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Find a Bijection [tex]\left[ 0,1 \right] \rightarrow \Re [/tex]

  1. Jan 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Find a Bijection [tex]\left( 0,1 \right) \rightarrow [/tex] R

    2. Relevant equations
    Detention of bijection


    3. The attempt at a solution

    let r be a number in the interval [tex]\left[ 0,1 \right] [/tex]

    r = a b1c1b2c2 ... bncn where a, b,c are digits between [0, 9]

    f(r) =
    { - b1b2b3 ...... bn . c1 c2 ..... cn if a<5
    {+ b1b2b3 ...... bn . c1 c2 ..... cn if a>=5

    i know its incorrect as it is not injective. any help in trying to get the plus minus part.
     
  2. jcsd
  3. Jan 4, 2009 #2

    HallsofIvy

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    Another problem with your function is that not every number in [0, 1] can be written that way!

    You need a one-to-one function that maps 0 into negative infinity and 1 into positive infinity.

    You might consider something like 1/(x-1)- 1/x.
     
  4. Jan 4, 2009 #3

    thank u for ur reply and understand how it will work.

    would : f(x) = 1/(2x-1) work

    as if x < 1/2 it will be -ve and if x > 1/2 it will be +ve
    as x --> 1/2 form 0 it will go to -ve infinity
    as x --> 1/2 form 1 it will go to +ve infinity

    or f(x) = arctanh(2x - 1)

    which will transform the arctanh(x) from the (-1, 1) interval to (0, 1) interval as required, and is a 1 to 1 and as you said a function that maps 0 into negative infinity and 1 into positive infinity is needed
     
    Last edited: Jan 4, 2009
  5. Jan 4, 2009 #4

    Dick

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    1/(2x-1) isn't onto. For example, it's never 0. The second example works fine.
     
  6. Jan 4, 2009 #5
    thank you for reply

    just for completeness the second part of the question was to: Find a Bijection [a,b] --> R where a, b are real numbers and b > a

    this can easily be done by modifying the second example as shown:
    let g(x) = x /(b-a) (to transform it into the inteval (0,1) )

    and f(x) = arctanh(2x - 1) (the first example)

    and have the final function as: f o g(x) ( f(g(x)) )
     
    Last edited: Jan 4, 2009
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