Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Goldbach's Conjecture (Theorem?)

  1. Jan 28, 2008 #1
    Goldbach's Conjecture (Theorem??)

    Hi guys,

    I just need to know if Goldbach's Conjecture has been demonstrated or not!

    Thx! :D
     
  2. jcsd
  3. Jan 28, 2008 #2
    No mathematical proof yet. It is very upsetting.
     
  4. Feb 8, 2008 #3
    are you trying to find a proof? :smile:
     
  5. Feb 8, 2008 #4
    what about if we write the even numbers as a sum of the form 3a + 5b? N = 3a + 5b

    consider gcd(3a,5b)=1, then "split" a and b obtaining a = a' + a", b = b' + b"

    N = (3a' + 5b') + (3a" + 5b")

    If we suceed to prove that one of these pairs (3a' + 5b') + (3a" + 5b"), when we test all the possible values for a', a", b' and b", are exactly two prime numbers, for every even number, the conjecture is proved.

    Of course this is far way of a proof for the conjecture, but should give us another angle to think about the problem.

    I made some observations but nothing pretty much beyond this point.
     
  6. Feb 12, 2008 #5

    HallsofIvy

    User Avatar
    Science Advisor

    In other words you have just changed to a slight variation, which looks to me to be more complicated and harder to prove than the original. You do understand, don't you, that testing "all the possible values for a', a", b', and b" " is no more possible than checking every even number?
     
  7. Feb 12, 2008 #6
    Let me try to explain a little bit better.

    What I meant saying "all the possible values for a', a", b', and b" " is: try to find some property for a', b' such that 3a' + 5b' is a prime number when 3a" + 5b" is a prime number too. Of course, as I said, this is far way a proof, wich means that perhaps this is a dead-end. But this is not the same thing as "checking every even number".

    Pick a even number, like 100.

    100 = 3*5 + 5*17, then a=5 and b=17, lets split a and b in a'=3, a"=2, b'=8, b"=9

    100 = (3*3 + 5*8) + (3*2 + 5*9)

    of course gcd(a',b') = gcd(a'',b'') = 1 when both expressions are prime numbers, but this is a necessary condition, not enough (I dont know if the correct word in english is this...)

    I was thinking on doing a kind of research to find the pairs of "a's" and "b's" of each expression such that their gcd() are both = 1, make a list, and try to find some pathern or property of the pairs with gcd()=1 when these pairs are prime numbers.

    I know this is a little bit vague, but could be a start... perhaps that other conjecture or mine about the numbers of the p + 2^n form, when proved, could be useful
     
  8. Feb 12, 2008 #7

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    "Necessary but not sufficient" was the phrase you were looking for. But I don't think that's true -- what about (3*2 + 5*4) + (3*2 + 5*1) = 37? gcd(4, 2) is not 1, but 37 is prime. So the condition appears to be neither necessary nor sufficient.
     
  9. Feb 12, 2008 #8
    sorry, my mistake, the correct is gcd(3a',5b') = gcd(3a'', 5b'') = 1, it is a necessary condition to both expression represents both prime numbers, but obviously it is not sufficient
     
  10. Feb 12, 2008 #9
    another point: what I mean wans't that (3a' + 5b') + (3a'' + 5b'') = prime, instead I meant that 3a' + 5b' = prime and 3a'' + 5b'' = prime each of then

    Call N = 3a + 5b an even number. When we split a and b, the question is: looking at the several pairs (3a' + 5b') + (3a'' + 5b'')*, such that gcd(3a',5b') = gcd(3a'', 5b'') = 1, obtained when we vary the "a's" and "b's", is there at least one of these pairs such that 3a' + 5b' and 3a'' + 5b'' are prime numbers?

    * I call a pair (3a' + 5b') + (3a'' + 5b'')
     
  11. Feb 12, 2008 #10

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    Ah, I misunderstood you, sorry.

