return to homepage THE SIMPLE PROOF OF
Take two facing walls in a room that are exactly the same size. Paint 10/59th of each wall yellow. That will stand for our primes. Paint 20/59th red. That will be the NPO's. Paint 29/59th blue. That will be the evens. With the blue, you must paint the left side of one wall and the right side of the other, like a mirror image, so that blue always meets blue. If we bring the walls together, blue will cancel itself out, and we can forget about it after this. The question, of course, is how the yellow meets the red. How much orange paint will be created when we move the walls together and have them touch eachother? We can have red meet red, yellow meet yellow or yellow meet red. Let us seek a minimum amount of red-red meeting. To do this we force as much red to meet yellow as possible. We forbid yellow from meeting yellow and then cover all available yellow with red (we are doing this to see if we can prevent yellow from meeting yellow, you see). The leftover red will have to meet red, and that will be our minimum amount. We have half as much yellow as red, so we must lose half our red in the meeting. What is our total density of each color now? We have 20/59 orange (red+yellow) and 10/59 red (red+red). But we must remember that 10/59 ≠ 5/30 or 1/6. Therefore, although our wall appears at first to corroborate our table, it does not. The table is a list of actual terms; the wall is a representation of densities. This is where the final and worst mistake is made, by those that may have made it this far. They have assumed that you could switch mid-problem to a density that expressed NPO's to odd numbers. That is, they assumed that we had a yellow density in the non-blue areas of 1/3 here, but we don't. We can't reduce our density and start talking of a density in the non-blue part of the wall. The density already applies to sub-parts of the wall as well as both walls together and each wall separately. So we cannot reduce it and find a sub-density of the odd parts of the two walls. You cannot reduce the density and find a sub-density. Or, you can, but you have to do it in the traditional way, by reducing the fraction in a legal way, such as 10/59 = x/30, and then cross multiplying to find the answer. You can't just say, "We have 30 odds and 10 primes, therefore the density of primes to odds is 1/3." The density of primes to odds does not generate the meetings, the overall density of primes does. Those who reduce in this way have made a terrible mathematical error: it is that simple.* We have half as much yellow as red, but we do not have a density of 1/3 yellow in the non-blue section or of 1/6 on each wall or both walls. Again, the density of yellow is 10/59, period, and it applies to the whole number line and every part of the number line. Both walls have that density and each wall has that density. Those who have made the mistake likely made it because they wrote the density (as I have) as the real number of primes over the real number of integers less than X. This fools them into thinking they can reduce that into sub-areas, but they can't. Once we define that fraction as a density, the numerator no longer stands for the actual number of primes in the table and the denominator no longer stands for the actual number less than X. We would be better off to write it as a density of 20/118 or something, so that we are no longer tempted to think of the numerator as the real number of primes and start reducing it in faulty ways. Now let us return to the solution. In the table, we found 1/3 of odds sums were NPO + NPO. But if we reduce our density fraction here (in the correct way) we find that 10/59 = 5.085/30. If we double that to represent a meeting or sum, we find 10.17/30 or 1.017/3. That is our minimum density of meetings or sums of NPO's given the density of primes. But 1/3 is less than that. It is therefore mathematically disallowed. The table we forced to work above fails for a mathematical reason. You cannot generate only five NPO-NPO meetings with 10/59 primes. Your number of NPO-NPO meetings must be greater than half your number of primes. This forces one of your necessary N + P meetings to fail, turning it into a P + P meeting. I will say that in a different way: If we divide all possible non-blue wall meetings into fifteen meetings (like fifteen equal size patches), at least six must be red-red, since that is the first available number that satisfies the requirement of being at least 1.017/3. If six out of fifteen are red-red, then some yellow will be uncovered by red, and it will have to meet yellow. A yellow-yellow meeting is a prime pair. Varying the local density does not change this fact, since a lower density of red paint on one given part of the wall must create a higher density at all other points. Less meetings of red-red in one vicinity must create more meetings in all other vicinities, in order to keep the overall density constant. Some will say that if you have considerably more red paint on one wall than the other wall the outcome must be affected. But this is not true. Even if we allow that the density of NPO's above the halfway mark of a given number must be greater than the density below, we cannot find a variation from my proof. Less red paint on one wall will give you more yellow paint on that wall, but the overall density of yellow paint is also constant for each given number, so more yellow on one wall means less on the other wall. This means that the total