# User talk:Musictheory2math

## Contents

## Goldbach's conjecture

**Main theorem**: Let $\mathbb{P}$ is the set prime numbers and $S$ is a set that has been made as below: put a point at the beginning of each member of $\Bbb{P}$ like $0.2$ or $0.19$ then $S=\{0.2,0.3,0.5,0.7,...\}$ is dense in the interval $[0.1,1]$ of real numbers.

$\,$This theorem is a base for finding formula of prime numbers, because for each member of $S$ like $a$ with its special and fixed location into $(0.1,1)$ and a small enough neighborhood like $(a-\epsilon ,a+\epsilon )$, but $a$ is in a special relation with members of $(a-\epsilon ,a+\epsilon )$ but there exists a special order on $S$ into $(0.1,1)$ and of course formula of prime numbers has whole properties related to prime numbers simultaneous. There is a musical note on the natural numbers that can be discovered by the formula of prime numbers. Musictheory2math (**talk**) 16:29, 25 March 2017 (CET)

- True, $S$ is dense in the interval $(0.1,1)$; this fact follows easily from well-known results on Distribution of prime numbers. But I doubt that this is "This theorem is a base for finding formula of prime numbers". Boris Tsirelson (talk) 22:10, 16 March 2017 (CET)

- Dear Professor Boris Tsirelson , in principle finding formula of prime numbers is very lengthy. and I am not sure be able for it but please give me a little time about two months for my theories expression. Musictheory2math (
**talk**) 16:29, 25 March 2017 (CET)

- Dear Professor Boris Tsirelson , in principle finding formula of prime numbers is very lengthy. and I am not sure be able for it but please give me a little time about two months for my theories expression. Musictheory2math (

- You mean, how to prove that $S$ is dense in $(0.1,1)$, right? Well, on the page "Distribution of prime numbers", in Section 6 "The difference between prime numbers", we have $ d_n \ll p_n^\delta $, where $p_n$ is the $n$-th prime number, and $ d_n = p_{n+1}-p_n $ is the difference between adjacent prime numbers; this relation holds for all $ \delta > \frac{7}{12} $; in particular, taking $ \delta = 1 $ we get $ d_n \ll p_n $, that is, $ \frac{d_n}{p_n} \to 0 $ (as $ n \to \infty $), or equivalently, $ \frac{p_{n+1}}{p_n} \to 1 $. Now, your set $S$ consists of numbers $ s_n = 10^{-k} p_n $ for all $k$ and $n$ such that $ 10^{k-1} < p_n < 10^k $. Assume that $S$ is not dense in $(0.1,1).$ Take $a$ and $b$ such that $ 0.1 < a < b < 1 $ and $ s_n \notin (a,b) $ for all $n$; that is, no $p_n$ belongs to the set

\[ X = (10a,10b) \cup (100a,100b) \cup (1000a,1000b) \cup \dots \, ; \]

- all $ p_n $ belong to its complement

\[ Y = (0,\infty) \setminus X = (0,10a] \cup [10b,100a] \cup [100b,1000a] \cup \dots \]

- Using the relation $ \frac{p_{n+1}}{p_n} \to 1 $ we take $N$ such that $ \frac{p_{n+1}}{p_n} < \frac b a $ for all $n>N$. Now, all numbers $p_n$ for $n>N$ must belong to a single interval $ [10^{k-1} b, 10^k a] $, since it cannot happen that $ p_n \le 10^k a $ and $ p_{n+1} \ge 10^k b $ (and $n>N$). We get a contradiction: $ p_n \to \infty $ but $ p_n \le 10^k a $.
- And again, please sign your messages (on talk pages) with four tildas: ~~~~. Boris Tsirelson (talk) 20:57, 18 March 2017 (CET)

**Theorem** $1$: For each natural number like $a=a_1a_2a_3...a_k$ that $a_j$ is $j$_th digit for $j=1,2,3,...,k$, there is a natural number like $b=b_1b_2b_3...b_r$ such that the number $c=a_1a_2a_3...a_kb_1b_2b_3...b_r$ is a prime number. Musictheory2math (**talk**) 16:29, 25 March 2017 (CET)

- Ah, yes, I see, this follows easily from the fact that $S$ is dense. Sounds good. Though, decimal digits are of little interest in the number theory. (I think so; but I am not an expert in the number theory.) Boris Tsirelson (talk) 11:16, 19 March 2017 (CET)

Now I want state philosophy of **This theorem is a base for finding formula of prime numbers**: However we loose the induction axiom for finite sets (Induction axiom is unable for discovering formula of prime numbers.) but I thought that if change space from natural numbers with cardinal $\aleph_0$ to a bounded set with cardinal $\aleph_1$ in the real numbers then we can use other features like axioms and important theorems in the real numbers for working on prime numbers and I think it is a better and easier way. Musictheory2math (**talk**) 16:29, 25 March 2017 (CET)

- I see. Well, we are free to use the whole strength of mathematics (including analysis) in the number theory; and in fact, analysis is widely used, as you may see in the article "Distribution of prime numbers".
- But you still do not put four tildas at the end of each your message; please do. Boris Tsirelson (talk) 11:16, 19 March 2017 (CET)

Importance of density in the Main theorem is similar to definition of irrational numbers from rational numbers.

