Bijective

[franz32]

Hello!

Are all rational expression of two linear expression, say f (x) =
(x + 3)/(x - 3), bijective? How will x = 3 affect the condition of a
bijective function?

[Muzza]

First thing's first, you have to define the domain and codomain of f... Let's use your example, f(x) = (x + 3)/(x - 3). It can't be function from R to R, since f(3) isn't defined (as you said). At best, it's a function from R \ {3} to R. But it isn't bijective, since there is no x such that f(x) = 1. (I found that value by considering the equation f(x) = y <=> (x + 3)/(x - 3) = y => x = 3(y + 1)/(y - 1), which isn't defined when y = 1).

But you can "make" it bijective if you define it to be a function from R \ {3} to R \ {1}, I think?

[HallsofIvy]

If (x+3)/(x-3)= (y+3)/(y-3), then (x+3)(y-3)= (y+3)(x-3) or xy+ 3y- 3x- 9= xy+ 3x- 3y-9. That reduces to 3y- 3x= 3x- 3y or 6y= 6x so x= y. Yes, the function (x+3)/(x-3) is bijective from R\{3} to R\{1}.

You should be able to do exactly the same thing for (ax+b)/(cx+d).,