Convert numbers between 0 and infinity to numbers between 0.0 and 1.0












3












$begingroup$


Can we arrange all numbers x; such that x lies between 0 and Infinity, between 0 and 1? The scale does not have to be linear, but for any a and b in x, where a <= b, then a' and b', the equivalent numbers on the new scale must also be such that: a' <= b'.



I am really curious about this as I need it for a ratings system.



Thanks!






real-numbers number-line







share|cite|improve this question











$endgroup$












  • $begingroup$
    "The scale does not have to be linear" how fortunate!
    $endgroup$
    – mdup
    53 mins ago
















3












$begingroup$


Can we arrange all numbers x; such that x lies between 0 and Infinity, between 0 and 1? The scale does not have to be linear, but for any a and b in x, where a <= b, then a' and b', the equivalent numbers on the new scale must also be such that: a' <= b'.



I am really curious about this as I need it for a ratings system.



Thanks!






real-numbers number-line







share|cite|improve this question











$endgroup$












  • $begingroup$
    "The scale does not have to be linear" how fortunate!
    $endgroup$
    – mdup
    53 mins ago














3












3








3





$begingroup$


Can we arrange all numbers x; such that x lies between 0 and Infinity, between 0 and 1? The scale does not have to be linear, but for any a and b in x, where a <= b, then a' and b', the equivalent numbers on the new scale must also be such that: a' <= b'.



I am really curious about this as I need it for a ratings system.



Thanks!






real-numbers number-line







share|cite|improve this question











$endgroup$




Can we arrange all numbers x; such that x lies between 0 and Infinity, between 0 and 1? The scale does not have to be linear, but for any a and b in x, where a <= b, then a' and b', the equivalent numbers on the new scale must also be such that: a' <= b'.



I am really curious about this as I need it for a ratings system.



Thanks!






real-numbers number-line




real-numbers number-line






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









J. W. Tanner

2,0341116




2,0341116










asked 9 hours ago









gbenrosciencegbenroscience

356




356












  • $begingroup$
    "The scale does not have to be linear" how fortunate!
    $endgroup$
    – mdup
    53 mins ago


















  • $begingroup$
    "The scale does not have to be linear" how fortunate!
    $endgroup$
    – mdup
    53 mins ago
















$begingroup$
"The scale does not have to be linear" how fortunate!
$endgroup$
– mdup
53 mins ago




$begingroup$
"The scale does not have to be linear" how fortunate!
$endgroup$
– mdup
53 mins ago










3 Answers
3






active

oldest

votes


















5












$begingroup$

$exp$ is a great tool, but there's also
$$
x mapsto frac{x^2}{1+x^2}
$$
which may be slightly easier to work with in some situations.



As @Servaes points out, you can also use
$$
x mapsto frac{x}{1+x}
$$
because you're working on the nonnegative reals rather than all reals.



And a personal favorite of mine is
$$
x mapsto frac{2}{pi} arctan(x).
$$






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    And in fact there is no need for the squares as the function is needed on the nonnegative reals only.
    $endgroup$
    – Servaes
    9 hours ago










  • $begingroup$
    Good point. Edited.
    $endgroup$
    – John Hughes
    8 hours ago



















3












$begingroup$

There are many options, one example is the function $f(x)=e^{-x}$. It maps the domain $(0,infty)$ onto the range $(0,1)$, though it reverses the ordering. That is, if $x<y$ then $f(x)>f(y)$. This is of course easily fixed by taking
$$g(x)=1-f(x)=1-e^{-x},$$
instead.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    $6$ seconds before me!
    $endgroup$
    – Peter Foreman
    9 hours ago










  • $begingroup$
    @PeterForeman Either this is a rather canonical answer, or you are very fast at paraphrasing.
    $endgroup$
    – Servaes
    9 hours ago






  • 1




    $begingroup$
    Are you sure this satisfies the second requirement?
    $endgroup$
    – Keatinge
    9 hours ago






  • 2




    $begingroup$
    Given that if $aleq b$ then $a'leq b'$ I think $1-e^{-x}$ should be what OP is looking for.
    $endgroup$
    – Infiaria
    9 hours ago






