Question

The function f: $N\to N$ defined by $f\left( x \right)=x-5\left[ \dfrac{x}{5} \right]$, where N is the set of natural numbers and $\left[ x \right]$ denotes the greatest integer less than or equal to x, is:
(a) one – one but not onto
(b) neither one – one nor onto
(c) onto but not one – one
(d) one – one and onto

Answer 1

Hint: We have given domain and range of a function $f\left( x \right)=x-5\left[ \dfrac{x}{5} \right]$ and we are asked to comment upon the nature of function like whether it is one – one or not one-one, onto or not onto. A function is one – one when every element of the range has at most one element in the domain. And a function is an onto function when for every element say “a” present in the range (or N) there exists an element say “b” in the domain (or N) such that $f\left( b \right)=a$ and also for each element in the domain there exists some element in the range. This is how you can check whether the function is one-one and onto.

Complete step-by-step solution:
We have given the following function:
$f\left( x \right)=x-5\left[ \dfrac{x}{5} \right]$
And the domain and range of the function is equal to:
f: $N\to N$
As you can see, the domain and range of the function is a set of natural numbers.
We know that a set of natural numbers is equal to:
$\left\{ 1,2,3,4...... \right\}$
Now, we have to comment on whether the function is one – one or onto.
We know that a function is one – one when the range of the function contains at most one element in the domain of the function.
The function which is given as:
$f\left( x \right)=x-5\left[ \dfrac{x}{5} \right]$
If we put $x=1$ because domain is a natural number then the value of f(x) is equal to:
$f\left( 1 \right)=1-5\left[ \dfrac{1}{5} \right]$
In the above equation, the greatest integer of $\left[ \dfrac{1}{5} \right]$ is 0 so substituting this value in the above we get,
$\begin{align}
  & f\left( 1 \right)=1-5\left( 0 \right) \\
 & \Rightarrow f\left( 1 \right)=1 \\
\end{align}$
Now, if we put $x=2$ in f(x) we get,
$f\left( 2 \right)=2-5\left[ \dfrac{2}{5} \right]$
The value of $\left[ \dfrac{2}{5} \right]$ is 0 so substituting this value in the above equation we get,
$\begin{align}
  & f\left( 2 \right)=2-5\left( 0 \right) \\
 & \Rightarrow f\left( 2 \right)=2-0 \\
 & \Rightarrow f\left( 2 \right)=2 \\
\end{align}$
For $x=3$ the value of f(x) is equal to:
$\begin{align}
  & f\left( 3 \right)=3-5\left[ \dfrac{3}{5} \right] \\
 & \Rightarrow f\left( 3 \right)=3-0 \\
 & \Rightarrow f\left( 3 \right)=3 \\
\end{align}$
For $x=4$ the value of f(x) is equal to:
$\begin{align}
  & f\left( 4 \right)=4-5\left[ \dfrac{4}{5} \right] \\
 & \Rightarrow f\left( 4 \right)=4-0 \\
 & \Rightarrow f\left( 4 \right)=4 \\
\end{align}$
For $x=5$ we get the value of f(x) is equal to:
$\begin{align}
  & f\left( 5 \right)=5-5\left[ \dfrac{5}{5} \right] \\
 & \Rightarrow f\left( 5 \right)=5-5\left[ 1 \right] \\
\end{align}$
We know that the value of $\left[ 1 \right]=1$ so substituting this value in the above we get,
$f\left( 5 \right)=5-5=0$
And we know that 0 is not a natural number. This means that when we put $x=5$ then what we get is an element which does not belong to range of the function
Hence, we can say that our function is not an onto function.
Now, we are going to check whether the function is one – one or not one – one.
If we take $x=6$ and substitute this value in f(x) we get,
$f\left( x \right)=6-5\left[ \dfrac{6}{5} \right]$
We know the value of $\left[ \dfrac{6}{5} \right]$ is 1 so putting this value in the above equation we get,
$\begin{align}
  & f\left( 6 \right)=6-5\left( 1 \right) \\
 & \Rightarrow f\left( 6 \right)=6-5=1 \\
\end{align}$
As you can see from the above that $f\left( 1 \right)=1$ and $f\left( 6 \right)=1$ and for one value of range (i.e. 1) there exists two values in the domain (1 and 6) which is violating the principle of one – one function. Hence, this function is not a one - one function.
Hence, our function is neither one – one nor onto.
Hence, the correct option is (b).

Note: The mistake that could happen in this problem is that generally, students mix natural numbers with whole numbers like you might think that 0 is a natural number which is not so make sure you know the clear definition of natural numbers. Apart from that, we have written the values for the greatest integer functions. In the below, we are going to show how to find the greatest integer value.
Let us suppose that we have to find the greatest integer value for $\dfrac{3}{2}$ then first of all draw a number line and then locate $\dfrac{3}{2}$ on that number line.

In the above figure, we have located the number $\dfrac{3}{2}$ by point A on the number line.