1、1此文发表在：Advances in Theoretical and Applied Mathematics (ATAM), ISSN 0793-4554, Vol. 7, 4, 2012, pp.417-424 Proving Goldbachs Conjecture by Two Number Axes Positive Half Lines which Reverse from Each Others DirectionsZhang TianshuNanhai west oil corporation,China offshore Petroleum, Zhanjiang city, G
2、uangdong province, P.R.China Email: tianshu_;AbstractWe know that every positive even number 2n(n3) can express in a sum which 3 plus an odd number 2k+1(k1) makes. And then, for any odd point 2k+1 (k1)at the number axis, if 2k+1 is an odd prime point, of course even number 3+(2k+1) is equal to the s
3、um which odd prime number 2k+1 plus odd prime number 3 makes; If 2k+1 is an odd composite point, then let 3PS. ProofOdd prime point PS represents odd prime number PS, it expresses the length of the line segment from odd prime point PS to point 0. Though lack the line segment from odd point 3 to poin
4、t 0 at the number axiss positive half line which begins with odd point 3, but odd prime point PS represents yet odd prime PS according to above-mentioned stipulation;Let OD PSF=PL 3Pb, odd prime point Pb represents odd prime number Pb, it expresses the length of the line segment from odd prime point
5、 Pb to point 0. Since PL 3Pb lack the segment from odd point 3 to point 0, therefore the integer which the length of PL 3Pb expresses is even number Pb-3, namely the integer which the length of OD PSF expresses is even number Pb-3. Consequently there is F=PS+(Pb-3), i.e. 3+F=odd prime PS + odd prime
6、 Pb . 5Theorem 2If 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.ProofSuppose the two odd prime numbers are Pb and Pd, then there be 3+F= Pb+Pd.
7、 It is obvious that there be OD 3F=PL 3Pb + OD PbF at the number axiss positive half line which begins with odd point 3.Odd prime point Pb represents odd prime number Pb according to above-mentioned stipulation, then the length of line segment Pb(3+F) is precisely Pd, nevertheless Pd expresses also
8、the length of the line segment from odd prime point Pd to point 0. Thereupon cut down 3 unit lengths of line segment Pb(3+F), we obtain OD PbF; again cut down 3 unit lengths of the line segment from odd prime point Pd to point 0, we obtain PL 3Pd, then there be OD PbF=PL 3Pd. Consequently there be O
9、D 3F=PL 3Pb + PL 3Pd.Theorem 3If the OD between odd point F and odd point 3 can 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. ProofSuppose one of the two PLS is PL 3PS, then there be FP S, and the OD be
10、tween odd point F and odd prime point PS is another PL. Consequently even number 3+F can express in a sum of two odd prime numbers according to theorem 1.6The ProofFirst let us give ordinal number K to from small to large each and every odd number 2k+1, where k1, then from small to large each and ev
11、ery even number which is not less than 6 is equal to 3+(2k+1). We shall prove this conjecture by the mathematical induction thereinafter.1When 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
12、express in a sum of two odd prime numbers. 2Suppose k=m, the even number which 3 plus m odd number makes, i.e. 3+(2m+1) can express in a sum of two odd prime numbers, where m4. 3Prove 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
13、 odd prime numbers.ProofIn 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 Pm, then the OD between odd prime
14、point Pm and odd composite point 2m+3 is either a PL or a CL. When the OD between odd prime point Pm and odd composite point 2m+3 is a PL, the even number 3+（2m+3）can express in a sum of two odd 7prime numbers according to theorem 1. If the OD between odd prime point Pm and odd composite point 2m+3
15、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 Pm(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
16、 others 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 IllustrationSuch a coincident line segment can shorten or elongate, namely
17、 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 directions half line. We call PLS wh
18、ich belong both in the leftward directions half line and in a coincident line segment “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 directions half line acts as the common right endmost point are writ
19、ten as RPLS2k+1, and RPL2k+1 denotes the singular, where k1. 8This 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 induct
20、ion. 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 pairs 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 RP
21、L2k+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 eith
22、er is equal to the half. A pair of PLS has a common ends 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+1 to become RPLS2k-y, the
23、n 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 . . 9Suppose
