State whether the function f is bijective. Justify your answer.

$f: \mathbf{N} \rightarrow \mathbf{N}$ is defined as $f(n)=\left\{\begin{array}{ll}\frac{n+1}{2}, & \text { if } n \text { is odd } \\ \frac{n}{2}, & \text { if } n \text { is even }\end{array}\right.$ for all $n \in \mathbf{N} .$

It can be observed that:

$f(1)=\frac{1+1}{2}=1$ and $f(2)=\frac{2}{2}=1 \quad$ [By definition of $\left.f\right]$

$\therefore f(1)=f(2)$, where $1 \neq 2$

∴ f is not one-one.

Consider a natural number (n) in co-domain N.

Case I: n is odd

$\therefore n=2 r+1$ for some $r \in \mathbf{N}$. Then, there exists $4 r+1 \in \mathbf{N}$ such that

$f(4 r+1)=\frac{4 r+1+1}{2}=2 r+1 .$

Case II: n is even

$\therefore n=2 r$ for some $r \in \mathbf{N}$. Then,there exists $4 r \in \mathbf{N}$ such that $f(4 r)=\frac{4 r}{2}=2 r$.

∴ f is onto.

Hence, f is not a bijective function.