证明哥德巴赫猜想的原文(英文)并附译后汉语

此文发表在：Advances in Theoretical and Applied Mathematics (A TAM), ISSN 0793-4554, V ol. 7, №4, 2012, pp.417-424
Proving Goldbach’s Conjecture by Two Number Axes’ Positive Half Lines which Reverse from Each Other’s Directions
Zhang Tianshu
Nanhai west oil corporation,China offshore Petroleum,
Zhanjiang city, Guangdong province, P.R.China
Email: tianshu_zhang507@;
Abstract
We know that every positive even number 2n(n≥3) can express in a sum which 3 plus an odd number 2k+1(k≥1) makes. And then, for any odd point 2k+1 (k≥1)at the number axis, if 2k+1 is an odd prime point, of course even number 3+(2k+1) is equal to the sum which odd prime number 2k+1 plus odd prime number 3makes; If 2k+1 is an odd composite point, then let 3<B<2k+1, where B is an odd prime point, and enable line segment B(2k+1) to equal line segment 3C. If C is an odd prime point, then even number 3+(2k+1) is equal to the sum which odd prime number B plus odd prime number C makes. So the proof for Goldbach’s Conjecture is converted to prove there be certainly such an odd prime point B at the number axis’s a line segment which take odd point 3 and odd point 2k+1 as ends, so as to prove the conjecture by such a method indirectly.
Keywords
Number theory, Goldbach’s Conjecture, Even number, Odd prime number, Mathematical induction, Two number axes’ positive half lines which reverse from each other’s direction, OD, PL, CL, and RPL.
Basic Concepts
Goldbach’s conjecture states that every ev en number 2N is a sum of two prime numbers, and every odd number 2N+3 is a sum of three prime numbers, where N≥2.
We shall prove the Goldbach’s conjecture thereinafter by odd points at two number axes’ positive half lines wh ich reverse from each other’s directions and which begin with odd point 3.
First we must understand listed below basic concepts before the proof of the conjecture, in order to apply them in the proof.
Axiom．Each and every even number 2n (n≥3) can express in a sum which 3 plus each odd number 2k+1 (k≥1) makes.
Definition 1．A line segment which takes two odd points as two ends at the number axis’s positive half line which begins with odd point 3 is called an odd distance. “OD” is abbreviated from “odd distance”.
The OD between odd point N and odd point N+2t is written as OD N(N+2t), where N≥3, and t≥1.
A integer which the length of OD between two consecutive odd points expresses is 2.
A length of OD between odd point 3 and each odd point is unique.
Definition 2．An OD between odd point 3 and each odd prime point at the number axis’s positive half line which begins with odd point 3, otherwise called a prime length. “PL” is abbreviated from “prime length”, and “PLS”
denotes the plural of PL.
An integer which each length of from small to large PL expresses be successively 2, 4, 8, 10, 14, 16, 20, 26. . .
Definition 3．An OD between odd point 3 and each odd composite point at the number axis’s positive half line which begins with odd point 3, otherwise called a composite length. “CL” is abbreviated from “composite length”.
An integer which each length of from small to large CL expresses be successively 6, 12, 18, 22, 24, 30, 32. . .
We know that positive integers and positive integers’ points at the number axis’s positive half line are one-to-one correspondence, namely each integer’s point at the number axis’s positive half line represen ts only a positive integer. The value of a positive integer expresses the length of the line segment between point 0 and the positive integer’s point here. When the line segment is longer, it can express in a sum of some shorter line segments; correspondingly the positive integer can also express in a sum of some smaller integers.
Since each and every line segment between two consecutive integer’s points and the line segment between point 0 and point 1 have an identical length, hence when use the length as a unit to measure a line segment between two integer’s points or between point 0 and any integer’s point, the line segment has some such unit length, then the integer which the line segment expresses is exactly some.
Since the proof for the conjecture relate merely to positive integers which are not less than 3, hence we take only the number axis’s positive half line which begins with odd point 3. However we stipulate that an integer which each integer’s point represents expresses yet the length of the line segment between the integer’s point and point 0. For example, an odd prime value which the right end’s point of any PL represents expresses yet the length of the line segment between the odd prime point and point 0 really.
We can prove next three theorems easier according to above-mentioned some relations among line segments, integers’ points and integers.
