Question

A function f from the set of natural numbers to integers defined by

\dfrac{{n - 1}}{2}\,\,{\text{if }}n{\text{ is odd}} \\\ \dfrac{{n - 1}}{2}{\text{if }}n{\text{ is even}} \\\ \end{gathered} \right.$$ A. Onto but not one-one. B. One-one and onto both. C. Neither one-one nor both. D. One-one but not onto.

Answer 1

In this problem, we need to use injectivity and surjectivity test to check whether the function is one-one or onto or both. A function is said to be an onto function, if co domain of the function is the same as the range of the function.

Complete step by step answer:
Injectivity:
Consider, A and B be any two elements in the domain of function f.
Case (i): Both A and B are even numbers.

$$\,\,\,\,\,f\left( A \right) = f\left( B \right) \\\ \Rightarrow \dfrac{{A - 1}}{2} = \dfrac{{B - 1}}{2} \\\ \Rightarrow A - 1 = B - 1 \\\ \Rightarrow A = B \\\$$

Case (ii): Both A and B are odd numbers.

$$\,\,\,\,\,f\left( A \right) = f\left( B \right) \\\ \Rightarrow \dfrac{{A - 1}}{2} = \dfrac{{B - 1}}{2} \\\ \Rightarrow A - 1 = B - 1 \\\ \Rightarrow A = B \\\$$

Case (iii): let, A is even B is odd.

$$\,\,\,\,\,f\left( A \right) = f\left( B \right) \\\ \Rightarrow \dfrac{{A - 1}}{2} = \dfrac{{B - 1}}{2} \\\ \Rightarrow A - 1 = B - 1 \\\ \Rightarrow A = B \\\$$

From, case (iii) it is clear that, the function $f$ has same value for different numbers, thus, the function $f$ is not one-one.
Surjectivity:
The co domain of the function $f$ is shown below.
{\text{Co domain of }}f\left( Z \right) = \left\\{ { \ldots , - 3, - 2, - 1,0,1,2,3, \ldots } \right\\}
Now, the range of function $f$ is shown below.

{\text{ Range of }}f = \left\\{ { \ldots ,\dfrac{{ - 2 - 1}}{2},\dfrac{{ - 1 - 1}}{2},\dfrac{{0 - 1}}{2},\dfrac{{1 - 1}}{2},\dfrac{{2 - 1}}{2},\dfrac{{3 - 1}}{2}, \ldots } \right\\} \\\ \Rightarrow {\text{Range of }}f = \left\\{ { \ldots , - 2, - 1, - \dfrac{1}{2},0,\dfrac{1}{2},1, \ldots } \right\\} \\\ \Rightarrow {\text{Co domain of }}f \ne {\text{Range of }}f \\\

So, the correct answer is “Option C”.

Note: A function is said to be one to one function if it maps distinct elements of its domain to distinct elements of its codomain. A function is said to be onto if every element of its codomain is the image of at least one element of domain.