4.2．数学的内在本质是含可逆的不可逆“非”运算

哥猜获证后可以为大力发展基础数学开道，可以进一步深层次第思考“非运算”了。即后继相邻的未知事件究竟会是什么呢？

何为可逆“非”运算与不可逆“非”运算；

不可逆“非”始终包含可逆“非”，而可逆“非”则无法囊括不可逆“非”。

不可逆“非”为基数表达的并运算，可逆“非”为基数表达的交运算；

相反，不可逆“非”为序数表达的交运算，可逆“非”为序数表达的并运算。

负数进行非运算变成正数，这是可逆非运算。

正数进行非运算重新变成负数，这是可逆非运算。

负数进行非运算变成2阶负数，这是不可逆非运算。

虚数进行非运算变成2阶虚数，这是不可逆非运算。

不是互补关系的越级非运算，都是不可逆非运算。

数学的突破性发展都是某个层面的不可逆非运算推动的。群论所描述的是对称互补关系，相对小范畴是不可逆运算，相对大范畴仍是可逆非运算，而相邻论所要表达的则是不可逆延伸关系。

群论表达了数学的形式结构；

而相邻论则表达了数学的内在本质。

研究和探索n阶负数、n阶虚数、n阶无理数的新数学时代即将来临。

不可逆“非运算”是中国乃至世界传统思想的核心，无论是老子的道可道非常道，还是佛陀的非想非非想，孔子的不偏不倚，耶稣的无知之幕，都将“非运算”伸向了极致，无穷延伸的素数就是不可逆“非运算”的象征。用倒金字塔的方式扩域，与戴德金分割的思想一致。布尔代数将所有的逻辑运算概括为“交”“并”“非”三种，其实还可以进一步概括为一个字就是“非”，有交运算性质（收敛）的“非”和并运算性质（发散）的“非”，有收敛才有等量关系，有发散才有序列关系，交集是公共混沌存在，不是根概念，根是指开放的全集。

基数1是交运算的最后结果；

序数1是并运算的最后结果。

世界传统思想的核心是，一切来自序数1，一无所有，应有尽有都是序数1派生出来的。序数1是不可量化的，一切都是序数1的子集。如果把序数1作为公理体系的话，该公理是开放的，它不受哥德尔定理的约束，它能证明的命题是发散的，而哥德尔宣告了现有的公理体系所能证明的命题是收敛的，图灵机会死机。哥猜获证所带来的思想收获是，图灵机就像薛定谔的猫，或生或死或不生不死。素数思想的突破将成为超弦理论以及凝聚态物理学的最佳数学工具。

无穷个离散量的自变量通过矩阵运算获得一个无穷连续量的值域，经证明该运算的幕后引擎其实质往往可归结为是由一个离散量的自变量决定的。高维度离散量的自变量通过积分运算获得一个高维连续量的值域，经证明该运算的幕后引擎其实质往往可归结为是由一维离散量的自变量决定的。也就是说自变量的第一个相邻关系至关重要。相邻论构建了一种包含同类集结且能异类区分的数学思想，建立了比集合论更基础的数学。如果自然数不能表达无限复杂的世界，那么有理数、实数、复数、代数数、超越数亦不能表达无限复杂的世界。真的应了克罗内克（Kronecker）的那句话：“上帝创造了正整数，其余都是人的事情。”而这样一个伟大判定，就基于哥德巴赫猜想获证后的一个简单推论。

即：首个相邻运算，决定了无穷无漏迭代运算；无穷无漏迭代运算，将归结为最后极限运算。

如此可将离散和连续融为一体。

前者是蛋生鸡，集合论是其基础，对象开放，是基数思维；

后者是鸡生蛋，相邻论是其基础，算法开放，是序数思维。

前者为还原论，后者为演生论。演生是更深刻的还原。

2013年12月3日于深圳

2018年11月8日修改

参考文献：

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附注1：

本文原稿关键部分发表于国家专业数学期刊《数学学习与研究》2013年第3期上，这里稍有改动。本文除了证明哥德巴赫猜想外，还用相邻论和重合法相继证明了孪生素数猜想、黎曼猜想等30多个国际数学界未解的相关猜想。该文2017年5月被收录于由海天出版社出版的《深圳基础理论原创文集》一书中。

附注2：

哥德巴赫猜想被相邻互素引理证明（中英文简报）

◎定理：自然数相邻互素（0除外）。（为相邻论核心内容）

证明：已知n、m是一对相邻自然数，即n+1=m，由于1与n互素，故n与m必互素。假如其中两项非互素，有公约数可约掉，就会产生整数与真分数相等，矛盾。故自然数是相邻互素的。