Theorem 1．If the OD which takes odd point F and odd prime point P S as two ends is equal to a PL, then even number3+F can express in a sum of two odd prime numbers, where F>P S.
Proof．Odd prime point P S represents odd prime number P S, it expresses the length of the line segment from odd prime point P S to point 0.
Though lack the line segment from odd point 3 to point 0 at the number axis’s positive half line which begins with odd point 3, but odd prime point P S represents yet odd prime P S according to above-mentioned stipulation;
Let OD P S F=PL 3P b, odd prime point P b represents odd prime number P b, it expresses the length of the line segment from odd prime point P b to point 0. Since PL 3P b lack the segment from odd point 3 to point 0, therefore the integer which the length of PL 3P b expresses is even number P b-3, namely the integer which the length of OD P S F expresses is even number P b-3. Consequently there is F=P S+(P b-3), i.e. 3+F=odd prime P S + odd prime P b .
Theorem 2．If even number 3+F can express in a sun of two odd prime numbers, then the OD which takes odd point 3 and odd point F as ends can express in a sun of two PLS, where F is an odd number which is more than 3.
Proof．Suppose the two odd prime numbers are P b and P d, then there be 3+F= P b+P d.
It is obvious that there be OD 3F=PL 3P b + OD P b F at the number axis’s positive half line which begins with odd point 3.
Odd prime point P b represents odd prime number P b according to above-mentioned stipulation, then the length of line segment P b(3+F) is precisely P d, nevertheless P d expresses also the length of the line segment from odd prime point P d to point 0. Thereupon cut down 3 unit lengths of line segment P b(3+F), we obtain OD P b F; again cut down 3 unit lengths of the line segment from odd prime point P d to point 0, we obtain PL 3P d, then there be OD P b F=PL 3P d.
Consequently there be OD 3F=PL 3P b + PL 3P d.
Theorem 3．If the OD between odd point F and odd point 3can express in a sum of two PLS, then even number 3+F can express in a sum of two odd prime numbers, where F is an odd number which is more than 3. Proof．Suppose one of the two PLS is PL 3P S, then there be F＞P S,and the OD between odd point F and odd prime point P S is another PL. Consequently even number 3+F can express in a sum of two odd prime numbers according to theorem 1.
The Proof
First let us give ordinal number K to from small to large each and every odd number 2k+1, where k≥1,then from small to large each and every even number which is not less than 6is equal to 3+(2k+1).
We shall prove this conjecture by the mathematical induction thereinafter.
1．When k=1, 2, 3 and 4, we getting even number be orderly 3+(2*1+1)=6=3+3, 3+(2*2+1)=8=3+5, 3+(2*3+1)=10=3+7 and 3+(2*4+1)=12=5+7. This shows that each of them can express in a sum of two odd prime numbers.
2．Suppose k=m, the even number which3 plus №m odd number makes, i.e. 3+(2m+1) can express in a sum of two odd prime numbers, where m≥4.
3．Prove that when k=m+1, the even number which 3 plus №(m+1) odd number makes, i.e. 3+(2m+3) can also express in a sum of two odd prime numbers.
Proof．In case 2m+3 is an odd prime number, naturally even number 3+（2m+3）is the sum of odd prime number 3 plus odd prime number 2m+3 makes.
When 2m+3 is an odd composite number, suppose that the greatest odd prime number which is less than 2m+3 is P m, then the OD between odd prime point P m and odd composite point 2m+3 is either a PL or a CL. When the OD between odd prime point P m and odd composite point 2m+3 is a PL, the even number 3+（2m+3）can express in a sum of two odd prime numbers according to theorem 1.
If the OD between odd prime point P m and odd composite point 2m+3 is a CL, then we need to prove that OD 3（2m+3）can express in a sum of two PLS, on purpose to use the theorem 3.
When OD P m(2m+3) is a CL, from small to large odd composite number 2m+3 be successively 95, 119, 125, 145. . .
First let us adopt two number axes’ positive half lines which reverse from each other’s directions and which begin with odd point 3.
At first, enable end point 3 of either half line to coincide with odd point 2m+1 of another half line. Please, see first illustration:
3 5 7 2m-3 2m+1
2m+1 2m-3 7 5 3
First Illustration
Such a coincident line segment can shorten or elongate, namely end point 3 of either half line can coincide with any odd point of another half line.
