Proof of Goldbach Conjecture

[Guest]

Keywords: Fundamental Theorem of Arithmetic (FTA) and Odd Prime-counting function π(*)

Note: We assume one is unit prime.

PROOF OF GOLDBACH CONJECTURE:

Goldbach Conjecture (GC) states every positive even integer is the sum of two prime numbers. (We count one as unit prime in the sense of additive number theory outside of the FTA.)

Proof of Goldbach Conjecture:
Suppose there exists a positive even integer, e > 4, that is not the sum of two odd prime numbers or 1. e ≠ p + q over S = {all odd prime numbers less than e} and where k = card(S)
= π(e).

Therefore,

e ≠ p + q over S, (p,q є S) , implies the following system of equations over S,

1 = e - [tex]n_{1 }[/tex] * [tex]q_{1 }[/tex];
3 = e - [tex]n_{2 }[/tex] * [tex]q_{2 }[/tex];
5 = e - [tex]n_{3 }[/tex] * [tex]q_{3}[/tex];
...
[tex]p_{k }[/tex] = e - [tex]n_{k }[/tex] * [tex]q_{k}[/tex],

according to the Fundamental Theorem of Arithmetic where [s]1 < [tex]q_{j}[/tex] ≤ [tex]\sqrt{n_{j} *q_{j}}[/tex] ≤ [tex]n_{j}[/tex]
for 1 ≤ j ≤ k where [tex]p_{j}[/tex], [tex]q_{j}[/tex] є Sand [tex]n_{j}[/tex] is a positive composite integer.

Note: If [tex]p_{j}[/tex]= 1, then [tex]n_{j}[/tex] є S, or [tex]n_{j}[/tex] is an odd prime less than e.

Therefore,
2. Probability(e ≠ p + q over S)= Prob(e ≠ p + q over S)
= [tex]\prod_{j=1}^{k\rightarrow \infty}[/tex][ Prob([tex]q_{j}[/tex] ≠ 1 | e - [tex]p_{j}[/tex] = [tex]n_{j}[/tex] * [tex]q_{j}[/tex] over S) * Prob(e - [tex]p_{j}[/tex] = [tex]n_{j}[/tex] * [tex]q_{j}[/tex]over S)]

= [tex]\prod_{j=1}^{k\rightarrow \infty}[/tex][π([tex]\sqrt{n_{j} *q_{j}}[/tex] - 1)/ π( [tex]\sqrt{n_{j} *q_{j}}[/tex] )] → 0.

Note: Prob(e - [tex]p_{j}[/tex]= [tex]n_{j }[/tex] * [tex]q_{j }[/tex] over S) = 1 for 1 ≤ j ≤ k.

This is an increasingly fast convergence for this almost everywhere monotonic
non-increasing expression. This implies that the expected value of
e ≠ p + q over S is practically zero,
or E[e ≠ p + q over S] = e * Probability(e ≠ p + q over S) ≈ 0 for all e ≥ 100.

Note: Probability (e - [tex]p_{j}[/tex]= [tex]n_{j}[/tex] * [tex]q_{j}[/tex]over S) = 1 for 1 ≤ j ≤ k.

In addition, empirical evidence has confirmed the validity of the conjecture for all positive even integers up to at least an order of [tex]10^{18}[/tex]. Therefore, we conclude the conjecture is true.

REFERENCES:
1. EMPIRICAL VERIFICATION OF THE EVEN GOLDBACH CONJECTURE, AND COMPUTATION OF
PRIME GAPS, UP TO 4 · 1018 by TOMAS OLIVEIRA E SILVA, SIEGFRIED HERZOG, AND SILVIO PARDI
http://www.ams.org/editflow/editorial/uploads/mcom/accepted/120521-Silva/120521-Silva-v2.pdf

2. The Exceptional Set In Goldbach Problem by Prof. HLM
http://matwbn.icm.edu.pl/ksiazki/aa/aa27/aa27126.pdf

*******
Author: David Cole
(aka primework123)

[Guest]

Correction! ... Note: If [tex]q_{j}[/tex]= 1, then [tex]n_{j}[/tex] є S, or [tex]n_{j}[/tex] is an odd prime less than e...

