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Homework Help: Proving a function is a one-to-one correspondence

  1. Oct 27, 2007 #1
    1. The problem statement, all variables and given/known data
    Determine whether each of the following functions is one-to-one, onto, neither or both.

    [tex] f : (2, \infty) \rightarrow (1, \infty)[/tex], given by [tex] f(x) = \frac{x}{x-2} [/tex]

    3. The attempt at a solution

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

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

    Proof: Assume [tex]f : (2, \infty) \rightarrow (1, \infty)[/tex], given by [tex] f(x) = \frac{x}{x-2} [/tex].
    First we must show that f is one-to-one.
    Let [tex] a,b \in (2, \infty)[/tex] such that [tex] f(a) = f(b) [/tex].
    Notice that
    [tex] f(a) = f(b) [/tex]
    [tex] \frac{a}{a-2} = \frac{b}{b-2} [/tex]
    [tex] a(b-2) = b(a-2) [/tex]
    [tex] ab - 2a = ab - 2b [/tex]
    [tex] -2a = -2b [/tex]
    [tex] a = b[/tex]
    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 [tex] b \in (1, \infty) [/tex] 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?
  2. jcsd
  3. Oct 27, 2007 #2
    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)!
  4. Oct 27, 2007 #3
    Thanks a lot.
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