**Goldbach's conjecture** is one of the oldest and best-known unsolved problems in number theory and all of mathematics. It states: Every even integer greater than $2$ can be expressed as the sum of two primes.

Assume $S_1=\{10^{-n}\cdot a\, |\, a\in S$ for $n=0,1,2,3,...\}$ & $L=\{(a,b)\,|\,a,b \in S_1,\, 0.01 \le a+b \lt 0.1,$ $\exists m \in \Bbb N,$ $a \cdot 10^m,\,b \cdot 10^m\in\Bbb P\setminus\{2\}\}$

**Theorem**: $S_1$ is dense in the interval $(0,1)$ and $S_1\times S_1$ is dense in the $(0,1)\times (0,1)$.

If $p,q$ are prime numbers and $n$ is the number of digits in $p+q$ and $m=$max(number of digits in $p$, number of digits in $q$), let $\varphi :\Bbb P\times\Bbb P\to \Bbb N,$ $\varphi ((p,q)) = \begin{cases} m+1 & n=m \\m+2 & n=m+1 \end{cases}$

**Theorem**: For each $p,q$ belong to prime numbers and $\alpha \in \Bbb R$ that $0 \le \alpha,$ now if $\alpha = q/p$ then $L \cap \{(x,y)\,|\,y=\alpha x \}=\{10^{-\varphi ((p,q))}(p,q)\}$ and if $\alpha \neq q/p$ then $L \cap \{(x,y)\,|\,y=\alpha x \}=\emptyset $ and if $\alpha = 1$ then $L \cap \{(x,x)\}$ is dense in the $\{(x,x)\,|\,0.005 \le x \lt 0.05 \}$.

**Definition**: Assume $L_1=\{(a,b)\,|\,(a,b) \in L$ & $b \lt a \}$ of course members in $L$ & $L_1$ are corresponding to prime numbers as multiplication and sum and minus and let $E=(0.007,0.005)$ (and also $5$ points to form of $(0.007+\epsilon _1,0.005-\epsilon _2)$ that $\epsilon _2 \approx 2\epsilon _1$) is a base for homotopy groups! and let $A:=\{(x,y)\,|\, 0 \lt y\lt x,$ $0.01 \le x+y \lt 0.1\}$ & $V:=\{(a+b)\cdot 10^m \,|$ $(a,b) \in ((S_1 \times S_1) \cap A) \setminus L,$ there is a least member like $m$ in $\Bbb N$ such that $(a+b)\cdot 10^m \in \Bbb N \}$ & $r:\Bbb N\to (0,1)$ is a function given by $r(n)$ is obtained as put a point at the beginning of $n$ like $r(34880)=0.34880$ and similarly consider $\forall k\in\Bbb N\cup\{0\}$ $r_k: \Bbb N \to (0,1)$ by $r_k(n)=10^{-k}\cdot r(n)$.

**Conjecture** $1$: For each even natural number like $t=t_1t_2t_3...t_k$, then $\exists (a,b),(b,a)\in L\,\cap$ $\{(x,y)\,|\, x+y=0.0t_1t_2t_3...t_k\}$ such that $0.0t_1t_2t_3...t_k=a+b$ & $10^{k+1} \cdot a,10^{k+1} \cdot b\in\Bbb P\setminus\{2\}$.

