Question

On the set of integers \[{\rm Z}\], define \[f:{\rm Z} \to z\] as
  \[f\left( n \right) = \left\{
  \dfrac{n}{2},\,n\,\,is\,\,even \\
  0,\,n\,\,is\,\,odd \\
  \right.\]
then \[f\] is
A. Injective but not surjective
B. Neither injective nor surjective
C. Surjective but not injective
D. Bijective

Answer 1

Hint: An injective function is also known as a one-one function and is defined as a function with one and only one image for each element and every element is associated with at least one other element. A function is said to be surjective if each element of the co-domain is mapped by at least one element of the domain

Complete step-by-step solution:
Given that \[f\left( n \right) = \left\{
  \dfrac{n}{2},\,n\,\,is\,\,even \\
  0,\,n\,\,is\,\,odd \\
  \right.\]
Now to check whether the given function is injective or not the following steps are followed:
Now it is given that \[n\] is even that means \[n = 2,4,6,8....\]
By substituting these values of \[n\] in the given function we get
When \[n = 2\]:
\[f\left( 2 \right) = \left\{ {\dfrac{2}{2}} \right. = 1\]
When \[n = 4\]:
\[f\left( 4 \right) = \left\{ {\dfrac{4}{2}} \right. = 2\]
When \[n = 6\]:
\[f\left( 6 \right) = \left\{ {\dfrac{6}{2} = 3} \right.\]
When \[n = 8\]:
\[f\left( 8 \right) = \left\{ {\dfrac{8}{2} = 4} \right.\]
Therefore, the given function has not a unique image.
 So, it is not injective.
Now it is given that every odd value \[n\]yields zeroes.
So, it is a many-one function.
Therefore, the ordered pair of the function is
\[\left\{ {\left( {0,0} \right),\left( {1,0} \right),\left( {2,1} \right),\left( {3,0} \right),\left( {4,2} \right),........} \right\}\]
That means co-domain = range
Thus, the given function \[f\left( n \right) = \left\{
  \dfrac{n}{2},\,n\,\,is\,\,even \\
  0,\,n\,\,is\,\,odd \\
  \right.\] is surjective but not injective.

Hence, option(C) is correct option

Note: A subjective function is also known as the onto function, while an injective function is known as a one-one function. If a function is both one-one and an onto and it is referred to as a bijective function.