Question

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

Answer 1

$\text{\bf{one - one function}}$ A function f : A --> B is said to be a one - one function or injective mapping, if different elements of A have different f images in B.

$\text{\bf{onto function}}$ If the function f :A → B is such that each element in B (codomain) is the f image of atleast one element in A. then we say that f is a function of A onto B.

given f : R → R, given by f (x) = [x] where [x] denotes greatest integer function.
We can see that f(1.2) = [1.2] = 1
f(1.9) = [1.9] = 1
⇒ f(1.2) = f(1.9), but 1.2 ≠ 1.9.
therefore, function f is not one- one.

Now, let us consider 0.6 ∈ R.
We know that f(x) = [x] is always an integer.
⇒ there does not exist any element x∈ R such that f(x) = 0.6
in simple,
codomain ∈ R but range ∈ I , where I is integers .e.g., codomain ≠ range .
therefore, function f is not onto.

Therefore, the greatest integer function is neither one-one nor onto