This proof will perform at some such coincident line segments. And for certain of odd points at such a coincident line segment, we use usually names which mark at the rightward direction’s half line.
We call PLS which belong both in the leftward direction’s half line and in a coincident line se gment “reverse PLS”. “RPLS” is abbreviated from “reverse PLS”, and “RPL” denotes the singular of RPLS.
The RPLS whereby odd point 2k+1 at the rightward direction’s half line acts as the common right endmost point are written as RPLS2k+1,and RPL2k+1 denotes the singular, where k>1.
This is known that each and every OD at a line segment which takes odd
point 2m+1 and odd point 3 as two ends can express in a sum of a PL and a RPL according to preceding theorem 2 and the supposition of №2 step of the mathematical induction.
We consider a PL and the RPL2k+1 wherewith to express together the length of OD 3(2k+1)as a pair of PLS, where k >1. One of the pair’s PLS is a PL which takes odd point 3 as the left endmost point, and another is a RPL2k+1 which takes odd point 2k+1 as the right endmost point. We consider the RPL2k+1 and another RPL2k+1 which equals the PL as twin RPLS2k+1.
For a pair of PLS, the PL is either unequal or equal to the RPL2k+1. If the PL is unequal to the RPL2k+1, then longer one is more than a half of OD 3(2k+1), yet another is less than the half. If the PL is equal to the RPL2k+1, then either is equal to the half. A pair of PLS has a common end’s point.
Since each of RPLS2k-1 is equal to a RPL2k+1,and their both left endmost points are consecutive odd points, and their both right endmost points are consecutive odd points too. So seriatim leftwards move RPLS2k+1to become RPLS2k-y, then part left endmost points of RPLS2k+1 plus RPLS2k-y coincide monogamously with part odd prime points at OD 3(2k+1), where y=1, 3, 5, ...
Thus let us begin with odd point 2m+1, leftward take seriatim each odd point 2m-y+2 as a common right endmost point of RPLS2m-y+2, where y= 1, 3, 5, 7 ... ỹ ...
Suppose that y increases orderly to odd number ỹ,and∑ part left endmost
points of RPLS2m-y+2(1≤y≤ỹ+2) coincide just right with all odd prime points at OD 3(2m+1) monogamously, then there are altogether(ỹ+3)/2 odd points at OD (2m-ỹ)(2m+1), and let μ=(ỹ+3)/2.
Let us separate seriatim OD 3(2m-y+2) (y=1, 3, 5,…) from each coincident line segment of two such half lines, and arrange them from top to bottom orderly. After that, put an odd prime number which each odd prime point at the rightward direction’s half line expresses to on the odd prime point, and put another odd prime number which each left endmost point of RPLS2m-y+2 at the leftward direction’s half line expresses to beneath the odd prime point.
For example, when 2m+3=95, 2m+1=93, 2m-1=91 and 2m-ỹ =89，μ=3. For the distributer of odd prime points which coincide monogamously with left endmost points of RPLS95, RPLS93, RPLS91, RPLS89 and RPLS87, please see second illustration:
OD 3(95)19 31 37 61 67 79
79 67 61 37 31 19
OD 3(93)7 13 17 23 29 37 43 53 59 67 73 79 83 89
89 83 79 73 67 59 53 43 37 29 23 17 13 7 OD 3(91) 5 11 23 41 47 53 71 83 89
89 83 71 53 47 41 23 11 5 OD 3(89)13 19 31 61 73 79
79 73 61 31 19 13
OD 3(87) 7 11 17 19 23 29 31 37 43 47 53 59 61 67 71 73 79 83
83 79 73 71 67 61 59 53 47 43 37 31 29 23 19 17 11 7
Second Illustration
Two left endmost points of twin RPLS2m-y+2 at OD 3(2m-y+2) coincide monogamously with two odd prime points, they assume always bilateral symmetry whereby the centric point of OD 3(2m-y+2)acts as symmetric
centric. If the centric point is an odd prime point, then it is both the left endmost point of RPL2m-y+2 and the odd prime point which coincides with the left endmost point, e.g. centric point 47 of OD3(91) in above-cited that example.
We consider each odd prime point which coincides with a left endmost point of RPLS2m-ỹ alone as a characteristic odd prime point, at OD 3(2m+1). Thus it can seen, there is at least one characteristic odd prime point at OD 3(2m-ỹ) according to aforesaid the way of making things, e.g. odd prime points 19, 31 and 61 at OD 3(89) in above-cited that example.
