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Math Help - Proving function is injective

  1. #1
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    Proving function is injective

    Hey guys.

    I've got the following problem:

    Let  U = (0, \infty ) \times R \subset R^2 og let f : U \rightarrow R^2 be defined by : f(x,y) = (x, y^3+xy)

    (a) Show that f(U) = U Done that already.
    (b) Show that f is injective
    (Hint: When is g: R \rightarrow R given by g(y)= y^3 + ay + b monotonic?)
    (c)... some more questions.

    I've already found the solution to (a), but I cant quite figure out how to solve (b). This is pretty much what I've tried so far:
    For f : U \rightarrow R to be injective, the following has to be true : q \neq p \Rightarrow f(q) \neq f(p) \forall q, p \in U.
    So, for proof by contradiction, we assume that for a pair q \neq p \in U we have that f(p) = f(q).
    Since f(q) = f(p) we have that x_q = x_p \wedge y_q^3+x_q y_q = y_p^3+x_p y_p (1).
    But since q \neq p we have that  x_q \neq x_p \vee y_q \neq y_p.
    According to (1) we know that x_q = x_p so y_q \neq y_p must be the case.

    From here I dont know exacly where to go. I can see that I'm not really using the hint.
    I know that all monotonic functions are injective, but how can I prove that f is monotonic?
    I can see by visualization that f will be monotonic, because x is defined to be positive, but how do I prove that rigorously?

    Thanks for your help guys.

    Morten
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  2. #2
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    Re: Proving function is injective

    the function is clearly continuous, since each component function is continuous. there is a theorem that says that a continuous real-valued function is injective if and only if it is strictly monotonic. but im not sure if there is an analogous theorem for \mathbb{R}^2
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  3. #3
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    Re: Proving function is injective

    Yes I understand that, but how do I prove that it is monotonic?
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