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Q.1 Show that the function $f: R_* \longrightarrow R_{*}$ defined by $f(x) = \frac{1}{x}$ is one-one and onto,
where R_∗ is the set of all non-zero real numbers. Is the result true, if the domain
R_∗ is replaced by N with co-domain being same as R_∗?

Answers (1)

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Given, $f: R_* \longrightarrow R_{*}$ is defined by $f(x) = \frac{1}{x}$ .

One - One :

$f(x)=f(y)$

$\frac{1}{x}=\frac{1}{y}$

$x=y$

$\therefore$ f is one-one.

Onto:

We have $y \in R_*$ , then there exists $x=\frac{1}{y} \in R_*$ ( Here $y\neq 0$ ) such that

$f(x)= \frac{1}{(\frac{1}{y})} = y$

$\therefore f is \, \, onto$ .

Hence, the function is one-one and onto.

If the domain R_∗ is replaced by N with co-domain being same as R_{∗ i.e.} $g: N \longrightarrow R_{*}$ defined by

$g(x)=\frac{1}{x}$

$g(x_1)=g(x_2)$

$\frac{1}{x_1}=\frac{1}{x_2}$

$x_1=x_2$

$\therefore$ g is one-one.

For $1.5 \in R_*$ ,

$g(x) = \frac{1}{1.5}$ but there does not exists any x in N.

Hence, function g is one-one but not onto.

Posted by

seema garhwal

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