Weak Goldbach Conjecture, SOLVED

Guess I will have to also show (since I posted on my facebook wall that I solved both the strong and the weak Conjecture) the solution for the weak Conjecture as well:

Christian of the Goldbach family said also that he regarded it as evident that every odd number greater than seven was the sum of three primes, something which, according to him, should be evident if every even number was the sum of two primes.
So, we now find out the truth, or rather, you do, since I would be lying if I said I did not know the answer already, right?
 It must be boring being a teacher, but some appear to make a living out of it, so, each to his own, I suppose.

Anyway, to the matter at hand, or rather, on the screen... this is also available on mobile, although it is not formatted differently from the computer/laptop version, and so you have to zoom in a lot, but, hey that is modern technology for you.
 
?!

Suppose one had an odd number θ and this odd number was so large that primes (p1;p2;p3;...;pn) existed where pn is the greatest prime just less than the root of θ ;-

Then 1<n<θ  could be written fully in terms of factor sets by:

θ = [Fp1] +[Fp2] +[Fp3] +...+[Fpn] +[Rpn]

(where the factor sets and remainder set are as previously defined in older posts, all up to the value of θ )

Now, since an odd number is of the form (2d-1), where d={1;2;3;...}, then it follows that there is likely to be, sometimes, an even number involved in the summation, and the ONLY even number that is prime is '2' and this means that, actually, even for the STRONG conjecture, the factor set CAN have ONE member involved in the pairing of the primes;- the factor prime.

However, that is not the point.

If the factor set primes are not factors of the odd number in question, then they will most likely (meaning DEFINITELY, since these are numbers, not rocket science or probability..whatever "probability" really means) going to pair up with other members of of other factor sets and then each  pair will pair with a remainder element according to a definite ratio, an INDISCRIMINATE, definite ratio.

In fact, one can casually say that a third of Rpn will be taken by the sharing with the factor set Z3, with a fifth of the remainder being claimed by Z5, and a seventh of whatever remains after that by Z7, an eleventh of whatever remains after Z7 by Z11, and whatever is left over after that by Z13, so that, up to the value of pn, the remainder after summation, φrpn  would be:

φrpn= [Rpn] - ([Fp1] +[Fp2]+[Fp3]+...+[Fpn])[Rpn],

            = [Rpn] {1- ([Fp1] +[Fp2]+[Fp3]+...+[Fpn])

This means that, this remainder will then have to be split three ways so that one gets the number of triple primes that give the sum of the odd number, θ , such that, in effect, 

p(triples)= (1/3)φrpn

Actually, if there are NO primes less than the odd number θ  ( and this applies for even Σ  as well in the strong conjecture) then there is no need for pn or such and the prime that is considered is just'1', such that Rpn ={1} only and therefore

p(triples)= (1/3).3  (since the odd number would be 3)

=> p(triples)=1

There would be 1 triple that gives a sum of the odd number.

So,one can consider the solution to the Goldbach Conjectures available, and fully solved. I can now post THIS on my facebook wall and dust my hands of it!

Enough already, LET US LOOK at

COUNT1NG NUMB3R5!!!

The Upcoming Book that will ROCK the WORLD!!!    

  

A case where a few words are worth a thousand

Let me give an example of what I mean by the solution of The Strong Goldbach Conjecture. Consider the case where e=44;




The prime just less than the root of '44' is 5, so only the factor sets Z3 and Z5 exist up to 44;

[Z3] = 1/6 (44) = 7 1/3. But since there can be no fraction of a number, the answer must be either 7 or 8. In this case, listing the numbers according to

Z3:=> (6d -3) ,1< n<44 gives, {3;9;15;21;27;33;39} with '7' as the number of elements of Z3.

[Z5]= 2/30 (44) = 3, Z5:=>(30d-25)^(30d-5) , 1<n< 45 gives {5;25;35}

The number of prime pairs, according to the solution for the STRONG Goldbach Conjecture, is given by;

p(pairs) =(1/2) (44/4) (1 - {[Z3] +[Z5]}[Zr5]) ,

where [Zr5] = (8/30 )(45)=12
 
But(1/6 + 2/30) =7/30.

