Question

On the set of integers $Z$. define $f: Z \to Z$ as $f(n) = \begin{cases} n/2 & \quad \text{if } n \text{ is even}\\\ 0 & \quad \text{if } n \text{ is odd}\\\ \end{cases}$ then $'f'$ is

• A. bijective
• B. injective but not surjective
• C. neither injective nor suijective
• D. surjective but not injective

Answer 1

Answer: (D)

surjective but not injective

Answer 2

Given, $f(n)=\begin{cases}\frac{n}{2}, & n \text { is even } \\\ 0, & n \text { is odd }\end{cases}$ Here, we see that for every odd values of $z$, it will give zero. It means that it is a many one function. For every even values of $z$, we will get a set of integers $(-\infty, \infty)$. So, it is onto. Hence, it is surjective but not injective.