Let $f : N \rightarrow N$ be defined by $f(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases}$ for all $n \in N$. State whether the function $f$ is bijective. Justify your answer.

(D) Given $f : N \rightarrow N$ is defined by $f(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases}$.
First,we check if $f$ is one-one:
$f(1) = \frac{1+1}{2} = 1$ and $f(2) = \frac{2}{2} = 1$.
Since $f(1) = f(2)$ but $1 \neq 2$,the function $f$ is not one-one.
Next,we check if $f$ is onto:
For any $n \in N$ (co-domain),we need to find $x \in N$ such that $f(x) = n$.
Case $I$: If $n$ is odd,let $n = 2r - 1$ for some $r \in N$. Then $f(4r - 3) = \frac{(4r - 3) + 1}{2} = 2r - 1 = n$.
Case $II$: If $n$ is even,let $n = 2r$ for some $r \in N$. Then $f(4r) = \frac{4r}{2} = 2r = n$.
Since for every $n \in N$,there exists a pre-image in the domain,$f$ is onto.
Conclusion: Since $f$ is not one-one,it is not a bijective function.