Easy Proof of Goldbach's Conjecture

Okay, when you say that an even number is a sum of two primes, all you're really saying is that there's some inbetween number between zero and that number, where from zero to the inbetween is a prime, and from the inbetween to the number is a prime.


So, call zero and the number each zero on two different number lines, with one counting from zero to the number starting at the left, and one counting backwards zero to the number starting at the right. For every prime hole on one number line, can we fill it with a frequency of a prime on the other?


Obviously, since no two prime numbers are an odd number apart, no frequency of a prime on one line - say 10, 15, 20, 25 - can fill more than one prime hole on the other line. The only exception woudl be 1-waves - which we're not allowed to use - and 2-waves - if we could stagger two of them or offset them an odd number.


But since we're only allowed even numbers from zero, we can't use two staggered frequencies of 2 to fill eveyr other hole on both lines. Moreover, since 2 is an even distance from zero - an exception to the even-distance from a prime rule - we can use it to fill two prime holes on the other line.


So some frequency or product of 2 on each line can fill 2 and 0 on the other, and then we need at least one different prime on each line to match every hole on the other in some product of itself. Problem is, 1's aren't available to use frequencies or products of on one line to fill holes on the other, but each line's 1-hole needs to be filled!


So, short a frequency, this leaves an unfilled hole somewhere on the opposite line, and there must be at least one such uniflled hole on each line. If they are in the same place, at the middle, they take care of each other, and present the most obvious Goldbach candidate.


If there is no middle, such as with the number 18, you might get a loose, say, 7 on both lines, but they're not on top of each other. So this creates a loose 11 which they are opposite. Only now we have an even number of holes called for, and we still must have that 1 extra. There must be an odd number of loose-end numbers other than 1.


More specifically, you have numbers like 17 whose products are never used, meaning there must be a naked prime on the other side. Theoretically, it seems you could usually just have, like, 17 and 1, leaving your 1's uncovered, and be done with it.


But then there are still other frequency primes whose products are never used, which must pair off symmetrically in even quantities. These are like numbers that are greater than half, and numbers like 7, where 14 - meaning 4 on the other side - is already called for by 2, and not prime, so the seven hangs useless, with 21/-3 being totally useless.


Did I miss anything?


eLROY

you should go collect your million dollah!


http://www.mscs.dal.ca/~dilcher/Goldbach/index.html


and what the hell is Goldbach's Conjecture anyway?


http://www.informatik.uni-giessen.de/staff/richstein/ca/Goldbach.html


verification:


http://www.ieeta.pt/~tos/goldbach.html

The simplest explanation of why Goldbach's conjecture cannot be violated is because of the way entropy is accounted for and lost in our logical product of two opposing number lines. The only way to fill every prime hole at the zero end of one number line using a prime frequency at the positive-even-number end of the other, is if you can describe as complex a space with the frequencies. But in our logical system, any time frequencies harmonize or synch up, entropy - meaning the power to describe as complex a sequence - is lost.


If we consider our number line as a series of 1's and 0's - 1's being frequencies and 0's being prime holes - 3 and 5 can be used to create two 1's total, 1 each at point 3 and point 5, or to create just one 1 at 15. In other words, if our number line could be described in greater resolution as a sum of frequencies than 0,1 - if 15 could be described as 0,1 or 2, for instance - we would not lose our entropy like this.


The total number of frequencies employed in both number lines is the number of primes available in a line of that length. But since our prime frequencies are in phase at our zero end, we harness their maximum entropy by using each to describe a unique spot. At the other end, no prime frequency even becomes useful, by not hitting a 1 already taken care of by a lower prime frequency, until it reaches its own square. 11 doesn't become useful until 121.


It is inevitable that, by the time our first, say, 4 primes all fall completely out of phase again - one hop to each side of being synched up - and become useful - we will perpetually already have a much greater supply of new, higher primes to account for at the zero end. The utilized primes at the zero end rises faster than the utilized frequencies at the positive, even end, in a tragic, bottomless way.


Therefore, using these axioms, the proof is simply a proof that there will be a prime number higher than any given prime number, but lower than its square, or you get the idea. For an illustration, consider the positive, even numbr 126, which makes separate use of 11,7,5,3, and 2 at its terminus. We already have the prime wave 13 available to describe this region, and we're not even talking half way to the midpoint!


Meaning, it has gotten to 126 and it hasn't even used 13 yet at any point back to 63. But we already get to use 53. They can never completely match us except by using primes, there will always be an AND hole. Their frequencies will stack up and lose resolution, and will always be less complex than our linearly-resolved primes.


eLROY

If you think about these frequencies, 2 has the most entropy. So much so, in fact, that if we were allowed to stagger our opposing number lines by an odd amount, we could use our prime frequencies in one to fill just about every prime hole in the other.


Actually, each time we add a higher prime to the base end, the entropy rises only slightly, but the entropy resolution rises enormously, right? With odd/staggered 2's, almost all the entropy resolution is blocked out by accounting losses. The 2's cover just the amount of entropy that isn't lost in the opposing 2's. Without two's, we need equally-fine numbers to resolve it.


But again, if we have to use out-of-synch combinations of odd numbers to cover up, exhaustively, an even sequence of opposing odds, that is impossible, unless we have every frequency, which isn't available to us in a sub-infinite stretch of number line. By climbing up a sequence of primes in the opposing line, you will inevitably uncover the frequencies we are missing.


Or they can be expressed as some combination of frequencies, by flippng holes on both sides until sort of a least-common-denominator resolutioon is reached. But no matter how you break it down, the frequency end will always have less entropy than the prime end, and it will have to be resolved somehow.


Put differently, since we are guaranteed to not make full use of our frequencies, by having some in synch short of their own square - and by always having a prime going in the other direction which is higher or finer than the one which has not yet reached its square - we are guaranteed to expose this mismatch of precisely some increment of unaccounted-for finess, in the amount or wavelength of at least one prime number.


We need merely to iterate through the wavelentghs to discover the unaccounted-for entropy. Moreover, depending on the chance harmonic characteristics of our frequencies at the frequency end, we may lose even more entropy, so that it can only be accounted for through offsetting combinations of missing holes, until a mismatch resolvable with the available entropy is the residual.


In fact, eLROY's conjecture is that the harmonic characteristics will never leave just one hole - meaning at least one sum of two primes to create any even number - but will have to resolve the mismatch through a horsetrade of opposing holes - meaning more than 1 set of two primes inside every even number. Now maybe some idiot with this much time on his hands will be talking about me in 250 years:)


eLROY

Here is the link to my current semi-final answer:


http://www.twoplustwo.com/cgi-bin/newforums/genpok.pl?read=31971


eLROY