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: To prove this we have to check the function for both one to one solution and onto solution. A function is said to be onto if there exists an x for every y. A function is said to be one-one if no element of B is the image of more than one element in A.

Complete step-by-step answer:
$f(x) = [x]$
Where [x] denotes that the greatest integer is less than equal to x
Let’s check if it is one-one or not
For f(1) =[1] = 1
f(1,1) =[1.1] = 1
f(1.2) =[1.2] = 1
f(1.8) =[1.8] = 1
f(1.999) =[1.999] = 1
Since different elements have the same image, the function is not one-one.
Let’s check if it is onto or not
Let y = f(x)
y = [x]
Hence, y is the greatest integer which is less or equal to x.
Therefore the value of y will always be an integer.
But, as y is a real number,
Hence f is not onto
Hence it is proved 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.

Note: A function is called one to one if no two elements in the domain of that function correspond to the same element in the range of that function which means each x in the domain has exactly one image in the range. This is also called injective function.
The mapping of f is said to be onto if every element of the Y is the f-image of at least one element of x. It is also called surjective function. A function f from a set X to set Y is surjective, if for every element y in the codomain Y of f, there is at least one element x in the domain X of f such that f(x)=y.
If a function is both one to one and onto, then it is called a bijective function.