[Guest]

A very simple proof of Golbach conjecture is available in the link below:
Manuscript Title : A Rigorous Proof for the Strong Goldbach Conjecture
Digital Library URI : IJCA - A Rigorous Proof for the Strong Goldbach Conjecture
ISBN : 973-93-80893-14-6

[leesajohnson]

Thanks for sharing the proof of Goldbach Conjecture.

[Guest]

FYI: Some Calculations based on the above proof of Goldbach Conjecture at the following link:

https://www.researchgate.net/post/Is_the_Goldbach_Conjecture_true

David Cole
aka primework123
https://www.linkedin.com/in/davidcole11

P.S. Please pray for me since my time on great Mother Earth is coming to an end. Best wishes to all! And may Lord GOD help us through these very difficult and challenging times.

[Guest]

There is another proof that's been published recently which I would like to share with you and to have your feedback and comments on. Thanks.

[Guest]

The proof is published in this paper : https://www.researchgate.net/publicatio ... Conjecture

[Guest]

Keywords: Fundamental Theorem of Arithmetic (FTA) and Odd Prime-counting function π(*)

Note: We assume one is unit prime.

PROOF OF GOLDBACH CONJECTURE:

Goldbach Conjecture (GC) states every positive even integer is the sum of two prime numbers. (We count one as unit prime in the sense of additive number theory outside of the FTA.)

Proof of Goldbach Conjecture:
Suppose there exists a positive even integer, e > 4, that is not the sum of two odd prime numbers or 1. e ≠ p + q over S = {all odd prime numbers less than e} and where k = card(S)
= π(e).

Therefore,

e ≠ p + q over S, (p,q є S) , implies the following system of equations over S,

1 = e - [tex]n_{1 }[/tex] * [tex]q_{1 }[/tex];
3 = e - [tex]n_{2 }[/tex] * [tex]q_{2 }[/tex];
5 = e - [tex]n_{3 }[/tex] * [tex]q_{3}[/tex];
...
[tex]p_{k }[/tex] = e - [tex]n_{k }[/tex] * [tex]q_{k}[/tex],

according to the Fundamental Theorem of Arithmetic where [s]1 < [tex]q_{j}[/tex] ≤ [tex]\sqrt{n_{j} *q_{j}}[/tex] ≤ [tex]n_{j}[/tex]
for 1 ≤ j ≤ k where [tex]p_{j}[/tex], [tex]q_{j}[/tex] є Sand [tex]n_{j}[/tex] is a positive composite integer.

Note: If [tex]p_{j}[/tex]= 1, then [tex]n_{j}[/tex] є S, or [tex]n_{j}[/tex] is an odd prime less than e.

Therefore,
2. Probability(e ≠ p + q over S)= Prob(e ≠ p + q over S)
= [tex]\prod_{j=1}^{k\rightarrow \infty}[/tex][ Prob([tex]q_{j}[/tex] ≠ 1 | e - [tex]p_{j}[/tex] = [tex]n_{j}[/tex] * [tex]q_{j}[/tex] over S) * Prob(e - [tex]p_{j}[/tex] = [tex]n_{j}[/tex] * [tex]q_{j}[/tex]over S)]

= [tex]\prod_{j=1}^{k\rightarrow \infty}[/tex]π([tex]\sqrt{n_{j} *q_{j}}[/tex] )/ (π( [tex]\sqrt{n_{j} *q_{j}}[/tex] ) + 1) → 0. ----- Corrected!

Note: π() is the odd prime-counting, so two is not counted. But one counts, and it is counted.

Note: Prob(e - [tex]p_{j}[/tex]= [tex]n_{j }[/tex] * [tex]q_{j }[/tex] over S) = 1 for 1 ≤ j ≤ k.

