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Old problem of infitesimals

  1. Aug 12, 2003 #1
    Here's a math problem which I think best represents the old problem of infitesimals. Do mathy guys accept there are infinately small numbers between 0 and finite numbers? I thought some famous maths guy said there wasn`t any. If so, how do you reslove this prob?...

    A rational number between 0 and 1, p ,is selected at random.
    As there are an infinite number of rationals between 0 and 1, it can be shown that the chance of any one rational being selected is 0. But we cannot deduce from this that it is impossible that a certain rational is selected, because it is possible. This has been proved for p, and as p is a variable, is prooved for all rationals between 0 and 1. So there is an infitesimal chance of selecting p. Constrast this with a number outside the boundary [0,1] being selected which really is zero chance.

    So 1 / infinity is greater than 0, yeah?
     
    Last edited by a moderator: Feb 5, 2013
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  3. Aug 12, 2003 #2

    Hurkyl

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    Mathematics would take that as proof that probability over finite spaces behaves slightly differently than probability over infinite spaces. :smile:


    Generally, probability is a measure on a space, meaning that it is an additive function that assigns a number to a class of subsets (they are measurable sets... aren't mathematicians creative? [:)). Like most measures, they are plagued with having subsets that are too small, and thus have measure 0... though over finite sets, this doesn't always happen. Thus, in general, "probability 0" is not synonymous with "cannot 'selected'" (whatever 'selected' means)

    Typically, over infinite spaces, one deals with intervals or similar things, not individual points. For example, one speaks of the probability that a measured quantity is 10 +/- 0.001, aka in the interval [9.999, 10.001]


    Incidentally, the rational numbers are more pathological in this respect than for what you give them credit. :smile: The rationals between [0, 1], as a countably infinite set, will have either measure 0 or measure ∞ no matter what homogenous metric you use, thus tremendously complicating any notion of having a uniform probability distribution over them.


    But, of course, it's easy enough to come up with nonuniform probability distributions... in fact, one virtue of being a countably infinite set is that there exist (nonhomogenous) probability distributions over the rationals between [0, 1] such that no individual point has probability 0 of being selected. For example, take the distribution:

    0 with probability 1/2
    1 with 1/4
    1/2 with 1/8
    1/3 with 1/16
    2/3 with 1/32
    1/4 with 1/64
    3/4 with 1/128
    1/5 with 1/256
    ...

    This covers all of the rationals in [0, 1] giving them a total probability of 1, but no individual point has probability 0.
     
  4. Aug 12, 2003 #3

    mathman

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    There is a subject called "non-standard analysis" developed by Abraham Robinson, which deals with these kinds of small numbers. Look it up in Wikipedia.
     
  5. Aug 21, 2003 #4
    Thanks for replys guys.
    Liked the non-uniform probabilty thing.
    So Hurkyl, you say it's mega hard to attach uniform probabilty to countably infinite sets, but is it possible with standard analysis? ( Don`t have to explain how ). Because it's surely a intuitively simple truth that uniform probabilty exists over such sets. If it's mega hard to proove with standard analysis, maybe mathy guys should go with non-stardard analysis.
     
  6. Aug 26, 2003 #5
    Hi meemoe_uk,

    Here is some non conventional point of view:


    {} = content does not exist = 0

    not{} = content exists = 1


    Any transformation from {} to not{} can't be anything but a phase transition from 0(= does not exist) to 1(=exists).

    So, what we have is a quantum-leap path between 0 to 1 ,with exactly 0 points in it.

    Through this point of view, all you have in this stage is a continuous smooth X-axis connector between 0 to 1.

    Any point needs at least two coordinates to exist, and X-axis alone is not enough.

    So, we put some y(=0)-axis on (not in) the X-axis and we get some x,0 point.

    Any x value determined by its relation to 0(={}).

    Through the “eyes” of 0, any non-0 = 1 (this is the reason why we can’t divide by zero), so to get a unique value to x, we must compare it to our first quantum-leap, which we call it '1'.

    If we find a ratio that can be expressed as a ratio between equaled quantum-leap sizes under '1', we call this x a Q member, and if not, it is an R member.

    Through this point of view, no x,0 point exists between 0 to 1, until we create it by putting some Y(=0)-axis on the X-axis.

    Therefore through this non conventional point of view, the ratio is
    1/0 and not 1/[oo].


    Yours,

    Doron
     
    Last edited by a moderator: Aug 26, 2003
  7. Aug 26, 2003 #6

    Hurkyl

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    Unfortunately, I don't have an advanced text on probability measures to which I can turn as a reference for your question... bit I'm pretty sure it is impossible without amending one of the axioms of a measure. (switching to the hyperreals is, of course, one possible amendment to the axioms of a measure ... I'm not sure if it would accomplish what is desired, though)
     
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