HOW TO UNDERSTAND RELATIONS AND FUNCTIONS ,INVERSE OF A FUNCTION

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                              Today we are going to  discuss Relations and Functions , "How to understand  Relations and  Functions, Inverse of a Function" under the topic  Relations and Functions.

Ordered-Pair Numbers :-

Ordered-pair number is written within a set of parentheses and separated by a comma.

For example, (5, 6) is an ordered-pair number; the order is designated by the first element 5 and the second element 6. The pair (3, 6) is not the same as (6,3) because they have different order. Sets of ordered-pair numbers can represent relations or functions.

Example of ordered pair :

(3,8),(2,1),(7,6)

Relation

A relation is a  set of ordered-pair numbers.

consider the following table

Number of Students	1	2	3	4	5	6
Marks Obtained	96	98	97	78	77	86

In the above table the numbers of students and marks obtained by them  is a relation and can be written as a set of ordered-pair numbers.

A= {(1, 96), (2, 98), (3, 97), (4, 88),(5,77),(6,86)}

When we collect all the elements written in 1st column of the ordered pairs and placed in a set then the set so formed is called  Domain of the relation.
The domain of A= {1, 2, 3, 4,5,6}

As all the elements written in 2nd column of the ordered pairs and placed in a set then the set so formed is called  Range of the relation.

The range of A = { 96, 98, 97,88 , 77, 86}

Function

A function is a relation in which every first element in ordered pairs have unique second element associated with them. Second  elements may or may not be same.

we can better understand this concept with the help of this video

Example

 {(1, 2), (2, 3), (3, 4), (4, 5),(5,6)}  is an example of function 

 { (1, 2), (2, 3), (3, 4), (4, 5),(5,6) } is a function because all the  first elements are different.

Example

{(1, 3), (3, 3), (2, 1), (4, 2)}  is an example of function 

 {(1, 3), (2, 3), (2, 1), (4, 2)}  is a function because all the first elements are different.

Example

{ (1, 6), (2, 5), (1, 9), (4, 3) }  is not an  example of function 

As in  {(1, 6), (2, 5), (1, 9), (4, 3)}  the element "1 "   appeared twice .

Example

{(2, 15), (3, 15), (4, 15), (5, 13),(6,18)}  is  an  example of function 

As in  {(2, 15), (3, 15), (4, 15), (5, 15)}   all the first elements are different.

Example

{(1, 1), (-1, 1),(2,4),(-2,4), (3, 9), (-3, 9),(4,16),(-4,16)}  is an  example of function although   the element "1" and "-1" ,"2" and "-2" , "3" and "-3"  ,"4", "-4" have same images. This is an example of many one function.

Question:-   Find x and y if: 

(i) (5x + 3, y) = (4x + 5,  2)

(ii) (x – y, x + y) = (8, 12)
(iii) ( 2x-y , y+5 ) = ( -2,3 )

Solution

(1)  Given  (5x + 3 , y) = (4x + 5, 2)
So By the equality of ordered pair elements,
1st element of the ordered number written on the left hand side will be equal to the 1st element of the ordered pair number written on the  right hand side . Therefore 

5x + 3 = 4x + 5   and y =  2 
5x-4x = 5 - 3   and y = 2 
x = 2 and y = 2

(ii) So By the equality of ordered pair elements
x – y = 8 and  x + y = 12

Solving these two equations for x and y 

 2x =20  and    10+ y =12 

x=10   y = 2

(iii) So By the equality of ordered pair elements
2x-y  =-2  , y+5 = 3 
2x = -2+y  , y = 3-5
2x = -2+y  , y = -2
Putting the value of y in 1st Equation ,we get
2x = -2 - 2
2x = -4
x = -2
so x= -2 and y =-2

Types of Relations

A relation R in a set A is called

(i) reflexive, if (a, a) ∈ R, for every a ∈ A,

(ii) symmetric, if (a, b) ∈ R implies that (a, b) ∈ R, for all a,b ∈ A.

(iii) transitive, if (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R, for all a, b,c ∈ A.

If you want to Read  HOW  TO  UNDERSTAND BINARY  OPERATIONS , RELATIONS  AND  FUNCTIONS || COMMUTATIVE || ASSOCIATIVE  click here 

Equivalence Relation

A relation R in a set A is said to be an equivalence  relation if R is reflexive, symmetric and transitive.

1 ) Let B be the set of all triangles in a plane with R a relation in B given by

R = {(T1, T2) : T1 is congruent to T2}. Then R is an equivalence relation.

2 ) Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7}  by

R = {(a, b) : both a and b are either odd or even}. Then R is an equivalence

one-one Function

A function f : X → Y is defined to be one-one (or injection ), if the images of distinct elements of X under f are distinct, i.e., for every x, y ∈ X, f (x) = f (y) implies x = y. Otherwise, f is called many-one.

Onto Function

A function f : X → Y is said to be onto (or surjective), if every element of Y is the image of some element of X under f, i.e., for every y ∈ Y, there exists an

element x in X such that f (x) = y.

Example

1   Function f : R → R, given by f (x) = 2x, is one-one and Onto As all the elements  have only one and uniqe image under f.

2  Function f : N → N, given by f (x) = 2x, is one-one but not onto.Because  the elements  have only one and unique image under f Therefore it is one one function .But not all elements of N have image under f 

e. g .  1,3,5,7... are not the image of any elements of N under f so it is not onto function

Example

The function f : N → N, given by f (1) = f (2) = 1 and f (x) = x – 1,

for every x > 2, is onto but not one-one.

Solution

Since f is Not one-one, as f (1) = f (2) = 1. 

But f is Onto, as given any y ∈ N, y ≠ 1,

Choose x = y + 1 s.t.

 f (y + 1) = y + 1 – 1

f (y + 1)  = y. 

Also for 1 ∈ N, 

we are given  f (1) = 1

Inverse of a Function

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

Example

Let S = {1, 2, 3}. Determine whether the functions f : S → S defined as below have inverses. Find f , if it exists.

(a) f = {(1, 1), (2, 2), (3, 3)}

(b) f = {(2, 2), (3, 1), (4, 1)}

(c) f = {(1, 5), (3, 4), (2, 1)}

Solution

(a) It is to  proved that  f is one-one and onto Hence f is invertible with the inverse f –1 of  f given by f –1 = {(1, 1), (2, 2), (3, 3)} = f.

(b) Since f (3) = f (4) = 1, f is not one-one, so that f is not invertible.

(c) Here  f   is one-one and onto, so that f is invertible with

 f –1 = {(5, 1), (4, 3), (1, 2)}.

Composition of Functions

Let f : A → B and g : B → C be two functions. Then the composition of f and g, denoted by gof, is defined as the function gof : A → C given by

gof (x) = g(f (x)), ∀ x ∈ A

Example

fof(x) = (16x + 12 + 18x -12 ) / ( 24x + 18 - 24x +16)

fof(x) = (34 x ) / ( 34)

fof(x) =  x  =  I(x)

Example

Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as f (2) = 3, f (3) = 4, f (4) = f (5) = 5 and g(3) = g(4) = 7 and g(5) = g(9) = 11. Find gof = ?

Solution

We are given

 gof (2) = g (f (2)) 

               = g(3) 

               = 7

 gof (3) = g(f (3)

             = g(4)

              = 7,

gof (4) = g(f (4)) 

           = g(5) 

             = 11 

and  gof (5) = g(f (5))

                   = g (5)                     

                    = 11

So gof ={(2,7),(3,7),(4,11),(5,11)} 

Example

conclusion

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