每个大于2的偶数都是两个素数之和

1、Any even number greater than 2 must be the sum of two prime numbers, i.e., 2N2 = Pa + PbAbstract hypothesis, simulate synchronous infinite basic arithmetic logic, judge reasoning and assuming that contradictionProof by contradiction (integer theory, philosophy) Wu Ye Tang Yiabcty111111

2、ORCID ID:/0000-0001-9904-7594Gantang Dongmen Village, Fuan City, Fujian Province, ChinaIntroduction: A prime number is a natural number greater than 1 and it is not divisible by natural numbers other than 1 and itself. In other words, a prime number has factors of 1 and itself only.

3、All other numbers are called composite numbers. According to the fundamental theorem of arithmetic, any integer greater than 1 is either a prime number itself or can be written as the product of a series of prime numbers. If the order of these prime numbers in the product is disregarded, the written

4、 form will be unique. The smallest prime number is 2.Abstract:Infinite arithmetic logic + infinite assuming that There can only be one truth: either 2N = Pa + Pb or 2N Pa + Pb.Assume that 2N Pa + Pb and its basic analog arithmetic logic diagram. According to the integer condition, the hypothesis is

5、refuted.For any problem, there can only be two judgments: the choice is either yes or no.Keywords: assuming that, Prime number, Composite number, Prime factorization,Cross-reference.Mathematical classification number: 03Axx(1Type one:simple analog infinite arithmetic logic (Computational diagram)(2)

6、 Type two: simple analog infinite arithmetic logic (Computational diagram)Cross reference（WY2）Computing logic obtains prime numbers，and（WY1）Computing logic obtains prime numbers，They have the same prime numbers ,Cross reference（WY2）and（WY1）Computing logic obtains prime numbers。（They have different p

7、rimes numbers。）（5）Suppose that：2NPaPb 。Its assumption is untenable where N is a natural number and N > 1.-0Any even number = 2N and any odd number = 2N1.Suppose that an even number Pa + Pb.Let N1 = even number = composite number and N2 = odd number = composite number.Then, N1 1= S1 = odd number a

8、nd N2 2 = S2 = odd number.(1)Type one:simple analog infinite arithmetic logic (Computational diagram)Extract the prime factors A1, B1, C1 using2N A1 = L2, assuming that L2 = Composite number = An2 2N B1 = H2, assuming that H2 = Composite number = Bn22N C1 = M2, assuming that M2 = Composite number =

9、Cn2Extract the prime numbers A2, B2, C2 using2N A2 = L3, assuming that L3 = Composite number = An3 2N B2 = H3, assuming that H3 = Composite number = Bn32N C2 = M3, assuming that M3= Composite number = Cn3Extract the prime numbers A3, B3, C3 using2N A3 = L4, assuming that L4 = Composite number = An4

10、2N B3 = H4, assuming that H4 = Composite number = Bn42N C3 = M4, assuming that M4 = Composite number =Cn4Extract the prime numbers A4, B4, C4 using2N A4 = L5, assuming that L5 = Composite number = An5 2N B4 = H5, assuming that H5 = Composite number = Bn52N C4 = M5, assuming that M5 = Composite numbe

11、r = Cn5L5, H5, M5 (prime number or composite number)Extract the prime numbers A5, B5, C5, . Analog infinite arithmetic logic (WY1).Suppose that 2N Pa + Pb. infinite arithmetic logic, it has two judgments:(A) Basic analog arithmetic logic, prime number loop(B) Basic analog arithmetic logic, infinite

12、calculation, infinite increase of prime numbers.Suppose that 2N Pa + Pb. Then, in the case of judgment (B) of basic analog arithmetic logic, infinite calculation and infinite increase of prime numbers, the hypothesis does not hold. Infinite calculation, it has an infinite increase of prime numbers d

13、iffering in quantity. Infinity > 2N The assuming that does not hold.If 2N Pa + Pb, then its choice is (A): basic analog arithmetic logic, prime number loop.Suppose that 2N Pa + Pb. Then, in the case of judgment (B) of basic analog arithmetic logic, infinite calculation and infinite increase of pr

