Question

A mapping $f : N \to N$, where $N$ is the set of natural numbers is defined as
$f(n) = n^2$ , for $n$ odd
$f(n) = 2n + 1$ , for $n$ even
for $n \in N $.
Then $f$ is

• A. surjective but not injective
• B. injective but not suijective
• C. bijective
• D. neither injective nor suijective

Answer 1

Answer: (B)

Since, $f(n) =
\begin{cases} n^2 & \text{if $n$ is odd} \\[2ex] 2n+1 & \text{if $n$ is even}
\end{cases}$

For even functions

$2n_1 + 1 = 2n_2 + 1$

$\Rightarrow 2n_{1} = 2n_{2}$

$\Rightarrow n_{1} = n_{2}$

and for two odd functions

$ n_{1}^{2} = n_{2}^{2} $

$ \Rightarrow n_{1}^{2} - n_{2}^{2} = 0$

$ \Rightarrow n_{1}-n_{2} = 0$ or $n_{1} +n_{2} = 0$
$ \Rightarrow n_{1} = n_{2} \left(\because n_{1} +n_{2} \ne0\right) $

$ \therefore f\left(n \right)$ is one-one.

But $f\left(n \right)$ is not onto.

$\therefore f\left(n \right)$ is injective but not surjective.