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.


closed as unclear what you're asking by Daniel, user147263, graydad, Matt Samuel, Strants Aug 30 '15 at 6:06

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    $\begingroup$ possible duplicate of Goldbach conjecture $\endgroup$ – Morgan Rodgers Aug 29 '15 at 12:29
  • $\begingroup$ i could not understand the question you mentioned. $\endgroup$ – Bibekpandey Aug 29 '15 at 13:13
  • $\begingroup$ There are like 30 "check my proof of the Goldbach conjecture" questions..... $\endgroup$ – Morgan Rodgers Aug 29 '15 at 14:39
  • $\begingroup$ @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. $\endgroup$ – Erick Wong Aug 30 '15 at 1:53
  • $\begingroup$ Use anything but $e$, because $e$ is already universally defined. $\endgroup$ – user253055 Sep 2 '15 at 0:08

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.

  • $\begingroup$ 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? $\endgroup$ – Bibekpandey Aug 29 '15 at 13:08
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    $\begingroup$ No. If you toss a coin three times, the number of heads will average to 3/2. But sometimes you will get no heads. $\endgroup$ – Empy2 Aug 29 '15 at 13:11
  • $\begingroup$ yeah, that's true. $\endgroup$ – Bibekpandey Aug 29 '15 at 13:15
  • $\begingroup$ and by the way, has it been proven that GC can't be proven or disproven?? $\endgroup$ – Bibekpandey Aug 29 '15 at 13:16
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    $\begingroup$ @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. $\endgroup$ – Erick Wong Aug 30 '15 at 1:49

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