Whereas there is not any such characteristic odd prime point in odd prime points which coincide monogamously with left endmost points of RPLS2m+1 plus RPLS2m-1 ... plus RPLS2m-ỹ+2.
In other words, every characteristic odd prime point is not any left endmost point of RPLS2m+1plus RPLS2m-1 ... plus RPLS2k-ỹ+2.
Moreover left endmost points of RPLS2m-y are №1 odd points on the lefts of left endmost points of RPLS2m-y+2monogamously, where y is an odd number≥1.
Consequently, №1 odd point on the left of each and every characteristic odd prime point isn’t a ny left endmost point of RPLS2m-1plus RPLS2m-3…plus RPLS2m-ỹ. — {1}
Since each RPL2m-y+2 is equal to a PL at OD 3(2m-y+2).
In addition, odd prime points which coincide monogamously with left endmost points of RPLS2m+1 plus RPLS2m-1 ... plus RPLS2m-ỹare all odd prime points at OD 3(2m+1).
Hence considering length, at OD 3(2m+1) RPLS whose left endmost points coincide monogamously with all odd prime points are all RPLS at OD 3(2m+1), irrespective of the frequency of RPLS on identical length. Evidently the longest RPL at OD 3(2m+1)is equal to PL 3P m.
When OD P m(2m+3) is a CL, let us review aforesaid the way of making thing once again, namely begin with odd point 2m+1, leftward take seriatim each odd point 2m-y+2 as a common right endmost point of RPLS2m-y+2, and∑ part left endmost points of RPLS2m-y+2(1≤y≤ỹ+2) coincide just right with all odd prime points at OD 3(2m+1) monogamously.
Which one of left endmost points of RPLS2m-y+2(1≤y≤ỹ+2) coincides first with odd prime point 3? Naturally it can only be the left endmost point of the longest RPL P m whereby odd prime point P m acts as the right endmost point.
Besides all coincidences for odd prime points at OD 3(2m+1) begin with left endmost points of RPLS2m+1, whereas left endmost points of RPLS2m-ỹare final one series in the event that all odd prime points at OD 3(2m+1) are coincided just right by left endmost points of RPLS.
Therefore odd point 2m-ỹ as the common right endmost point of RPLS2m-ỹcannot lie on the right of odd prime point P m, then odd point 2m-ỹ-2 can only lie on the left of odd prime point P m . This shows that every RPL2m-ỹ-2 at OD 3(2m-ỹ-2) is shorter than PL 3 P m.
In addition, №1odd point on the left of a left endmost point of each and every RPL2k-ỹ is a left endmost point of RPLS2k-ỹ-2.
Therefore each and every RPL2m-ỹ-2can extend contrary into at least one RPL2m-y+2, where y is a positive odd number ≤ỹ+2.
That is to say, every left endmost point of RPLS2m-ỹ-2 is surely at least one left endmost point of RPLS2k-y+2.
Since left endmost points of RPLS2m-ỹ-2 lie monogamously at №1 odd point on the left of left endmost points of RPLS2m-ỹ including characteristic odd prime points.
Consequently, №1 odd point on the left of each and every characteristic odd prime point is surely a left endmost point of RPLS2m-y+2, where 1≤y≤ỹ+2.—— {2}
So we draw inevitably such a conclusion that №1 odd point on the left of each and every characteristic odd prime point can only be a left endmost point of RPLS2m+1 under these qualifications which satisfy both above-reached conclusion {1}and above-reached conclusion {2}.
Such being the case, let us rightwards move a RPL2m+1 whose left endmost point lies at №1odd point on the left of any characteristic odd prime point to adjacent odd points, then the RPL2m+1 is moved into a RPL2m+3. Evidently the left endmost point of the RPL2m+3is the characteristic odd prime point, and its right endmost point is odd point 2m+3.
So OD 3(2m+3) can express in a sum of two PLS, and the common
endmost point of the two PLS is exactly the characteristic odd prime point. Thus far we have proven that even if OD P m(2m+3) is a CL, likewise OD 3(2m+3) can also express in a sum of two PLS.