 This is the portion of Zr5 that is taken over by the factor sets Z3 and Z5 in sharing; The portion left over is therefore (23/30)

(23/30)(11) = 8.433
                     
=> 8<(answer)<9

                     

=> p(pairs) = (1/2)(8) or (1/2)(9)
                         =4 or 4 1/2

                     
The pairs are, actually, (from Zr5:=> (3od-r5) ; r5 ={ 29;23;19;17;13;11;7;1}
 which gives,
 for 1<n<44, {1;7;11;13;17;19;23;29;31;37;41;43})

{(43+1);(41+3);(37+7);(31+13)}


Found to be 4, which raises questions about where the so-called 'Simultaneous equations' got their origins.

Solution to the Goldbach Conjecture

This should not take that long: I will drop this quickly so I can go back to being bored:


If one has been following my previous posts, then one will know the terms by now; if not, then I suggest that one reads the previous posts for definition of terms.

About even numbers; let Σ  be the set of all even numbers; it follows then that, if  'e' the element of Σ is sufficiently large that there are primes less than e such that the greatest prime, pn, is the one just less than the square root of  'e' then the following factor sets FULLY define the even number 'e' ( in the sense that every number from 1 to e is accounted for... a bit tricky but you had better get this before you proceed, because if you are lost here you are lost for good)

e = [Fp1] + [Fp2] +[Fp3] +...+[Fpn] +[Rpn]

The sizes of the factor sets and the remainder set are from '1' to 'e' as I have mentioned.

Now, it is a known fact that every even number is 'double' some number, or that there are numbers less than the evn number that when added will give the sum of the even number. The greater the even number, the more the possible pairs, both even and odd, yes?

The number of possible pairs for the even number are exactly half the value of the evn number, and, if one removes the even number pairs...the pairs due tyo even numbers pairing with other even numbers less than the even number in question, then the remaining half is also halved... work that out.

Suppose the even number has other factors that are greater than two, and up to pn, for example. These factors will have their multiples pairing with other multiples which have THOSE factors as common, when THOSE multiples add up to give the sum of 'e'.

The point? Only NON- factors of the particular even number will  have their factors  pair with the elements of Rpn, which will contain only primes, in case you haven't noticed.

So, to find the NUMBER of prime pairs that, per even number will give a sum of the even number, factor out the factor sets of the evn number, remove them from the above equation, then use what is left to find the "proportion" which are claimed by the factor sets which have their primes as non-factors of 'e'. The remainder in the remainder will be the number of prime pairs which give the sum of the even. You can then identify the primes using the laborious method that I used in proving that the primes are infinite.

If you want an equation, then it will be something like



p(pairs) =  (1/2)(e/4)(1 - { [Fp2] +[ Fp3] +[Fp4]+...+[ Fpn]} [Rpn])) 

(where p(pairs) are the number of prime pairs that give the sum of e for every even number)


provided that none of the primes, p2;p3;p4;...;pn are factors of e, in which case they will not be included in the equation. Note the factor sets have an "upper boundary" of 'e'.

Come on, this is NOT that difficult to prove!!!

Now, give me a REAL challenge!!!

Part two; Even Theory

This is a brief overview of what is coming next:-
 

The postulate is that for every even number there exist AT LEAST ONE pair of primes which when added together will give a sum of the evn number;


the point is to show just how many such pairs there are for every even number, and that is ONLY done by coming up with an equation/statement that satisfies for every even number (..easy peazzy) ; as well as showing WHICH pairs we are talking about, i.e.,  come up with aFOOLPROOF method for unerring identifying a particular number and its complement in the sum as not THE prime pair, or one of a known number of pairs.