24、that y increases orderly to odd number , and part left endmost points of RPLS2m-y+2 (1y+2) coincide just right with all odd prime points at OD 3(2m+1) monogamously, then there are altogether (+3)/2 odd points at OD (2m-)(2m+1), and let =(+3)/2. Let us separate seriatim OD 3(2m-y+2) (y=1, 3, 5,) from
25、 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 directions half line expresses to on the odd prime point, and put another odd prime number which each left endmost point
26、of RPLS2m-y+2 at the leftward directions half line expresses to beneath the odd prime point.For example, when 2m+3=95, 2m+1=93, 2m-1=91 and 2m- =89，=3.For the distributer of odd prime points which coincide monogamously with left endmost points of RPLS95, RPLS93, RPLS91, RPLS89 and RPLS87, please see
27、 second illustration: OD 3(95) 19 31 37 61 67 79 79 67 61 37 31 19OD 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 7OD 3(91) 5 11 23 41 47 53 71 83 8989 83 71 53 47 41 23 11 5OD 3(89) 13 19 31 61 73 79 79 73 61 31 19 13OD 3(87) 7 11 17 19 23 29 31 37 43 47 53
28、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 Illustration10Two 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
29、 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 OD 3(91) in above-cited that example.We consider each odd prime point which coincides with a left en
30、dmost point of RPLS2m- 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-) 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 t
31、here 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-+2. In other words, every characteristic odd prime point is not any left endmost point of RPLS2m+1 plus RPLS2m-1 . plus RPLS2k-+2. More
32、over left endmost points of RPLS2m-y are 1 odd points on the lefts of left endmost points of RPLS2m-y+2 monogamously, where y is an odd number1.Consequently, 1 odd point on the left of each and every characteristic 11odd prime point isnt any left endmost point of RPLS2m-1 plus RPLS2m-3 plus RPLS2m-.
33、 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- are all odd prime points at OD 3(2m+1). Hence considering length, at OD 3(2m+1) RPLS whose left endmost points coincid
34、e 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 3Pm. When OD Pm(2m+3) is a CL, let us review aforesaid the way of making thing once again, namely begin with odd poin
35、t 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 (1y+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 (1y+2) coincides first with o
36、dd prime point 3? Naturally it can only be the left endmost point of the longest RPLPm whereby odd prime point Pm 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- are final o
37、ne series in the event that all odd prime points at OD 3(2m+1) are 12coincided just right by left endmost points of RPLS. Therefore odd point 2m- as the common right endmost point of RPLS2m- cannot lie on the right of odd prime point Pm, then odd point 2m-2 can only lie on the left of odd prime poin
38、t Pm . This shows that every RPL2m-2 at OD 3(2m-2) is shorter than PL 3 Pm. In addition, 1 odd point on the left of a left endmost point of each and every RPL2k- is a left endmost point of RPLS2k-2.Therefore each and every RPL2m-2 can extend contrary into at least one RPL2m-y+2, where y is a positiv
39、e odd number +2. That is to say, every left endmost point of RPLS2m-2 is surely at least one left endmost point of RPLS2k-y+2. Since left endmost points of RPLS2m-2 lie monogamously at 1 odd point on the left of left endmost points of RPLS2m- including characteristic odd prime points. Consequently,
40、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 1y+2. 2So 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
41、 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 1 odd point on the left of any characteristic odd prime point 13to adjacent odd points, then the RPL2m+1 is moved
42、 into a RPL2m+3. Evidently the left endmost point of the RPL2m+3 is 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
43、 have proven that even if OD Pm(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 conclusi
44、on 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 k1. To wit every even number 2N can express in a sum
45、 of two odd prime numbers, where N2.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 N2.Since every odd number 2N+3 can express in a sum which a prime number 14plus the even number makes, consequ
46、ently every odd number 2N+3 can express in a sum of three prime numbers, where N2.To sum up, we have proven that two propositions of the Goldbachs conjecture are tenable, thus Goldbachs 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_摘 要我们知道,依次增大的每一个正偶数2n(n3)可以表示成3分别与依次增大的一个奇数2k+1（k1）之和.于是,对于数轴上的任意一个奇数点2k+1（k1）,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 32时，每一个偶数2N都可以表示成两个奇素数之和. 24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 N2.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 N2.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.