Consequently even number which 3 plus №(m+1)odd number makes, i.e. 3+(2m+3) can also express in a sum of two odd prime numbers according to aforementioned theorem 3.
Proceed from a proven conclusion to prove a larger even number for each once, then via infinite many an once, namely let k to equal each and every natural number, we reach exactly a conclusion that every even number 3+(2k+1) can express in a sum of two odd prime numbers, where k≥1.
To wit every even number 2N can express in a sum of two odd prime numbers, where N>2.
In addition let N =2, get 2N=4=even prime number 2+even prime number 2. Consequently every even number 2N can express in a sum of two prime numbers, where N≥2.
Since every odd number 2N+3 can express in a sum which a prime number plus the even number makes, consequently every odd number 2N+3 can express in a sum of three prime numbers, where N≥2.
To sum up, we have proven that two propositions of the Goldbach’s conjecture are tenable, thus Goldbach’s conjecture holds water.
附，翻译成汉语：
利用互为反向数轴的正射线证明哥德巴赫猜想
张天树
Tianshu_zhang507@
摘要
我们知道,依次增大的每一个正偶数2n(n≧3)可以表示成3分别与依次增大的一个奇数2k+1（k≧1）之和.于是,对于数轴上的任意一个奇数点2k+1（k≧1）,如果2k+1是一个奇素数点,当然,偶数3+(2k+1）可等于奇素数2k+1 与奇素数 3 之和;如果2k+1是一个奇合数点,那么,取3<B<2k+1,这里,B是一个奇素数点,且使线段B(2k+1)等于线段3C. 如果C是一个奇素数点,那么, 偶数3+（2k+1）等于奇素数B与奇素数C之和.于是对哥德巴赫猜想的证明就变换成了去证明：在数轴的以奇数点3和2k+1为端点的线段上，总是存在着这样的奇素数点B，以此方法来间接地证明哥德巴赫猜想.
关键词
数论、哥德巴赫猜想、数学归纳法、偶数、奇素数、互为反方向的数轴的正射线、奇距、素长、合长和反向素长.
基本慨念
哥德巴赫猜想表为:每一个偶数2N都是两个素数之和,每一个奇数2N+3都是三个素数之和,这里N≧2.
本文将用互为反方向的两条数轴的从奇数点3开始的正方向射线上的奇数点来证明这个猜想.
在证明这个猜想之前，我们先要熟知下述的基本慨念，以便在证明的过程中应用它们.
公理.依次增大的每一个正偶数2n(n≧3)可以表示成3分别与依次增大的一个奇数2k+1（k≧1）之和.
定义1. 在数轴的从奇数点3开始的正方向射线上，以任意两个奇数点为端点的线段,称为这两个奇数点之间的距离，简称奇距．我们用符号“OD”表示奇距.
奇数点N与奇数点N+2t之间的奇距,写成OD N(N+2t)，这里N≧3,t≧1. 相邻两个奇数点之间的奇距是2，奇数点3与各个奇数点之间的奇距长度,都是唯一的.
定义2.在数轴的从奇数点3开始的正方向射线上，以任意一个奇素数点和奇数点3为端点的奇距，也被称为这个奇素数点的素长，并用符号“PL”表示一条素长，和符号“PLS”表示至少两条素长，即素长的复数.
从小到大的素长依次是: 2、4、8、10、14、16、20、26. . . . . .
定义3.在数轴的从奇数点3开始的正方向射线上，以任意一个奇合数点和奇数点3为端点的奇距，又被称为这个奇合数点的合长，并用符号“C L”表示一条合长.
从小到大的合长依次是: 6、12、18、22、24、30、32. . . . . .
我们知道，在数轴的正方向射线上的整数点与正整数是一一对应的.也就是说，在数轴的正方向射线上，每一个整数点只代表一个整数，这个整数的值在这里表示这个整数点距0点的线段长度. 当这条线段较长时，它可表为若干条线段之和. 因此，这个整数也可表为若干个整数之
和. 且因为0点与1点、以及相邻两个整数点之间的线段长度都是相等的,当我们把这些相等线段每条的长度作为1个长度单位去度量任意一个整数点距0点的线段或任意两个整数点之间的线段长度时,该线段有多少个这样的长度单位，它就代表多大的整数.
