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Math Help - Proving a function is a one-to-one correspondence

  1. #1
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    Proving a function is a one-to-one correspondence

    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}



    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 in to show that it gives me f(a) = b?
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  2. #2
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    That works. Also the derivative is everywhere negative.
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  3. #3
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    How do i prove that it's onto though?
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  4. #4
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    f(x) = \frac{x}{{x - 2}}\quad  \Rightarrow \quad f^{ - 1} (x) = \frac{{2x}}{{x - 1}}<br />
    This works as I said above.
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  5. #5
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    gotcha, thanks.
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