Relation and Function Exercise 1.2 Solutions

17. Show that the function f: R* → R* defined by f(x) = 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 the co-domain being same as: R*?

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9 months ago

The $f x$ n is $f (x) = \frac{1}{x}$ , which is a $f : R_{*}$ $\to$ $R_{*}$ and $R_{*}$ is set of all non-zero real numbers

For, $x_{1}, x_{2} \in R_{*}, f (x_{1}) = f (x_{2})$

$\Rightarrow \frac{1}{x_{1}} = \frac{1}{x_{2}}$

$\Rightarrow x_{1} = x_{2}$ So, $f$ is one-one

For, $y \in R_{*},$ $x = \frac{1}{f (x)} = \frac{1}{y}$ such that

So, $f (x) = y$

So, every element in the co-domain has a pre-image in $f$

So, $f$ is onto

If $f : N \to R_{*}$ such that $f (x) = \frac{1}{x}$

For, $x_{1}, x_{2} \in N,$ $f (x_{1}) = f (x_{2})$

$\Rightarrow \frac{1}{x_{1}} = \frac{1}{x_{2}}$

$\Rightarrow x_{1} = x_{2}$ So, $f$ is one-one

For, $y \in R_{*}$ and $f (x) = y$ we have $x = \frac{1}{y} \notin N$

Eg., $3 \in R_{*}$ so $x = \frac{1}{3} \notin N$

So, $f$ is not onto

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