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

$\left(a\right)\left(i\right)f\left(x\right)=\frac{1}{x}$

$f\left(x\right)=f\left(x\right)$

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

$\Rightarrow x_1=x_2$

R has a unique being in co-domain

there,for t is one-zero

(ii) for each y belonging co-domain

then $y=\frac{1}{x}or,x=\frac{1}{y}$

there is a unique pre-being of y

there,for t is one-zero

(b) when domain R is replaced by N.

co-domain R remcoining the same then

$f:N\to R$

$if\left(x_1\right)=f\left(x_2\right)$

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

$\Rightarrow x_1=x_2$

there,for f is one- zero

but for every x ect number belonging to co-domain may not have a pre-being in N

$i.e\frac{1}{2}=\frac{3}{2}\notin N$