  • 1




    $begingroup$
    Wow, I . am impressed with the dexterity and eagerness here. Thanks so much for helping out! I have to accept the other answer by @John Hughes, though. Because I need the answers not to converge too quickly towards 1. Thanks all the same!
    $endgroup$
    – gbenroscience
    8 hours ago



















1












$begingroup$

The expression you are looking for is a one-to-one mapping from positive reals into $[0,1]$. Consider the exponential mapping $f_k(x) = exp(- (x^k))$. Other people suggested $f_2$. There exist other mappings.






share|cite|improve this answer











$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    $exp$ is a great tool, but there's also
    $$
    x mapsto frac{x^2}{1+x^2}
    $$
    which may be slightly easier to work with in some situations.



    As @Servaes points out, you can also use
    $$
    x mapsto frac{x}{1+x}
    $$
    because you're working on the nonnegative reals rather than all reals.



    And a personal favorite of mine is
    $$
    x mapsto frac{2}{pi} arctan(x).
    $$






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      And in fact there is no need for the squares as the function is needed on the nonnegative reals only.
      $endgroup$
      – Servaes
      9 hours ago










    • $begingroup$
      Good point. Edited.
      $endgroup$
      – John Hughes
      8 hours ago
















    5












    $begingroup$

    $exp$ is a great tool, but there's also
    $$
    x mapsto frac{x^2}{1+x^2}
    $$
    which may be slightly easier to work with in some situations.



    As @Servaes points out, you can also use
    $$
    x mapsto frac{x}{1+x}
    $$
    because you're working on the nonnegative reals rather than all reals.



    And a personal favorite of mine is
    $$
    x mapsto frac{2}{pi} arctan(x).
    $$






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      And in fact there is no need for the squares as the function is needed on the nonnegative reals only.
      $endgroup$
      – Servaes
      9 hours ago










    • $begingroup$
      Good point. Edited.
      $endgroup$
      – John Hughes
      8 hours ago














    5












    5








    5





    $begingroup$

    $exp$ is a great tool, but there's also
    $$
    x mapsto frac{x^2}{1+x^2}
    $$
    which may be slightly easier to work with in some situations.



    As @Servaes points out, you can also use
    $$
    x mapsto frac{x}{1+x}
    $$
    because you're working on the nonnegative reals rather than all reals.



    And a personal favorite of mine is
    $$
    x mapsto frac{2}{pi} arctan(x).
    $$






    share|cite|improve this answer











    $endgroup$



    $exp$ is a great tool, but there's also
    $$
    x mapsto frac{x^2}{1+x^2}
    $$
    which may be slightly easier to work with in some situations.



    As @Servaes points out, you can also use
    $$
    x mapsto frac{x}{1+x}
    $$
    because you're working on the nonnegative reals rather than all reals.



    And a personal favorite of mine is
    $$
    x mapsto frac{2}{pi} arctan(x).
    $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 8 hours ago

























    answered 9 hours ago









    John HughesJohn Hughes

    64.1k24191




    64.1k24191








    • 2




      $begingroup$
      And in fact there is no need for the squares as the function is needed on the nonnegative reals only.
      $endgroup$
      – Servaes
      9 hours ago










    • $begingroup$
      Good point. Edited.
      $endgroup$
      – John Hughes
      8 hours ago














    • 2




      $begingroup$
      And in fact there is no need for the squares as the function is needed on the nonnegative reals only.
      $endgroup$
      – Servaes
      9 hours ago










    • $begingroup$
      Good point. Edited.
      $endgroup$
      – John Hughes
      8 hours ago








    2




    2




    $begingroup$
    And in fact there is no need for the squares as the function is needed on the nonnegative reals only.
    $endgroup$
    – Servaes
    9 hours ago




    $begingroup$
    And in fact there is no need for the squares as the function is needed on the nonnegative reals only.
    $endgroup$
    – Servaes
    9 hours ago












    $begingroup$
    Good point. Edited.
    $endgroup$
    – John Hughes
    8 hours ago




    $begingroup$
    Good point. Edited.
    $endgroup$
    – John Hughes
    8 hours ago











