On the set of integers $Z$, define $f:Z ightarrow Z$ as $f(n) = \left\{ \begin{array}{r} \frac{n}{2},\ n\ is\ even \ 0,\ \ \ n\ \ is\ odd \end{array} ight\}$, then $'f^{'}$is

Question

On the set of integers $Z$, define $f:Z ightarrow Z$ as $f(n) = \left\{ \begin{array}{r} \frac{n}{2},\ n\ is\ even \ 0,\ \ \ n\ \ is\ odd \end{array} ight\}$, then $'f^{'}$is

Dr. Abubacker · Accepted Answer

(c)
Given, $f(n) = \left\{ \begin{array}{r}
\frac{n}{2},\ n\ is\ even \
0,\ \ \ n\ \ is\ odd
\end{array} ight.\$
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.

Q. On the set of integers $Z$, define $f:Z \rightarrow Z$ as $f(n) = \left\{ \begin{array}{r} \frac{n}{2},\ n\ is\ even \\ 0,\ \ \ n\ \ is\ odd \end{array} \right\}$, then $'f^{'}$is

Explanation

Reviewed by Dr. Abubacker

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