Question

If $f : \mathbb{N} \rightarrow \mathbb{W}$ is defined as \[ f(n) = \begin{cases} \frac{n}{2}, & \text{if } n \text{ is even} \\ 0, & \text{if } n \text{ is odd} \end{cases} \] then $f$ is :

• A. injective only
• B. surjective only
• C. a bijection
• D. neither surjective nor injective

Hint:

To check if a function is injective, verify if distinct inputs map to distinct outputs. To check for surjectivity, ensure every element in the codomain has a pre-image in the domain.

Answer 1

The Correct Option is D

- For $f$ to be injective, each element in the domain should map to a unique element in the codomain. However, both even and odd numbers in the domain are mapping to different values, making $f$ not injective. - For $f$ to be surjective, every element in the codomain must be mapped from an element in the domain. But $f$ cannot map to all elements in the codomain $\mathbb{W}$, specifically, it cannot map to all odd numbers, making it not surjective. Thus, $f$ is neither injective nor surjective.