◎推论：相邻偶数约掉因子2后定是互素的，即2n+2=2m, n与m必互素。

证明：相邻偶数2n与2m约掉2因子后是一对相邻自然数，据上文已证定理，n与m一定是互素的。

◎定义：两个任意奇素数p与q互异相加所得到的所有偶数2m（其中m＞3）叫可表偶数，也叫基础偶数。与之互异的所有偶数2h叫例外偶数，也叫非基础偶数，是不含基础偶数的通解偶数。

◎推论：2m与p、q每次三元互素，但三元累积解集一定是非互素的。即m必定蕴含所有素因子。

证明：令每次p>q，且皆为任意奇素数，p、q的累积解集由定义可知是相同的，皆蕴含所有奇素数，故p、q二元累积解集非互素。现假设三元累积解集m欠缺奇素数r，那么这会与p、q含所有奇素数的推论相矛盾。该推论是，根据p、q皆为奇素数全集，可令其子集不含奇素数r, r＜p, r就是p、q的真子集在全集奇素数上的补元，而p+q能生成互素补元因子r，包括2因子，这是由三元互素方程p+q=2m的性质决定的。可令方程两边的解集互素，左边的素数解集互素生成了右边的素数解集，即左边为任意奇素数，可取r解集生成右边的非r解集，反过来，也可取非r的解集生成右边的r解集。

根据三项互素方程性质，若p+q=∏p1p2p3...pn（不含素数r因子），则一定p、q∈r；

同理，若p, q不含素数r, p+q=∏pk+1pk+2pk+3...pk+n，则一定pk+1, pk+2, pk+3, ..., pk+n∈r.

因左边p, q为全集奇素数，故右边m欠缺r因子就会与该结论相矛盾。因此m必定蕴含所有素因子的推论就获得了证明。

以下提供该推论的第二种证明方法。

假如可表偶数中不含奇素因子r，其中r＜p，由伯特兰-切比雪夫定理得到，大于6的所有偶数可用三项互素方程表达，即2n=p+kq, p＞n，当且仅当k≠1, gcd（r, p）=1，则2r是例外偶数，2r=p+kq必每次三项互素。例外偶数存在本原解，才有更多通解，于是我们来考察有三项互素性质的本原解方程。

为何可表偶数的本原解方程可以每次三元互素但累积解集非三元互素呢？是因为定义允许p与q解集相等。而例外偶数的本原解方程则要求，不但（2m、p、kq）每次三元互素，且kq的每次解还必须与q的所有解互异（k≠1），因互异必互素（kq满足乘法交换律）。由于可表偶数是任意两奇素数的和，说明kq同p、q的所有累积解集仍三元互素。

于是可推得p不但同2r与kq每次互素，且会累积解集三元互素，而生成元p解集是奇素数全集，既然r p，那r就是奇素数空集，既然k、qp，那k、q就是奇素数空集。可见本原解非可表偶数2r不存在，其通解自然不存在，于是反证了可表偶数必囊括全部奇素数因子及2素数因子。

还有第三种证明方法。

根据伯特兰-切比雪夫定理，2n=p+kq，其中每次p＞kq, p、q是奇素数，k是正整数，n是大于3的所有自然数，由于p＞n, p不能整除2n，可知方程必三元互素。根据可表偶数2m=p+q，可判定向量（1, k）的特征值c作用2m会等于向量（1, k）内积于p+q，即会共同等于2n。如此可知，不小于8的所有偶数2n就是可表偶数2m的数乘值2mc，当且仅当有理数c≠1, k≠1时，2mc=p+kq就是例外偶数。

为何c是有理数？因为2m2n，根据唯一析因定理，在2m的基础上通过作用c=x/y≠1，即通过添加分子x因子或删减分母y因子就能得到全部2n，故构造例外偶数的c是非1有理数。而2m是从所有奇素数中两两相加构造得到的，每次都会产生互素因子，且至少有一个奇素因子或至少有两个2因子，总之m不会等于1。

关键点是，c=x/y≠1必须同大于中值数的某一奇素数p每次互素，且须同任意奇素数p累积互素，有一次非互素都会c=1, k=1, q=p，这与例外偶数的定义矛盾。既然c同任意奇素数p累积互素，那么m与任意奇素数p也须累积互素，在例外偶数2mc中，m与c满足乘法交换律。可见所有的2p与例外偶数2mc在奇素因子种类上须累积互异。