Johnny Nash said it could not be done, and they gave him a prize and said he had a "Beautiful Mind"(seen the movie?). Well, I DO the impossible, because they can not be done, so, I will shortly show the solution.

Goldbach Conjecture... Prime Analysis [Contd...ii]

So, if Zr3:=> ( 6d -5)^(6d -1)


then the first element in the remainder greater than 1 but less than the square of the preceding prime is a prime, and will be used as the next prime factor in the next factor set.


When d=1, (6d -5) = 1, (6d -1) = 5.


so, '5' is the next prime and p3 is '5'.


Z5 is therefore the factor set that has every fifth element from 5 in Zr3 as a factor, and this can be represented mathematically as;


Z5:=> 5[ (6d -5)^ (6d -1)]
       = (30d -25)^ (30d -5).


for successive values of 'd' then Z5 has the elements given by

Z5 ={ 5; 35;65;95;...}^{25;55;85;115;...} et.c., which can just be written
Z5 ={5;25;35;55;65;85;95;115;...}

and one gets the trend as the values of 'd' increase.


Back to the factor equations give us Rp3 as being of  size:

[Rp3] = (p1 -1)(p2 -1)(p3 -1) N
                            p1.p2.p3

 but since p1=2; p2=3;p3=5, then

[Zr5] =  (8/30). N


This implies that Zr5 will be represented by '8' subsets, and these will be determined by looking at the remainder when d=1 for Z5, after, of course, the preceding factor sets have taken their elements;

The elements of Σ are, {2;4;6;8;10;12;14;16;18;20;22;24;26;28;30}, and they are 15, as can be shown from the factor set equations. The elements of Z3 are {3;9;15;21;27} and these are 5 as can also be shown. Those of Z5 are {5;25} which means the remainder up to 30 is given by

Zr5= {1;7;11;13;17;19;23;27;29} and these 8 make up the full complement of 30 numbers from1 up to 30.

Obviously, this means that the number in Zr5 that is nearest to '1' is the next prime, and this would be 7. But before determining how Z7 looks we will need to find the nature of Zr5, then i will do Z7 and the rest I am sure you will know how to figure out, as I move on to the next part of the solution of Goldbach's Conjecture.

Zr5's elements can be represented as, in terms of d, ( from when d=1 in the case above)

(30d -29); (30d -23); (30d- 19); (3od -17); (30d -13);(30d -11); (30d -7); (30d -1) so that

Zr5:=> [ 30d- r5], r5 ={29;23;19;17;13;11;7;1}


This means that Z7 has elements determinable by


Z7:=> 7[ 30d- r5]

       = (210d- f7) , f7 ={ 203; 161; 133; 119;77; 49; 7}

And all the members of Z7 at d=1, and for all values of d, actually, are the exclusive members of the factor set with '7' as the least common factor.

 THIS is how one finds primes, continuously and unimaginatively and in a practical manner. The next primes are easy to find, from a continued use of the equations of the factor sets; In mathematics, it is considered sufficient to show from three successive examples the trend, and then extrapolate from there, since there will not be anything new. This is why I have limited myself to just a few examples and whenever I mentioned that , for example, n={1;2;3;...} I was merely showing that the trend continues after n=3 in the same manner as for n=1, n=2, n=3; with just an increase of '1' every time till infinity.


Once that sinks in, I am sure you can then go on and find the rest of the primes.


I have just shown a way to "practically" prove that primes can be found continuously up to infinity, and that, if 1 keeps on recurring as a remainder and is not claimed as a multiple by any factor set, then there is no reason why '1' should not be considered a prime. So, based on prime analysis, I declare that the number '1" is indeed a prime.


therefore,


N= PUM , P^M={}, [P] =(infinity)


QED!

More counting numbers...

Prime Analysis, Continued...

