Goldbach's Conjecture - Proof (better link)

  1. jcsd
  2. Can you upload better images so I can print it out to read?
     
  3. Better yet, make it LaTeX and upload it to arxiv.
     
  4. CRGreathouse

    CRGreathouse 3,682
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    Unsurprisingly, the proof is flawed. On the last page, you make the assumption that the probabilities are 1/3, 1/3, and 2/3, respectively. First, this assumption is unjustified. Second, depending on how you define 'probability' in this context, it's probably false. Third, this assumption is not sufficient to prove the conjecture, even were it true.
     
  5. I have the last three pages on facebook now... I originally had duplicates of the first three.
     
  6. CRGreathouse

    CRGreathouse 3,682
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    Ah, I see where you're going.

    This approach is flawed. It is true that, for any bound B, a positive fraction of integers are relatively prime to all integers 2, 3, ..., B. It is even true that, for a given N, a positive fraction of integers are such that both n and N - n are relatively prime to all numbers 2, 3, ..., sqrt(N). But you need more than that: in particular, you need to show that n < N/2, or else the second term will be negative.

    Let me put this another way, considering the simpler problem of finding primes in a range. Asymptotically, you expect at least 1/2 of numbers to be good mod 2, at least 2/3 of numbers to be good mod 3, at least 4/5 to be good mod 5, and at least 6/7 to be good mod 7. 1/2 * 2/3 * 4/5 * 6/7 = 8/35. So you might expect that out of
    200, 201, 202, 203, 204, 205, 206, 207, 208
    about 9 * 8/35 = 2.05... would be prime to 2, 3, 5, and 7. But in fact none are. This is because you can't guarantee that (out of N consecutive integers) N/2 numbers are odd; rather, you show that at least (N-1)/2 are. Likewise, not N/3 but at least (N-3)/3 are prime to 6, and so on. You haven't kept track of these additive portions, and you must!

    In fact, if you properly track them, you'll see that as the number of primes you consider increases, your ranges quickly shrink to nothing. The approach doesn't work. I encourage you to work this out in detail so you can see this for yourself!
     
  7. CRGreathouse

    CRGreathouse 3,682
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    Here's an example. Suppose S is a finite set of primes. Then, asymptotically,
    [tex]k=\prod_{s\in S}1-\frac1s[/tex]
    of numbers are relatively prime to all the primes in S ("candidates"). To get a nonasymptotic result, it's easy to see that every
    [tex]P=\prod_{s\in S}s[/tex]
    consecutive numbers give Pk candidates. Thus, at least
    [tex]\lceil k(N-P-1)\rceil[/tex]
    out of N consecutive numbers are candidates. Now, let S = {2, 3, 5, ..., sqrt(N)} so that the candidates, if less than N, are prime. Then
    [tex]k=\prod_{p\le\sqrt N}1-\frac1p\approx\frac{2}{e^\gamma\log N}[/tex]
    [tex]P=\prod_{p\le\sqrt N}p\approx\exp\left(\sqrt N\right)[/tex]
    and so you have "at least" approximately
    [tex]2\frac{N-\exp\left(\sqrt N\right)}{e^\gamma\log N}[/tex]
    primes below N. But this is clearly negative!

    So there's no obvious way to use this method to prove that there are any primes up to N/2, let alone primes that satisfy additional conditions.

    You may object that the square root in the equation above could be replaced by something like a fourth root if the Riemann hypothesis holds. But (1) the RH hasn't been proven, and (2) the term is still far too large to work. (It looks like the Riemann hypothesis might be able to prove that at least 100-200 primes exist... but not even infinitely many primes, a truly ancient result!)
     
    Last edited: Jan 4, 2010
  8. Thanks for the response. Concerning your example for the integers 201, 202, ... ,208:
    I don't think it is necessarily relevant. My proof only requires that for any three consecutive odd integers, at least one will not be divisible by 3; for any 5 consecutive odd integers, at least one will not be divisible by 3 and 5; for any n consecutive odd integers, at least one will not be divisible by 3, 5, ..., n.

