# Functions of Set

MathematicsComputer EngineeringMCA

A Function assigns to each element of a set, exactly one element of a related set. Functions find their application in various fields like representation of the computational complexity of algorithms, counting objects, study of sequences and strings, to name a few. The third and final chapter of this part highlights the important aspects of functions.

## Function - Definition

A function or mapping (Defined as f: X → Y) is a relationship from elements of one set X to elements of another set Y (X and Y are non-empty sets). X is called Domain and Y is called Codomain of function ‘f’.

Function ‘f’ is a relation on X and Y such that for each x ∊ X, there exists a unique y ∊ Y such that (x,y) ∊ R. ‘x’ is called pre-image and ‘y’ is called image of function f.

A function can be one to one or many to one but not one to many.

## Injective / One-to-one function

A function f: A → B is injective or one-to-one function if for every b ∊ B, there exists at most one a ∊ A such that f(s) = t.

This means a function f is injective if a1 ≠ a2 implies f(a1) ≠ f(a2).

## Example

• f: N → N, f(x) = 5x is injective.

• f: N → N, f(x) = x2 is injective.

• f: R → R, f(x) = x2 is not injective as (-x)2 = x2

## Surjective / Onto function

A function f: A → B is surjective (onto) if the image of f equals its range. Equivalently, for every b ∊ B, there exists some a ∊ A such that f(a) = b. This means that for any y in B, there exists some x in A such that y = f(x).

## Example

• f : N → N, f(x) = x + 2 is surjective.

• f : R → R, f(x) = x2 is not surjective since we cannot find a real number whose square is negative.

## Bijective / One-to-one Correspondent

A function f: A → B is bijective or one-to-one correspondent if and only if f is both injective and surjective.

### Problem

Prove that a function f: R → R defined by f(x) = 2x – 3 is a bijective function.

Explanation − We have to prove this function is both injective and surjective.

If f(x1) = f(x2), then 2x1 – 3 = 2x2 – 3 and it implies that x1 = x2.

Hence, f is injective.

Here, 2x – 3 = y

So, x = (y+5)/3 which belongs to R and f(x) = y.

Hence, f is surjective.

Since f is both surjective and injective, we can say f is bijective.