    3












    $begingroup$

    There are many options, one example is the function $f(x)=e^{-x}$. It maps the domain $(0,infty)$ onto the range $(0,1)$, though it reverses the ordering. That is, if $x<y$ then $f(x)>f(y)$. This is of course easily fixed by taking
    $$g(x)=1-f(x)=1-e^{-x},$$
    instead.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      $6$ seconds before me!
      $endgroup$
      – Peter Foreman
      9 hours ago










    • $begingroup$
      @PeterForeman Either this is a rather canonical answer, or you are very fast at paraphrasing.
      $endgroup$
      – Servaes
      9 hours ago






    • 1




      $begingroup$
      Are you sure this satisfies the second requirement?
      $endgroup$
      – Keatinge
      9 hours ago






    • 2




      $begingroup$
      Given that if $aleq b$ then $a'leq b'$ I think $1-e^{-x}$ should be what OP is looking for.
      $endgroup$
      – Infiaria
      9 hours ago






    • 1




      $begingroup$
      Wow, I . am impressed with the dexterity and eagerness here. Thanks so much for helping out! I have to accept the other answer by @John Hughes, though. Because I need the answers not to converge too quickly towards 1. Thanks all the same!
      $endgroup$
      – gbenroscience
      8 hours ago
















    3












    $begingroup$

    There are many options, one example is the function $f(x)=e^{-x}$. It maps the domain $(0,infty)$ onto the range $(0,1)$, though it reverses the ordering. That is, if $x<y$ then $f(x)>f(y)$. This is of course easily fixed by taking
    $$g(x)=1-f(x)=1-e^{-x},$$
    instead.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      $6$ seconds before me!
      $endgroup$
      – Peter Foreman
      9 hours ago










    • $begingroup$
      @PeterForeman Either this is a rather canonical answer, or you are very fast at paraphrasing.
      $endgroup$
      – Servaes
      9 hours ago






    • 1




      $begingroup$
      Are you sure this satisfies the second requirement?
      $endgroup$
      – Keatinge
      9 hours ago






    • 2




      $begingroup$
      Given that if $aleq b$ then $a'leq b'$ I think $1-e^{-x}$ should be what OP is looking for.
      $endgroup$
      – Infiaria
      9 hours ago






    • 1




      $begingroup$
      Wow, I . am impressed with the dexterity and eagerness here. Thanks so much for helping out! I have to accept the other answer by @John Hughes, though. Because I need the answers not to converge too quickly towards 1. Thanks all the same!
      $endgroup$
      – gbenroscience
      8 hours ago














    3












    3








    3





    $begingroup$

    There are many options, one example is the function $f(x)=e^{-x}$. It maps the domain $(0,infty)$ onto the range $(0,1)$, though it reverses the ordering. That is, if $x<y$ then $f(x)>f(y)$. This is of course easily fixed by taking
    $$g(x)=1-f(x)=1-e^{-x},$$
    instead.






    share|cite|improve this answer











    $endgroup$



    There are many options, one example is the function $f(x)=e^{-x}$. It maps the domain $(0,infty)$ onto the range $(0,1)$, though it reverses the ordering. That is, if $x<y$ then $f(x)>f(y)$. This is of course easily fixed by taking
    $$g(x)=1-f(x)=1-e^{-x},$$
    instead.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 1 hour ago









    J. W. Tanner

    2,0341116




    2,0341116










    answered 9 hours ago









    ServaesServaes

    25.4k33996




    25.4k33996








    • 1




      $begingroup$
      $6$ seconds before me!
      $endgroup$
      – Peter Foreman
      9 hours ago










    • $begingroup$
      @PeterForeman Either this is a rather canonical answer, or you are very fast at paraphrasing.
      $endgroup$
      – Servaes
      9 hours ago






    • 1




      $begingroup$
      Are you sure this satisfies the second requirement?
      $endgroup$
      – Keatinge
      9 hours ago






    • 2




      $begingroup$
      Given that if $aleq b$ then $a'leq b'$ I think $1-e^{-x}$ should be what OP is looking for.
      $endgroup$
      – Infiaria
      9 hours ago






