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.