因为本文的证明仅仅用到不小于3的整数，所以，我们只取数轴的从奇数点3开始的正方向射线，但我们规定：在这射线上的每个整数点代表的整数值仍然是指这个整数点距0点的线段长度. 例如，在这射线上任意一条素长的右端点所代表的奇素数值，实际上是指这个奇素数点距0点的线段长度.
根据以上所述，我们很容易地证明以下三条定理:
定理1.如果以奇数点F和奇素数点P S 为端点的奇距等于一条素长，那么，偶数3+F可表为两个奇素数之和，这里F＞P S.
证明：奇素数点P S代表奇素数P S，P S的值表示奇素数点P S到0点的长度.在数轴的从奇数点3开始的正方向射线上虽然缺少从奇数点3到0点的一段，但是按照前面的规定，奇素数点P S仍然代表奇素数P S；
令OD P S F=PL 3P b，奇素数点P b代表奇素数P b，P b的值表示奇素数点P b 到0点的长度. 但是，PL 3P b缺少从奇数点3到0点的一段，因此，PL 3P b 的长度表示的整数是偶数P b-3，即OD P S F的长度表示的整数是偶数P b-3. 所以, F=P S+(P b-3)，即3+F=奇素数P S+奇素数P b.
定理2. 如果偶数3+F可表为两个奇素数之和，那么，以奇数点F和奇数点3为端点的奇距可表为两条素长之和，这里F是一个大于3的奇数.
证明：假设这两个奇素数为P b和P d，则有3+F= P b+P d. 显然，在数轴的从奇数点3开始的正方向射线上，OD 3F=PL 3P b + OD P b F. 根据前面
的规定，奇素数点P b代表奇素数P b，那么，线段P b(3+F)的长度就是P d，该长度是指从奇素数点P d到0点的线段长度.在线段P b(3+F)上去掉3个单位长度，得到OD P b F；在从奇素数点P d到0点的线段上去掉3个单位长度，得到PL 3P d，于是，OD P b F=PL 3P d.所以, OD 3F=PL 3P b + PL 3P d.
定理3.如果奇数点F与奇数点3之间的奇距可表为两条素长之和，那么，偶数3+F可表为两个奇素数之和，这里F是一个大于3的奇数.
证明：假设这两条素长中的一条是PL 3P S，那么，F＞P S，且奇数点F与奇素数点P S 之间的奇距是另一条素长. 根据定理1，偶数3+F可表为两个奇素数之和.
证明
首先,让我们以自然数k给每一个依次增大的奇数2k+1编上序号，这里k≧1,那么，每一个不小于6的依次增大的偶数等于3+(2k+1). 在下文中, 我们将运用数学归纳法来证明这个猜想.
1.当k= 1、2、3和4时，我们依次得到的偶数: 3+（2×1+1）=6=3+3，3+（2×2+1）=8=3+5，3+（2×3+1）=10=3+7或5+5，3+（2×4+1）=12=5+7都可表为两个奇素数之和.
2.假设当k=m时，3加上第m个奇数所得的偶数、即3+（2m+1）可表为两个奇素数之和，这里m≧4.
3.证明:当k=m+1时，3加上第m+1个奇数所得的偶数、即3+（2m+3）也可表为两个奇素数之和.
证明如果2m+3是一个奇素数，当然，偶数3+（2m+3）可表为奇素数3与奇素数2m+3之和.
当2m+3是一个奇合数时，假设小于2m+3的最大奇素数为P m，那么，奇合数点2m+3与奇素数点P m之间的奇距或是一条素长，或是一条合长.
当奇合数点2m+3与奇素数点P m之间的奇距是一条素长时，根据定理1，偶数3+（2m+3）可表为两个奇素数之和.
如果奇合数点2m+3与奇素数点P m之间的奇距是一条合长，我们则需要证明OD 3（2m+3）可表为两条素长之和，以便应用定理3.
当OD P m(2m+3)是一条合长时，2m+3从小到大的值依次是95、119、125、145. . . . . .
首先，让我们采用两条互为相反方向的数轴的从奇数点3开始的正方向射线，最初,让一条射线上的奇数点3与另一条射线上的奇数点2m+1重合. 请参看第一图：
3 5 7 2m-3 2m+1
2m+1 2m-3 7 5 3
第一图
这两条射线互相重合的线段能够缩短或伸长，也就是说，一条射线的端点3能够重合另一条射线的任意一个奇数点.本文的整个证明将在这样互相重合、且可伸缩的线段上施行，并且，对这线段上互相重合后的一个奇数点，我们主要使用它在向右方向射线上的名称.