This is an increasingly fast convergence for this almost everywhere monotonic
non-increasing expression. This implies that the expected value of
e ≠ p + q over S is practically zero,
or E[e ≠ p + q over S] = e * Probability(e ≠ p + q over S) ≈ 0 for all e ≥ 100.

Note: Probability (e - [tex]p_{j}[/tex]= [tex]n_{j}[/tex] * [tex]q_{j}[/tex]over S) = 1 for 1 ≤ j ≤ k.

In addition, empirical evidence has confirmed the validity of the conjecture for all positive even integers up to at least an order of [tex]10^{18}[/tex]. Therefore, we conclude the conjecture is true.

REFERENCES:
1. EMPIRICAL VERIFICATION OF THE EVEN GOLDBACH CONJECTURE, AND COMPUTATION OF
PRIME GAPS, UP TO 4 · 1018 by TOMAS OLIVEIRA E SILVA, SIEGFRIED HERZOG, AND SILVIO PARDI
http://www.ams.org/editflow/editorial/uploads/mcom/accepted/120521-Silva/120521-Silva-v2.pdf

2. The Exceptional Set In Goldbach Problem by Prof. HLM
http://matwbn.icm.edu.pl/ksiazki/aa/aa27/aa27126.pdf

*******
Author: David Cole
(aka primework123)

[Guest]

Keywords: Fundamental Theorem of Arithmetic (FTA) and Odd Prime-counting function π(*)

Note: We assume one is unit prime.

PROOF OF GOLDBACH CONJECTURE:

Goldbach Conjecture (GC) states every positive even integer is the sum of two prime numbers. (We count one as unit prime in the sense of additive number theory outside of the FTA.)

Proof of Goldbach Conjecture:
Suppose there exists a positive even integer, e > 4, that is not the sum of two odd prime numbers or 1. e ≠ p + q over S = {all odd prime numbers less than e} and where k = card(S)
= π(e).

Therefore,

e ≠ p + q over S, (p,q є S) , implies the following system of equations over S,

1 = e - [tex]n_{1 }[/tex] * [tex]q_{1 }[/tex];
3 = e - [tex]n_{2 }[/tex] * [tex]q_{2 }[/tex];
5 = e - [tex]n_{3 }[/tex] * [tex]q_{3}[/tex];
...
[tex]p_{k }[/tex] = e - [tex]n_{k }[/tex] * [tex]q_{k}[/tex],

according to the Fundamental Theorem of Arithmetic where [s]1 < [tex]q_{j}[/tex] ≤ [tex]\sqrt{n_{j} *q_{j}}[/tex] ≤ [tex]n_{j}[/tex]
for 1 ≤ j ≤ k where [tex]p_{j}[/tex], [tex]q_{j}[/tex] є Sand [tex]n_{j}[/tex] is a positive composite integer.

Note: If [tex]p_{j}[/tex]= 1, then [tex]n_{j}[/tex] є S, or [tex]n_{j}[/tex] is an odd prime less than e.

Therefore,
2. Probability(e ≠ p + q over S)= Prob(e ≠ p + q over S)
= [tex]\prod_{j=1}^{k\rightarrow \infty}[/tex][ Prob([tex]q_{j}[/tex] ≠ 1 | e - [tex]p_{j}[/tex] = [tex]n_{j}[/tex] * [tex]q_{j}[/tex] over S) * Prob(e - [tex]p_{j}[/tex] = [tex]n_{j}[/tex] * [tex]q_{j}[/tex]over S)]

= [tex]\prod_{j=1}^{k\rightarrow \infty}[/tex]π([tex]\sqrt{n_{j} *q_{j}}[/tex] )/ (π( [tex]\sqrt{n_{j} *q_{j}}[/tex] ) + 1) → 0. ----- Corrected!

Note: π() is the odd prime-counting function, so two is not counted. But one counts, and it is counted.

Note: Prob(e - [tex]p_{j}[/tex]= [tex]n_{j }[/tex] * [tex]q_{j }[/tex] over S) = 1 for 1 ≤ j ≤ k.

