Question

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

Answer 1

$f(n) = \begin{cases} \frac{n+1}{2} & \quad \text{if } n \text{ is odd}\\\ \frac{n}{2} & \quad \text{if } n \text{ is even} \end{cases}$ for all $n∈N$.
It can be observed that:
$f(1)=\frac{1+1}{2}=1\text{ and }f(2)=\frac{2}{2}=1.$
$\therefore f(1)=f(2), \text{ where } 1\neq2.$
∴ f is not one-one.

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

Case I: n is odd
∴$n = 2r + 1$ for some $r ∈ N$.
Then, there exists $4r + 1∈N$ such that
$f(4r+1)=\frac{4r+1+1}{2}=2r+1$.

Case II: n is even
∴$n = 2r$ for some $r ∈ N$.
Then,there exists $4r ∈N$ such that $f(4r)=\frac{4r}{2}=2r$
∴ f is onto.

Hence, f is not a bijective function.