Proving a function is a one-to-one correspondence

[Jacobpm64]

Homework Statement

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

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

The Attempt at a Solution

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

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

Proof: Assume $$f : (2, \infty) \rightarrow (1, \infty)$$, given by $$f(x) = \frac{x}{x-2}$$.
First we must show that f is one-to-one.
Let $$a,b \in (2, \infty)$$ such that $$f(a) = f(b)$$.
Notice that
$$f(a) = f(b)$$
$$\frac{a}{a-2} = \frac{b}{b-2}$$
$$a(b-2) = b(a-2)$$
$$ab - 2a = ab - 2b$$
$$-2a = -2b$$
$$a = b$$
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 $$b \in (1, \infty)$$ 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 into show that it gives me f(a) = b?

 

[ZioX]

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

I think what you're getting at is since a,b in (2,infty) then a/(a-2) make sense (the denominator is nonzero).

For ontoness (not a word) solving for b=a/(a-2) is the way to go. A lot of times you don't get a very nice function where you can't solve and so you'll have to show some kind of existence. But this is a nice function, not only can you show that there must exist an a mapping to b, but you can actually find the value of a.

NB: You must demonstrate that a is in the domain of the function. If you keep that same function and fudge around with the domain you can lose ontoness. eg if you define f the same, but change the domain to (3,infty) you get a very different function: the image of f under (3,infty) is (1,3) which is missing quite a bit of (1,infty)!

 

[Jacobpm64]

Thanks a lot.