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Math Help - Bijections

  1. #1
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    Bijections

    Show that the function f(x)=x/(x^2-1) is a bijection between (-1,1) and the real numbers.

    I was able to show that the function is one-to-one and y=x, but I can't figure out how to show that it is onto.

    How can I also show that given any a and b, which are elements of the real numbers with a<b, there is a bijection between (a,b) and (-1,1).
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  2. #2
    Member kalyanram's Avatar
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    Re: Bijections

    Quote Originally Posted by lovesmath View Post
    I was able to show that the function is one-to-one and y=x
    This statement is not clear. what do you mean by y=x.

    Quote Originally Posted by lovesmath View Post
    I can't figure out how to show that it is onto.
    Use the definition that for every \alpha \in \mathbb{R} \exists \beta \in (-1,1) \ni f(\beta) = \alpha
    Spoiler:
    Consider
    \frac{x}{x^2-1} = \alpha \implies f(x) = \alpha x^2 - x -\alpha = 0 when \alpha = 0, x= 0 \alpha \neq 0 we solve for x we have x = \frac{1 \pm \sqrt{1 + 4 \alpha^2}}{2\alpha}. Now we are not sure as to which root to select.
    Bijections-parabola.png
    Consider \alpha > 0 figure 1 shows thw plot of the quadratic where we need the roots in (-1,1) so this requires f(-1)> 0 , f(1)> 0. but we have f(1) = -1 and f(-1)=1. \therefore we choose \beta = \frac{1 - \sqrt{1 + 4 \alpha^2}}{2\alpha} \ni \beta \in (-1,1) , f(\beta) > 0 and f(\beta) = \alpha. Similarly you can argue from figure 2 that for \alpha < 0 choose \beta = \frac{1 + \sqrt{1 + 4 \alpha^2}}{2\alpha}. Hence it is onto.
    Last edited by kalyanram; September 11th 2012 at 09:02 AM.
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  3. #3
    Member kalyanram's Avatar
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    Re: Bijections

    Quote Originally Posted by lovesmath View Post
    How can I also show that given any a and b, which are elements of the real numbers with a<b, there is a bijection between (a,b) and (-1,1).
    Bijections-ab-11.png
    Can you construct such a bijective map between (a,b) and (-1,1)?
    Spoiler:

    Consider f(c) = 2.\frac{c-a}{b-a}-1, \forall a<c<b
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  4. #4
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    Re: Bijections

    Quote Originally Posted by lovesmath View Post
    Show that the function f(x)=x/(x^2-1) is a bijection between (-1,1) and the real numbers.

    I was able to show that the function is one-to-one and y=x, but I can't figure out how to show that it is onto.

    How can I also show that given any a and b, which are elements of the real numbers with a<b, there is a bijection between (a,b) and (-1,1).
    To show "1 to 1" let y be any real number and show there exist at least one x such that x/(x^2- 1)=y. You can do that by solving for x. You will, of course, need to show that the value of x you get is between -1 and 1.
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