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

$f:R \to R,\;\;f(x) = [x],$

Injectivity
We have, for $1 \le x < 2,$
$f(x) = 1$
Thus, f is not one-one.

Surjectivity
$f:R \to R$ does not attain non-integral values.
Therefore, Non-integer points in R do not have their pre images in the domain.
Therefore, f is not onto.

Hence, f is neither one-one nor onto.