Download Proof of Goldbach hypothesis

Each even number 2x≥2 can be expressed with the sum of two primes
Zhang xi-wen
(CHINA)
Abstract
By the method of mathematical expectation to discuss P2x(1,1) and its infimum, here we
show 2x≡p1﹢p2 , (x≥1).
Key words: Goldbach hypothesis [1], prime, infimum.
MSC(2010)11-xx.
Theorem 1. The distribution of prime numbers in natural numbers:
(x／㏒ x)·1＜π(x)≤(x／㏒ x) ㏒ ymax , (x≥a).
(1)
π(x)／x
Proof. If y﹦x
, then π(x)﹦(x／㏒ x) ㏒ y ,
∵ lim π(x)／x ﹦lim 1／㏒ x, (x→∞).
[2]
π(x)／x
1／㏒ x
We have lim x
﹦lim x
,
(x→∞).
1／㏒ x
π(x)／x
∵ x
﹦e . ∴ lim x
﹦e ﹦ymin ,（x→∞). ∴ ㏒ ymin﹦1.
When x≥a, we have ymin＜y≤ymax .
∴ ⑴ is obtained.
Theorem 1 is proved.
Theorem 2. P2x(1,1)≥[((2x﹣1)／㏒(2x﹣1)﹣((x﹣1)／㏒(x﹣1))㏒ ymax )(x／㏒ x)／
((x﹢1)／2)]﹢1﹦[k(x)]﹢1, (a≤x ﹦2n﹣1).
(2)
P2x(1,1)≥[((2x﹣1)／㏒(2x﹣1)﹣(x／㏒ x)㏒ ymax )((x﹣1)／㏒(x﹣1))／
(x／2)]﹢1﹦[f(x)]﹢1,
(a≤x ﹦2n).
⑶
Proof. Let P2x(1,1) be the number of primes p1 or p2, (2＜p1≤p2) which suit
2x﹦p1﹢p2, when the natural number x＞2 is given.
∵ 2＜p1≤p2 , 4＜2p1≤p1﹢p2 , ∴ 2＜p1≤x.
P2x(1,1) ﹦∑(π(p2)﹣π(p2﹣1)),
(2＜p1≤p2 ﹦2x﹣p1).
﹦∑(π(2x﹣p1)﹣π(2x﹣p1﹣1)),
(2＜p1≤x).
⑷
①．If x﹦2n﹣1, writing down the folding expression of sequence of odd
number as the following: （the sum of two meeting numbers﹦2x）
︱x ,x﹢2,… … …,2x﹣3,2x﹣1︱
︱x ,x﹣2,… … …, 3 , 1 ︱
The upper row contains p2, the lower row contains p1.
The number of odd number between both vertical lines﹦(x﹢1)／2.
The mean of meeting number of p1,p2:
EP2x(1,1)﹦(π(2x﹣1)﹣π(x﹣1))(π(x))／((x﹢1)／2).
⑸
By ⑴, transforming ⑸ into the greatest lower bound of meeting number
of p1,p2 . ⑵ is obtained.
②. If x﹦2n, writing down the folding expression of sequence of odd number
as the following：
（the sum of two meeting numbers﹦2x）
︱x﹢1,x﹢3,… … …,2x﹣3,2x﹣1︱
︱x﹣1,x﹣3,… … …, 3 , 1 ︱
The upper row contains p2, the lower row contains p1.
The number of odd number between both vertical lines﹦x／2.
The mean of meeting number of p1,p2:
EP2x(1,1)﹦(π(2x﹣1)﹣π(x))(π(x﹣1))／(x／2).
⑹
By ⑴, transforming ⑹ into the greatest lower bound of meeting number
of p1,p2 . ⑶ is obtained.
1
Theorem 2 is proved.
Theorem 3. The infimum of P2x(1,1) exists. *
P2x(1,1)≥[((2x﹣1)／㏒(2x﹣1)﹣((x﹣1)／㏒(x﹣1))λ)(x／㏒ x)／((x
﹢1)／2)]﹢1﹦[k(x)]﹢1≥1, (5≤x﹦2n﹣1 or 2n＜∞).⑺
30／113
Proof. Ⅰ. If π⑴﹦0，then x≥a﹦11, ㏒ ymax﹦㏒ 113
﹦μ.
P2x(1,1)≥[((2x﹣1)／㏒(2x﹣1)﹣((x﹣1)／㏒(x﹣1))μ)(x／㏒ x)／((x
﹢1)／2)]﹢1﹦[k(x)]﹢1,
(11≤x ﹦2n﹣1＜∞).
(8)
P2x(1,1)≥[((2x﹣1)／㏒(2x﹣1)﹣(x／㏒ x)μ)((x﹣1)／㏒(x﹣1))／((x
／2)]﹢1﹦[f(x)]﹢1,
(12≤x ﹦2n＜∞).
(9)
From (8), When x﹦19, P2x(1,1)﹦2, [k(x)]﹢1﹦2,
When x﹦199, P2x(1,1)﹦7, [k(x)]﹢1﹦8,
From (9),When x﹦34, P2x(1,1)﹦2, [k(x)]﹢1﹦3
When x﹦64, P2x(1,1)﹦3, [k(x)]﹢1﹦4
When x﹦166, P2x(1,1)﹦6, [k(x)]﹢1﹦7
When x﹦316, P2x(1,1)﹦10, [k(x)]﹢1﹦11,
When x﹦496, P2x(1,1)﹦13, [k(x)]﹢1﹦14,
When x﹦1336, P2x(1,1)﹦28, [k(x)]﹢1﹦30,
30／113
㏒ ymax﹦㏒ 113
is useless, as everyone knows,
it is a man-made error, π⑴﹦0 is wrong.
