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How to prove 'infinite primes' kind of problem

  1. Jun 18, 2011 #1
    Other than the Eucledean method
    (1+p1p2p3...)and so on

    other than this, how do we prove that there are infinte prime numbers?
    in algebreic way (excluding analyses)
     
    Last edited: Jun 18, 2011
  2. jcsd
  3. Jun 18, 2011 #2

    HallsofIvy

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    The Euclidean proof is so neat and simple, I cannot imagine why you would want a different proof.
     
  4. Jun 18, 2011 #3
    indeed

    but I was just wondering if there is any other technique to prove there are infinite primes so it can be applied to prove twin prime conjecture, mersenne prime conjecture, or sophie-germain conjecture.
     
  5. Jun 18, 2011 #4
    There are many many proofs of the infinitude of the primes

    http://primes.utm.edu/notes/proofs/infinite/

    And about 1/4 of the way down through this
    http://en.wikipedia.org/wiki/Prime_number

    and many more, but I hesitate to claim an infinite number of them.

    However thus far nobody no one has seen how to use these to prove any of the conjectures you mention.

    On a brighter and prime related note, there is now a non-crank proposed proof of Goldbach that has been submitted. It would be interesting to see if that could be explained in a form that those not in the "bright student" group could understand.
     
  6. Jun 19, 2011 #5
    One proof due to the German mathematician Kummer:

    Assumption:
    Let P[itex]_{r}[/itex]:={2,3,5,7,...,p[itex]_{r}[/itex]} the finite set of all primes

    form the prime factorial p[itex]_{r}[/itex]#:=2*3*5*...*p[itex]_{r}[/itex]
    and the number q:=-1+ p[itex]_{r}[/itex]#

    q is not in P[itex]_{r}[/itex], that means, q is composite, with prime
    factors from P[itex]_{r}[/itex]; let p[itex]_{k}[/itex] be one of those factors

    Now: p[itex]_{k}[/itex] divides both p[itex]_{r}[/itex]# and q, hence it
    must divide their difference: but this difference is 1
    and p[itex]_{k}[/itex] does NOT divide 1.

    Therefore our assumption must be false
     
  7. Jun 19, 2011 #6

    disregardthat

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    The above proof is essentially equivalent to euclids proof.
     
  8. Jun 20, 2011 #7
    I think not.

    Recall the original version of Euclids proof:

    Let Q[itex]_{r}[/itex]:={q[itex]_{1}[/itex], q[itex]_{2}[/itex],...,q[itex]_{r}[/itex]} be the set a r distinct
    prime numbers, and set q:=1+q[itex]_{1}[/itex]*...*q[itex]_{r}[/itex]

    (Note: the r primes must NOT be the first r primes 2,3,..,p[itex]_{r}[/itex] !)

    Then q is NOT divisible by any of the q[itex]_{i}[/itex] and must therefore be an new
    prime or at least a composite number formed with additional primes.

    Therefore: from any Q[itex]_{r}[/itex] you can construct a Q[itex]_{r+1}[/itex], and that proves the infinitude of the primes

    Now compare this to the Kummer proof I cited.
     
  9. Jun 20, 2011 #8
    The following proff is due to Polya

    Let e:=2[itex]^{k}[/itex] and F[itex]_{k}[/itex]:=1+2[itex]^{e}[/itex] be the Fermat numbers

    with F[itex]_{0}[/itex]:=3, F[itex]_{1}[/itex]:=5, F[itex]_{3}[/itex]:=17. F[itex]_{4}[/itex]:=257, F[itex]_{5}[/itex]:=65537, F[itex]_{6}[/itex]:=4294967297, etc

    we have the recurrence relation F[itex]_{k+1}[/itex]:=2+[itex]\prod[/itex][itex]^{k}_{i=0}[/itex]F[itex]_{i}[/itex]

    From the recurrence relation we have the following

    Lemma: Any two F[itex]_{i}[/itex] and F[itex]_{j}[/itex] for distinct i, j are co-prime,
    i.e. have no factors in common


    Now let F:={F[itex]_{0}[/itex],F[itex]_{1}[/itex],F[itex]_{2}[/itex],...} be the set of all Fermat numbers and this set is countable,
    i.e. has the cardinality of N:={1,2,3,4,...}, the set of the natural numbers

    Therefore no finite P[itex]_{r}[/itex]:={2,3,5,...,p[itex]_{r}[/itex]} can meet the requirements
    of our Lemma, there would simply be not enough primes.
     
  10. Jun 20, 2011 #9
    Let P:={p[itex]_{i}[/itex]} = {2,3,5,7,...} and
    E[itex]_{k}[/itex]:=1+[itex]\prod[/itex][itex]^{k}_{i=1}[/itex]p[itex]_{i}[/itex] be the Euclid numbers with E[itex]_{1}[/itex]:=3, E[itex]_{2}[/itex]:=7, E[itex]_{3}[/itex]:=31. etc

    We have the following

    Lemma All E[itex]_{i}[/itex] are odd, i.e. 2 is not a factor of E[itex]_{i}[/itex].

    and it is not difficult to prove the following

    Theorem Any two E[itex]_{i}[/itex], E[itex]_{j}[/itex] for different i,j are co-prime, i.e. have no common factors.

    Now supose, P:={2,3,5,7,...,p[itex]_{r}[/itex]}, then the corresponding {E[itex]_{i}[/itex]} must be the numbers {2,3,5,7,...,p[itex]_{r}[/itex]}
    in some order (our Theorem), but that is impossible (our Lemma)

    This proof was inspired by the Polya proof
     
  11. Jun 20, 2011 #10
    From the Euler [itex]\zeta[/itex]-function [itex]\zeta[/itex](s):=[itex]\sum[/itex][itex]^{\infty}_{i=1}[/itex]([itex]\frac{1}{i}[/itex])[itex]^{s}[/itex]

    we know, that [itex]\zeta[/itex](2) is [itex]\frac{\pi^{2}}{6}[/itex]

    This quantity is irrational, since [itex]\pi[/itex] itself is irrational

    From the Euler product formula, we have

    [itex]\zeta[/itex](2):=[itex]\prod[/itex](1 - [itex]\frac{1}{p^{2}}[/itex])[itex]^{-1}[/itex], where the product is taken over all primes

    If there were only a finite number of primes, this product would
    be a rational number, in contradiction to the irrational nature of [itex]\frac{\pi^{2}}{6}[/itex],
    which concludes the proof.
     
  12. Jun 20, 2011 #11

    disregardthat

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    Yes, it is not equivalent, but "essentially" equivalent. It doesn't contain any new ideas.
     
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