14、ime numbers, the hypothesis does not hold. Infinite calculation, it has an infinite increase of prime numbers differing in quantity. Infinity > (2N + 1) The hypothesis does not hold.If 2N Pa + Pb, then its choice is (A): Basic analog infinite arithmetic logic, prime number loop（2） Type two: simpl

15、e analog infinite arithmetic logic (Computational diagram)Extract all the prime numbers, A1, A2, A3, B1, B2, B3, C1, C2, of the analog arithmetic logic (WY1) using the following expressions: 2N 2 × A1 = 2 × T1, assuming that T1 = Composite number = Mn1 2N 2 × B1 = 2 × T2, assumin

16、g that T2 = Composite number = Mn22N 2 × C1 = 2 × T3, assuming that T3 = Composite number = Mn3 2N 2 × A2 = 2 × T4, assuming that T4 = Composite number = Mn4 2N 2 × B2 = 2 × T5, assuming that T5 = Composite number = Mn5 2N 2 × C2 = 2 × T6, assuming that T6 = C

17、omposite number = Mn6 2N 2 × A3 = 2 × T7, assuming that T7 = Composite number = Mn7 2N 2 × B3 = 2 × T8, assuming that T8 = Composite number = Mn82N 2 × C3 = 2 × T9, assuming that T9 = Composite number = Mn9 。Extract the prime factors, M1, M2, M3, M4, M5, M6, M7, M8, , ,

18、 using the following expressions:2N Mn1 = T1, assuming that T1 = Composite number = Mn1 2N Mn1 = T2, assuming that T2 = Composite number = Mn22N Mn1 = T3, assuming that T3 = Composite number = Mn3 2N Mn1 = T4, assuming that T4 = Composite number = Mn4 2N Mn1 = T5, assuming that T5 = Composite number

19、 = Mn5 2N Mn1 = T6, assuming that T6 = Composite number = Mn6 2N Mn1 = T7, assuming that T7 = Composite number = Mn7 2N Mn1 = T8, assuming that T8 = Composite number = Mn8 2N Mn1 = T9, assuming that T9 = Composite number = Mn9 。Extract the prime factors：S1 、S2、S3、S4、S5、S6、S7、S8、using。.Analog infinit

20、e arithmetic logic（WY2）There are only two choices for its judgment:Either 2N = Pa + Pb or 2N Pa + Pb：Suppose that 2N Pa + Pb. infinite arithmetic logic, it has two judgments:(A) Basic analog infinite arithmetic logic, prime number loop(B) Basic analog infinite arithmetic logic, infinite calculation,

21、 infinite increase of prime numbers.Suppose that 2N Pa + Pb. Then, in the case of judgment (B) of basic analog arithmetic logic, infinite calculation and infinite increase of prime numbers, the hypothesis does not hold. Infinite calculation, it has an infinite increase of prime numbers differing in

22、quantity. Infinity > (2N + 1) The assuming that does not hold.If 2N Pa + Pb, then its choice is (A): Basic analog infinite arithmetic logic, prime number loop（3）Cross reference（WY2）Computing logic obtains prime numbers，and（WY1）Computing logic obtains prime numbers，They have the same prime numbers

23、 ,Suppose that：（WY2），and，（WY1）Computing logic obtains prime numbers。Same prime numberset up：Its prime number，A1、A2、A3、B1、B2、B3、C1、C2、C3、2NA1=L1=Compound number=A2n 2N2CC=2TT=Compound number=A2A 2N2naE=2naAE A1+2CC=A2nA2A A1+2naE=A2n2naA2E （WY2），and，（WY1），The calculation logic obtains the same prime

24、number.（WY1）2NA1=L1，Suppose that：L1=Compound number=A2n 2NB1=M1，Suppose that：M1=Compound number=B2n 2NC1=H1，Suppose that：H1=Compound number=C2n 2NA2=L2，Suppose that：L2=Compound number=A3n 2NB2=M2，Suppose that：M2=Compound number=B3n 2NC2=H2，Suppose that：H2=Compound number=C3n 2NA3=L3，Suppose that：L3=

25、Compound number=A4n 2NB3=M3，Suppose that：M3=Compound number=B4n 2NC3=H3，Suppose that：H3=Compound number=C4n （WY1）Suppose that：They get prime numbers，A1、A2、A3、B1、B2、B3、C1、C2、C3、（WY2）2N2A1=2T1,Suppose that：T1=Compound number=M1n 2N2A2=2T2,Suppose that：T2=Compound number=M2n 2N2A3=2T3,Suppose that：T3=C

