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)

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.

 

 

 

 

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