我们把既属于向左方向射线，又在两条射线互相重合的线段上的素长
称为“反向素长”. 一条反向素长用符号“RPL”表示，它的复数表为“RPLS”. 以向右方向射线上的奇数点2k+1作为共同右端点的至少两条反向素长被写作RPLS2k+1,以及RPL2k+1表示其中的一条，这里k＞1.
根据前列的定理2和数学归纳法中第二步的假设，我们知道，在以奇数点2m+1和奇数点3为端点的线段上的各条奇距，都能够表为一条素长与一条反向素长之和. 我们把表示OD 3(2k+1)为两条素长之和的两条素长看作是一对素长，这里k＞1. 在这对素长中，一条是以奇数点3为左端点的素长，另一条则是以奇数点2k+1为右端点的反向素长. 我们把这对素长中的反向素长和在长度上等于这对素长中素长的另一条以奇数点2k+1为右端点的反向素长看作是孪生的反向素长.
对于共表OD 3(2k+1)长度的一对素长，如果它们不等，那么，较长的一条比OD 3（2k+1）的一半长，而另一条则比这一半短. 如果这两条素长相等，那么，每一条都等于OD 3（2k+1）的一半.共表OD 3（2k+1）长度的一对素长，它们有一个共同的端点.
因为在长度上每一条RPL2k-1 等于一条RPL2k+1. 并且，它们的左端点是相邻的奇数点，右端点也是相邻的奇数点. 于是,逐点向左移动RPLS2k+1使之成为RPLS2k-y,那么，RPLS2k+1和RPLS2k-y 的部分左端点就一对一地重合OD 3（2k+1）上的奇素数点，这儿y=1, 3, 5, …...
因此,让我们在数轴的从奇数点3开始的正方向射线上，从奇数点2m+1开始，向左依次取每一个奇数点2m-y+2作为反向素长的共同右端点，这里y=1、3、5、7、...ỹ、...
假设y依次增大到奇数ỹ，且RPLS2m+1、RPLS2m-1 ... RPLS2m-ỹ+2和
RPLS2m-ỹ的部份左端点一对一地恰好重合完OD 3（2m+1）上的全部奇素数点，那么，在OD（2m- ỹ）（2m+1）上共有（ỹ+3）/2个奇数点，并且令μ =（ỹ+3）/2.
然后，我们按照从长到短的次序，从两条射线的重合段中依次分离出OD 3（2m-y+2），并把它们从上到下依次排列起来，还把向右方向射线上表示奇素数点的奇素数放在这个奇素数点的上方，把重合这奇素数点的表示反向素长左端点的奇素数放在它的下方.
例如:当2m+3=95时，2m+1=93, 2m-1=91和2m-ỹ=89，μ=3. 对于OD 3(95)、OD 3(93)、OD 3(91) 、OD 3(89) 和OD 3(87)上反向素长左端点重合奇素数点的分布情况，如下图：
OD 5(95)19 31 37 61 67 79
79 67 61 37 31 19
OD 5(93) 7 13 17 23 29 37 43 53 59 67 73 79 83 89
89 83 79 73 67 59 53 43 37 29 23 17 13 7
OD 5(91)5 11 23 41 47 53 71 83 89
89 83 71 53 47 41 23 11 5
OD 5(89) 13 19 31 61 73 79
79 73 61 31 19 13
OD 5(87) 7 11 17 19 23 29 31 37 43 47 53 59 61 67 71 73 79 83
83 79 73 71 67 61 59 53 47 43 37 31 29 23 19 17 11 7
第二图
在OD 3（2m-y+2）上，孪生反向素长的两个左端点一对一重合的两个奇素数点，以该奇距的中点为对称中心左右对称. 如果该奇距的中点是一个奇素数点，那么,它既是反向素长的左端点，又是这个左端点重合的奇素数点，例如，在上面引用例子中，OD 3（91）的中点47.
在OD 3（2m+1）上，我们仅仅把唯一由RPLS2m-ỹ的左端点重合的奇素数点，即在RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ+2的左端点一一重合的
奇素数点中没有的奇素数点看作是特有的奇素数点. 那么，根据上述的作法，在RPLS2m-ỹ的左端点一一重合的奇素数点中，有至少一个特有的奇素数点.例如，上面所举那个例子中，在OD 3(89)上的奇素数点19、31和61.
