Question

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

Answer 1

It is given that $f:R_{\ast }\to R_{\ast }$ is defined by $f\left(x\right)=\frac{1}{x}$
For one-one, $f\left(x\right)=f\left(y\right)\Rightarrow \frac{1}{x}=\frac{1}{y}\Rightarrow x=y$
Therefore, f is one-one.
For onto,
It is clear that $y\in R_{\ast }$. there exists $x=\frac{1}{y}\in R_{\ast }$. such that $f\left(x\right)=\frac{1}{\left(\frac{1}{y}\right)}=y$
Therefore, f is onto. Thus, the given function (f)is one-one and onto.
Now, consider function $g:N\to R_{\ast }$ defined by $g\left(x\right)=\frac{1}{x}.$
We have, $g\left(x_1\right)=g\left(x_2\right)\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}\Rightarrow x_1=x_2$
Therefore, g is one-one.
Further, it is clear that g is not onto as for $1.2\in R_{\ast }$,there does not exist any x in N such that $g\left(x\right)=\frac{1}{1.2}$
Hence, function g is one-one but not onto.