26、ompound number=M3n 2N2B1=2T4,Suppose that：T4=Compound number=M4n 2N2B2=2T5,Suppose that：T5=Compound number=M5n 2N2A3=2T6,Suppose that：T6=Compound number=M6n 2N2C1=2T7,Suppose that：T7=Compound number=M7n 2N2C2=2T8,Suppose that：T8=Compound number=M8n。Suppose that：They get prime numbers，M1Mn，All the sa

27、me。A1、A2、A3、B1、B2、B3、C1、C2、C3、（WY2），and，（WY1）The calculation logic obtains the same prime number（4）,Cross reference（WY2）and（WY1）Computing logic obtains prime numbers。（They have different primes numbers。）Suppose that：Its prime number，A1、A2、A3、B1、B2、B3、C1、C2、C3、Judgment reasoning：Suppose that：2NB1=TA1

28、 Get the same prime number2N2B1=TA1B1 2N3B1=TA12B1 2N4B1=TA13B1 2N5B1=TA14B1 。Or，2N2A1=B1+（T2）A1 2N3A1=B1+（T3）A1 2N4A1=B1+（T4）A1 2N5A1=B1+（T5）A1 。Suppose that：（WY1）Prime number obtained by computing logic。从小依次排列，A1，A2，A3，AnMinimum prime number，A1>2 Arbitrary prime number：【（A2，A3，An）+A3，An】>2A1

29、如：2NA3=SAA 2N(A3AA )=SAA AA set up：(A3AA )÷2=2A6 2N2A6=SAA AA Minimum prime numbers，A12N2A1，It does not have the same integer solution as other prime numbers。2N2A1=B1+（T2）A1 【B1+（T2）A1】÷（B1,A1）=fraction （WY），and，（WY1）Computing logic obtains prime numbers（They have different prime numbers）I

30、ts assumption is untenable（5），Suppose that：2NPaPb 。Its assumption is untenable（WY），and（WY1）Computing logic obtains prime numbers（They have different prime numbers）set up：（WY）The calculation logic is obtained, and the prime number is newly added，Pa，2NPaPaSIt must add new prime numberAgain according t

31、o：（WY1）The calculation logic obtains different prime numbers, and the prime number loops。Suppose that：according to（WY1）The calculation logic is obtained, and the prime number is newly added，从小依次排列，Pa，Pb,Pc,Pd，according to（WY2）Computational logic, minimum prime number， Pa，2N2Pa，It adds new primesInfi

32、nite synchronous computing。（WY2），（WY1），（WY2），（WY1），It infinitely increases different prime number。2N=Limitedinfinite>2N Suppose that：2NPaPb 。Its assumption is untenableReferences:EuclidThesis topic Abstract Hypothesis, Basic Analog Arithmetic Logic, Judgment Inference and Hypothesis Contradiction

33、Philosophy, Proof by contradiction (its condition: integer) Proven by the contradiction that the number of primes is infiniteAbstract hypothesis:Suppose that the number of primes is finite.They are permuted in the ascending order as P1, P2, P3.PnBasic analog arithmetic logic: multiplication in ascen

34、ding orderP0×P1×P2×P3×.×Pn=N 2×3×5×7×.×Pn=N Then, N+1 is either prime or not prime. N+1>PnJudgment inference:If N+1 is a composite number,suppose that N+1=WLet W = P1, P2, P3.Pn (any prime number)(N + 1) ÷ W N ÷ W = Integer and 1 ÷

35、W = Fraction (N + 1) ÷ W = Fraction The propositional condition is an integer (definition of a prime number) N + 1 is either a composite or a prime number.The prime factors obtained by the factorization of the integer N + 1 are definitely not within the assumed P1, P2, P3.Pn.There are other pri

36、mes besides the finite number of primes assumed. Hence, the original hypothesis does not hold. In other words, there exist infinitely many prime numbers.Hypothesizing that the infinite number of primes belongs to the unconditional mathematical theory.Then, there is an equation N + 1 = W × YnLet

