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A representation using reciprocal primes

  1. Nov 23, 2012 #1

    RobertCairone

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    Is this interesting?

    I have a way to represent all of the real numbers with a subset of the open interval (0,1). I write as a binary decimal x = .a[itex]_{0}[/itex]a[itex]_{1}[/itex]a[itex]_{2}[/itex]a[itex]_{3}[/itex]... where x = [itex]\sum -1^{a_{0}+1}/p_{i}[/itex] where p[itex]_{i}[/itex] is the i[itex]^{}th[/itex] prime.

    Since the reciprocals of the primes diverges, in this notation, .[itex]\overline{0}[/itex] diverges to negative infinity and .[itex]\overline{1}[/itex] diverges to positive infinity. By changing the digits, this forms a conditionally convergent series that can converge to any number in between. For example, √2 = .11111111010011010011001... We could pick [itex]\pi[/itex], but as the series diverges very slowly that would lead to a long string of ones before the first zero. Still, in principal any number can be achieved.

    There are an infinite number of ways to converge to any number, but some are more efficient than others. If we impose a simple rule that if the partial sum is less than the targeted value, the next a[itex]_{i}[/itex] is 1, and if the partial sum exceeds the targeted value then the next a[itex]_{i}[/itex] is zero. If perchance the value is exactly equal, the next digit is a one. This representation is unique, and I call it a well formed number. A test can show if any partial sum steps outside the bounds of a well formed envelope, that is, after a finite number of steps any sequence can be shown to be ill formed, or to convergent to a decreasing range of possible values. The examples here should be well formed.

    In this notation, since every prime must contribute something to the value, either as a positive or negative element, all numbers are transcendental. There are no numbers that end in ..[itex]\overline{0}[/itex] or ..[itex]\overline{1}[/itex] I don't think any numbers can be repeating decimals, but I'm not certain of that. A rational number like 5/6 does not have the representation .11, but .110110101010110..., dancing around the value as it converges once again.

    In a properly formed number, there can be only so many consecutive digits, and once the value is approached closely enough, I think at most two consecutive digits can appear before it flips. But every so often there must be at least one such a double, as a number that terminated in the repeating decimal ..[itex]\overline{01}[/itex] or ..[itex]\overline{10}[/itex] must diverge. I don't know if I could say the same for ..[itex]\overline{011}[/itex] or ..[itex]\overline{01011}[/itex] but I suspect it is true

    Anyway, that's the general idea if I've expressed it well enough.
     
  2. jcsd
  3. Nov 23, 2012 #2
    Hi Robert,

    My first question about this is: What is the significance of prime numbers here? Could you just as well use any increasing sequence of natural numbers for which the sum of the reciprocals diverges?
     
  4. Nov 23, 2012 #3

    RobertCairone

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    Yes, the natural numbers would work as well, but I chose the primes as a smaller set of basis elements. I was looking for some kind of "minimalist" way of approximating the reals. The slowness by which the reciprocals of the primes diverges made them seem natural. Does anything diverge more slowly?
     
  5. Nov 23, 2012 #4

    haruspex

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    Treating this as a question about slowly diverging series of reals, and reinterpreting as integrals, the following all diverge, steadily more slowly:
    ∫1/x, ∫1/(x ln(x)), ∫1/(x ln(x) ln(ln(x))), ...
    The primes, of course, would correspond to the second of those.
     
  6. Nov 24, 2012 #5

    RobertCairone

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    So if the first corresponds to the integers, and the second to the primes, is there a naive interpretation for the third of these? Does this say there is a definable subset of the primes that still diverges?
     
  7. Nov 24, 2012 #6

    haruspex

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    That's a tantalizing question.
     
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