Question

Prove that the Greatest Integer Function \[f:R \to R\;\] given by \[f(x) = [x]\], is neither one - one nor onto, where \[[x]\;\] denotes the greatest integer less than or equal to \[x\].

Answer 1

Hint: - Here, We go through with the properties of greatest integer function. It always returns an integer as output.

Complete step-by-step solution -
Here $f(x) = [x]$
The Greatest Integer Function is denoted by ${\text{y = }}\left[ {\text{x}} \right]$. For all real numbers \[x\].
The greatest integer function returns the largest less than or equal to \[x\].
In essence, it rounds down a real number to the nearest.
By above definition it is seen that,
$
  f(1.2) = [1.2] = 1,f(1.9) = [1.9] = 1. \\
  \therefore f(1.2) = f(1.9) = 1 \\
 $
But ${\text{1}}{\text{.2}} \ne {\text{1}}{\text{.9}}$
$\therefore f $ is not one - one $\because $ (if each element of one set mapped with a unique element of another set is known as one-one function. But here for two values we get only one result)
Now, consider $0.7 \in R$
It is known that $f{\text{(}}x{\text{) = [}}x{\text{]\;}}$ is always an integer. Thus, there does not exist any element $x \in R\;$such that $f(x) = 0.7$
$\therefore f $ is not onto.
Hence, The greatest integer function is neither one - one nor onto.

Note: - Whenever we face such types of problems like greatest integer function. First of all we should recall the property of that function. Then check if the property of that function is matching with the given statement or not to verify the answer.