    I think this is a perfectly valid statement.

    Andy Lee
     
  9. CRGreathouse

    CRGreathouse 3,682
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    1. You didn't prove this, so even if it was true your proof would fail.
    2. It's false. Here are 47 consecutive odd numbers, each divisible by one of 3, ..., 47:
    Code (Text):
    6002108856728919 = 3^3 * 11 * 17 * 1188771807631
    6002108856728921 = 31 * 2711 * 71418817681
    6002108856728923 = 29 * 1301 * 159084758587
    6002108856728925 = 3 * 5^2 * 80028118089719
    6002108856728927 = 37 * 181^2 * 4951593611
    6002108856728929 = 7 * 857444122389847
    6002108856728931 = 3 * 2000702952242977
    6002108856728933 = 13 * 461700681286841
    6002108856728935 = 5 * 647 * 1855365952621
    6002108856728937 = 3^2 * 9719 * 50441 * 1360367
    6002108856728939 = 43 * 870097 * 160423409
    6002108856728941 = 11 * 71 * 127 * 9721 * 6224983
    6002108856728943 = 3 * 7 * 1009 * 2003 * 8293 * 17053
    6002108856728945 = 5 * 89 * 13487885071301
    6002108856728947 = 23 * 701 * 372269978089
    6002108856728949 = 3 * 83 * 24104854846301
    6002108856728951 = 19 * 61 * 1861 * 2782749149
    6002108856728953 = 17 * 101 * 5441 * 642472949
    6002108856728955 = 3^2 * 5 * 74717 * 1785138547
    6002108856728957 = 7 * 109 * 7907 * 8761 * 113557
    6002108856728959 = 13 * 461700681286843
    6002108856728961 = 3 * 97 * 199 * 367 * 282417587
    6002108856728963 = 11 * 317 * 3217 * 535057997
    6002108856728965 = 5 * 1117 * 3209 * 6277 * 53353
    6002108856728967 = 3 * 2000702952242989
    6002108856728969 = 41 * 1171 * 125015285179
    6002108856728971 = 7 * 7927 * 36721 * 2945659
    6002108856728973 = 3^4 * 1153 * 64267224061
    6002108856728975 = 5^2 * 467 * 91939 * 5591743
    6002108856728977 = 47 * 139 * 23627 * 38885047
    6002108856728979 = 3 * 67 * 3457 * 8637905147
    6002108856728981 = 29 * 100673 * 2055856793
    6002108856728983 = 31 * 157 * 1233225571549
    6002108856728985 = 3 * 5 * 7^2 * 11 * 13^3 * 919 * 367687
    6002108856728987 = 17 * 727 * 436061 * 1113713
    6002108856728989 = 19^2 * 653 * 25461470633
    6002108856728991 = 3^2 * 666900984080999
    6002108856728993 = 23 * 59 * 19069 * 231950921
    6002108856728995 = 5 * 53 * 1709 * 5813 * 2279899
    6002108856728997 = 3 * 26357 * 75907840507
    6002108856728999 = 7 * 2144509 * 399832373
    6002108856729001 = 37 * 151 * 797 * 3673 * 366983
    6002108856729003 = 3 * 69593 * 28748623457
    6002108856729005 = 5 * 1873 * 73303 * 8743279
    6002108856729007 = 11 * 251 * 2173889480887
    6002108856729009 = 3^2 * 11261 * 17551 * 3374291
    6002108856729011 = 13 * 149939 * 3079256773
    I can go higher if you like.
     
  10. The 47 consectutive numbers would have to appear before 47^2 for this to be a counter-example.

    Andy Lee
     
  11. CRGreathouse

    CRGreathouse 3,682
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    No, they wouldn't; that wasn't a part of your claim.

    Why don't you write up precisely what you mean, prove it, then use that proof in your full proof of GC.
     
