Let $f: N \rightarrow N $ defined by $f(N) = \begin{equation}
f(x)=\begin{cases}
\dfrac {n+1}{2} , & \text{if n is odd}.\\
\dfrac {n}{2} , & \text{if n is even}.
\end{cases}
\end{equation}$

then $f$ is

The correct answer is onto but not one-one

Given that $f: N \rightarrow N $ defined by
$f(N) = \begin{equation}
f(x)=\begin{cases}
\dfrac {n+1}{2} , & \text{if n is odd}.\\
\dfrac {n}{2} , & \text{if n is even}.
\end{cases}
\end{equation}$


For n = 1, we have $f(1) = \dfrac {1+1}{2} = 1$

For n = 2, we have $f(2) = \dfrac {2}{2} = 1$

So $f(1) = f(2)$, but $1 \ne 2$

Therefore, $f(x)$ is not one-one


$f(x) = \dfrac {n+1}{2}$ if n is odd

If $y = \dfrac {n+1}{2}$, the $n = 2y - 1$ $ \forall y$


Also, $f(x) = \dfrac {n}{2}$ if n is even

If $y = \dfrac {n}{2}$, the $n = 2y$ $ \forall y$


Therefore $f(x)$ is onto