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;
Zr5:=> [ 30d- r5], r5 ={29;23;19;17;13;11;7;1}
This means that Z7 has elements determinable by
Z7:=> 7[ 30d- r5]
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= {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!
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!
