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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.
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