RD-BH
Human
[TYPO] The last image should show sigma_(x=0)^(infinity) instead of sigma_(n=0)^(infinity)
For consideration:
Integers:
2*I are integers divisible by 2.
2*I +- 2^0 are integers indivisible by 2.
2*I +- 2^0 + K*(2) are integers indivisible by 2.
3*(2*I +- 2^0) are integers divisible by 3 and indivisible by 2.
3*(2*I +- 2^0) +- 2^J for J>0 are integers indivisible by 2 or 3.
3*(2*I +- 2^0) +- 2^J + K*(2*3) for J>0 are integers indivisible by 2 or 3.
...
5*3*(2*I +- 2^0) +- 2^J + K*(2*3*5) for J>0 are integers indivisible by 2, 3, or 5.
7*5*3*(2*I +- 2^0) +- 2^J + K*(2*3*5*7) for J>0 are integers indivisible by 2, 3, 5, or 7.
... P[n] = nth Prime
P[n]*...*7*5*3*(2*I +- 2^0) +- 2^J + K*(2*3*5*7*...*P[n]) for J>0 are integers indivisible by Primes P[1] to P[n].
If f(x) = product_(n=1)^(x){P[n]} then:
... (f(x)/2)*(2*I+-2^0) +- 2^J + K*f(x) are integers indivisible by P[1] to P[x]
... there remain infinite indivisible integers as x approaches infinity
f'(x) = (f(x)/2)*(2*I+-2^0) + K*f(x)
Therefore there are infinite indivisible twin pairs of integers such that:
... f'(x) - 4, f'(x) - 2, f'(x) + 2, and f'(x) + 4 are indivisible by x primes as x approaches infinity.
Assume finite twin primes:
1) there must be a finite number of indivisible twin pairs
2) there must exist an x (ie P[n]) such that any integer greater than x (ie P[n] + 2) is divisible by at least two primes
1) is false, infinite indivisible twin pairs
2) is false, infinite indivisible twin pairs and infinite primes
... consider P[y] such that P[x] < P[y] < f"(x) = product_(n=1)^(x){P[n]}
... ... P[y] = (2*I +- 2^0)*f"(x)/2 + K*f"(x) +- 2^J for J>0
... ... example: P[3] < P[10] < f"(3) {ie 5 < 29 < 30 }
... ... 29 = (2*0+2^0)*2*3*5/2 + (1)*2*3*5 - 2^4 {ie 29 = (1)*15 + (1)*30 - 16 }
Indivisible twin pairs:
... ITP(x) = product_(n=1)^(x){P[n]-2} as x approaches infinity
Form of those pairs:
... f'(x) - 4, f'(x) - 2, f'(x) + 2, and f'(x) + 4 as x approaches infinity
1) a Prime is an Integer > 1 and only divisible by 1 and itself
2) there are infinite Primes indivisible by 2
Therefore:
3) there are infinite Twin Primes
... Quantity: ITP(x) = product_(n=1)^(x){P[n]-2} as x approaches infinity
... Form: f'(x) +- 2^J for J=1 and 2, as x approaches infinity
You have three points on a wave:
Zero: 0 ... evenly divisible by every Prime
Midpoint: 2*3*5*7*...*P[x]/2 divisible by every odd Prime
Endpoint: 2*3*5*7*...*P[x] evenly divisible by x Primes
It follows:
midpoint - 4 is indivisible by x Primes
midpoint - 2 is indivisible by x Primes
midpoint + 2 is indivisible by x Primes
midpoint + 4 is indivisible by x Primes
These points have a period = Endpoint, defining infinite series:
... midpoint - 4 + K*Endpoint
... midpoint - 2 + K*Endpoint
... midpoint +2 + K*Endpoint
... midpoint +4 + K*Endpoint
As x (number of Primes) approaches infinity, these series remain infinite.
Zero +- 1 are indivisible by all Primes (Primes are Integers > 1)
Endpoint +- 1 + K*Endpoint defines an infinite series of integers indivisible by x Primes.
P[n] = nth Prime
1) There are infinite Primes, therefore no set of Primes can divide an infinite series.
... P[n]/P[n+1] are not integers
2) There are infinite Primes, therefore no set of Primes can divide two infinite series.
... for P[n+1]-P[n]>2, P[n]/P[n+1] and (P[n]+2)/P[n+1] are not integers
3) There are infinite Primes, therefore no set of Primes can divide infinite number of infinite series of indivisible twin pairs.
... For n = 1 to x, as x approaches infinity (midpoint +-3 +-1 + K*Endpoint)/P[n] are not integers
4) Therefore, there must be infinite Twin Prime pairs.
Integer I, Prime P:
... remainder of I/P has P possible results ranging from 0 to (P-1)
... I/2 has remainders of 0,1
... I/3 has remainders of 0,1,2
... I/5 has remainders of 0,1,2,3,4
... I/7 has remainders of 0,1,2,3,4,5,6
... etc ...
... I/P has remainders of 0,1,...,(P-1)
Code:
Integers (0..29): 012345678901234567890123456789
Remainders (2): 010101010101010101010101010101
Remainders (3): 012012012012012012012012012012
Remainders (5): 012340123401234012340123401234
Remainders (7): 012345601234560123456012345601Zeroes appear where an integer is divisible by a Prime.
Vertically (first 4 primes):
Code:
2357...etc...
========
0000...etc...
1111...etc...
222...etc...
33...etc...
44...etc...
5...etc...
6...etc...Combinations of remainders form unique Integers.
Total combinations formed by set of x primes:
... rCombos(x) = product_(n=1)^(x){P[n]} for P[n] = nth Prime
Strike the zeroes (ie combos divisible by set of x primes).
Code:
2357...etc...
========
1111...etc...
222...etc...
33...etc...
44...etc...
5...etc...
6...etc...Total combinations indivisible by set of x primes:
... rIndivisibleCombos(x) = product_(n=1)^(x){P[n]-1} for P[n] = nth Prime
How many indivisible twin pairs remain?
... I/2 leaves (1) infinite series, remainders: (1,1)
... I/3 leaves (1) infinite series, remainders: (2,1)
... I/5 leaves (3) infinite series, remainders: (1,3),(2,4),(4,1)
... I/7 leaves (5) infinite series, remainders: (1,3),(2,4),(3,5),(4,6),(6,1)
...etc...
... I/P leaves (P-2) infinite series for P > 2, eliminated remainders: (0,2),((P-2),0)
Indivisible Twin Pair Combos:
... ITPCombos(x) = product_(n=1)^(x){P[n]-2} for P[n] = nth Prime
Spoiler :
Conclusion:
... There are infinite Primes and therefore infinite Twin Prime Pairs.
end For Consideration.
is divergent.
