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

[thinhnghiem]

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

1.       Prove that the conjecture is correct with n=2.

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

1.       With n = 2 => 2n= 4 = 2+2.

The conjecture has been ascertained as correct in this case.

2.       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

[uncool]

that's not how it works

that's not how any of this works

Edited June 26, 2021 by uncool

[thinhnghiem]

On 6/25/2021 at 10:18 PM, uncool said:

that's not how it works

that's not how any of this works

Could you explain more?

[uncool]

Let's start with this:

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

Your definition of B was: "B is also the set of the odd and prime numbers y which qualify the expression y < 2n". How does the conjecture relate to the nonemptiness of B (which, if anything is obviously true - there is always an odd number below 2*n)?

 

It sounds like you're trying to define B based on summing, but you have not done so.