- Conjecture $1$ is an equivalent to Goldbach's conjecture, this conjecture has two solutions $1)$ Homotopy groups $\pi _n(X)$ (by using cognition $L_1$ from homotopy groups this conjecture is solved of course we must attend to two spheres because $S^2$ minus the tallest point in north pole as topological and algebraic is an equivalent with plane $\Bbb R^2$ (except $\infty$) and also every mapping is made between these two spheres easily if these spheres aren't concentric.), using homotopy theory linear problems like Goldbach can be solved and $2)$ Algebraic methods.
**Assuming**conjecture $1$, it guides us to finding formula of prime numbers at $(0,1) \times (0,1),$ in natural numbers based on each natural number is equal to half of an even number so in natural numbers main role is with even numbers but when we change space from $\Bbb N$ to $r(\Bbb N)$ then main role will be with $r( \{2k-1\, | \, k \in \Bbb N \} )$, because $r(\{2k-1\,|\,k\in \Bbb N\})\subset r(\{2k\,|\,k\in \Bbb N\})$ or in principle $r(\Bbb N)=r(\{2k\,|\,k\in \Bbb N\})$ for example $0.400=0.40=0.4$ or $0.500=0.50=0.5$ but however a smaller proper subset of $r( \{2k-1\, | \, k \in \Bbb N \} \cup \{2\} )$ namely $S$ is helpful, but for finding formula of prime numbers we need to all power of Main theorem not only what such that is stated in above conjecture namely for example we must attend to the set $V$ too!**Conjecture**$2$: For each even natural number like $t=t_1t_2t_3...t_k,$ $\exists x\in \{\alpha (a^2+b^2)^{0.5}\,|$ $(a,b) \in L,$ $\alpha \in (1,\sqrt 2] \} \cap r_1(\{2k\,|\, k\in \Bbb N \} )$ such that $t=10^{k+1} x$.- Conjecture $2$ is an equivalent to conjecture $1$, because $\forall t=t_1t_2t_3...t_k \in \Bbb N$ that $t$ is even, $\forall (a,b)\in \{(x,y)\,|\, x+y=0.0t_1t_2t_3...t_k,\, 0\lt y\le x \}$ we have: $(a^2+b^2)^{0.5}\lt 0.0t_1t_2t_3...t_k\le \sqrt 2\cdot (a^2+b^2)^{0.5}$ so by intermediate value theorem we have $0.0t_1t_2t_3...t_k=\alpha (a^2+b^2)^{0.5}$ that $1\lt \alpha \le \sqrt 2$. But now if $a=10^{-k-1} p,b=10^{-k-1} q$ for $p,q$ belong to prime numbers we have:
_{$$\alpha = \frac{t}{\sqrt {p^2+q^2}}$$}

**Theorem**: $\forall p,q,r,s$ belong to prime numbers & $q \lt p$ then $(p,q)$ is located at the direct line contain the points $(0,0),10^{-\varphi ((p,q))}(p,q)$ and if $(r,s)$ is belong to this line then $p=r$ & $q=s$.

Let $S^2_2$ be a sphere with center $(0,0,r_2)$ and radius $r_2$ and $S^2_1$ be a sphere with center $(0.007,0.005,c)$ and radius $r_1$ such that $S^2_1$ is into the $S^2_2$ now suppose $f_1,f_2$ are two mapping from $A$ to $S^2_2$ such that $1)$ if $x \in A,$ $f_1 (x)$ is a curve on $S^2_2$ that is obtained as below: from $x$ draw a direct line that be tangent on $S^2_1$ and stretch it till cut $S^2_2$ in curve $f_1 (x)$ and $2)$ if $x \in A,$ $f_2 (x)$ is a curve on $S^2_2$ that is obtained as below: from $x$ draw a direct line that be tangent on $S^2_1$ and then in this junction point draw a direct line perpendicular at $S^2_2$ till cut $S^2_2$ in curve $f_2 (x)$.

Let $f_3: L_1 \to S^1$ is a mapping with $f_3 ((a,b)) = (a^2+b^2)^{-0.5}(a,b)$ and $f_4:L_1 \to S^2$ is a mapping with $f_4 ((a,b)) = (a^4+a^2b^2+b^2)^{-0.5}(a^2,ab,b) $

**Guess** $1$: $f_3 (L_1) $ is dense in the $S^1 \cap \{ (x,y)\,|\, 0 \le y$ & $2^{-0.5} \le x \}$.

Let $U: S^2_2 \setminus \{(0,0,2r_2)\} \to \{(x,y,0)\,|\, x,y \in \Bbb R \}$ is a mapping such that draw a direct line by $(0,0,2r_2)$ & $(x,y,z)$ till cut the plane $\{(x,y,0)\,|\, x,y \in \Bbb R \}$ in the point $(x_1,y_1,0)$. Now must a group be defined on the all the points of $S^2_2 \setminus \{(0,0,2r_2)\}$.

