## Sunday, January 24, 2010

### Goldbach's Conjecture : a proof.

Goldbach's Conjecture : a proof.
Andre Willers
24 Jan 2009

Synopsis :
A simple proof for Goldbach's conjecture using iterative systems and infinite descent .

Discussion:
Goldbach's conjecture : any positive even number can be written as the sum of two prime numbers .

Part 1
Assume the conjecture is true .
2n = p1 + p2 , where p is an odd prime number . ….(1)
p=2n+1 , where n is a positive integer >=0 ……….(2)

Iterate expression (2) into (1)

Every (2n) part can be expressed as another sum of p .

Iterate this m times until at least one p=3 (ie n=1) for the first time .
Let s be the number of terms terms of p equal 3 at this iteration .
Then
2n = sum(p) + m(m+1) - 2s ….(3)
where s must be equal to m(m+1)/2 for a true tautology .
This can only be true if m=1 if s=1 (see comments below . S is dependant on the number base , and can always be set to 1 .)
This means that there is always a sum of two primes equal to 2n .

Part 2
The inverse :
This is where we use the properties of prime numbers .
If even only case can be found where the sum of primes is equal to an even number , then the theorem must be true for all primes . (eg 7+11=18) . It sounds trivial , but it is true nonetheless .

1.Notice how the proof sidesteps the definition of primes except in Part 2. Some values can be adduced using the algorithm . (See equation (3) ) .

2.The problem with this sort of proof is always the infinite search space . What we have done is use a form of metalogic to show the general case .
Eq (3) can then be written in an iterative form
2(n-s) = sum(p) + m(m+1) ….(4)
The rest follows .

3.Note that sum(m) = m(m+1)/2 , as expected . There is a deep underlying logic to the iterative metasteps .

4.Expressing numbers in terms of different bases illustrates this
Why m(m+1) –2s =0 must be true if only one instance of the theorem is true .
S is dependant on the number base , which we can vary at will , but m is a pure counter (a metacounter , if you will) . We can always set s=1 by choosing the number base .

5. Notice the use of Infinite Descent .
A general expression is described in terms of constraints to a particular instance .
What is sauce for the goose must then be sauce for the gander .

6. Other uses .
Only beth(0) (ie aleph(0) ) infinite descent was used above .
The same technique can be used fruitfully to probe infinite feedback processes of higher orders of randomness .
See http://andreswhy.blogspot.com "Negative Evolution" et al

7.Other proofs:
Longer but easier proofs can easily be adduced .

Hint : Use the transformation 0 -> i0 in the definition of number .

Keep track of those pesky transfinitesimal zeros . You've only had them for 1500 years . Losing them is sheer carelessness .

Tora! Tora! Tora!

Andre .