37、 W<Pn , Y<PnThe equation (N + 1) ÷ W =Yn (Y<Pn) must be established. The theorem of infinite primes does not hold.Any rigorous theory proves that one is tenable and the other is not tenable2N A1 = L2, assuming that L2 = Composite number = Bn × Cn × Dn2N B = E, assuming that E

38、 = Composite number = Bn12N C = F, assuming that F = Composite number = Cn12N D = G, assuming that G = Composite number = Dn 12N B1 = E1, assuming that E1 =Composite number = B × C2N C1 = F1, assuming that F1 = Composite number = CC × D2N D1 = G1, assuming that G1 = Composite number = DD2N

39、 B = E, assuming that E = Composite number = Bn12N C = F, assuming that F = Composite number = Cn12N D = G, assuming that G = Composite number = Dn 1Prime number loop (Computational diagram)12863=65=5×13Extract the prime numbers 5,13 using1285=123=3×4112813=115=5×23Extract the prime n

40、umbers 3,5,41,23 using1283=125=5×5×512841=87=3×2912823=105=3×5×7Extract the prime numbers 3,5,29,7 using12829=99=3×3×111287=121=11×11Extract the prime numbers 3,11 using12811=117=3×3×13.Infinite assuming that , infinite computation（2）Type two：Extract

41、 the prime factors,3,5,7,11,13,41,23，using the following expressions:1282×5=2×591282×3=2×611282×7=2×57=2×3×19 1282×11=2×591282×13=2×511282×23=2×411282×29=2×35=2×5×71282×41=2×23（It is an unknown number

42、: if it is a prime, suppose it is a compound number。）12859=69=3×23 12861=67 , assuming that 67=Composite number。12819=109,assuming that 109=Composite number。.Infinite assuming that , infinite computation抽象假设，模拟同步无穷基本算术逻辑，判断推理和假设矛盾任意大于2偶数必定是俩个素数的和2N=PaPb 反证法（整数论、哲学）引言：素数，一个大于1的自然数，除了1和它本身外，不能被其他

43、自然数整除，换句话说就是该数除了1和它本身以外不再有其他的因数;否则称为合数。根据算术基本定理，每一个比1大的整数，要么本身是一个质数，要么可以写成一系列质数的乘积;而且如果不考虑这些质数在乘积中的顺序，那么写出来的形式是唯一的。最小的素数是2摘要：无限基本计算逻辑，无穷假设，无限增加素数。关键词：假设，素数，合数，相互参考。数学分类号: 03Axx无穷大的偶数，是否存在偶数（1）介绍设：2NPaPb ,第一种、模拟无穷同步计算逻辑。假设它们得到全部都是合数。（计算示意图）计算逻辑两种选择：A无穷计算逻辑，素数循环。B无穷计算逻辑，无穷增加不同素数。（2）介绍设：2NPaPb,第二种、模拟无穷

44、同步计算逻辑。（计算示意图）它有两个选择：A无穷算术逻辑，素数循环。B无穷算术逻辑，无穷增加不同素数。（3）（WY2）计算逻辑获得素数，和（WY1）计算逻辑获得素数.，它们有相同的素数，相互参照（WY2）和（WY1）计算获得素数。（4）（WY2）计算获得素数和（WY1）计算获得素数,(它们有不相同素数。（5）假设：2NPaPb 。它的假设站不住脚设：2NPaPb （自然数N）N>2任意：偶数=2N任意：奇数=2N1 设：2NPaPbN1=偶数，N2=奇数 N1=偶数，N11=S1=奇数 N2=奇数，N22=S2=奇数，假设：2NS1=W1=合数=A1n×B1n×.&#

45、215;C1n2NS2=W2=合数=A1n×B1n×.×C1n 模拟计算逻辑：如果余项等于素数，则：2N=PaPb假设：2NPaPb 。使用抽象模拟计算逻辑。第一种：模拟无穷,同步计算逻辑。（模拟同步计算，示意图）抽取素因数、A1，B1，C1,使用2NA1=L1，假设：L1=合数=A2n 2NB1=M1，假设：M1=合数=B2n 2NC1=H1，假设：H1=合数=C2n 抽取素因数、A2、B2、C2,使用2NA2=L2，假设：L2=合数=A3n 2NB2=M2，假设：M2=合数=B3n 2NC2=H2，假设：H2=合数=C3n 抽取素数：A3，B3，C3,使用。2