It has been discovered that p1 is '2' and p2 is '3' and that Σ is the set of all even numbers with a defining statement for every member , e, of the set given by

e:=> (2d) , d={1;2;3;...}

and that the odd set θ is also defined for every element  θ in it by

θ :=> (2d -1) , d ={1; 2; 3;...}

and the values of d for both (θ ,Σ ) are the same in every instance so that when 'd' has any specific value, it is the same for both sets. This should be borne in mind, at all times, as it will apply to all sets and their corresponding remainder sets.


For convenience, all the factor sets not either the odd or the even set will henceforth be designated 'Z'. So, Fp2 is actually Z3 and Rp2 is Zr3 , et.c. .


Now, if Z3 is the factor set Fp2, then it is, as has been shown, of magnitude 1/6(N).


The defining statement of all the members of Z3 is to be found by the following;

 

Z3 claims a third of θ , and that means every third  member of the odd set will, from the number '3' itself onwards, be claimed by Z3, and this can be mathematically shown as

z3:=> 3( 2d-1)

      = (6d-3), d={1;2;3;...}

These are natural numbers, so there can not be any number less than '1' in subtraction...and still have a member of N as the result..., so , it is important to point out that 'd' for Z3 is not the same as for (Σ , θ ) but is unique...and it is a variable anyway so there is no need to specify what it is for every factor set/ remainder set pair. (The value of 'd' is the same only for a specific factor set and its remainder set).

When Z3 begins to exist as 'n' the element of N increases in value, it follows that when d= 1, then
z3= [6(1) -3] =3

[but, as I have pointed out before, the value of 'd'  at this point is not the same for Z3 and  Σ , since they are not related]

But, according to the factor form equations, where we have our Fp2 being given its size by

[Fp2] = (p1-1). N
                    p1.p2

and [Rp2] = (p1 - 1)(p2 -1) N
                            p1.p2
 then, technically, it follows that, Rp2, that is Zr3, is practically defined as having "two members in every six numbers" in N. This is because

 (p1-1)(p2-1)=2  [from p1=2, p2=3] and  p1.p2.=6, which is where the conclusion of the member representation can be drawn from.

For, d=1 in z3:=>(6d-3) there should therefore be two members of Zr3 which are left after both Σ AND Z3 take their elements;

Z3 can only begin at (n=6) since that is where (p1.p2) can have any meaningful value, so up to n=6, there are

1/2(6)= 3 members of  Σ , which are {2;4;6}, from e:=> 2d, d={1;2;3}

there is 1/6(6) = 1 member of Z3, which is [6(1)-3] =3

there should be 2/6 (6)= 2 members of Zr3, and these would be the remainder that fully account for every element of N from '1' up to '6'

So, the missing numbers are {1;5}

1 = (6d-5) while

5= (6d-1) when d=1 for Z3

these two then are the factor forms of the remainder set, Zr3;

Zr3 is fully defined by:

zr3:=> (6d-5) ^ (6d-1)

 Think about that for the day!!!

Prime Analysis Continued...ENGINEERING Numbers

The question of whether or not the number 1 is a prime or not means that  I can not say, with a clear conscience, that N= PUM, P^M= 0 is really a fact without going all out to prove it, at least for this ONE number, since by logical extrapolation ANYONE can see that there are no OTHER numbers that can be classified as neither prime nor multiple, or for that matter EITHER one or the other, or BOTH.

So, the point is to have to take the analysis to a different level;- How would one go about creating factor sets to begin with, and what numbers would one have to use to have these factor sets? I will now have to REALLY count numbers...

Counting Numbers; Prime Engineering:

It is a given that p1 is the least number that can be used as a factor for numbers greater than itself, which, obviously rules out the number 1, so the next number in line in N will have to be the first [note the following qualification] substantial prime.

That means p1 is the number '2'.