    What are you saying then? Are you trying to use this to disprove the Goldbach conjecture by showing that there is an even number that cannot be split into coprime pairs (a, b), (a', b')? Are you trying to lessen the search space for experimental verification of a finite portion of the conjecture?
     
  12. Feb 12, 2008 #11
    I don't believe that the conjecture is wrong. If it is wrong, showing that there is an even number such that it cannot be split into coprime pairs might prove that the conjecture is false, yes.

    Unfortunally, I don't believe that such even number exist.

    I was thinking on another aproach, a little vague yet...
     
  13. Feb 12, 2008 #12

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    I'm just saying that a proof requires something that's sufficient, not something that's necessary. Any further thoughts on the problem?
     
  14. Feb 12, 2008 #13
    Ok, lets work a little bit more on our definitions.

    if a given number n =3a+5b is such that gcd(3a,5b)=1, we cannot say that n is a prime number, right? for instance,

    a=1 and b=11, gcd(3,55)=1 and n = 58,
    a=12 and b=11, gcd(36,55)=1 and n = 91 = 7*13

    then gcd(3a,5b)=1 is a necessary condition to n be a prime number, but not sufficient

    Given a even number written of the form N = 3a + 5b

    lets split this number like I said before in two other numbers m = 3a'+5b' and k = 3a''+5b'' such that N = m+k

    if we prove that for an any N there is no m+k pairs such that gcd(3a',5b')=1 AND gcd(3a'',5b'')=1 then this is a proof that goldbach's conjecture is FALSE, since gcd(3a',5b')=1 AND gcd(3a'',5b'')=1 is a necessary condition to m AND k be prime numbers, since there is no prime numbers represented by 3a+5b such that gcd(3a,5b)>1 (note that all numbers >7 can be expressed in the 3a+5b form)

    so this particular proof, in this particular case, requires a necessary condition, not a sufficient
     
  15. Feb 12, 2008 #14
    Has it not been proven that Goldbach's conjecture can not be proven in Peano arithmetic?
     
  16. Feb 13, 2008 #15
    I don't know about it... you mean that someone have proved that Goldbach's conjecture is undecidable? If yes, the conjecture is true.

    Could you give the link?
     
  17. Feb 13, 2008 #16
    I'm only guessing. There have been theorems about the natural numbers which are not provable in PA, but are provable in ZFC.

    What do you mean by "If yes, the conjecture is true"?
     
  18. Feb 13, 2008 #17

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    I imagine the meaning was thus, as I was going to post a similar note:

    Suppose the proposition "There exists no positive number n such that P(n)" for some predicate P. If the proposition is independent of Peano arithmetic (or some other framework, as long as the framework is strong enough to state the predicate P and express all natural numbers), then it cannot be disproved in that system by definition. But no natural number n exists that makes P(n) true, since if it did then the proposition could be disproved with the counterexample P(n).
     
  19. Feb 13, 2008 #18
    I did search my notes about it, but unfortunally I lost the thing... I was just working on a approach about these ideas I had put here.

    I noticed that almost all numbers such that gcd(3a,5b)=1 are primes or semi-primes (meaning two prime factors), but I don't know if I am wrong or not

    This is important if there is some pattern, even statistical pattern, such that prime numbers are more closer (or near?) to semiprime numbers than other numbers, but I can't explain why without my notes

    This is pretty much vague... if you have some insight about it reading this please post here (or tell me if you think that this will lead us to nowhere)
     
  20. Feb 13, 2008 #19

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    An attractive thought, but it doesn't work out.

    The most obvious problem is that "Goldbach's conjecture is undecidable in PA" means that "PA + Goldbach's Conjecture is false" is a consistent theory!


    The more subtle problem is that the following could be possible:
    ZFC is able to prove a particular finite ordinal x is a counterexample to Goldbach's conjecture.
    ZFC is able to prove that PA is incapable of proving that x is a counterexample.

    or similarly
    ZFC is able to prove a particular finite ordinal x is a counterexample to Goldbach's conjecture.
    ZFC is able to construct another model of PA where the corresponding x is not a counterexample to Goldbach's conjecture.

