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I have seen the following in my observation. I am so sorry, if I am wrong. I do not know how far I am right. But, I feel that, it might be correct. If correct, Please let me know that way to proceed further to prepare a proof.

every number can be expressible either, prime + prime or prime + non prime or non prime + non prime or non prime + prime as shown below. To support GC, I would like to take an example to show these combination. For example: number 10 (>2) and consider that non prime as NP and prime is P.

10 = 1 + 9 or 2 + 8 or 3 + 7 or 4 + 6 or 5 + 5

I am sure there is prime + prime in every even integer.

We can notice that every integer > 2 can be expressible any one of the above form. If P + P is not taken place in the above example, then there is no question of discussion of GC. Now, find the number of primes in fraction. If the fraction is > ¼, then we need not to discuss further proof of GC. No matter what fraction we have taken, we find the same. If the fraction of primes is 1/8, then the probability of finding N + P are 6/16, but we require 8/16 to avoid any P + P sums. If the fraction of primes is 1/10, then the probability of finding N + P are 8/25, but we require 10/25 to avoid any P + P sums. The probability is always strongly in favor of a P + P. So, we can conclude the GC.

Here: n is non prime and p is prime

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3 Answers 3

Your idea is interesting, but sadly enough we can prove it is no good for a large enough integer $N$. It is very well know that the prime-counting function, $\pi(x)$, which counts how many primes there are less than $x$, is asymptotic to $x/\log(x)$, i.e. $$ \lim_{n \to \infty} \frac{\pi(n)}{n/ \log(n)} = 1. $$ This means that $$ \lim_{n\to \infty} \frac{ \pi(n)}{n} = \lim_{n \to \infty} \frac 1{\log(n)} \frac{\pi(n)}{n/\log(n)} = 0. $$ Therefore, the "fraction" you speak of can be put arbitrarily small, since this fraction goes to $0$ as $n$ goes to infinity. This technique of using a pigeonhole principle argument (using the fact that there are enough primes to counter-balance the non-primes) is therefore quite useless.

In some way, this means that the Goldbach conjecture would hold not because there is enough primes in probability, but because the primes are positioned "at the right spots", in the sense that if we take $N$ large enough and consider a set of $\pi(N)$ numbers (a subset of the first $N$ integers) as special numbers, the probability that two special numbers sum up to $N$ is zero, because $\pi(N)/N \to 0$ as $N$ goes to infinity.

Hope that helps,

(P.S. I don't blame you for thinking about it, it is a cool thing to do, I'm just telling you this is not a good direction. =D )

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3  
As Henry's answer indicates, quite the opposite is true: Goldbach's conjecture could easily be true "by accident" for no special reason, simply because there are enough primes. This has been discussed on MO as well: mathoverflow.net/questions/73651/… –  Alex B. Dec 26 '11 at 16:22
    
Thank you so much –  Baba BITS Dec 27 '11 at 6:31

The problem with Goldbach's conjecture is that observationally it is extremely likely to be true, but that does not prove it. Your observation does not provide a proof as there are not enough primes to ensure that there must be a pair, even though there are enough to suggest it is probable there is a pair.

You might imagine that since the number of primes less than $n$ is about $\frac{n}{\log_e n}$, the number of pairs of primes adding up to an even $n$ might be about $\frac{n}{2(\log_e n)^2}$ and all the evidence is that this gives the correct order of magnitude, though some $n$ have higher numbers than $2$ and other lower. The following chart shows for even numbers up to $20000$ the number of prime pairs which add up to $n$ multiplied by $\frac{(\log_e n)^2}{n}$

enter image description here

but while this may suggest Goldbach's conjecture is going to be true, it too is not a proof.

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How did you come up with the $n/(2 \log(n)^2)$? –  Patrick Da Silva Dec 26 '11 at 13:22
    
@Partrick: In a handwaving sense, there are about $n/2$ pairs of positive integers adding up to $n$, and a proportion of about $\frac{1}{\log_e n}$ of each of these individual numbers are prime, so about $\frac{1}{(\log_e n)^2}$ of the $n/2$ pairs might be both be primes. –  Henry Dec 26 '11 at 14:58
    
Oh, I see. I just didn't see your way of handwaving. =P –  Patrick Da Silva Dec 26 '11 at 23:20
    
Thank you so much for comments and positive responses. –  Baba BITS Dec 27 '11 at 6:30

Moreover, I think it would be sufficient to establish rigorously that the lim inf of the function plotted by Henry is twice the twin prime constant to prove the extended Goldbach's conjecture of Hardy and Littlewood.

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