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!
