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
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
Updated On: May 14, 2024
- bijective
- injective but not surjective
- neither injective nor suijective
- surjective but not injective
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The Correct Option is D
Solution and Explanation
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.
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Concepts Used:
Types of Functions
Types of Functions
One to One Function
A function is said to be one to one function when f: A → B is One to One if for each element of A there is a distinct element of B.
Many to One Function
A function which maps two or more elements of A to the same element of set B is said to be many to one function. Two or more elements of A have the same image in B.
Onto Function
If there exists a function for which every element of set B there is (are) pre-image(s) in set A, it is Onto Function.
One – One and Onto Function
A function, f is One – One and Onto or Bijective if the function f is both One to One and Onto function.
Read More: Types of Functions