an equivalence relation.

Problems for Practice
Q1. Let T be the set of all triangles in a plane with R a relation in T given by
R = {(T1, T2) : T1 is congruent to T2}. Show that R is an equivalence relation.
Q2. Let L be the set of all lines in a plane and R be the relation in L defined as
R = {(L1, L2) : L1 is perpendicular to L2}. Show that R is symmetric but neither
reflexive nor transitive.
Q3. Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by
R = {(a, b) : |a – b| is a multiple of 4} is an equivalence relation.
Find the set of all elements related to 1
Q4. Let R be a relation on the set A of ordered pairs of positive integers
defined by (x, y) R (u, v) if and only if xv = yu. Show that R is an equivalence relation.
Q5. Given a non empty set X, consider P(X) which is the set of all subsets of X.
Define the relation R in P(X) as follows:
For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation
on P(X)? Justify your answer.
Q6*. In the set of natural numbers N, define a relation R as follows: ∀ n, m ∈ N, nRm if on
division by 5 each of the integers n and m leaves the remainder less than 5, i.e. one of the
numbers 0, 1, 2, 3 and 4. Show that R is equivalence relation. Also, obtain the pairwise
disjoint subsets determined by R.

FUNCTIONS
Function:-A function f from set A to set B, associates
each element of set A to a element of set B.
we write this as f : AB.
If an element a Є A is associated to an element b Є B
then b is called “the f – image of a “ and “a is called
pre image of b”and it is written as b=f(a).

1
2
3
4
a
b
c
d
e
f1
A B A B
a
b
c
d
e
1
2
3
4
f2
1
2
3
4
a
b
c
d
e
A B
f3
EXAMPLE:---
1.Let A={1,2,3,4} and B={a, b, c, d, e} be two sets and let f1,f2,f3, and f4 be rules
associating elements B as shown in the following figures. Identify which of these
are functions from A to B
Solution:
 f1 is not a function as there is one element left (2) in set A which is not
mapped to any of the element in set B .
 f2 is also not a function as element “3” in set A does not have a unique image
in set B.
f3 is a function as each element of set A has a unique image in set B.

Domain of a Function
Let f : AB .then the set A is called the domain of f.
The set B is called the co domain of the function f.
Range of a Function
The set of all the f-images is called the range of f and is denoted by f
(A).
1
2
3
4
a
b
c
d
e
A B
f From the adjoining figure, we observe
that f is a function from A to B
Domain is A
Co- domain is B
Range is {a, b, c, d}

1. One-One function or injective function :
A function f : A  B is said to be one-one function or an
injection if different element s of A have different
images in B.
Thus f: AB is one-one iff a ≠ b ⟹ f(a) ≠f(b) for all a ,b єA
or f(a) = f(b)⟹a=b for all a ,b єA
Example: Here f : AB is one –one function
A B
Types of Functions
a
b
c
d
1
2
3
4
5

Note: In order to define one-one functions from A to B,
n(A)≤ n(B)
Further Let n(A) = m and n(B) = n. Then number of 1-1
functions from A to B is P(n, m)
A function f : A to B is not one-one if there exist x1, x2Є A such
that x1≠x2 but f(x1) = f(x2)
Example:
f: R R defined by f(x) = x2 is not one-one as
-1 ≠ 1 but f(-1)=f(1) = 1

Onto functions or Surjective functions :
A function f: A B is said to be an onto function if every element of B is
the f-image of some element of A i.e. range of f is the Co domain of f .
In other words
A function f from AB is an onto function iff for each b є B there exist
a є A s.t f(a) = b.
a
b
c
d
1
2
3
f
Here f is onto as range of f = co-domain of f

A function f : A to B is not onto if there exist some b Є B such that it
has no pre- image in A.
a
b
c
d
1
2
3
4
f
A B
Here f is not onto because
There exist 4 Є B which has no pre-image
Note: In order to define one-one functions from A to B, n(A)≥ n(B)

