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
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 $k^{VI}$.