    I imagine even more pathological things could happen.
     
    Last edited: Feb 13, 2008
  21. Feb 13, 2008 #20

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    Of course the reason I didn't post my original thoughts was that I was unable to properly cover all of the pathologies, and as the complexity rose I had trouble justifying it for a one-off answer to the question.

    But I will admit it's hard to imagine these pathologies. It's like nonprinciple ultrafilters -- sure, ZFC has them, but you'll never run into one.
     
  22. Feb 13, 2008 #21

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    I guess the only rigorous thing I can pull out of my statement is that if the Goldbach conjecture is independent of PA then no consistent system can (1) construct an explicit natural number counterexample to the Goldbach conjecture, or even (2) construct an explicit upper bound for such a counterexample. In the latter case PA could take this upper bound and with finite effort check all partitions of all natural numbers up to it.
     
  23. Feb 14, 2008 #22
    some ideas may help.

    ok,here is an idea may help,
    even number ends with alast digit from the left,(0,2,4,6,8).prime number ends with(1,3,7,9) except,5.now the sum of any 2 prime numbers must end with(0,2,4,6,8).the two statements are equivalent to each others.
    another idea.the even numbers can be expressed as 2(p+q+1),p&q.are positive integers.now,prime numbers are odd and odd numbers can be expressed as,2p+1 or 2q+1,we get even numbers can be expressed as asum of two prime numbers.you may say that prime number cannot be expressed as,2p+1,because it includes all of the odd numbers.but the real question is,if the set,A,belongs to set,B,then not all the judgements that can be applied on,B,are applicable on,A,like countablity.if you can prove that the statement"ALL ODD NUMBERS CAN BE EXPRESSED AS,2p+1"can be applied on the prime numbers set.then the conjecture would be clear.i think you will have aproblen because our unwell defined statements seem to be tricky,for example,when i say,all even numbers can be expressed as 2n,i can also say,not all even numbers can be expressed as 2n.some of even numbers can be expressed as,3n+1.end of talk.
     
  24. Feb 14, 2008 #23

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    The rightmost digit of an even in base 10 is one of {0, 2, 4, 6, 8}, yes. For the primes you forgot to exclude 2: prime numbers other than 5 and 2 have a rightmost digit in base 10 of one of {1, 3, 7, 9}.

    The statement that the sum of any two primes greater than 2 is even follows, but that doesn't show Goldbach's (strong) conjecture, which is that any even number can be additively split into two primes.

    For example, consider n = 10^1000. n - 2 is composite, n - 3 is composite, n - 5 is composite, ..., n - (10^1000 - 1769) is composite. If all the omitted numbers in the ... are composite as well, then Goldbach's conjecture fails. Sure, all the numbers in the middle (all numbers except n - 2) are odd, but that doesn't help.

    Okay, all even numbers greater than 4 can be expressed in this form, sure.

    Backward again. Yes, each odd prime can be expressed in the form 2n+1 for some positive integer n, but that doesn't mean that n = p for the p you've already used. It could be a different positive integer.

    No. All primes greater than 2 can be expressed in the form 2n+1 for some positive integer n, but that doesn't imply the Goldbach conjecture -- it's not nearly a proof.

    All even numbers can be expressed as 2n for some integer n. Some even numbers can also be expressed as 3n+1 for some integer n -- these are precisely the integers congruent to 4 mod 6.
     
    Last edited: Feb 14, 2008
  25. Feb 14, 2008 #24
    ZFC could prove that such a counterexample exists without constructing it. In terms of finite sets/numbers, how strong is ZFC compared to PA? I'm trying to remember the theorem that was proved in ZFC but now PA; it involved constructing a sequence by taking the previous number, using it to get an expression involving towering powers of 2, then adding 1 -- I think -- and showing that the sequence eventually ends at 1.
     
  26. Feb 14, 2008 #25

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    http://en.wikipedia.org/wiki/Goodstein's_theorem
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook