Show that the function f : R − {3} → R − {2} given by

Solution:

$f: R-\{3\} \rightarrow R-\{2\}$ given by

$f(x)=\frac{x-2}{x-3}$

Injectivity:
Let x and y be any two elements in the domain (R − {3}), such that f(x) = f(y).

$f(x)=f(y)$

$\Rightarrow \frac{x-2}{x-3}=\frac{y-2}{y-3}$

$\Rightarrow(x-2)(y-3)=(y-2)(x-3)$

$\Rightarrow x y-3 x-2 y+6=x y-3 y-2 x+6$

$\Rightarrow x=y$

So, f is one-one

Surjectivity:
Let y be any element in the co-domain (R − {2}), such that f(x) = y for some element x in R − {3} (domain).

$f(x)=y$

$\Rightarrow \frac{x-2}{x-3}=y$

$\Rightarrow x-2=x y-3 y$

$\Rightarrow x y-x=3 y-2$

$\Rightarrow x(y-1)=3 y-2$

$\Rightarrow x=\frac{3 y-2}{y-1}$, which is in $R-\{3\}$

So, for every element in the co-domain, there exists some pre-image in the domain.

$\Rightarrow f$ is onto.

Since, f  is both one-one and onto, it is a bijection.