Given that f : R -->R and greatest integer function f(x) = [x]

Now let x = 1.2

f(1.2) = [1.2] = 1  ..........1

Again let x = 1.9

f(1.9) = [1.9] = 1 ...........2

from equation 1 and 2, we get

f(1.2) = f(1.9)

But 1.2 ≠ 1.9

So f(x) = [x] is not one-one.

Again let 0.7 ∈ R

But it is known that f(x) = [x] is always an integer.

So there does not exist any element x ∈ R such that f(x) = 0.7

So the greatest integer function f(x) = [x] is neither one-one nor onto.

Hense proved.