If p1 is '2', then, 1/2 of all numbers, starting with 2 itself, can be claimed as members of Fp1, [or F2, or Σ, the set of even numbers ],  which means the size of the even set in N is given by;
[Σ ] = 1/2(N)

 Therefore, the set of all odd numbers, θ is also half the size of N in magnitude, as can be easily seen, although it should be noted that for the statement on size to make sense, there should be an upper limit set in N which would, for example, be 10, in which case if 10 is the limit, 10 would replace N in the equation for both the even and the odd factor sets:

[Σ ],(1<n<1o) = 1/2 (N)

                            = 1/2(10)
                            
                            = 5

Therefore the number of 'members' of Σ up to 10 is 5. Technically, we are yet to IDENTIFY the numbers in question, so we may for the moment assume that they are unknowns and take the analysis step by step.

The remainder θ is the set that would have been called Rp1 and as such it is equal to the difference between the 'number of numbers' between 1 and 10 that are in N and the 'number of numbers' between 1 and 10 that are claimed by the even set Σ , as per the initial analysis with factor sets.

[θ ], (1<n<10)= 1/2(N)

               =5


The above was just an illustration, to give an example of what is going on.


However, since the first substantial prime has been identified, it makes sense to identify the defining characteristics of every member of the first factor set, and thus be able to find the definition of the first remainder set, and thus be able to identify other numbers, and so on.

Fact: every number in Σ is divisible by 2.

So, if 'n' is any number in N, then n is a member of Σ if

n/2= d                 where ( n, d) C N i.e.,( 'n' and 'd' are both natural numbers)

=> n= 2d,     d= {1;2;3;...}

So, the set Σ  is fully defined by the statement that every member 'e' in it has to be of the form given by

e:=> 2d ; d= {1;2;3;...}

For 'd' with successive values of {1;2;3;...} then 'e' has values; {2;4;6;...} which means now ALL the numbers in N which belong to Σ can be identified.


If the limit is set to n<10, then the members of  the remainder set can be found, for example, as { 1;3;5;7;9}, while the members of the even set are {2;4;6;8;10}, which means that up to 10, N is fully defined in terms of the even set and its remainder complement.

I hope that is clear now.

Now, that being the case, how is the set θ defined?

θ is a remainder set, so a look at the first remainder is illuminating:

e:=> 2d, and when d=1,  e=2, and θ =1, so it follows that ALL θ are of the form given by

θ  :=> (2d -1),  [since 2-1= 1, then 2d-1 = θ , would work for all members of θ if the same value of 'd' used to find 'e' was used to find the corresponding remainder, θ . When d=24, for example, e=48, θ = 47,  using (2d) and (2d -1) respectively for the factor set notation and the remainder set as well]

It is from the equation of the remainder set, as well as the members of the remainder themselves, that the next factor set gets its own definition.

Remember, the number 1 can not be used in ANY meaningful factorisation, so it is not considered when finding a substantial prime factor for the next factor set.
 

Looking at the next least remainder, gives 3 as the number, so 3 must be a prime, since it is not divisible by any number greater than '1'. In fact, since '3' is less than the second member of Σ , which is the square of '2', that is, '4', it follows that when one wants to look for the most likely primes at any stage, even with the greater factor sets, than one must consider the numbers in the remainder set which are less than the square of the functional prime in the  factor  set, in the remainder.

In this case, the square is 4; look at the numbers less than 4 but greater than 2 and you have the prime that should be used in the next factor set.

Therefore p2 =3

From [Fp2] = 1/p2 ( p1-1)/p1 .N

one gets that the factor set with '3' as functional prime, which we will call Z3, is of the size

[Z3] = (2-1) N

             2.3

           =1/6 .N


More about this in the next post!!

Digest this one first.

The Goldbach Conjecture; Part 1: Prime Analysis [full theory]

It was postulated that, every even number is the sum of two primes, originally, but this later became, every even number greater than two is the sum of two primes, since, based on the laws of Number Theory, the number 1 could not be taken as prime.

The first part of solving the Conjecture is, therefore, based on Primes themselves, to find out whether, just as, according to Number Theory, even numbers are infinite;- primes can also be said to be infinite.