因为在RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ+2的左端点一一重合的奇素数点中，没有一个特有的奇素数点. 换言之，每一个特有的奇素数点都不是RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ+2的左端点.
又RPLS2m-y+2的左端点左边的第一个奇数点是RPLS2m-y的左端点，这里y是≧1的正奇数.
所以,在每一个特有奇素数点左边的第一个奇数点不是RPLS2m-1、RPLS2m-3、...和RPLS2m-ỹ的左端点.--- {1}
因为在OD 3(2m-y+2)上，每一条反向素长等于一条素长. 加之，RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ的左端点一一重合的奇素数点是OD 3（2m+1）上全部的奇素数点.
因此，仅以长度而言，在OD 3(2m+1)上，其左端点一对一地重合完全部奇素数点的反向素长是OD 3(2m+1)上全部的反向素长，而不与具有同一长度的反向素长的条数有关.
显然，在OD 3(2m+1)上的全部反向素长中，最长的反向素长等于PL 3P m.
当奇合数点2m+3与奇素数点P m之间的奇距是一条合长时，让我们回顾前述的作法，即从2m+1开始，向左依次取每一个奇数点作为反向素长的共同右端点，并且，RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ的左端点一对一重合的奇素数点是OD 3（2m+1）上全部的奇素数点.
究竟哪一条反向素长的左端点首先重合奇素数点3呢？显然，它仅仅能够是以奇素数点P m为右端点的最长的反向素长的左端点.
又从RPLS2m+1的左端点开始，RPLS2m-ỹ的左端点是重合完OD 3（2m+1）上全部奇素数点的最后一组.
所以，作为RPLS2m-ỹ的共同右端点的奇数点2m-ỹ不能位于奇素数点P m的右边，于是，奇数点2m-ỹ-2位于奇素数点P m的左边. 由此可知，在OD 3（2m- ỹ-2）上的每一条反向素长都比PL 3P m短.
又因为在RPLS2m-ỹ的左端点左边的第一个奇数点全部是RPLS2m-ỹ-2的左端点.
所以，每一条RPL2m-ỹ-2都能反向延长成为至少一条RPL2m-y+2. 那就是说，RPLS2m-ỹ-2的每个左端点都一定是RPLS2m-y+2的至少一条的左端点，这里y是≦ỹ+2的正奇数.
因为RPLS2m-ỹ-2的左端点一对一地全部在RPLS2m-ỹ左端点（包括特有奇素数点）左边的第一个奇数点上，于是，在特有奇素数点左边的第一个奇数点一定是RPLS2m-y+2的左端点，这里y是≦ỹ+2的正奇数. ---{2}
所以,在既要满足结论{1}、又要满足结论{2}的情况下，我们只能够得出: 在特有奇素数点左边的第一个奇数点是RPLS2m+1的左端点.
既然如此，让我们向右移动任意一条这样的PLS2m+1到相邻的奇数点，那么,这条PLS2m+1被移动成了一条PLS2m+3. 显然，这条PLS2m+3的左端点重合这个特有奇素数点，而它的右端点是奇数点2m+3.
这样，OD 3（2m+3）可以被表为两条素长之和，这两条素长的共同端点就是这个特有奇素数点.
到此为止，我们已经证明了: 当OD P m（2m+3）是一条合长时，OD 3（2m+3）也能表为两条素长之和.
那么，根据定理3，3加上第m+1个奇数所得的偶数,即3+(2m+3)可以表示成两个奇素数之和.
每次都从已证得的结论出发，就可以推出:当k等于每一个自然数时，每一个偶数3+(2k+1)都可以表示成两个奇素数之和. 即当N>2时，每一个偶数2N都可以表示成两个奇素数之和.
又当N=2时,有2N=4=偶素数2+偶素数2.
所以,每一个偶数2N都可以表示成两个素数之和，这里N≧2.
因为每一个奇数2N+3等于偶数2N加上奇素数3，因此，每一个奇数2N+3都可以表示成三个素数之和，这里N≧2.
综上所述，我们已经证明了哥德巴赫猜想的两个命题都是站得住脚的，因此，哥德巴赫猜想成立.。