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FYI: Proofs of Goldbach Conjecture and Riemann Hypothesis
PROOF OF GOLDBACH CONJECTURE:
Goldbach Conjecture (GC) states every positive even integer is the sum of two prime numbers. (We count one as prime in the sense of additive number theory outside of the FTA.)
Proof of Goldbach Conjecture:
Suppose there exists a positive even integer, e > 4, that is not the sum of two odd prime numbers or 1. e ≠ p + q
over S = {all odd prime numbers less than e} and where k = card(S) = π(e).
Therefore,
e ≠ p + q over S, (p,q є S) , implies the following system of equations over S,
1 = e - n1 * q1, 3 = e - n2 * q2, ..., pk = e - nk * qk,
according to the Fundamental Theorem of Arithmetic where 1 < qj ≤ (nj * qj)^.5 ≤ nj for 1 ≤ j ≤ k
where pj, qj є S and nj is a positive integer.
Note: If qj = 1, then nj є S, or nj is an odd prime less than e.
Therefore,
2. Probability(e ≠ p + q over S) = ∏(j=1 to k→ [tex]\infty[/tex])of [Probability (qj ≠ 1 | e - pj = nj * qj over S) * Probability (e - pj = nj * qj over S)]
= ∏(j=1 to k→ [tex]\infty[/tex]) of (π( (nj * qj)^.5) - 1)/π( (nj * qj)^.5) → 0.
(This is an increasingly fast convergence for this almost everywhere monotonic non-increasing expression. This implies that the expected value of
e ≠ p + q over S is practically zero, or E[e ≠ p + q over S] = e * Probability(e ≠ p + q over S) ≈ 0 for all e ≥ 100.)
Note: Probability (e - pj = nj * qj over S) = 1 for 1 ≤ j ≤ k.
In addition, empirical evidence has confirmed the validity of the conjecture for all positive even integers up to at least an order of 10^18. Therefore, we conclude the conjecture is true.
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PROOF OF RIEMANN HYPOTHESIS:
The Riemann Hypothesis states that all non-trivial zeta zeros of the Riemann zeta function or ζ(s) = Σ(k=1 to [tex]\infty[/tex]) of 1/k^s) have a real part equal to one-half.
Fact I: The real part of all non-trivial zeros of the Riemann zeta function are located in the critical strip, [0, 1], according to a Riemann Theorem.
Fact II: There are infinitely many non-trivial zeros of the Riemann zeta function whose real part equals one-half according to a Hardy Theorem.
Fact III: The sum of the complex conjugate pairs, s and s', respectively, of non-trivial zeros,
s = a + bi and s' = c + di (Note: a = c and b = -d)
where
ζ(s) = Σ(k=1 to [tex]\infty[/tex]) of 1/k^s = 0 and ζ(s') = Σ(k=1 to [tex]\infty[/tex]) of 1/k^s' = 0,
of the Riemann zeta function equals one according to the Fundamental Theorem of Arithmetic and the Harmonic Series (H):
Note: Euler and others have proven that there exists an infinitely many primes in H. And the divergence of H is the key reason for that result.
H = Σ(k=1 to [tex]\infty[/tex]) of 1/k = Σ(k=1 to [tex]\infty[/tex]) of (1/k^s)(1/k^s') = Σ(k=1 to [tex]\infty[/tex]) of 1/k^(s+s') = [tex]\infty[/tex].
Therefore, according to Facts I, II, and III, we have the following properties for all non-trivial zeros of the Riemann zeta function:
s + s' = 1 which implies a + c = 1 and b + d = 0 such that 0 ≤ a ≤ 1 and 0 ≤ c ≤ 1.
And of course, k^s = k^(a + bi) implies k^(s-bi) = k^a, and k^s' = k^(c + di) implies
k^(s-di) = k^c.
Fact IV: For all k > 1, k is a positive integer, there exists a prime number, p, so that p|k such that p = k or p ≤ k^(1/2).
Therefore, according to Facts I, II, III, and IV, we have:
k^(1/2) ≤ k^a ≤ k, k^(1/2) ≤ k^c ≤ k, and a + c = 1.
Hence, k^a = k^c = k^(1/2) which implies a = c = 1/2 which is the exponent of k^(1/2).
Thus, Riemann Hypothesis is true! Riemann was right!
Note the Importance of the Harmonic Series (H) with regards to prime numbers:
H = Σ(k=1 to [tex]\infty[/tex]) of 1/k = 2Σ(k=1 to [tex]\infty[/tex]) of [1/(2k) = 1/(q - p)] = [tex]\infty[/tex]
where q > p, and p,q are odd prime numbers.
H1 = 2( 1/(5-3) + 1/(11-7) + ...) = [tex]\infty[/tex];
There is a infinite set, P1, of primes, p, generated from H1, and there is a infinite set, Q1, of primes, q,
generated from H1.
H2 = 2( 1/(7-5) + 1/(17-13) + ...) = [tex]\infty[/tex];
There is a infinite set, Q2, of primes, q, generated from H2.
H3 =2( 1/(13-11) + 1/(23-19) + ...) = [tex]\infty[/tex];
There is a infinite set, Q3, of primes, q, generated from H3.
...
There is a infinite set (Q∞) of primes, q, generated from H∞.
H∞ = 2(...) =[tex]\infty[/tex];
Therefore, the infinite set of all odd primes or ℙ\{2} = P1 ∪ Q1 ∪ Q2 ∪ Q3 ∪ ... ∪ Q∞.
Note: If we accept one as prime in the sense of additive number theory outside of FTA, this inclusion of 1 as a prime will change H1 through H∞, slightly. For example,
H1 = 2( 1/(3-1) + 1/(11-7) + ...) = [tex]\infty[/tex], and
H2 = 2( 1/(5-3) + 1/(17-13) + ...) = [tex]\infty[/tex], ...,H∞ = 2(...) = [tex]\infty[/tex].
Thank Lord GOD!
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David Cole
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