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;...}
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;
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;...}
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
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
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
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!!!
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!!!
