The proving of the GoldBach conjecture by using the Zermelo Fraenkel axioms

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thomasnghiem

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Goldbach conjecture can be proved by using the Zermelo Fraenkel axioms in set theory.
We apply the inductive method to prove that this conjecture is correct with all even numbers 2n with n ∈ IN. The process includes two steps
  • Prove that the conjecture is correct with n=2.
  • Assume that the conjecture is correct with a certain value n, prove that the conjecture is also correct with (n+1).
We proceed with our process in sequence as below
  • With n = 2 => 2n= 4 = 2+2.
  • The conjecture has been ascertained as correct in this case.
  • Now we assume that the conjecture is correct with n, the next step is we have to prove that it is also correct with (n+1).
  • Here is our solution

  • Let’s define A as the set of the odd and prime numbers x which qualify the expression x < 2n.
  • B is also the set of the odd and prime numbers y which qualify the expression y < 2n
  • S(x,y) is the statement to express the equation (x+y) = 2n.

  • By the axiom schema of Replacement, there is the existence of the set B to ensure that S(x,y) is correct with at least one element x ∈ A.

  • Since the conjecture is correct with n, we have B ≠ Ø.

  • Let’s define A+ as the set of the odd and prime numbers x which qualify the expression x < 2(n+1).
  • B+ is the set of the odd and prime numbers y with y < 2(n+1) and S(x,y) is correct with at least one element x ∈ A+.

  • By the axiom of Infinity, A+ and B+ are the next items of A and B relevantly in the infinite sets.
  • Since both A and B are ≠ Ø, then we can utilize the axiom of infinity to declare the existing of A+ and B+. That means A+ ≠ Ø and B+ ≠ Ø.

  • The conjecture has been finally proven as correct with (n+1). By the inductive method, it is determined as correct with every n ∈ IN.

  • Thomas Nghiem (Ontario – Canada)
Reference:
https://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Lian.pdf
 
I am not sure I follow your proof.

Your set A includes all odd primes < 2n. So does set B. In other words, they are the same set. The whole distinction between A and B seems unnecessary.

Moreover, your case of 2 + 2 = 4 = 2n is irrelevant because 2 is not an odd prime. Now, I agree that it is not hard to show that for n = 3, that 3 + 3 = 6 = 2n. So I grant your base case, but this irrelevancy does not give me confidence in your logic.

So we assume that there exists integer k such that Ak contains all the odd primes < 2k
and that there exist xk and yk in Ak such that xk + yk = 2k.

When we move on to k + 1, we are dealing with a completely different set Ak+1. Now I admit that xk and yk are members of Ak+1, but they self evidently do not sum to 2(k + 1).

I also admit that (xk + 2) + yk = xk + (yk + 2) = 2k + 2 = 2(k + 1),
but there is no assurance that either xk + 2 or yk + 2 is prime.
 
JeffM said:
I am not sure I follow your proof.

Your set A includes all odd primes < 2n. So does set B. In other words, they are the same set. The whole distinction between A and B seems unnecessary.

Morover, your case of 2 + 2 = 4 = 2n is irrelevant because 2 is not an odd prime. Now, I agree that it is not hard to show that for n = 3, that 3 + 3 = 6 = 2n. So I grant your base case, but this irrelevancy does not give me confidence in your logic.

So we assume that there exists integer k such that Ak contains all the odd primes < 2k
and that there exist xk and yk in Ak such that xk + yk = 2k.

When we move on to k + 1, we are dealing with a completely different set Ak+1. Now I admit that xk and yk are members of Ak+1, but they self evidently do not sum to 2(k + 1).

I also admit that (xk + 2) + yk = xk + (yk + 2) = 2k + 2 = 2(k + 1),
but there is no assurance that either xk + 2 or yk + 2 is prime.
Click to expand...
Hi JeffM,

Thank you for your feedback. It is very valuable for me to improve my research.

your case of 2 + 2 = 4 = 2n is irrelevant because 2 is not an odd prime
[Thomas] The origin statement of GoldBach conjecture said that "every even whole number greater than 2 is the sum of two prime numbers". It does not mention the parity of these prime. 2 is the only even prime number. So 4 is the special case here.

So we assume that there exists integer k such that Ak contains all the odd primes < 2k
and that there exist xk and yk in Ak such that xk + yk = 2k.

When we move on to k + 1, we are dealing with a completely different set Ak+1. Now I admit that xk and yk are members of Ak+1, but they self evidently do not sum to 2(k + 1).

I also admit that (xk + 2) + yk = xk + (yk + 2) = 2k + 2 = 2(k + 1),
but there is no assurance that either xk + 2 or k + 2 is prime.
[Thomas] Yes you are right. If we use number theory like this we are never able to prove this conjecture successfully and we will get stuck.

That's the reason why I refused to dip into the format of the even numbers. Instead, I explored indirectly with algebra as shown above.
 
My point about 2 + 2 not being an apposite case for the base of an argument from induction still stands. The number two is not relevant to the conjecture with respect to any even number greater than 4. Because it is quite obvious that if an even number greater than four is expressible as the sum of two primes, the two primes must both be odd. Now I agree that you can find a valid base case, namely six. Therefore, your use of a fallacious base case is not an insuperable objection, but it does not give me confidence in your work.

Your calling the same set by two different names does not give me confidence. I wonder whether you are using the axiom of replacement to say the existence of A entails the existence of A.

Your failure to distinguish the relevant sets when you move from k to k + 1 does not give me confidence.

Moreover, your failure to specify what property you are using to justify the axiom of replacement between the relevant set in the case of k and the relevant set in case k + 1 does not give me confidence. It is obvious that the property

[MATH]x_k + y_k = 2k[/MATH] does not entail that [MATH]x_k + y_k = 2(k + 1).[/MATH]
And finally I wonder whether (even if all your set theory is correct) whether what you have proved is that in every set of odd primes, the sum of any two of them is an even number, which is not the Goldbach conjecture.

You have to prove that in set Ak+1
there exists prime xk+1 such that 2xk+1 = 2(k + 1) or
there exist primes xk+1 and yk+1 such that xk+1 + yk+1 = 2(k + 1).
 
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