amount of red paint that can meet yellow will stay the same. The minimum red-red events therefore cannot change, no matter what you do. This is because to minimize your red-red meetings, you force as much red to meet yellow as possible. You cover all yellow with red. But if your total amount of yellow on both walls does not change, you cannot change your minimum number of red-red meetings. So having a higher density of NPO's at higher numbers does not affect my proof. Others will say that when I was calculating probabilites, I jettisoned the evens, since they skewed the problem. But now I am doing the opposite. Yes, I am. I am forbidding you from jettisoning the evens, since the density must include them. We aren't concerned with variance now, since we aren't calculating probabilities. Our fraction is not a probability, it is now a density. The density must include all numbers less than X, since that is how it is defined. Notice that you cannot say that the density of evens is 100%, allowing it to be ignored. That doesn't make any sense. Nor can you find a density of primes relative to odds. That doesn't make any sense either, since there isn't any way to subtract out the evens. Every potential spread of primes will create different gaps, and these gaps will be changed in different ways by filtering out evens. So, not only is it impossible to achieve, it is both senseless and counterproductive. It is not how the numbers actually match up. The primes and NPO's match up around the evens, therefore the density must express this. The density we have does that, so it would be absurd to propose filtering evens to solve a problem we do not have. Filtering would actually unsolve a problem we have just solved.Now, if you go back to our table (the number 60) and apply this fact to the list, you find that to create another NPO + NPO sum, you must switch a P with an N. You cannot do so without creating a P + P sum. What all this means is that my new rule utterly overthrows the tacit assumptions of the example table and every possible table or even number. My minimum can be calculated in the same way for any given even number or prime density, and in every case it forces the appearance of at least one prime pair. We had assumed that we could put an N wherever there was a space for it in a list, but we can't. We have to be scrupulously careful with how many of each sort of sum we create, since many possible combinations will contradict the given prime density. A prime density of 10/59 at the number 60 requires that we find at least one prime pair, no matter where the primes turn up in the sequence. Real positions and densities will create other possible numbers of prime pairs, but I have now proven that no possible density or position can create zero prime pairs. Lower and lower densities of primes will create higher and higher densities of NPO's, and this will always force the N + N sums to invade the expected N + P sums, turning at least one of them into a P + P. One thing this does is match our findings from the probability calculations, and shows that I have simply extended those odds into a non-probability solution. Our probabilities told us that no matter how low we took our density of primes, the odds were very strongly in favor of finding at least one P + P. This was because the odds were even more strongly against covering all primes with non-primes. At any rate, I turned the probability fractions into usable fractions by looking at densities and minimum values. These minimum values provide a general proof without having to look at actual sums or the real distribution of primes. I could avoid all contact with prime generators, as well as esoteric fields or non-linear maths. Although it is possible there are other more complex solutions, I believe this is the simplest possible solution to Goldbach's Conjecture. So, the general rule is that your total density of NPO + NPO sums to all sums cannot be less than half your total density of NPO's. It therefore cannot be less than the prime density. But to cover all primes, you must create a fraction of NPO + NPO sums that is less than the prime density. This being disallowed, one of the NPO + P sums must switch to an NPO + NPO sum. Since the total number of NPO's and P's is fixed for each given situation, to create the extra NPO + P sum requires the creation of a P + P sum.
*Some readers have claimed that there is an apparent contradiction in my pointing at the "worst mistake" of switching mid-problem from a fraction of terms to a fraction of odds, and then doing it myself at the beginning of my proof. I show that my example table must create 1/3 NPO + NPO sums as a fraction of odd sums. Haven't I just broken my own rule, by talking of sums as a fraction of odd sums? No I haven't. I warn against switching mid-problem to a fraction to odds only when I am talking about densities of primes and NPO's. You can't switch because you have a denominator that is always an odd number--in this case the number 59. You can't divide that in half and find a round number that stands for your fraction of odd terms. But that only applies when you are talking about terms; it does not apply when you are talking about sums. The fraction of sums If this paper was useful to you in any way, please consider donating a dollar (or more) to the SAVE THE ARTISTS FOUNDATION. This will allow me to continue writing these "unpublishable" things. Don't be confused by paying Melisa Smith--that is just one of my many |