Question

Let $f:N\to N$ be defined by $f\left(n\right)=\{\begin{array}{cc}\frac{n+1}{2},& ifnisodd\\ \frac{n}{2},& ifniseven\end{array}$
For all $n\in N$ state whether the function f is onto, one-one or bijective. Justify your answer.

Answer 1

Here, $f:N\to N$ is defined as, $f\left(n\right)=\{\begin{array}{cc}\frac{n+1}{2},& ifnisodd\\ \frac{n}{2},& ifneven\end{array}$

for all $n\in N$. It can be observed that $f\left(1\right)=\frac{1+1}{2}=1$ and $f\left(2\right)=\frac{2}{2}=1$ [By definition of f]
$\therefore f\left(1\right)=f\left(2\right)$, where $1\ne 2.$
Therefore, f is not one-one. Consider a natural number n in co-domain N.
Case 1: When n is odd.
Therefore, n=2r +1 for some $r\in N.$
Then, there exists 4r+1 $\in$ N such that $f(4r+1)=\frac{4r+1+1}{2}=2r+1.$
Therefore, f is onto.
Case 2: When n is even
Therefore, n=2r for some $f\in N$
Then, there exists $4r\in N$ such that $f\left(4r\right)=\frac{4r}{2}=2r$
Therefore, f is onto. Hence, f is not a bijective function.
If f is one-one and onto, then we say that f is bijective function.