Question

A mapping from $N$ to $N$ is defined as follows
$f:N⟶N$
$f\left(n\right)={(n+5)}^2,n\in N$
($N$ is the set of natural numbers). Then,

• A. $f$ is not one to one
• B. $f$ is onto
• C. $f$ is both one to one and onto
• D. $f$ is one to one but not onto

Answer 1

The correct option is D $f$ is one to one but not onto

Given, $f:N\to N$, $f\left(n\right)={(n+5)}^2$
For one to one
$f\left(n_1\right)=f\left(n_2\right)$
$\Rightarrow {(n_1+5)}^2={(n_2+5)}^2$
$\Rightarrow (n_1-n_2)(n_1+n_2+10)=0$
$\Rightarrow n_1=n_2$
$\therefore$ $f$ is one to one.
When we put $n=1,2,3,4,\dots \infty$, we will get
$f\left(1\right)=36,f\left(2\right)=49,f\left(3\right)=64,f\left(4\right)=81,\dots$
Here, we see that we do not get any pre-images of $1,2,3$ etc.
Hence, $f$ is not onto.