Prime Theory

If  P is the set of all primes in N, the set of all natural numbers, and M is the set of all Multiples among the naturals, then the three sets are related by;

N = P U  M ,  P ^ M = { }, I P I = (infinity)


Proof;


*** Note*** The statement above, in plain English, reads; N is the sum of all the numbers found in P and in M, and no number found in P is found in M, and the size of P is infinite, i.e., the primes do not end.


Let p1 be the first prime that can be used as a factor of another number greater than itself . That means if all the numbers in N were to be grouped in a set according to each having p1 as the least number they can be divided by, then every p1th number from p1, including p1 itself, would be in the set. Let this set be called Fp1.


The size of Fp1, given by [Fp1] (meaning the number of numbers...think counting numbers... that would be found if all the numbers in Fp1 were counted, in the set N), would therefore be given by


[Fp1] = 1/p1 (N)



If this set were to be isolated in N, then the Remainder set, (R), after Fp1 would be the set of ALL numbers that do NOT have p1 as a factor, nor are divisible by p1. This set is Rp1, and its magnitude is given by


[Rp1] = N - [Fp1]


            = N - 1/p1(N)


            =N (1- 1/p1)


             =(p1 - 1) N

                     p1



Of this remainder, the prime in succession to p1, that is, p2, would claim all those numbers which have p2 as the least common factor, and these would also be in their own set, Fp2, whose size would be given by



[Fp2]= 1/p2 [Rp1]



            = 1/p2 (p1 -1) N

                              p1


           = (p1 - 1) N

                  p1 .p2

The next remainder after that would be the set Rp2, whose own size would be

[Rp2] = [Rp1] -  [Fp2]

             = {(p1 -1) - (p1 -1) } N
                       p1         p1.p2
         ={(p1 -1) (p2-1)} N

                       p1.p2

 

The next set with a claim to the all the numbers which have a least common factor as prime would be Fp3 , which would have its size given by

[Fp3] = 1/p3 [Rp2]

            = (p1 -1) (p2 -1) N
                   p1.p2.p3

So that, even [Rp3] can be found to be given by;

[Rp3] = (p1 -1) (p2 -1) (p3 -1) N
                      p1.p2.p3



So, in general, if pn is the nth prime that can be considered as the least factor of any other numbers in n, after p1; p2; p3;...; p(n-1) have appeared in N, then, the size of the factor set Fpn is given by

[Fpn] = (p1 -1)(p2 -1)(p3 -1)...(pn-1  -1) N

                        p1.p2.p3. ... .pn

And


[Rpn] = (p1 -1)(p2-1)(p3-1)...(pn -1) N
                       p1.p2.p3. ... .pn


This means that, technically, the Remainder set after each prime is NOT less than the factor set that precedes it, so much so that for every given factor set Fpn that exists in N, there will always be a LARGER remainder, which gives more room for a prime greater than pn, i.e. pn+1 , to have numbers which will have the prime pn+1 as least factor.


To make it simple; Factor sets give rise to more factor sets which, because the remainder does not get less than the factor set in size, means that there will always be another factor set that can exist by using the numbers in the remainder set generated by the preceding factor set.

Rp2 makes Fp3 exist, which gives rise to Rp3 when Fp3 is isolated from Rp2, and since [Rp3] - [ Fp3] =[Rp4]; where [Rp4] > [Fp3] ( if any limit is set in N, as both sets are infinite and would thus not be able to be compared for size to infinity otherwise)... then it means that there WILL ALWAYS  be room for another prime to have its own factor set.

From pure analysis, therefore, it follows that the set of Primes, P, is infinite.

From definition,it also follows that, since a prime is a number that can be used as a least factor and is itself not factorisable, while a multiple is a number that can be factorised, then no number can be both a prime and a multiple.

Which leaves the number 1.

The square of 1 is 1, and 1 is not divisible by any number less than itself in N, since the least natural number is 1. So, is 1 prime or a multiple, or neither?

If N is fully defined by N=PUM, then where does 1 lie, or is the theory void.

Check the next post!