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probability that a natural number is prime

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  • Kermit Rose
    2a. Are Big Numbers More Likely To Be Prime? Posted by: Chester Elders chesterelders28@yahoo.com chesterelders28 Date: Sun May 23, 2010 2:17 am ((PDT)) ...
    Message 1 of 1 , May 23, 2010
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      2a. Are Big Numbers More Likely To Be Prime?
      Posted by: "Chester Elders" chesterelders28@... chesterelders28
      Date: Sun May 23, 2010 2:17 am ((PDT))

      > I have no idea whether this proof is unique to me; but, I think I can
      > prove that the probability that a natural number is prime is directly
      > proportional to it's size.


      How do you calculate the probability that a number of a given size
      be prime?

      By my own intuition, I would calculate it as follows.

      Probability of N being prime = (Number of positive primes =< N) / N

      which by the prime number theorem,

      for large N,

      is approximately,

      (N/log(N)) / N = 1/log(N)

      which approaches 0 as N increases without bound.


      > Proof:

      > Let A be a natural number.

      > Let S be the set of all prime numbers such that any element of S is
      > less than the step function of the square root of A.


      Why not say that S is the set of prime numbers < square root of A.

      S = { p such that p is prime and p**2 < A }



      > Define the probability that A is a prime number as the quotient
      > (A-S)/A.

      A is a number.
      S is a set.

      What do you mean by A - S?



      > Since the quotient of the square root of a natural number divided by
      > the number decreases in proportion to the size of that number, it
      > follows that the limit of (A-S)/A approaches 1 as A gets arbitrarily
      > large.

      If S were a number, not a set, this would have a better chance of making
      sense.

      Suppose we let S be the number of primes < square root of A.

      Then your ratio (A - S)/A is more related to the probability that
      A is not prime, than it is to the probability that A is prime.

      If we took S to be the number of primes < A, then

      the ratio (A - S)/A would be the probability that a positive number
      less than A is composite.

      Perhaps you might want to argue that the probability that a number < A
      being composite is the same as the probability that A is composite.




      > This is tantamount to saying that the probability that A is prime
      > increases as A increases. QED.

      Did you come close to proving that the probability that A is composite
      increases as A increases?







      > Wouldn't this imply that the prime numbers get arbitrarily
      > dense when one considers extremely large natural numbers?

      :) Composite numbers do get arbitrarily dense in sets of
      extremely large natural numbers.


      > What really has me worried is the seemingly inescapable conclusion
      > that, for sufficiently large natural numbers, wouldn't primes tend to
      > become sequential? I know that's not possible because the
      > distribution of prime numbers is random.

      Distribution of primes is not random. The distribution is directly
      determined by smaller primes.


      > I assume that probability holds uniformly across the number line.


      Not sue what you have in mind when you say "probability holds uniformly"



      > I would appreciate some feedback.

      > Sincerely,
      > Chester


      :) You got it.

      Kermit Rose
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