Let $f:R-\left\{n\right\}\to R$ be a function defined by $f\left(x\right)=\frac{x-m}{x-n}$ such that $m\ne n$ then

Given $f\left(x\right)=\frac{x-m}{x-n}$

where $m\ne n,\forall x\in R-\left\{n\right\},\text{\hspace{0.33em}Let\hspace{0.33em}}{x}_1,x_2\in R$

$\therefore f\left(x_1\right)=f\left(x_2\right)\Rightarrow \frac{x_1-m}{x_1-n}=\frac{x_2-m}{x_2-n}\Rightarrow x_1=x_2$

$\therefore f$ is one-one

Let $\lambda \in R$ such that $f\left(x\right)=\lambda ,\therefore \frac{x-m}{x-n}=\lambda$

$\therefore x=\frac{m-n\lambda }{1-\lambda }$

x is not defined for $\lambda =1,$

$\therefore f\left(x\right)$ is not onto function.

If a function is not onto it refered that it is into function. Hence f is one-one into function.