We must repeal them as a heap of rubbish.
9／19
Ⅱ. If π⑴﹦1，then x≥a﹦2, ㏒ ymax﹦㏒ 19 ﹦λ.
P2x(1,1)≥[((2x﹣1)／㏒(2x﹣1)﹣((x﹣1)／㏒(x﹣1))λ)(x／㏒ x)／((x
﹢1)／2)]﹢1﹦[k(x)]﹢1,
(7≤x ﹦2n﹣1＜∞). (10)
P2x(1,1)≥[((2x﹣1)／㏒(2x﹣1)﹣(x／㏒ x)λ)((x﹣1)／㏒(x﹣1))／((x
／2)]﹢1﹦[f(x)]﹢1,
(8≤x ﹦2n＜∞).
(11)
From (10), When x﹦19, P2x(1,1)﹦2, [k(x)]﹢1﹦2,
When x﹦199, P2x(1,1)﹦7, [k(x)]﹢1﹦6,
From (11), When x﹦34, P2x(1,1)﹦2, [k(x)]﹢1﹦2,
When x﹦64, P2x(1,1)﹦3, [k(x)]﹢1﹦3,
When x﹦166, P2x(1,1)﹦6, [k(x)]﹢1﹦5,
When x﹦316, P2x(1,1)﹦10, [k(x)]﹢1﹦8,
When x﹦496, P2x(1,1)﹦13, [k(x)]﹢1﹦11,
When x﹦1336, P2x(1,1)﹦28, [k(x)]﹢1﹦23,
9／19
㏒ ymax﹦㏒ 19
is useful, π⑴﹦1 is right, mathematicians all over the
world must support it for science and for human.
The characteristics of the infimum of P2x(1,1):
ⅰ. Uniformly continuous.
k(x),f(x) are elementary function,theirs interval of definition [7,x﹦2n
﹣1],[8, x﹦2n] are closed, thus, k(x),f(x) are uniformly continuous. [3]
∵ When n≥4, [k(x)]≥[f(x)]≥0,
Deleting the lower value function [f(x)].
P2x(1,1)≥[((2x﹣1)／㏒(2x﹣1)﹣((x﹣1)／㏒(x﹣1))λ)(x／㏒ x)／((x
﹢1)／2)]﹢1﹦[k(x)]﹢1,
(5≤x﹦2n﹣1 or 2n＜∞).
When x﹦6, P2x(1,1)﹦[k(x)]﹢1﹦1, critical point.
2
When x﹦19, P2x(1,1)﹦[k(x)]﹢1﹦2, critical point.
When x﹦34, P2x(1,1)﹦[k(x)]﹢1﹦2, critical point.
When x﹦64, P2x(1,1)﹦[k(x)]﹢1﹦3, critical point.
ⅱ. Monotone increasing.
Differentiating the function k(x):
k′(x) ﹦(A(BC﹢DE)﹣BD)F，（5≤x＜∞).
A﹦x﹢1＞0，
B﹦(2x﹣1)／㏒(2x﹣1)﹣(x﹣1)λ／㏒(x﹣1)＞0，
2
C﹦(㏒ x﹣1)／(㏒ x) ＞0，
D﹦x／㏒ x＞0，
2
2
E﹦2(㏒(2x﹣1)﹣1)／(㏒(2x﹣1)) ﹣(㏒(x﹣1)﹣1)λ／(㏒(x﹣1)) ＞0，
2
F﹦2／(x﹢1) ＞0, (5≤x＜∞).
2
﹣2
﹣2
When x≥5, A(BC﹢DE)﹣BD＞F, (x (㏒ x) ＞x ); F＞0, (5≤x＜∞).
∴ k′(x)＞0, (5≤x＜∞). k(x) is monotone increasing in [5,x].
P2x(1,1)≥[k(x)]≥1,（5≤x＜∞）.
∴ ⑺ is obtained.
Theorem 3 is proved.
Theorem 4. Each even number 2x≥2 can be expressed with the sum of two primes.
Proof. From Theorem 3 , P2x(1,1)≥1, (5≤x＜∞).
∵ 1 is a prime number,
x﹦4, 2x﹦8﹦1﹢7﹦3﹢5,
x﹦3, 2x﹦6﹦1﹢5﹦3﹢3,
x﹦2, 2x﹦4﹦1﹢3,
x﹦1, 2x﹦2﹦1﹢1,
∴
P2x(1,1)≥1, (1≤x≤4).
∴
P2x(1,1)≥1, (1≤x＜∞).
Theorem 4 is proved.
*
The supremum of P2x(1,1) exists.
P2x(1,1)≤π(2x﹣1)﹣π(x﹣1), (x﹦2n﹣1).
P2x(1,1)≤π(2x﹣1)﹣π(x),
(x﹦2n).
Reference literature
[1]. Goldbach, on 7 June 1742, the mathematician Christian Goldbach wrote a letter
to Leonhard Euler in which he proposed the following conjecture:
Every even number which can be written as the sum of two primes, can also be written as
the sum of as many primes as one wishes, until all terms are 1.
e.g.
4﹦1﹢3
6﹦1﹢5
﹦1﹢1﹢2
﹦1﹢2﹢3
﹦1﹢1﹢1﹢1.
﹦1﹢1﹢1﹢3
﹦1﹢1﹢1﹢1﹢2
﹦1﹢1﹢1﹢1﹢1﹢1.
He considered 1 to be a prime number, a convention subsequently abandoned.
But this conventional abandon is an aberration obviously.
[2]. Hadamard ＆ De La Vall′ee Poussin, Prime number Theorem. 1896.
[3]. Cantor, Cantor Theorem about uniformly continuous. 1872.
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