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I conducted a study on the Goldbach binary conjecture and now I will show my proposal to demonstrate this conjecture, where P{2} represents the odd prime numbers set. I would like to know if you think this demonstration is correct, thank you very much.

So I will formulate the conjecture in mathematical terms not considering the number 4, since this natural number is the only writable as the sum of 2 even prime numbers.

Strong Goldbach conjecture statement:

               ∀ k ∈ N*\{1,2} ∃ p,q ∈ P\{2} | 2k = p + q 

Proof

               ∀ k’∈ N*|k’< k → k’ = k - k°      k° < k, k° ∈ N*

               2k’ = 2(k - k°) = 2k - 2k°

An even number can be written as a sum of two odd numbers hence:

               2k’ = (2k” + 1) + (2k’’’ + 1)         k”, k’’’ ∈ N* 

We know that every odd number can be written as a difference between a prime and an even number, therefore:

               2k” + 1 = p – 2k^IV

               2k”’ + 1 = q – 2k^V                   k^IV, k^V ∈ N,  p, q ∈ P\{2}

               2k’ = (p – 2k^IV) + (q – 2k^V)   

               2k’ = p + q – 2(k^IV + 2k^V)   

               2k’ = p + q – 2k^VI                   k^VI ∈ N*          

              *2k - 2k° = p + q – 2k^VI

               p, q ∈ 2Z+1 → 2n = p + q              n ∈ N*\{1,2}    

               2k - 2k° = 2n – 2k^VI 

             ** k - k° = n – k^VI 

Continuous approach. We can express a real number as a function of every other real number, hence (given that N ⊂ R):

               k - k° = n – k^VI       n = f(k) 

               k - k° = f(k) – k^VI    f(k) = g*k,  g ∈ R|g*k ∈ N*\{1,2}        

               k - k° = g*k – k^VI                  g ∈ G, G ⊂ R

               2n = 2f(k) = 2g*k → 2g*k = p + q   

               2k = p*g^-1 + q*g^-1 = (p + q)*g^-1  

               lim(g → 1)⁡2k = p + q  

Otherwise we can say that g = 1 guarantees that (p + q) is an integer greater than 4 while the fact that both p and q are odd numbers guarantees that the parity of (p + q) is in agreement with 2k parity.

Discrete approach. We can also say:

  *2k = p + q – 2(k^VI - k°) where if k° = k^VI ⇒ 2k = p + q, k ∈ N*\{1,2}, p, q ∈ P\{2}  

or

       **n = k ± k°’,  k°’∈ N hence 2n = 2(k ± k°’) = p + q where k°’ = 0 ⇒ 2k = p + q
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closed as off-topic by Andrés E. Caicedo, egreg, ahulpke, caverac, Mark Bennet Dec 12 '17 at 16:56

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  • "This question is not about mathematics, within the scope defined in the help center." – Andrés E. Caicedo, egreg, Mark Bennet
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  • $\begingroup$ Why is this proof written with so many symbols? Proofs are about ideas, which are best expressed in English (or some other natural language). Can you distill the basic idea of your proposed proof into simple sentences? $\endgroup$ – G Tony Jacobs Dec 12 '17 at 13:36
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    $\begingroup$ This is very hard to follow. You appear to be trying to give an explicit construction. That is, given $2k$ you appear to construct primes $p,q$ with $p+q=2k$. But do you? If $2k=24=15+9$ we write $15=17-2,9=11-2$. Then $p=17, q=11$. Yes? But $17+11\neq 24$. So what are you saying? If I have misunderstood your construction (high probability) then please go through an explicit numerical example in your post. $\endgroup$ – lulu Dec 12 '17 at 13:45
  • $\begingroup$ In your example 2k = 24, k = 12; so we have to find p and q such that 2k =24 = p + q We take, for example, K’ = k -2 =10 and so 2k’ = 10*2 = 20 20 = 24 – 4, 20 = 11 + 9 24 – 4 = 11 + 9 = (13 – 2) + (11 -2) = (13 + 11) – (2 + 2) --> 24 – 4 = (13 + 11) – 4 from which it derives that 24 = 13 + 11 But I think the heart of the demonstration is when we arrive to say that k - k ° = f (k) - k ^ VI from which derives this important statement : 2n = 2f(k) = 2g * k → 2g * k = p + q $\endgroup$ – Daniele Bertaggia Dec 12 '17 at 15:31
  • $\begingroup$ @DanieleBertaggia How did it so conveniently happen that, when you chose $k'=k-2$, and then chose $20=11+9$ - both $11+2$ and $9+2$ turned out to be prime? (I know, some other pairs would do in your proof, but isn't finding such pairs actually equivalent to Goldbach's conjecture?) $\endgroup$ – user491874 Dec 12 '17 at 16:11
  • $\begingroup$ As I said above, I humbly think that we must arm ourselves with patience and follow the demonstration from the beginning to the end. My approach is of the arithmetic-algebraic type and examples that confirm the passage in question are always possible. But I repeat, in my opinion, the proof must be analyzed from the beginning to the end from a logical-set point of view. I would like to understand in which single algebraic-arithmetic step I'm doing wrong ... $\endgroup$ – Daniele Bertaggia Dec 12 '17 at 16:24
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I did not have time to look at your continuous proof, but the discrete one has a flaw. Your $p$ and $q$ depend on the choice of $k^0$ but then later you forget about that and just set $k^0$ to be equal to $k^{VI}$.

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  • $\begingroup$ And that is quite an elementary flaw: making the same flaw, you can conclude that every number is $0$: start with $a\in\mathbb R$, split it as a sum $a=b+c$ and then let's choose $c$ to be $-b$ so $a=b+(-b)=0$. $\endgroup$ – user491874 Dec 12 '17 at 13:52
  • $\begingroup$ In your example, it seems to me that there is an initial defect, you want to prove a untruth: to show that all the numbers are 0 (which finds countless counterexample that deny it, rather exist only a circumstance that confirm it: c=-b ). On the contrary, I try to prove a conjecture that at the moment does not yet present a counterexample that denies it. For do this I rely on the Euclid's theorem on the infinity of prime numbers, which allows me to always identify a prime number greater than an odd number... $\endgroup$ – Daniele Bertaggia Dec 12 '17 at 14:44
  • $\begingroup$ @DanieleBertaggia, I think you should reply to the comments directly below your question, in particluar, to the one by lulu. $\endgroup$ – G Tony Jacobs Dec 12 '17 at 15:01
  • $\begingroup$ @DanieleBertaggia My point is that, with flawed reasoning as in your proof, you can purportedly prove untruths. Similarly, Goldbach's conjecture may still be false and you may still think that you've proven it. Please have a look at my answer and try to understand the reason your 'discrete proof' is flawed. $\endgroup$ – user491874 Dec 12 '17 at 15:49

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