例外偶数的素因子个数因互异至少要大于2个奇素数或3个偶素数或更多个，那所有奇素数的两倍2p就不可能是例外偶数。同样例外偶数的素因子个数因互异至少要小于2个奇素数或3个偶素数或更少个。那所有奇素数的两倍2p就不可能是例外偶数。可见2p与例外偶数2mc在素因子个数上须累积互异。这样可表偶数2m就包含了所有2p，当然也就蕴含了所有的素数因子。

素数二元相加再分解运算在所有奇素数因子集上封闭，这一结论还有第四种证明。

若互异型可表偶数2m=p+q, p、q为互异奇素数，则rad（p+q）中的所有奇素数因子集合，与p、q中的所有奇素数因子集合等价。证明如下：

令2m（含2p亦含2w）为互异型可表偶数，它是能用两互异奇素数之和表达的偶数，2p＇为例外偶数，它是不能用两互异奇素数之和表达的偶数，p、p＇为互异奇素数，它们的并集须囊括所有奇素数q。那么必有2p＇-2p=2t, p＇与p因互异而互素，根据三元方程若两元互素必三元两两互素的性质，p与t必互素，p＇与t必互素。

由于构造t的素因子始终要与p及p＇互素，其累积结果，导致要与所有的奇素数互素。如此t就没有奇素因子可构造，加上2p＇-2w=2t, t与偶素数2也互素，故例外偶数2p＇不存在。因为所有奇素数q 的2倍，定是互异型可表偶数2p以及互异型非可表偶数2p＇的并集，意味着t要始终与所有的奇素数及偶素数互素。因此2t就不存在，故2p=2q,2q必为互异型可表偶数。如此就证明了互异型可表偶数2m包含了2倍的所有奇素数q。素数二元相加再分解运算在所有奇素数因子集上封闭得证。

也就是说，例外的（2r）都不存在（心外无物），存在的（2p）都不例外（法不离心），无时间必无空间，有空间必有时间，无空间不一定无时间。

◎定理：例外偶数是空集。（为相邻论核心内容——单向绵延的一种根本改变）

证明：与可表偶数互异存在的例外偶数，因互异而至少有例外偶数首项生成元与可表偶数相邻，例外偶数与可表偶数之间以及例外偶数与不同例外偶数之间，因须首项偶数相邻互素，故始终没有非2公约数，例外偶数首项生成元与可表偶数因互异而必有首项相邻，因相邻而必须m与h互素（自然数相邻互素定理已证）。

而上文还证明，可表偶数2m中的m蕴含所有素因子，h既然要与所有的素因子累积互素，h就不存在能超越素数全集的新增素数因子，故首项例外偶数2h中的h无素数因子可构造，因此首项例外偶数2h是空集，偶数0不是空集，既然无首项例外偶数，当然也就不存在后继例外偶数。故例外偶数是空集。

总结下就是，因两类偶数互异，c不等于1，导致例外偶数与可表偶数不但会在素数个数上无穷无漏互异，还会在素数种类上无穷无漏互异。例外偶数通过与可表偶数在两类性质上区分，从而被判定为空集。

◎定理：不小于8的所有偶数皆可表为两互异奇素数之和。（为重合法核心内容）

证明：既然例外偶数2h是空集，根据不小于8的所有偶数2n等于可表偶数2m与例外偶数2h的两类偶数并集，可推得不小于8的所有偶数2n与可表偶数2m是无缝重合，是完全同构的，故不小于8的所有偶数2n也就同可表偶数2m一样与两互异奇素数之和p+q同构，互异版哥德巴赫猜想获证。

补上非互异版的3+3=6，欧拉版哥德巴赫猜想也就获证。

Note 2:

The Goldbach Conjecture is Proved by the Adjacent Mutually Prime Lemma

(briefing in Chinese and English)

◎ Theorem: The adjacent natural number are mutually prime (0 except). (For the core content of the adjacent theory)

Proof: It is known that n, m is a pair of adjacent natural numbers, that is, n+1=m.Since 1 and n are mutually prime, n and m must be mutually prime.If two of them are non-mutually prime, there are common factors that cancel out, and the integer and the proper fraction will be equal so contradictory. Therefore, the adjacent natural numbers are mutually prime to each other.

◎ Inference: The adjacent even number of cancel out factors 2 will be mutually prime. That is,2n+2=2m, n and m must be mutually prime.

Proof: After the adjacent even number 2n and 2m is cancel out factors 2, it is a pair of adjacent natural numbers. According to the above theorem, n and m must be mutually prime.

◎ Definition: All even number 2m obtained by adding two arbitrary odd prime numbers p and q are called the basic even number(where m>3). All distinct even numbers 2h from it are called the non-basic even number. That is to say, it is general even number without the basic even number.

