Analysis:
Venn diagram: Used to show the overlapping relationship between elements.
Morgan's law:
The so-called prime number distribution problem in the additive relation a+b refers to the problem of the number of any sufficiently large positive integer M expressed as the sum of two positive integers. Because when x→∞, the addition relation can only give the limit of ∞+∞=2∞. Therefore, the study of the distribution of prime numbers in the additive relation a+b can only be carried out in the interval (0,2 ∞). There are:
2 ∞ =1+(2 ∞-1) = 2+(2 ∞-2) = ... = ∞ +∞ Obviously, in the addition relation a+b, when a→∞, then B can only be expressed as ∞+1 and beyond the natural number. Therefore, in the addition relation a+b, its cardinality has surpassed that of natural number set N. It is concluded that the elements in the given addition relation a+b of M are *** G, and like natural number set N, the elements in *** G are transitive. ② Trigeminal nerve. ③ For each element a+b, as long as it is within the interval (1, ∞), it must be a successor. (4) good foundation. Therefore, the addition relation a+b conforms to zermelo-fraenkel's axiom of sum regularity, because the value of B in the infinite *** G element is originally an extension of natural numbers.
Obviously, Eratosthenes screening method can't be used to sequence the infinite *** G well. Because Eratosthenes screening method is only for natural sequences, its p=x-H is only applicable to the property that the investigated element has only one natural number. In the natural sequence, screening out any natural number will not affect the existence of other natural numbers. However, this is not the case in the addition relation a+b, because the elements in *** G are composed of the sum of two natural numbers, and screening out any natural number will inevitably affect the existence of another natural number. From quantitative change to qualitative change, the law obtained in natural sequence does not apply to the additive relationship A+B.
In this paper, we investigate the properties of the sum of two positive integers in the addition relation a+b, including prime plus prime, prime plus sum, and sum plus sum (excluding the addition with 1 here). So, in *** G, according to the principle of completeness, there are:
Prime number plus prime number =G- prime number plus complex number-complex number plus complex number is represented by symbols, including
P (1, 1) = g-{(p, h)+h (1, 1)} This formula is the famous Morgan's law in * * * theory: A ~ ∩ B ~ = (A ∨.
Because in the addition relation a+b, let m be the value, there are elements m =1+(m-1) = 2+(m-2) =1... = M/2+m/2 * * and m/2. Morgan's law is applied to the addition relation a+b: in the interval (1, m/2), all elements a+b with complex number properties are summarized as * * * * A；; ; Reset in the interval [M/2, M], where the property that a+b has a complex number is summarized as * * * * B；; ; Then A∪B=(p, H)+H( 1, 1) and.
(a ∪ b) ~ = g-(p, h)-h (1, 1) and the complement set A~ of *** A is the * * of all the elements with prime numbers in the interval (1, M/2); The complement set B~ of *** B is the sum of * * of all elements with prime number property in the interval [M/2, M]. So there is A~∩B~=p( 1, 1)
To sum up, there are
A ~ ∩ b ~ = p (1, 1) = g-(p, h)-h (1, 1) = (a ∪ b) ~ Morgan's law tells us that there are
Since it is an addition relationship, we must apply the formula in the addition ring. When m is set as the selected value, according to the unique decomposition theorem:
M = (p _ I) α * (p _ j) β *...* (p _ k) γ has.
M=np=(n-m)p+mp From this formula, we can know that the sum of prime factors with m is always added to the same element with another sum of prime factors with m. A+b determined by the unique decomposition theorem is called eigenvalue. Since multiples of p are always added to the same element, every other value of p will have a multiple of p plus an element. Therefore, in M=a+b, if the occurrence probability of multiples of the eigenvalue p is 1/p, then the occurrence probability of elements that are coprime with it is (1- 1/p).
In addition, according to the residual class ring
From the formula M=nq+r=(n-m)q+mq+r, we can see that any prime number Q is not a multiple of the prime factor of M, and it can never be added to the composite number with the prime factor Q in the same element, and R is the difference between the two. In order to distinguish it from eigenvalue, we call it residual value, which is obtained from residue ring according to it. Because each value of r < q and q has two elements with prime factor q, one in A and the other in B, in M=a+b, the occurrence probability of multiples of residual value q is 2/q, and the occurrence probability of elements that are coprime with it is (1-2/q).
For the coefficient (1- 1/p) which is coprime with the eigenvalue p, we can know from Euler function ψ(N) that the coefficient in the eigenvalue p is an integrable function: m/2 {∏ p | m} (1-1/). So, is the coefficient of residual Q also an integrable function? Because the coefficient (1-2/q) which is coprime with residual value has never been involved before, it is my initiative, so it is necessary to demonstrate whether it is an integrable function.