Procedure to verify whether f from A to B is onto or not
Method 1
Find the range of f by putting f(x) = y
and obtaining x in terms of y
If Range = B(co-domain) then
conclude that f is onto.
Method 2
1. Begin with arbitrary element of B say y
Let y = f(x)
2. Obtain the value of x in terms of y
3. Verify that x so obtained exist in
A(domain) for every y in B
4. In case we can get some y in B for which
x does not exist in A then conclude that f is
not onto.
Example : Check whether f from R R s.t. f(x) = x2+2 is onto or not
Method 1
Let y = x2 +2
⟹ x =
For range y-2≥0 i.e [2, ∞ ) ≠ R(co-
domain)
So f is not onto
2 y
Method 2
Let y ЄR (co domain)
Let f(x) = y
i.e. x2 +2 = y ⟹ x =
taking y = 0 ЄR we get
x = ∉ R (domain)
So f is not onto
2 y
2

Bijective Function
A function f : A B is a bijection if it is one -one as well as onto i.e.
1. f(x) = f(y) ⇒x=y or all x, y є A
2. for all y є B, there exist x є A s.t f(x)=y
Here f : A—> B is a bijection
Note: In order to define bijective function from A to B, n(A) = n(B)
Further number of bijective functions from A to B is n! where n(A)
= n(B) = n

Example(HOTS)
Solution:
When x and y are of opposite sign
Let x is positive and y is negative then
f(x) is +ve but f(y) is –ve so f(x) ≠ f(y)
Similarly for other case f(x) ≠ f(y)
So f is one-one
Case i) let -1<y <0
Let f(x) = y i.e = y
Clearly x < 0
∴ = y ⇒ x = ∈R
||1 x
x

x
x
1 y
y
1
Case ii) let 0<y <1
Let f(x) = y i.e = y
Clearly x > 0
∴ = y ⇒ x = ∈R ∴ f is onto
||1 x
x

x
x
1 y
y
1
Hence f is bijective
bijective

Q1.Show that the function f : R → R defined by f (x) is neither one-one nor onto.
Q2. Let the function f : R → R be defined by f (x) = cosx, ∀ x ∈ R. Show that f is
neither one-one nor onto.
Q3. Let A = R – {3}, B = R – {1}. Let f : A → B be defined by f (x) =
Show that f is bijective.
3
2


x
x
Q4.
Q5.
Questions for Practice

Composition of Functions
Definition : Let f: AB and g : B C be two functions. Then a
function gof from AC is defined by
(gof) (x) = g (f (x) ) for all x єA
is called the composition of f and g .
X f(x) g f (x)
f g
gof gof
Note: i) In order to define composition gof range of f must be subset of domain of g
ii) |||ly In order to define composition fog range of g must be subset of domain of f

Example: If f from R R defined by f(x) = x² and g : RR defined by g(x) = 2x+1 then find fog
and gof .
Solution : Here range of f is a subset of domain of g and range of g is a subset of domain of f.
∴ fog and gof are defined
(fog)(x) = f ( g (x) )
= f (2x+1)
= (2x+1)² = 4x2+4x+1
and (gof)(x) = g(f(x))
= g (x*2)
= 2x2+1
Clearly in this case fog is not equal to gof.
Note : It is not necessary that both fog and gof exists and if they exists they may not be equal.
Example : If the function f and g are given by f = {(1,2), (3,5), (4,1)} and g = {(2,3), (5,1),(1,3)},
Find fog and gof if they exist.
Solution: Range of f = {2,5,1} domain of g ={2,5,1} so gof exist
range of g = {3,1} domain of f = {1,3,4} so fog exist
hence gof = {(1,3),(3,1),(4,3)} Similarly fog = {(2,5), (5,2), (1,5)}
gof(1) =g(f(1)) = g(2) = 3. similarly we can find other values

A function f : X → Y is defined to be invertible, if there exists a function
g : Y → X such that gof = IX and fog = IY. The function g is called the inverse of f and
is denoted by f –1.
Invertible Functions
Note: If f is invertible, then f must be one-one and onto and conversely, if f is
one-one and onto, then f must be invertible
i) f1 is not onto so we cannot
define g1 from X2 toX1
ii) F2 is neither 1-1 nor onto so we
cannot define g2 from X2 to X1
iii) f3 is not 1-1 so we cannot define
g3 from X3 to X1
iv) f4 is both 1-1 and onto so we
can define g4 from X4 to X1
which will also be 1-1 and onto

Example
For 1-1
f(x) = f(y)
9x2+6x-5=9y2+6y-5
9(x+y)(x-y)+ 6(x-y) = 0
(x-y)(9x+9y+6) = 0
X-y = 0 or x+y = -2/3 rejected as x, y ∈ R+
So f is 1-1

Relation and function_xii

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