Question

The mapping $f : N \to N$ given by $f(n) = n^3 + 3, n \in N$ where $N$ is the set of natural number, is

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

Answer 1

Answer: (B)

One to one but not onto

Answer 2

The correct option is(B): One to one but not onto.

We have, $f : N \to N$ given by $f (n)=n^{3}+3$
Let $f (n_{1})=f (n_{2})$
$\Rightarrow n^{3}_{1}+3=n^{3}_{2}+3$
$\Rightarrow n^{3}_{1}=n^{3}_{2}$
$\Rightarrow n_{1}=n_{2}$
So, $f (n)$ is one to one mapping
Let $y=f (n) = n^{3}+3$
$\Rightarrow n=(y-3)^{1/3}$
Now, $\forall y \in\,N, n \notin\,N$
so, $f (n)$ is not onto