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...
That means p1 is the number '2'.
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)
[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
= 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.
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;...}
=> 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;...}
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.
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.
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.
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.
