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# 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 $x$.

Toolbox:
• A function $f: X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function.
• A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Given $f: R \rightarrow R$ defined by $f(x)=[x],$ where $[x]$ denotes the greatest integer $leq x$:
Step1: Injective or One-One function
For example, if we consider $x=1.1$, we see that $f(1.1) = 1$.
Similarly, if we consider $x = 1.9$, we see that $f(1.9) = 1$.
$\Rightarrow [1.1]=[1.9]$, but given $1.1 \neq 1.9$,
$f: R \rightarrow R$ defined by $f(x)=[x],$ is not one-one.
Step 2: Surjective or On-to function:
Let us consider $x = 0.7 \in R \rightarrow [x] = 0$.
Since it is given that $[x]$ denotes the greates integer $\leq x$, there can be no element such that $f(x) = 0.7$.
Therefore, $f: R \rightarrow R$ defined by $f(x)=[x],$ is not onto.
Solution: $f: R \rightarrow R$ defined by $f(x)=[x],$ is neither one-one nor onto.

edited Mar 20, 2013 by meena.p