Question

The mapping $f:N\to N$ given by $f\left(n\right)=1+n^2,n\in N$ where N is the set of natural numbers, is

• A. One-to-one and onto
• B. Onto but not one-to-one
• C. One-to-one but not onto
• D. Neither one-to-one nor onto

Answer 1

The correct option is C One-to-one but not onto

Since, $f\left(n\right)=1+n^2$
For one-to-one
$1+n_1^2=1+n_2^2$
$\Rightarrow n_1^2=n_2^2$
$\Rightarrow (n_1-n_2)(n_1+n_2)=0$
$\Rightarrow n_1=n_2(\because n_1+n_2\ne 0)$
$\therefore f\left(n\right)$ is one-to-one.
But $f\left(n\right)$ is not onto.
Hence, $f\left(n\right)$ is one-to-one but not onto.