Determine whether each of the following functions is one-to-one, onto, neither or both.

$\displaystyle f : (2, \infty) \rightarrow (1, \infty)$, given by $\displaystyle f(x) = \frac{x}{x-2} $

So, I think this is one-to-one and onto. So i need to prove it.

Claim: If $\displaystyle f : (2, \infty) \rightarrow (1, \infty)$, given by $\displaystyle f(x) = \frac{x}{x-2} $, then f is a one-to-one correspondence.

Proof: Assume $\displaystyle f : (2, \infty) \rightarrow (1, \infty)$, given by $\displaystyle f(x) = \frac{x}{x-2} $.

First we must show that f is one-to-one.

Let $\displaystyle a,b \in (2, \infty)$ such that $\displaystyle f(a) = f(b) $.

Notice that

$\displaystyle f(a) = f(b) $

$\displaystyle \frac{a}{a-2} = \frac{b}{b-2} $

$\displaystyle a(b-2) = b(a-2) $

$\displaystyle ab - 2a = ab - 2b $

$\displaystyle -2a = -2b $

$\displaystyle a = b$

Hence, f is one-to-one. Here, do I need to use the fact that the codomain is (1, infinity) or the domain is (2, infinity)?

Now we need to show that f is onto. Let $\displaystyle b \in (1, \infty) $ and take a = ....

Here I would have solved b = a / (a-2) for a, but I do not know how to solve this. Is there any other way to do this proof besides solving for a then substituting back in to show that it gives me f(a) = b?