  12. You are correct, it was not part of my claim in the earlier post. But it most certainly is evident in the proof.

    Andy Lee
     
  13. CRGreathouse

    CRGreathouse 3,682
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    Your proof doesn't even state the required condition, let alone prove it. You need more specifics and less handwaving.

    (Sorry, this comes off as harsh. But you need to do it for a valid proof!)
     
  14. Hey, CRGreathouse, can you translate his proof into English rather than picture? I have no clue what A1, A2,... B1, B2,... C1, C2,... are, and I believe it would easier to discuss the argument, and possibly any flaws, if it was written like this.

    Cheers,

    Rofler
     
  15. CRGreathouse

    CRGreathouse 3,682
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    If someone translated his whole argument into valid, gapless mathematics then the flaws would be obvious. But it's not easy to do that translation!
     
  16. Wow, that's a pretty hostile comment! It also does not make a lot of sense, when you think about it. After one week and I have one person on this board who states the proof is flawed. Any other comments?
     
  17. Well, could you explain what A1, A2, B1, B2, etc. are? I can't comment on the quality of the proof if I don't understand/can't read/ dont want to turn my head 90 degrees to read the second page.

    Cheers,

    Rofler

    By the way, saying that there is no set of n consecutive numbers UNDER n^2 each containing an odd factor less than n is quite a strong statement, and may even be stronger than the Goldbach conjecture itself. Of course the numbers n!+k each contain k as a factor, but n^2 is significantly stronger than n!. Or it may just be wrong, but I won't comment any further until I am able to understand the proof.

    And CRGreathouse, despite apparently having valid arguments, I can't support either one of you until I see the original proof. You don't need to be so annoyed by this, at least he tried something, and apparently asides from omitting his endpoints, the proof from what you described would have been more or less correct.

    Of course, the last sentence was more towards CRGreathouse's attitudes, I am always suspicious at purported "easy" proofs.
     
    Last edited: Jan 12, 2010
  18. CRGreathouse

    CRGreathouse 3,682
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    I spent the time to look over your proof and found numerous mistakes and gaps. When I finally understood your method and saw that the method cannot be used to prove the conjecture, I stopped looking further.

    Most real mathematicians would refuse to even look at your proof -- there are lots of crackpots out there, and you would need to do something to demonstrate that you're not to convince them to spend their valuable time looking over the proof. Even going as far as I did was a gesture of good faith.

    Perhaps if you clean up the proof you can get rofler to look it over. He's clearly not a mathematician, but he may have useful input to give you.
     
  19. CRGreathouse

    CRGreathouse 3,682
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    If you're not willing to turn your head 90 degrees to read the purported proof, I think you'll understand if I'm not willing to attempt to translate it for you. :)

    I also don't see an easy way to show this. I'm not even sure if it's true, though I would suspect so.

    I'm not in the business of convincing you. I'd rather let you make your own decision, frankly. I looked the purported proof over and came to my own conclusion, and I chose to share that with this forum in case others were curious.

    You may decide not to read it, in which case there's really not a whole lot to say.

    You may read it and not be able to follow it; this really wouldn't be surprising given how it's written.

    You may read it and conclude that it's wrong. In that case we agree and, again, there's not much to say.

    *If* you chance to read it and, flabbergastingly decide that it's correct, perhaps I'll be interested enough to go back to the proof and point out the flaws in more detail. But I think I was explicit enough in post #4. Was that post not understandable, or do you not see why it applies?
     
  20. CRGreathouse

    CRGreathouse 3,682
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    I did want to comment on this, at least. It is ABSOLUTELY not true. This may seem like a small point in the proof, but it actually causes the proof to fail. In particular, the error term becomes larger than the main term, so instead of proving that there are at least f * N Goldbach partitions, where f tends to 0 but f*N increases without bound, it shows that at least f * N - O(stuff) Goldbach partitions, where O(stuff) actually becomes larger than f * N (and indeed, larger even than N).

    It is a fatal flaw, not an accounting error.
     
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