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  4. On the set of integers Z. define f: Z → Z as f(n) = begincases n/2 textif n text is even 0 textif n text is odd endcases then 'f' is

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Q. 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

KCETKCET 2009Relations and Functions - Part 2

Solution:

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.