LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

RELATION & MAPPING - FUNCTION & RELATION

**Function & Relation -**

**We have already discussed earlier about the relation & mapping but now we would like to discussed about the function & relation. There are some example are given below**

__This is to be remembered that –__

**1) All
functions are relations but all relations are not functions.**

**The domain of
relation (1) = {1, 2, 3, 4, 5} = set A and the domain of relation (5) = {1, 2,
3, 4, 5} = set A**

**2) The domain
of a mapping or function from set A to set B = set A**

**The range of
relation (1) = {a, b, c, d, e} = set B, while the range of relation (5) = {a,
b, c, d, e} ⊂ set B**

**Next,
consider the relation “is half of” from set A = {3, 5, 7, 9} to set B = {6, 10,
14, 18}, illustrated by the arrow diagram alongside.**

**It is clear
that, 2 R 4, 5 R 10, 7 R 14,
9 R 18, the relation is a function because each element of A is
associated with only one member of B. If we denote this relation by ‘f ’ then
we can represent the function ‘f’ from set A to set B as f : A → B. **

**We can write f(3)
= 6, similarly f(5) = 10, f(7) = 14, and f(9) = 18. In other words the
images of 3, 5, 7 & 9 are 6, 10, 14,
& 18 respectively. We can represent f in the roster form as –**

** f = {(3, 6), (5, 10), (7, 14), (9, 81)}
where 3, 5, 7, 9 belongs to set A and 6, 10, 14, 18 belongs to set B**

**or, f = {(x,
y) : x Є A, y Є B and y = 2x}**

**if ‘x’ is an
element of A then the image of ‘x’ is 2x, i.e., f(x) = 2x**

**so, f = {(x,
f(x)) such that x Є A} = {(x, f(x) : x Є A}**

**we can also
write, f : A → B such that f(x) = 2x, x Є A **

**this is the equation form of the functions, The set A is called the domain
of the function ‘f ’, the set B is called the co-domain of the function ‘f’ and the set of
all the images of the elements of A is called the range
of the function ‘f’.**

**Next,
consider the relation “is double of” from set A = {49, 64, 81, 121} to set B =
{7, 8, 9, 11}, illustrated by the arrow diagram alongside.**

**It is clear
that, 49 R 7, 64 R 8, 81 R 9,
121 R 11, the relation is a function because each element of A is
associated with only one member of B. If we denote this relation by ‘f ’ then
we can represent the function ‘f’ from set A to set B as f : A → B. **

**We can write
f(49) = 7, similarly f(64) = 8, f(81) = 9, and f(121) = 11. In other words
the images of 49, 64, 81 & 121 are
7, 8, 9, & 11 respectively. We can represent f in the roster form as –**

** f = {(49, 7), (64, 8), (81, 9), (121, 11)}
where 49, 64, 81, 121 belongs to set A and 7, 8, 9, 11 belongs to set B**

**or, f = {(x,
y) : x Є A, y Є B and 2x = y}**

**if ‘x’ is an
element of A then the image of ‘x’ is 2x, i.e., f(2x) = x**

**so, f = {(x,
f(2x)) such that x Є A} = {(x, f(2x) : 2x Є A}**

**we can also
write, f : A → B such that f(2x) = x, 2x Є
A **

**this is the equation form of the functions, The set A is called the domain
of the function ‘f ’, the set B is called the co-domain of the function ‘f’ and the set of
all the images of the elements of A is called the range
of the function ‘f’.**