Prime n/2n Proof
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Proof that prime numbers continue to occur along the howling infinite of the number sequence.
The following proof is rather technical, but here is an aid to visualize the core idea. In order to find the prime numbers from a certain number (9) to twice that same number (2*9 = 18) we only need the prime numbers from 2 to 9. Map those primes on the vertical and the number sequence on the horizontal. Mark every multiple of the listed primes (for 2 that's 2 and 2+2 and 2+2+2 etc). When all prime multiples are charted look for marks in every vertical column from 10 to 18. Those columns without any marks are the prime numbers from 10 to 18.
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | |
| 2 | ||||||||||||||||||
| 3 | ||||||||||||||||||
| 5 | ||||||||||||||||||
| 7 | ||||||||||||||||||
| => | => |
Calculating the amount of primes between N/2 and N.
Let a0 be 1.
Let a1 be the value of the first prime number (=2).
Let a2 be the value of the second prime number (=3).
Let ap be the value of prime number p.
Let Ai be the amount of numbers identified to be not primes by the isofactor of i's.
A1 = |N/a0*a1|
A2 = |N/a0*a2| - |N/a1*a2|
A3 = |N/a0*a3| - |N/a1*a3| - |N/a2*a3| + |N/a1*a2*a3|
Etc. up to Ap.
Amount of primes between N/2 and N = N-1-A total.
If this amount is to be zero (that is after the final prime number) then 0 = N-1-A. In this case A = N-1. When the prime numbers become very large, N becomes very large but the a*a-phrase becomes larger faster. The number A will never be large enough to be N-1.
