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Math Help - Bijection

  1. #1
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    Bijection

    Show that f:\mathbb R\to(-1,1) where f(x)=\dfrac x{1+|x|} is a bijection.

    Wonder if there's a way using not so advanced analysis tools to prove this, and not to do it by the old way by proving first that is injective and surjective.
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    Quote Originally Posted by Connected View Post
    Show that f:\mathbb R\to(-1,1) where f(x)=\dfrac x{1+|x|} is a bijection.
    Wonder if there's a way using not so advanced analysis tools to prove this, and not to do it by the old way by proving first that is injective and surjective.
    And what is wrong with that?
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  3. #3
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    Haha, sorry, it's not wrong actually.

    Well I'm having problems to prove surjectivity, can you help me with that?
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Connected View Post
    Haha, sorry, it's not wrong actually.

    Well I'm having problems to prove surjectivity, can you help me with that?

    We can write f in other form.

    h:R-->(-1,1)

    h(x)={x/(1+x) for x>0; 0 for x=0; x/(1-x) for x<0}

    f=h for all x in R.
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  5. #5
    Senior Member Tinyboss's Avatar
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    Quote Originally Posted by Connected View Post
    Haha, sorry, it's not wrong actually.

    Well I'm having problems to prove surjectivity, can you help me with that?
    What AsZ said is simpler, but here's another way to think of it.

    Its limit at infinity is 1 and its limit at -infinity is -1. It's continuous on a connected set, so its image is connected, i.e. an interval. So its image contains (-1, 1).
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Tinyboss View Post
    What AsZ said is simpler, but here's another way to think of it.

    Its limit at infinity is 1 and its limit at -infinity is -1. It's continuous on a connected set, so its image is connected, i.e. an interval. So its image contains (-1, 1).
    And to prove that f is 1-1 you need to use Rolle's theorem.
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