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Keywords: Prime Counting Function is π[*]; P\{2}, the set of all odd prime numbers, N is the set of all positive integers; Euclid's Theorem; and the Fundamental Theorem of Arithmetic (FTA); and sqrt[*] is the square root function; and card[A] = |A| is the number of elements in a set A; ∩[*] is the intersection function; and ∏ [*] is the product function; NextPrime[ x ] is the function which generates the nearest prime number greater than x where x>0, and min[*] is t'he minimum function.
Proof of Polignac's Conjecture:
Let us assume there exists a prime gap of size 2i = 2 * i for some positive integer, i, that does not occur infinitely many times in the natural sequence of odd prime numbers. And if we let
S = { s ∈ P\{2} | s + 2i = NextPrime[s] for some i ∈ N}, then 0 ≤ card[S] = |S| < ∞.
Now, if we construct an exceptional set, E, which has a strictly monotonic increasing sequence such that:
E = { e ∈ P\{2} | e + 2i < NextPrime[e] for some i ∈ N where min[e] ≥ π[e] * i / b for some b ≥ 1 },
then card[E] = |E| = ∞ according to Euclid's Theorem which states that there are infinitely many prime numbers, and according to the fact that prime numbers generate the positive even integers so efficiently that gaps between two consecutive prime numbers increase without bound.
Thus, e ≡ -2i (mod r) over E where r ∈ P\{2} such that 1 < r ≤ sqrt[m* r] ≤ m ∈ N.
From e ≡ -2i (mod r), we generate a system of infinite equations over E according to FTA:
e1 + 2i = m1* r1; e2 + 2i = m2* r2; e3 + 2i = m3* r3; ..., where e1 < e2 < e3 < ... such that 1 < rk ≤ sqrt[ mk * rk ] ≤ mk ∈ N for all k ∈ N.
Note: If rk = 1, then mk =NextPrime[ek]. And this result is a contradiction!
So, the Probability[ the prime gap, 2i, does not occur infinitely many times in the natural sequence of odd prime numbers ]
= Probability[ e ≡ -2i (mod r) over E] = Prob[ e ≡ -2i (mod r) over E]
= Prob[ ∩[ from k = 1 to ∞ of rk ≠ 1] ] = ∏ [ from k = 1 to ∞ of Prob[ rk ≠ 1] ]
= ∏ [ from k = 1 to ∞ of Prob[ rk ≠ 1 | ek + 2i = mk * rk ] * Prob[ ek + 2i = mk * rk ] ]
= ∏ [ from k = 1 to ∞ of Prob[ rk ≠1 | ek + 2i = mk * rk ] ]
since Prob[ ek + 2i = mk * rk ] = 1 for all k ∈ N.
∏ [ from k = 1 to ∞ of Prob[ rk ≠ 1 | ek + 2i = mk * rk ] ]
= ∏ [ from k = 1 to ∞ of ( π[ sqrt[ mk * rk] ] − 1 ) / π[ sqrt[ mk * rk ] ] ].
Note: 0 < ( π[ sqrt[ mk * rk ] ] − 1) / π[ sqrt[ mk * rk] ] < 1 for all k ∈ N.
Note: The possibility of rk = 1 cannot be ignored.
Thus, 1. ∏ [ from k = 1 to ∞ of ( π[ sqrt[ mk * rk ] ] − 1 ) / π[ sqrt [mk * rk ] ] ] = 0.
This is a contradiction! And thus, the Polignac's Conjecture is true. Q.E.D.
Note:
Why is equation 1:
∏ [ from k = 1 to ∞ of ( π[ sqrt[ mk * rk ] ] − 1 ) / π[ sqrt [mk * rk ] ] ] = 0?
Big Hint: Given an open interval, ( p, q ) where q = NextPrime[ p ] for some odd prime prime, p, in
the natural sequence of odd primes, the metric d(p, q) = q - p is arbitrarily long according to PNT
(Prime Number Theorem) or according to PPL (Prime Parity Law).
So, for all integers, jk ∈ (p, q), with integral index, k,
π[ j1] = π[ j2 ] = ... = π[ jm] where p < j1 < j2 < ... < jm < q, and π[*] is the prime-counting function.
I hope this hint helps you. удачи!
*Note: We do not count the prime, 2 (it's even!), and count the controversial prime, 1, in our prime count for the left side of equation one.
*****An Important Observation from the Proof of Polignac's Conjecture*****
Recall:
E = { e ∈ P\{2} | e + 2i < NextPrime[e] for some i ∈ N where e = π[e] * i / b for some b ≥ 1}.
e = π[e] * i / b implies more generally:
∑[(π[e] / e) *(1/b) over all primes e] = ∑ [ 1 / i over all integers i ≥ 1 ] = [tex]\infty[/tex].
*****
P.S. Cheers and Happy Birthday (on September 17)! Georg Friedrich Bernhard Riemann!
Thank you!
(https://en.wikipedia.org/wiki/Bernhard_Riemann).
And of course, we thank and cheer Gauss and Euler too!
They got so much, so right, so early, and their great works shall last an eternity. Thank Lord GOD!
*****
http://biblia.com/verseoftheday/image/Ro8.28
