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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
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