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Prove or disprove that there is a rational bijective function f : R to (0; 1)

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove or disprove that there is a rational bijective function f : R to (0; 1)

    2. Relevant equations

    i found a bijective map from (0,1) to R (y=(2x-1)/(2x^2-2x)



    3. The attempt at a solution

    Im just stuck and i was thinking since it has to be a rational function, denominaotor should be defined on all R ... so nothing funky in the denominator..
    Somehow i have to play with the numerator to limit the map to (0,1) which i think is impossible...
     
  2. jcsd
  3. Apr 14, 2012 #2
    I don't really know the answer to this, but if you consider a function such as:

    [itex]f \left( x \right) = \frac{x^2}{x^2+1}[/itex]

    The range is [itex]\left[ 0, 1 \right)[/itex], but it is not bijective.
     
  4. Apr 14, 2012 #3

    morphism

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    A rational function is the ratio of two polynomials. You observed if there's anything "funky" in the denominator, then certainly R won't get mapped to (0,1). Now just think about what "funky" really means.. Remember, the denominator is a polynomial.

    [STRIKE]Edit: Sorry--that's not really helpful.[/STRIKE] Actually, I think it can be made to work. One thing the denominator can't be is a polynomial of odd degree (why?). Keeping this in mind, consider the derivative of the rational function, and try to think about turning points. You will also need to think about the degree of the numerator.

    Edit2: There's a much easier approach: consider limits as ##x\to\pm\infty##.
     
    Last edited: Apr 14, 2012
  5. Apr 14, 2012 #4
    What i mean by funky is that the denominator should be defined on the enitre R.(for example, x-2 wont be defined at x=0, we dont want the denominator to have any roots in R)
    The function that im trying to find has to be continuous so that as x gets sent to infinity, it has to go to either 0 or 1

    I dont get why you said the denominator cannot be of odd degree. odd degree polynomial means one end is up and the other end is down.
     
  6. Apr 14, 2012 #5

    morphism

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    A polynomial of odd degree always has a real root, and the denominator can't have a root if we want the rational function to be defined on all of R.
     
  7. Apr 14, 2012 #6
    i have plugged in some random rational functions with the denominator beign a polynomial of degree 2. I can get it to be restricted between 0 and 1 but it is not injective.
    I dont think you can get it to be injective.
    In order for one end to go up (approaching 1) and for the other to go near 0, the numerator has to be odd. But then f will be 0 as x goes to -,+ infinity
     
  8. Apr 14, 2012 #7

    morphism

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    Sounds like you've pretty much got it. Care to summarize your argument in one post, so we can make sure there are no gaps?
     
  9. Apr 14, 2012 #8
    The reason why i think it is impossible to find a bijective function going from R to (0,1) is
    1, the denominator has to be a polynomial of even
    2, one end of the function has to approach 1 asymptotically and the other end has to approach 0 and you can only achieve something like this and this means the numerator has to be a polynomial of odd
    3, in order to restrict it in the interval (0,1) , the degree of my numerator cannot exceed the degree of the numerator
    4, if it is the same degree (even), then it will both approach positive 1 and will not be injective
    5, if the degree of my numerator is less than that of the denominator, it will approach 0 both ways and will not be injective.
     
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