    • 1




      $begingroup$
      Wow, I . am impressed with the dexterity and eagerness here. Thanks so much for helping out! I have to accept the other answer by @John Hughes, though. Because I need the answers not to converge too quickly towards 1. Thanks all the same!
      $endgroup$
      – gbenroscience
      8 hours ago














    • 1




      $begingroup$
      $6$ seconds before me!
      $endgroup$
      – Peter Foreman
      9 hours ago










    • $begingroup$
      @PeterForeman Either this is a rather canonical answer, or you are very fast at paraphrasing.
      $endgroup$
      – Servaes
      9 hours ago






    • 1




      $begingroup$
      Are you sure this satisfies the second requirement?
      $endgroup$
      – Keatinge
      9 hours ago






    • 2




      $begingroup$
      Given that if $aleq b$ then $a'leq b'$ I think $1-e^{-x}$ should be what OP is looking for.
      $endgroup$
      – Infiaria
      9 hours ago






    • 1




      $begingroup$
      Wow, I . am impressed with the dexterity and eagerness here. Thanks so much for helping out! I have to accept the other answer by @John Hughes, though. Because I need the answers not to converge too quickly towards 1. Thanks all the same!
      $endgroup$
      – gbenroscience
      8 hours ago








    1




    1




    $begingroup$
    $6$ seconds before me!
    $endgroup$
    – Peter Foreman
    9 hours ago




    $begingroup$
    $6$ seconds before me!
    $endgroup$
    – Peter Foreman
    9 hours ago












    $begingroup$
    @PeterForeman Either this is a rather canonical answer, or you are very fast at paraphrasing.
    $endgroup$
    – Servaes
    9 hours ago




    $begingroup$
    @PeterForeman Either this is a rather canonical answer, or you are very fast at paraphrasing.
    $endgroup$
    – Servaes
    9 hours ago




    1




    1




    $begingroup$
    Are you sure this satisfies the second requirement?
    $endgroup$
    – Keatinge
    9 hours ago




    $begingroup$
    Are you sure this satisfies the second requirement?
    $endgroup$
    – Keatinge
    9 hours ago




    2




    2




    $begingroup$
    Given that if $aleq b$ then $a'leq b'$ I think $1-e^{-x}$ should be what OP is looking for.
    $endgroup$
    – Infiaria
    9 hours ago




    $begingroup$
    Given that if $aleq b$ then $a'leq b'$ I think $1-e^{-x}$ should be what OP is looking for.
    $endgroup$
    – Infiaria
    9 hours ago




    1




    1




    $begingroup$
    Wow, I . am impressed with the dexterity and eagerness here. Thanks so much for helping out! I have to accept the other answer by @John Hughes, though. Because I need the answers not to converge too quickly towards 1. Thanks all the same!
    $endgroup$
    – gbenroscience
    8 hours ago




    $begingroup$
    Wow, I . am impressed with the dexterity and eagerness here. Thanks so much for helping out! I have to accept the other answer by @John Hughes, though. Because I need the answers not to converge too quickly towards 1. Thanks all the same!
    $endgroup$
    – gbenroscience
    8 hours ago











    1












    $begingroup$

    The expression you are looking for is a one-to-one mapping from positive reals into $[0,1]$. Consider the exponential mapping $f_k(x) = exp(- (x^k))$. Other people suggested $f_2$. There exist other mappings.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      The expression you are looking for is a one-to-one mapping from positive reals into $[0,1]$. Consider the exponential mapping $f_k(x) = exp(- (x^k))$. Other people suggested $f_2$. There exist other mappings.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        The expression you are looking for is a one-to-one mapping from positive reals into $[0,1]$. Consider the exponential mapping $f_k(x) = exp(- (x^k))$. Other people suggested $f_2$. There exist other mappings.






        share|cite|improve this answer











        $endgroup$



        The expression you are looking for is a one-to-one mapping from positive reals into $[0,1]$. Consider the exponential mapping $f_k(x) = exp(- (x^k))$. Other people suggested $f_2$. There exist other mappings.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 1 hour ago









        J. W. Tanner

        2,0341116




        2,0341116










        answered 9 hours ago









        PackSciencesPackSciences

        67216




        67216






























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