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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 \(x\).

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  • 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.


answered Mar 8, 2013 by thagee.vedartham
edited Mar 20, 2013 by meena.p

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