Let N=nq+r=(n-m)q+mq+r, and change mq+r into a multiple of p, that is, mq+r=kp. We can see that "q is not divisible by kp, so the number of (q- 1) is: p, 2p, ..., (q-65438). Because k < q, in M=a+b, the multiple of p added to the same element starts from M=(n-m)q+kp. If pq is added or subtracted continuously, there will be m = (n-m-IP) q+(k+IQ) p; Every pq value appears 1≤i≤M/pq.
Therefore, in M=a+b, the multiples of Q and the coprime of P should not only screen out the elements of the prime factor of P in (n-m)q itself, but also screen out the composite number of the prime factor of P in the composite number of its constituent elements pair mq+r=kp. Therefore, in M=a+b, the number of coprime with p in element A+b composed of multiples of q is M/q( 1-2/p).
In M=a+b, if p⊥M and q⊥M (where the symbol ⊥ indicates inseparability), the elements a+b that are coprime with p and q are: M/2( 1-2/p) and m/2 (65433) respectively.
m/2( 1-2/p)-m/q( 1-2/p)=(m/2-m/q)( 1-2/p)= m/2( 1-2/p)。 In other words, in M=a+b, the coprime coefficient whose prime number is not more than √M is calculated by the principle of gradual elimination, and the eigenvalues and residuals are integrable functions.
Through analysis, we know that in M=a+b, both eigenvalue and non-eigenvalue are integrable functions. Therefore, in M=a+b, the number of coprime with prime number less than √M is:
P ( 1， 1)= m/2 {∏p | m }( 1- 1/p){∏p⊥m }( 1-2/p)。 From the coefficients of the formula, we can clearly see the function of Morgan's law: using prime numbers not greater than √M as a sieve, for multiples of prime numbers that are prime factors of m, the screened coefficient is (1-1/p); For multiples of prime numbers with prime factors other than m, the screening coefficient is (1-2/p).
When m is odd, because prime number 2 is not an eigenvalue, we can know that there is a zero factor from the coefficient of the remainder: (1-2/2)=0, so when m is odd, the sum of two odd prime numbers is zero.
Therefore, in the addition relation a+b, the number of p (1, 1) is required, and the value of m must be even, that is, prime number 2 must be an eigenvalue to get the number of p (1, 1). It can be seen from (1-1/p) > (1-2/p) that if there are other prime numbers not greater than √M as eigenvalues, the coefficient can't be minimum. So only when m = 2 n will there be a minimum coefficient, p (1,1) = m/4 ∏ (1-2/p) = m/4 ∏ ({p-2}/p).
According to the arrangement order of prime numbers in natural sequence, the difference between two adjacent prime numbers is greater than 2, at least not less than 2, so there is (p_n)-2≥(p_{n- 1}). (2) Substitute the conclusion of inequality (2) into (1). We can get p (1,1) ≥ m/4 (1/p) ≥ m/4 (1/√ m) = √ m/4, and when M→∞, there is √M/4→∞. In other words, if the big even table is the sum of two odd prime numbers, then its number will not be less than √M/4. Therefore, if m is an even number, it is called Goldbach conjecture, and when a→∞, Goldbach conjecture holds.
Because the general solution is obtained when m is infinite, there will be some errors when m is finite. Even so, the coefficient can well reflect the law that a big even number is the sum of two odd prime numbers. Because from the coefficient analysis: for the m with the same eigenvalue, the bigger m is, the more p( 1, 1): p( 1, 1)≥Lim(√N/4)→∞.
For N with different eigenvalues, the smaller the eigenvalue, the more P (1, 1): if P < q, then (1-kloc-0//p) (1-2/q) > (1)
The more eigenvalues, the more p( 1, 1):
( 1- 1/p)>( 1-2/p).
Of course, these three factors must be organically combined to truly reflect the number of p( 1, 1).
Regarding the subset of residual class values in H( 1, 1) which have the same occurrence probability but do not intersect with each other, there are:
φ，H(f，e)，H(g，e)，...，H(α，e)，H(β，e)，H(γ，e)， ...
H(e，f)，φ，H(g，f)，...，H(α，f)，H(β，f)，H(γ，f)， ...
H(e，g)，H(f，g)，φ，...，H(α，G)，H(β，G)，H(γ，G)， ...
......
H(e，α)，H(f，α)，H(g，α)，...，φ，H(β，α)，H(γ，α)， ...
H(e，β)，H(f，β)，H(g，β)，...，H(α，β)，φ，H(γ，β)， ...
H(e，γ)，H(f，γ)，H(g，γ)，...，H(α，γ)，H(β，γ)，φ， ...
Where e < f < g