◎ Inference: 2m and p, q each time be ternary mutually prime, but the ternary cumulative solution must be non-mutually prime. That is, m must contain all prime factors.

Proof: Let each time p>q, and all are any odd prime numbers. The cumulative solution set of p and q can be known by definition, all the odd prime numbers are included. Therefore, the p and q of cumulative solution sets are non-mutually prime. Now suppose that the ternary cumulative solution set m lacks the odd prime number r, then this will contradict the inference that p and q contain all odd prime numbers. The inference is that, according to p and q are the complete sets of odd prime numbers, the subset can be free of odd prime numbers r, r < p, r is the complement of the true subset of p and q on the all odd prime numbers of the complete set, and mutually prime complement set factor r is generated with p+q, including factors 2, which is determined by the nature of the ternary prime equation p+q=2m. The solution of the two sides of the equation can be mutually prime, and the prime solution of the left side solves the prime solution of the right side, that is, the left side is any odd prime number, and the r solution set can generate the non-r solution set on the right side, and vice versa.

According to the properties of the ternary mutually prime equation，

p+q=∏p1 p2p3...pn(No prime factor r), then certain p, q∈r, empathy，

p, q (No prime number r), p+q=∏ pk+1 pk+2pk+3...pk+n, then certain pk+1, pk+2, pk+3, ..., pk+n∈r.

The non-r solution set will generate the r solution set on the right side, so the lack of the r factor on the right side will contradict this conclusion. Therefore, the inference that m must contain all prime factors is proved.

The second proof of this inference is provided below.

If the basic even number factor 2m does not contain the odd prime factor r, each time r<p, from Bertrand-Chebyshev＇s theorem,2n=p+kq is obtained, if and only if k≠1, gcd(r, p)=1,2 r is the non-basic even number,2r=p+kq must be ternary mutually prime each time. The non-basic even number only have primitive solutions, there are more general solutions. So we will study the primitive solutions equation with ternary mutually prime properties.

Why primitive solutions equation of the basic even number be ternary mutually prime each time, but the cumulative solution set is not ternary mutually prime? Because the definition allows p to be equal to q solution set. But primitive solution equations of the non-basic even number that not only (2m, p, kq) is ternary mutually prime, but also each solution of kq must be different from all solutions of q (k≠1), because it be different from each other, so be mutual prime(kq satisfies the law of multiplicative exchange). Since the basic even number is the sum of any two odd primes, it is shown that all the cumulative solutions of kq with p and q are still ternary mutually prime.

So we can deduce that p is not only mutually prime with 2 r and kq each time, but also accumulates solution set be ternary mutually prime. The generator p solution set is the universal set of odd prime numbers.Since r p, r is the empty set of odd prime numbers. Since k, q p, k and q are the empty set of odd prime numbers. It can be seen that the primitive solution of the non-basic even number 2r does not exist, and its general solution naturally does not exist, thus proving that the basic even number must include all odd prime factors and 2 prime factors.

There is also a third proof.

According to the Bertrand-Chebyshev theorem,2n=p+kq, where p>kq, p, q are odd prime numbers, k is a positive integer, n is all natural numbers greater than 3,2n cannot be divisible by p, due to p>n, and we know that the equation must be ternary mutually prime. According to the basic even number 2m=p+q, it can be determined that, the eigenvalue c of the vector (1, k) acts 2m, equal to the inner product of the vector (1, k) and p+q, which is all equal to 2n. Thus, all even numbers 2 n not less than 8 are the multiplication value 2 mc of the basic even number 2 m. If and only if the rational number c≠1, k≠1,2 mc is the non-basic even number.

Why is c a rational number? Because 2m2n, according to the unique factorial theorem, All even numbers can be obtained by multiplying the basic even number 2m with the rational number c（c=x/y). That is to say, all 2n can be obtained by adding molecular x factor or deleting denominator y factor. Therefore, the factor c that generates the non-basic even number is a non-1 rational number. While 2m is the sum of two arbitrary odd prime numbers, mutual prime factors are produced every time, and there is at least one odd prime factor or at least two factors 2. In short, m will not equal to 1.

It can be seen that the number of prime factors with The non-basic even number is at least more than two odd prime numbers or three even prime numbers or more, because two kinds of even numbers are different from each other, The 2p obtained by multiplying any odd prime number p by 2 cannot be the non-basic even number. Similarly, the number of prime factors with the non-basic even number is at least less than two odd prime numbers or three even prime numbers or less, because two kinds of even numbers are mutual difference, The 2p obtained by multiplying any odd prime number p by 2 cannot be the non-basic even number. It can be seen that the number of prime factors of 2 p and the non-basic even number 2 mc must be cumulatively different from each other.

