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So the Goldbach conjecture says 'Every even integer greater than 2 can be written as sum of two primes'. Here is what I have roughly done to verify it, using probability. I don't say it is correct but I just want to show it.

Let a be an even integer greater than 2. The prime counting function gives number of primes n as n ~ a/ln a. Now, for every prime less than a, we generate odd numbers(except for 2)

a-p(i) where, p(i) is the ith prime below a. Thus, we have approximately n odd numbers below a.

We have n/a = 1/lna which gives the probability of prime number below a.

If we consider only the odd numbers below a, the probability becomes 2/lna, which means 2 out of lna odd numbers below a are prime numbers.

But we have generated approximately(which just excludes case for 2 and is negligible for large numbers) a/lna odd numbers.

So, from unitary method, 2/lna out of 1 odd numbers below a are prime, which yields for our case, $ 2 a/(lna)^2$ prime numbers out of a/lna odd numbers generated.

And we can see that, 2 a/(lna)^2 is obviously greater than 1 which even grows when a gets larger.

This shows that, among our generated odd numbers(a-p(i)) there is at least one prime number q giving,

a-p(i) = q

or, p(i) + q = a. where p(i) and q are primes, verifying Goldbach Conjecture.

Edit: I just wanted to know what is wrong with this approach(I knew there was). And I got my answer too. Thank you.

Bibekpandey
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  • i could not understand the question you mentioned. – Bibekpandey Aug 29 '15 at 13:13
  • @BeWakePandey It isn't clear what your question is. You give a crude heuristic which could be refined to a sharp heuristic, and though neither is proven. What type of answer are you expecting that you'd accept?? As it stands, there isn't any question in your question. – Erick Wong Aug 30 '15 at 01:53
  • Use anything but $e$, because $e$ is already universally defined. –  Sep 02 '15 at 00:08
  • @AaronThompson, I have corrected that. Thank you. – Bibekpandey Sep 02 '15 at 01:33

1 Answers1

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On average, you would expect around $2e/(\log e)^2$ primes. But the $e/\log e$ numbers are a small proportion of the numbers below $e$, and it might happen, for one particular $e$, that none of them are prime.

Empy2
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  • but 2e/(log e)^2 is greater than 1 always, should not that mean by plain probability that there is at least 1 prime favouring GC? – Bibekpandey Aug 29 '15 at 13:08
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    No. If you toss a coin three times, the number of heads will average to 3/2. But sometimes you will get no heads. – Empy2 Aug 29 '15 at 13:11
  • yeah, that's true. – Bibekpandey Aug 29 '15 at 13:15
  • and by the way, has it been proven that GC can't be proven or disproven?? – Bibekpandey Aug 29 '15 at 13:16
  • I don't know. Here is an earlier question. http://math.stackexchange.com/questions/864149/could-it-be-that-goldbach-conjecture-is-undecidable – Empy2 Aug 29 '15 at 13:38
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    @BeWakePandey If it had been proven that GC can't be disproven, that would amount to a proof that it's true, since any counterexample would become a trivial disproof. – Erick Wong Aug 30 '15 at 01:49