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From: Archimedes Plutonium on 12 Aug 2010 01:05 Archimedes Plutonium wrote: (snipped) > > The more I think about this, the more I realize I do not need the > Algebra of interchange > between multiplication and addition, where both require a minimum of > two primes for every > Even Number >2. > > Let me crudely write out the proof, continually improving it. > > PROOF: > (1) Every Even Natural >2 has at minimum two prime factors in a > decomposition. For example 6 = 2 x 3 > > (2) Hypothetically assume there is a Even Natural >2, call it K, that > has no two prime summands which added together equals K. > > (3) Now K has at least two prime factors in a decomposition of > multiplication > > (4) Now let me use an example to guide this proof of that of 12 which > to Goldbach would > be 7 + 5. But as an example, say it only had one prime such as 10 + 2. > > (5) Now can I achieve a contradiction > > (6) I think I can, and maybe I do not even need the multiplication > lemma that every even number >2 has two prime divisors. > > (7) Without loss to argument take 2 as the singlet prime in Goldbach > then we have the Goldbach pairs as (K-2, 2), such as the (10,2) for > 12. > > (8) But then 10 or K-2 has two prime summands. And in this case they > are (5,5) > > (9) This is almost looking like a Ferrmat's Infinite Descent or > Mathematical Induction. > > (10) So we have the decomposition of K into (K-2, 2) and the > decomposition of K-2 > into (p_1, p_2) > > (11) Now, all I need is the idea that if I add 2 to that of either p_1 > or p_2, I end up with > two prime summands. > > (12) looking good and shaping up good and nicely, because it looks > like a mathematical > induction for a Goldbach proof, where the idea is that if Goldbach > breaks down somewhere > it is a even number called K and we can then utilize K-2, and 2 as > summands and that we > know K-2 obeys Goldbach, that all I have to retrieve is the adding of > 2 to either the p_1 or > p_2 yields two prime summands. Maybe, or maybe not, the multiplication > lemma comes in > handy. tired now and will continue later.... > Maybe I need that multiplication lemma. For I notice that if K is 12 we would have (K-2, 2) would be (10,2) and where the 10 is (5,5) and adding 2 to 5 we have (7,5) Now if we go to the next higher even number 14 we have (12,2) which is (7,5) for the 12 and adding 2 to 5 is (7,7). So what I need is a guarantee that the adding of the 2 to the previous Goldbach lower even number prime summands is going to insure a new set of Goldbach prime summands. Now maybe the multiplication lemma is that insurance because I notice that in the case of 12 we have 5x5 = 25 and 5x7 = 35 with a difference of 10 from adding the 2 to the 5. Then in the case of 14 we had 7x5 = 35 for the 12 and then for the 14 we had 7 x 7 = 49 for a difference of 14 once we added the 2 to the 5. So maybe the guarantee of turning the Goldbach singlet prime cell into a dual prime cell by adding the 2 to one of the K-2 primes is guaranteed by the Multiplication Lemma. Getting closer.... Obviously the Algebra solution is the easiest, for all I have to note is that interchanging multiplication for addition preserves the minimum requirement of two primes. But maybe I can do this proof without that Algebra Interchange. It reminds me of Projective Geometry where a proof of lines becomes a proof of points and vice versa. So if Galois Algebra does not have the interchangeability, then perhaps I need to use Projective Geometry where I call multiplication that of lines and addition that of points. Archimedes Plutonium http://www.iw.net/~a_plutonium/ whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies |