Show that the function `f: R-{3}->R-{1}` given by `f(x)=(x-2)/(x-3)` is a bijection.

Explanation

One-one:

`x, y in R-{1}`

`f(x)=f(y)`

`frac{x-2}{x-3}=frac{x-3}{y-3}=>y=x` Then the function is on-one.

Onto:

`f(x)=frac{x-2}{x-3}`

`y=frac{x-2}{x-3}=>x(y-1)-2+3y`

`x=frac{-2+3y}{y-1}`
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