Question

Let \( f : \mathbb{N} \rightarrow \mathbb{N} \) 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 \( n \in \mathbb{N} \). Then \( f \) is _____

• A. One-one and onto
• B. Many-one and onto
• C. One-one but not onto
• D. Neither one-one nor onto

Hint:

Check both even and odd inputs carefully in piecewise functions.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A function is One-one if different inputs give different outputs. It is Onto if every element in the codomain ($N$) has a corresponding input in the domain ($N$).
Step 2: Formula Application:
Check for One-one: $f(1) = \frac{1+1}{2} = 1$ $f(2) = \frac{2}{2} = 1$ Since $f(1) = f(2)$, the function is Many-one.
Step 3: Explanation:
Check for Onto: If we want an output $y$, can we find $n$? If $n = 2y$, then $f(n) = \frac{2y}{2} = y$. Since $2y$ is always an even natural number for any $y \in N$, every $y$ is reached. Thus, the function is Onto.
Step 4: Final Answer:
The function is many-one and onto.