46、NA3=L3，假设：L3=合数=A4n 2NB3=M3，假设：M3=合数=B4n 2NC3=H3，假设：H3=合数=C4n 抽取素数：A4，B4，C4，使用。2NA4=L4，假设：L4=合数=A5n 2NB4=M4，假设：M4=合数=B5n 2NC4=H4，假设：H4=合数=C5n .抽取素数：A5、B5、C5、.模拟无穷假设，计算逻辑（WY1）。要么，2N=PaPb 。要么,2NPaPb假设：2NPaPb 。根据无穷计算两种选择：（A）模拟无穷，同步计算逻辑，获得素数计算循环。（B）模拟无穷，同步计算逻辑，获得素数计算循环不循环。则，无穷假设计算，无穷增加不相同素数。B假设：2NPaPb ,

47、无穷假设计算，无穷增加不相同素数。 无穷同步计算，无穷增加不相同素数。 无穷>2N 这个假设不成立。假设： 2NPaPb 。选择（A）模拟无穷，同步计算逻辑，素数循环。（2）转换第二种：模拟无穷计算逻辑。（WY2）（模拟同步计算，示意图）提取（WY1）计算获得全部素数。A1、A2、A3、B1、B2、B3、C1、C2、C3、使用。 2N2A1=2T1,假设：T1=合数=M1n 2N2A2=2T2,假设：T2=合数=M2n 2N2A3=2T3,假设：T3=合数=M3n 2N2B1=2T4,假设：T4=合数=M4n 2N2B2=2T5,假设：T5=合数=M5n 2N2A3=2T6,假设：T6=

48、合数=M6n 2N2C1=2T7,假设：T7=合数=M7n 2N2C2=2T8,假设：T8=合数=M8n。提取素因数：M1 、M2、M3、M4、M5、M6、M7、M8、使用。2NM1=Y1,假设：Y1=合数=S1n 2NM2=Y2,假设：Y2=合数=S2n2NM3=Y3,假设：Y3=合数=S3n 2NM4=Y4,假设：Y4=合数=S4n 2NM5=Y5,假设：Y5=合数=S5n2NM6=Y6,假设：Y6=合数=S6n 2NM7=Y7,假设：Y7=合数=S7n 2NM8=Y8,假设：Y8=合数=S8n 。根据（WY1）计算逻辑，假设素数又循环。提取素因数：S1 、S2、S3、S4、S5、S6、S

49、7、S8、使用。.模拟无穷计算逻辑。（WY2）要么，2N=PaPb 。要么，2NPaPb 。假设：2NPaPb 。根据模拟计算两种选择：（A）模拟无穷同步计算逻辑，素数循环。（B）模拟无穷同步计算逻辑，素数不循环。则，无穷假设计算，无穷增加不相同素数。假设：2NPaPb,无穷计算，无穷增加不相同素数。 无穷假设计算，无穷增加不相同素数。 无穷>2N 这个假设不成立。设： 2NPaPb 。它的选择（A），模拟无穷假设，同步计算逻辑，素数循环假设：2NPaPb。 第一种，和第二种计算逻辑，同时选择（A）（3）相互参照（WY2）计算逻辑获得素数，和（WY1）计算逻辑获得素数。（WY2）计算逻辑

50、获得素数，和（WY1）计算逻辑获得素数.，它们有相同的素数，假设：（WY2）和（WY1）计算逻辑获得素数相同。设：它们里面有S个素数，A1、A2、A3、B1、B2、B3、C1、C2、C3、2NA1=L1=合数=A2n 2N2CC=2TT=合数=A2A 2N2naE=2naAE A1+2CC=A2nA2A A1+2naE=A2n2naA2E （WY2）和（WY1），计算逻辑获得相同素数。（WY1）2NA1=L1，假设：L1=合数=A2n 2NB1=M1，假设：M1=合数=B2n 2NC1=H1，假设：H1=合数=C2n 2NA2=L2，假设：L2=合数=A3n 2NB2=M2，假设：M2=合数=B3n 2NC2=H2，假设：H2=合数=C3n 2NA3=L3，假设：L3=合数=A4n 2NB3=M3，假设：M3=合数=B4n 2NC3=H