This is an increasingly fast convergence for this almost everywhere monotonic
non-increasing expression. This implies that the expected value of
e ≠ p + q over S is practically zero,
or E[e ≠ p + q over S] = e * Probability(e ≠ p + q over S) ≈ 0 for all e ≥ 100.

Note: Probability (e - [tex]p_{j}[/tex]= [tex]n_{j}[/tex] * [tex]q_{j}[/tex]over S) = 1 for 1 ≤ j ≤ k.

In addition, empirical evidence has confirmed the validity of the conjecture for all positive even integers up to at least an order of [tex]10^{18}[/tex]. Therefore, we conclude the conjecture is true.

REFERENCES:
1. EMPIRICAL VERIFICATION OF THE EVEN GOLDBACH CONJECTURE, AND COMPUTATION OF
PRIME GAPS, UP TO 4 · 1018 by TOMAS OLIVEIRA E SILVA, SIEGFRIED HERZOG, AND SILVIO PARDI
http://www.ams.org/editflow/editorial/uploads/mcom/accepted/120521-Silva/120521-Silva-v2.pdf

2. The Exceptional Set In Goldbach Problem by Prof. HLM
http://matwbn.icm.edu.pl/ksiazki/aa/aa27/aa27126.pdf

*******
Author: David Cole
(aka primework123)

[Guest]

Relevant Reference Link:

'Proof of Goldbach Conjecture (GC)',

https://www.researchgate.net/publication/310845846_Proof_of_Goldbach_Conjecture.

Remarks:. Our proof of GC depends on the distribution of odd primes along the natural number line or the prime-counting function. And we assume RH. And therefore, the error associated the prime-counting function is known, and it does not violate our proof.

Dave.

[Guest]

Relevant Reference Link:

'Proof of Goldbach Conjecture (GC)',

https://www.researchgate.net/publication/310845846_Proof_of_Goldbach_Conjecture.

Remarks: Our proof of GC depends on the distribution of odd primes along the natural number line or the prime-counting function. And we assume RH. And therefore, the error associated with the prime-counting function is known, and it does not violate our proof.

Dave.

[Guest]

The proof, by induction, states: Let P(n) be the statement that "every even number greater than two is the sum of two prime numbers".

Base Case: P(2) is true, since 2 is the sum of the two prime numbers 1 and 1.

Inductive Step: Suppose that P(k) is true for some k greater than two.

We must prove that P(k + 2) is true.

Let k + 2 be an even number greater than 2. If k + 2 is itself prime, then P(k + 2) is true, with the two prime numbers being k + 2 and 0.

Otherwise, k + 2 = c + d, where c and d are integers with c and d both greater than 1. By the inductive hypothesis, c and d are both the sum of two primes, say c = p1 + p2 and d = q1 + q2. Thus, k + 2 = p1 + p2 + q1 + q2 where p1, p2, q1, and q2 are all prime. Therefore, P(k + 2) is true.

By induction, every even number greater than 2 is the sum of two primes.

[Guest]

The proof, by induction, states: Let P(n) be the statement that "every even number greater than two is the sum of two prime numbers".

Base Case: P(2) is true, since 2 is the sum of the two prime numbers 1 and 1.

Inductive Step: Suppose that P(k) is true for some k greater than two.

We must prove that P(k + 2) is true.

Let k + 2 be an even number greater than 2. If k + 2 is itself prime, then P(k + 2) is true, with the two prime numbers being k + 2 and 0.

Otherwise, k + 2 = c + d, where c and d are integers with c and d both greater than 1. By the inductive hypothesis, c and d are both the sum of two primes, say c = p1 + p2 and d = q1 + q2. Thus, k + 2 = p1 + p2 + q1 + q2 where p1, p2, q1, and q2 are all prime. Therefore, P(k + 2) is true.

By induction, every even number greater than 2 is the sum of two primes.

Nonsense!!!