Functions

posted on: 21 Mar, 2012 | updated on: 27 Aug, 2012

Algebra functions are one of the most important concepts of mathematics. A function can be considered as a rule which produces new elements out of some given elements. There are many terms such as 'map' to denote a function.
A relation ‘f’ from a Set ‘A’ to a set ‘B’ is said to be a function, if every element of set ‘A’ has one and only one image in set ‘B’.
In other words Algebra of functions is a relation from a non-empty set ‘A’ such that the Domain of ‘f’ is ‘A’ and no two distinct ordered pairs in ‘f’ have the same first element.
If ‘f’ is a function from ‘A’ to ‘B’ and (a, b) belongs to ‘f’, then f (a) = b, where ‘b’ is called the image of ‘a’ under ‘f’ and ‘a’ is called the pre image of ‘b’ under ‘f’.
The algebra function ‘f’ from ‘A’ to ‘B’ is denoted by f: A → B,
A function which has either ‘R’ or one of its Subsets as its range is called a real value function. Further, if its domain is also either ‘R’ or a subset of ‘R’, than it is called a real function.
Addition of two real functions,
Suppose f: X→ R and g: X→ R be any two real functions, where X ⊂ R. Then (f + g): X → R by
(f + g) (x) = f(x) + g(x), for all ‘x’ belongs to ‘X’.
Subtraction of two real functions
Suppose f: X→ R and g: X→ R be any two real functions, where X ⊂ R. Then (f - g): X → R by
(f - g) (x) = f(x) - g(x), for all ‘x’ belongs to ‘X’.
Multiplication by a Scalar of a function
Suppose f: X→ R be a real valued function and a be a scalar. Here by scalar we Mean a real number. Then the product of is a function from ‘f’ to ‘R’ defined by (af) (x) = a f(x) where ‘x’ belongs to ‘X’.
Multiplication of two real functions,
The product of two real functions f: X→ R and g: X→ R is a function fg: X→ R, defined by (fg) (x) = f(x) g(x) for all ‘x’ belongs to ‘X’.
Quotient of two real functions,
Let ‘f’ and ‘g’ be two real functions defined from X→ R where X ⊂ R. The quotient of ‘f’ and ‘g’ denoted by f/g is a function defined by,
(f/g) (x) = f(x) / g(x), provided by g(x), is not equals to zero. [‘x’ belongs to ‘X’].

Let ‘x’ be the set of all 60 students of a class in a central school. Let f: A →N be a function defined by F(x) = roll number of student ‘x’, show that ‘f’ is one-one but not onto?

Here ‘f’ stands for each student to his (her) roll number. So there are no two students of the class who can have the same roll number.
Therefore we can say that ‘f’ is one- one,
We can see that
F(x) = range of f = 1, 2, 3,… 57, 58, 59, 60 ≠ N.
Therefore we can say that range of f is not same as its co-domain. So the function f is not Onto function.

Show that the function f: N →N, given by f(x) = 3x, is one – one but not onto function.

The function f: N →N, given by f(x) = 3x, here we observe all the properties of ‘f’:
First we check for injectivity: let the function x_1,x₂ ∊N such that f(x₁) = f (x₂) then we see that,
f (x₁) = f (x₂),
3x₁= 3x_2,
x₁= x_2,
So ‘f’ is one- one function.
And in case of subjectivity it is clear that the function ‘f’ takes even values and therefore no odd natural number in N (co –domain) have its pre- image in Domain.

Prove that f: N →N, it is given by f(x) = 3x; is one –one and onto function?

The function f: N →N , given by f(x) = 3x, here we observe all the properties of ‘f’:
Firstly we check for injectivity: let the function x_1,x₂ ∊N such that f(x₁) = f (x₂) then we see that
          f(x₁) = f (x₂),
          3x₁= 3x_2,
x₁= x₂,
So ‘f’ is one- one function.
Now we check for subjectivity: let any number ‘y’ be any real number in ‘R’ (co-domain). Then we will see that
           F(x) = y = 3x;
X = y/3;
Clearly, y/2 ∊ R for y ∊ R such that f(y/3) = 3(y/3) = y.
Thus, for each y ∊ R (co-domain) there exist x = y/2 ∊R (domain) such that f(x) = y
This means that the element in co-domain has its pre-image in Domain
So the function f: R →R is Onto.
Hence f: R →R is a bijection function.

Prove that f: R →R, it is given by f(x) = x⁴; is a bijection function?

The function f :R →R , given by f(x) = x⁴, here we observe all the properties of function ‘f’:
First we check for injectivity: Let the function x_,y ∊ R such that f(x) = f (y) then we see that,
          f(x) = f (y),
          x⁴= 3y⁴,
           x= y,
So f :R →R is one- one function.
Now we check for subjectivity: Let any number ‘y’ be any real number in ‘R’ (co-domain). Then we will see that
           F(x) = y = x⁴;
                    X = y^1/4;
Clearly, y^1/4∊ R for y ∊ R such that f(x) = y^1/4 ∊ R.
Thus, for each y ∊ R (co-domain) there exist x = y^1/4∊ R (domain) such that f(x) = y.
This means that the element in co-domain has its pre-image in Domain,
So the function f: R →R is Onto.
Hence f: R →R is a bijection function.

Show that the function f: R →R, it is given by f(x) = px + q; where p and q ∊ R, and p ≠ 0 is a bijection function?

The function f: R →R, it is given by f(x) = px + q; where ‘p’ and ‘q’ ∊ R, and p ≠ 0 is a bijection function.
Firstly we check for injectivity: Let the function x_,y ∊ R be any two real number then
          f (x) = f (y) ⇒ px + q = py + q ⇒ py ⇒ x = y then,
          f(x) = f (y) ⇒ x = y for all x, y ∊ R (domain).
So f: R →R is injection function.
Now we check for subjectivity: Let any number ‘y’ be any arbitrary element ‘R’ (co-domain). Then we will see that
           F(x) = y = px + q = y ⇒x = (y – q/ p);
                    X = y – q/ p ∊ R;
Clearly, y – q/ p ∊ R for all y ∊ r (co-domain).
Thus, for each y ∊ R (co-domain) there exist x = (y – q/ p) (domain) such that,
f(x) = f(x = (y – q/ p) = p(x = (y – q/ p) + q = y,
This means that the element in co-domain has its pre-image in Domain
So the function f: R →R is surjection.
Hence f: R →R is a bijection function.

Let A = R – 2 and B = R – 1, if f: A →B is a mapping defined by f(x) = (x – 1/ x – 2), show that the function is a bijection?

A = R – 2 and B = R – 1, if f: A →B is a mapping defined by
f(x) = (x – 1/ x – 2), show that the function is a bijection,
Firstly we check for injectivity: Let the function x_,y ∊R be two element Numbers then
f(x) = f (y),
(x – 1/ x – 2) = (y – 1/ y – 2),
On solving the equation we get.
(x – 1) (y – 2) = (x – 2) (y – 1).
Now solving the equation
xy – y – 2x + 2 = xt – x – 2y + 2,
⇒ x = y.
Thus f(x) = f(y) ⇒ x = y for all x, y ∊A,
So f: R →R is injection function.
And in case of subjectivity: Let ‘y’ be an arbitrary element of ‘B’ then,
F(x) = y,
(x – 1) / (x – 2) = y,
(x – 1) = y (x - 2),
Then we find the value of ‘x’
x = (1- 2y) / (1 – y),
Clearly, x = (1 – 2y) / (1 – y) is real number for all y ≠1.
This means that the element in co-domain has its pre-image in Domain
So the function f: R →R is surjection.
Hence f: R →R is a bijection function.