The key point is that c=x/y≠1 must be mutually prime every time with an odd prime number p greater than the median number, and must accumulate mutually prime with any odd prime number p. Once non-mutually prime will be c=1, k=1, q=p, which contradicts the definition of The key point is that c=x/y must be mutually prime every time with an odd prime number p greater than the median number, and must accumulate mutually prime with any odd prime number p. Once non-mutually prime will be c=1, k=1, q=p, which contradicts the definition of the non-basic even number. Since c accumulates mutually prime with any odd prime number p, m and any odd prime number p must also accumulate mutually prime. In the non-basic even number 2 mc, m and c satisfy the law of multiplicative exchange. It can be seen that the types of prime factors of 2p and the non-basic even number 2 mc must be cumulatively different from each other. In this way, the basic even number 2m contains 2p, of course, it also contains all prime factors.

The prime binary addition and decomposition operations are closed on all the odd prime factor sets, There is a fourth proof of this conclusion.

If the basic even number 2m=p+q, p and q are the odd prime numbers, also different from each other, then all the odd prime number factor sets in rad (p+q) are equivalent to all the odd prime number factor sets in p and q. The proof is as follows:

Let 2m (including 2p and 2w) be the basic even number of different from each other type, It is an even number that can be expressed by the sum of two different odd prime numbers,2p＇be the non-basic even number of different from each other type, It is an even number that can not be expressed by the sum of two different odd prime numbers, p、p＇are mutually different odd prime numbers. Their union must include all odd prime numbers q. Then there must be 2p＇-2p=2t, p＇and p are coprime, because they be mutually different. According to the properties of ternary equation, if two elements are mutually prime, three elements are pairwise mutually prime, p and t must be mutually prime, p＇and t must be mutually prime.

Because the prime factor of constructing t always has to be inter prime with p and p＇, its cumulative result leads to inter prime with all odd prime numbers. So t has no odd prime factor to construct, In addition,2p＇-2w=2t, t and even prime number 2 are also mutually prime, so the non-basic even number 2p＇does not exist. Because the two times of all odd prime numbers q, it must be the union of the basic even of different from each other type 2p and the non-basic even of different from each other type 2p＇, which means that t should always be coprime with all odd prime numbers and even prime numbers. Therefore,2t does not exist, so 2p=2q,2q must be the basic even of different from each other type. Thus, the basic even number of different from each other type 2m contains two times all odd prime q. The prime binary addition and decomposition operations are closed on all the odd prime factor sets, has been proved.

That is to say, all exceptional（2r）does not exist（nothing outside the mind）, all existing （2p）must is no exception（all methods are not centrifugal）, if there is no time then without space, but if there is no space then still be time.

◎ Theorem: The non-basic even number is an empty set.（For the core content of the adjacent theory -a fundamental change in one-way stretch）

Proof: There are at least the non-basic even number first generating element is adjacent to the basic even number, because their solution sets complement each other, so there must be adjacent elements between the non-basic even number and the basic even number. There is no non-2 common number between the non-basic even number and the basic even number. Because there must be adjacent elements between complementary solution sets, and the m and h must be mutual prime to each other (the theorem of adjacent prime of natural number has proved).

It is also proved that m in the basic even number 2m contains all prime factors. Since h is to be mutually prime with all prime factors, there is no new prime factor that can surpass the complete set of prime numbers. Therefore, the first the non-basic even number 2h cannot be constructed, so the non-basic even number 2h is an empty set, successive even numbers of the non-basic even number are also empty sets. Even number 0 is not an empty set. Since there is not first the non-basic even number, of course, there is no successor the non-basic even number. So the non-basic even number is empty set.

To sum up, because two kinds of even numbers are mutually different, c is not equal to 1, so the non-basic even number and the basic even number will not only differ in arbitrary prime factors quantity, but also in arbitrary prime factors types. The non-basic even number are determined as empty sets by distinguishing them from the two properties of the basic even number.

◎ Theorem: All even numbers not less than 8 can be expressed as the sum of two mutually odd prime numbers. (for the core content of the Coincidence Method)

Proof: Since The non-basic even number 2h is an empty set, according to all even numbers 2n not less than 8 equal to the union of the basic even number 2m and the non-basic even number 2h, it can be inferred that all even numbers 2n and the basic even number 2m of not less than 8 are seamlessly coincident, completely Isomorphic, so all the even 2n of not less than 8 is the same as the even number of 2m and the sum of the two mutually different odd primes p+q, and Goldbach Conjecture of the original solution version is proved.

Complementing the non-original solution version with 3+3=6, Goldbach Conjecture of the Euler version is proved.