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# Injective, Surjective and Bijective Functions

## Injective / One-to-one function

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

This means a function **f** is injective if $a_1

e a_2$ implies $f(a1)

e f(a2)$.

### Example

$f: N \rightarrow N, f(x) = 5x$ is injective.

$f: N \rightarrow N, f(x) = x^2$ is injective.

$f: R\rightarrow R, f(x) = x^2$ is not injective as $(-x)^2 = x^2$

## Surjective / Onto function

A function $f: A \rightarrow B$ is surjective (onto) if the image of f equals its range. Equivalently, for every $b \in B$, there exists some $a \in A$ 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

$f : N \rightarrow N, f(x) = x + 2$ is surjective.

$f : R \rightarrow R, f(x) = x^2$ is not surjective since we cannot find a real number whose square is negative.

## Bijective / One-to-one Correspondent

A function $f: A \rightarrow B$ is bijective or one-to-one correspondent if and only if **f** is both injective and surjective.

### Problem

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

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

If $f(x_1) = f(x_2)$, then $2x_1 – 3 = 2x_2 – 3 $ and it implies that $x_1 = x_2$.

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**.