Mathematics for Biotechnology and Data Science
Chapter 3: Application of Set Theory
Chapter 6: MAPPING OR FUNCTIONS
Chapter 8: Inverse of a Function
Chapter 6: MAPPING OR FUNCTIONS
6.1: Function:
A function is defined as
a relation between a set of inputs having one output each. In simple words, a
function is a relationship between inputs where each input is related to
exactly one output. Every function has a domain and co-domain or
range. A
function is generally denoted by 
where 
is the
input. The general representation of a function is 
.
Definitions:
1. Write the definition of a mapping.
Let A and B be two non-empty sets. A relation f
from A to B, i.e., a subset of 
, is called a function
(or a mapping or a map) from A to B if
I.
For each 
there exists 
such that 
,
II.
and 
Thus, a non-void subset f of is a function from A to B if each element of A
appears in some ordered pair in f and no two ordered pairs in f have the same
first element.
2. What is image?
Ans. If
, then b is called the
image of an under f.
3. Give one example of an image or not an image
Ans. Let 
and 
be three subsets of as given below
Then 
is a function from A to B but 
and 
are not functions from A to B. is not a function from A to B, because 
has two images 2 and 3 in B and is not a function from A to B because 
has no image in B.
4. What is the domain and range of a function?
Ans. If a function f is expressed as the set of ordered pairs, the domain of f is the set of all first components of members of f, and the range of f is the set of second components of members of f i.e.,
Domain of 
Range of 
Problems with solutions:
1. Express the following function as sets of ordered pairs and determine their ranges
Solution:
We have, 
So, 
So, the Range of 
2.
Find the domain for which the functions 
are equal
Solution.
We have 
Thus, f(x) and g(x) are equal on the set
.
3.
Is 
a function? If this is described by the
formula, 
,
then what values should be assigned to 
.
Solution.
Since no two ordered pairs in g have the same first component. So g is a
function such that
.
It is given that
. Therefore
4.
Let f be a relation on the set N of
natural numbers defined by
.
Is f a function from N to N? If so, find the range of f.
Solution.
Since for each
, there exists a unique
such that
. Therefore, f is a
function from N to N.
Now, the range of 
5.
Let 
be a function described by the formula 
for some integers a, b. Determine a, b.
Solution.
We have 
It is given that
. Therefore
Thus, 
. Clearly, 
are true.
Hence a=2, b=-1
6.
Let f be a subset of 
defined by
.
Is f a function from Z into Z? Justify your answer
Solution. We observe that:
So, f is not a function from Z to Z.
Exercise problem
1.
Let, 
and 
be a function defined by 
.
Find (a) Range of A,
(b) Preimages of 6, -3 and 5
Ans.
(a) 
(b) 
2.
If a function 
be defined by
Find 
Ans.
3.
Let 
is defined by
. Determine
(a) Range of f
(b) 
(c) 
Ans.
(a) 
(b) {-2, 2}
(c) 
4.
Let and 
be two functions defined as 
. Are they equal
functions?
Ans. No, since the domain of f is not equals to the domain of g
5.
If 
are three functions defined from R to R as
follows:
(i) 
Ans.
6.
Let 
and be a function given by f(x)=highest prime
factor of x. Find the range of f.
Ans.
5.2: ONE-ONE, ONTO MAPPINGS
Definitions:
5.2.1: One-one
function (Injection): A function 
is said to be a one-one function or an
injection if different elements of A have different images in B.
Thus, is one-one 
Illustrations:
1. A function which associates to each country in the world, its capital is one-one because different countries have their different capitals.
2. Let
We have, 
Clearly different elements in A have different images under
function f. So is an injection.
Algorithm to check the injectivity of a function
Step I: Take two arbitrary elements x, and y (say) in the domain of f.
Step II: Put f(x) = f(y)
Step
III: Solve f(x) = f(y). If f(x) = f(y) gives x=y only, then is a one-one function (or an injection).
Otherwise not.
Note: Let and let 
Then
is always true from the definition. But 
is true only when f is one-one.
5.2.2:Onto
Function (Surjective): A function is said to be an onto function or a surjective
if every element of B is the f-image of some element of A i.e., f (A)=B or
range of f is the co-domain of f.
Thus, is a surjective iff for each
.
Algorithm to check the surjectivity of a function
Let be the given function.
Step I: Choose an arbitrary element y in B.
Step
II: Put
Step
III: Solve the equation for 
and obtain in terms of
. Let 
Step
IV: If
for all values of
, the values of obtained from are in A, then f is onto. If there are some, for which
, given by is not in A, then f is not onto.
is a bijection if it is one-one as well as
onto.
In other words, a function is a bijection if
(i) It is one-one i.e., 
(ii) It is onto i.e., for all 
, there exists 
Problem with solution:
1. Find whether the following function is one-one or not?
Solution: Let x, y be two arbitrary elements of R such that f(x) = f(y). Then,
Hence, f is one-one function from R to itself.
2. Discuss the surjectivity of the function
Solution: Let y be an arbitrary element of Z (c0-domain). Then
Clearly, if 
, then 
. Thus, 
does not have its pre image in Z (domain).
Hence f is not an onto function.
3.
Show that the function 
defined by 
is a bijection.
Solution: For injectivity, let x, y be any two elements of R (domain).
Then, 
Thus, 
So, f is an injective map.
For subjectivity, let y be an arbitrary element of R (co-domain). Then,
Thus we find that for all
Such that
This shows that every element in the co-domain has its pre-image in the domain. So, f is a surjective. Hence, f is bijective.
4.
Show that the function 
is neither one-one nor onto.
Solution:
Injectivity:
We know that 
So, different elements in R may have the same
image. Hence, f is not an injective.
Surjective: Since, the values of 
lie between
, it follows that the
range of f(x) is not equal to its co-domain. So, f is nor surjective.
5.
Let A and B be two sets. Show that 
defined by 
is a bijective.
Solution:
Injectivity, let 
Thus, f is an injective map
5.2.4: Surjectivity: Let (b,a) be an arbitrary element of 
. Then 
Thus for all 
So, f is an onto function
Thus, f is bijective.
6.
Let 
and 
.
If 
is a mapping defined by 
,
show that f is bijective.
Solution: Injectivity, Let x, y be any two elements of A. Then,
Thus, . So f is an injective
map.
Surjectivity, let y be an arbitrary element of B. Then
Clearly, 
, then we get 
, which is wrong.
Thus, every element 
in B has its pre-image in A given by
.
So, f is a surjective map.
Hence f is a bijective map.
Exercise problem:
1. Which of the following functions from A to B are one-one and onto?
I.
II.
III.
Ans.
2. Give an example of a map
I. Which is one-one but not onto.
II. Which is not one-one but onto.
III. Which is neither one-one nor onto.
3.
Prove that the function
, defined by 
is one-one but not onto.
4.
Show that the function 
given by 
is a bijection.
5.
Let
. Show that 
is neither one-one nor onto.
6.
Let 
be defined by
Show that f is a bijection.
7.
If is an injection such that range of
. Determine the number
of elements in A.
Ans. 1