Let $G=S^2_2 \setminus \{(0,0,2r_2)\}$ be a group by operation $g_1 + g_2 = U^{-1} (U(g_1)+U(g_2))$ that second addition is vector addition in the vector space $(\Bbb R^2,\Bbb Q,+,.)$ and now we must attend to subgroups of $G$ particularly $y=\pm x,\,y=0,\,x=0$

**Theorem**: Let $K_3 =\{p+q+r\,|\, p,q,r \in \Bbb P \}$ then $r(K_3)$ is dense in the interval $(0.1,1)$ of real numbers. Proof from Goldbach's weak conjecture

**Guess** $2$: Let $K_2 =\{p+q\,|\,p,q \in \Bbb P \}$ then $r(K_2)$ is dense in the interval $(0.1,1)$ of real numbers.

Let $F= \Bbb Q$ so what are Galois group of polynomials $x^4+b^2x^2+b^2$ and $(1+a^2)x^2 +a^4$.

**Theorem** $2$: If $(a,b),(c,d)\in \{(u,v)\,|\, u,v\in S_1$ & $0.01\le u+v\lt 0.1$ & $0\lt v\lt u \}$ and $(a,b),(c,d),(0,0)$ are located at a direct line then $(a,b)=(c,d)$.

- Proof: Suppose $A_1=\{(x,y)\,|\, y \lt x\lt 0.01,\, x+y\ge 0.01\}$ & $A_2=\{(x,y)\,|\, y\lt x\lt 0.1,$ $x+y\ge 0.1\}$ so $\forall (x,y) \in A_2 :\,\, 0.1(x,y)\in A_1$ & $\forall (x,y) \in A_1 :\,\, 10(x,y)\in A_2$ so theorem can be proved in $A_3=\{(x,y)\,|\, 0\lt y\lt x\lt 0.1,\, x\ge 0.01\}$ instead $A$, but in $A_3$ we have: $\forall (x_1,y_1),(x_2,y_2)\in A_3\cap (S_1\times S_1)$ so $x_1=10^{-r_1}p_1,\, y_1=10^{-s_1}q_1,$ $x_2=10^{-r_2}p_2,$ $y_2=10^{-s_2}q_2$ and if ${{y_1}\over {x_1}}$=${{y_2}\over {x_2}}$ then ${{10^{-s_1}q_1}\over {10^{-r_1}p_1}}$=${{10^{-s_2}q_2}\over {10^{-r_2}p_2}}$ so $p_1=p_2,\, q_1=q_2$ so $x_1=x_2$ so $y_1=y_2$ therefore $(x_1,y_1)=(x_2,y_2)$.

Let $Y=\{(a,b)\,|\, (a,b)\in (S_1\times S_1)\setminus L,\, 0.01\le a+b\lt 0.1\}$ & $\forall i\in \Bbb N,$ $E_i=\{(a,b)\,|$ $(a,b)\in S_1\times S_1,$ $a+b=r_1(2i)\}$ & $O_i=\{(a,b)\,|\, (a,b)\in S_1\times S_1,\, a+b=r_1(2i-1)\}$.

In theorem $2$ I obtained a cognition to $(S_1\times S_1)\cap A$ from $(0,0)$ but now I want do it from $\infty$; in the trapezoid shape with vertices $\{(0.1,0),(0.01,0),(0.05,0.05),(0.005,0.005)\}$ intersection of two direct lines contain points $\{(0.1,0),(0.01,0)\}$ & $\{(0.05,0.05),(0.005,0.005)\}$ is $(0,0)$ so we can describe $(S_1\times S_1)\cap A$ from $(0,0)$ but when we look at two parallel lines contain points $\{(0.1,0),(0.05,0.05)\}$ & $\{(0.01,0),(0.005,0.005)\}$ there isn't any point as a criterion for description of $Y$ or $L$ only inaccessible $\infty$ remains to description or the same these parallel lines contain points $\{(0.1,0),(0.05,0.05)\}$ & $\{(0.01,0),(0.005,0.005)\}$.

**Theorem** $3$: $1)\, \forall x\in [0.01,0.1)\setminus r_1 (\Bbb N),\,\, \{(u,v)\,|\, u+v=x\}\cap (S_1\times S_1)=\emptyset ,\,\,$ $2)\, \forall i\in \Bbb N,$ $E_i\subsetneq L,\, O_i\cap L\neq \emptyset \neq O_i\cap Y\neq O_i,\qquad Y=(\bigcup _{i\in \Bbb N} O_i )\setminus L,$ $L=(\bigcup _{i\in \Bbb N} E_i)\cup (\bigcup _{i\in \Bbb N} (O_i\cap L)),\,\,$ $3)\, \forall i\in \Bbb N,$ cardinal$(O_i \setminus L)\in \Bbb N$.

- Proof: $1,2)\, \forall (u,v)\in \{(x,y)\,|\, 0\lt y,x,\, 0.01\le x+y\lt 0.1\}$ be aware to summation $u+v$ at the lines $x+y=c$ for $0.01\le c\lt 0.1 \,\,\,3)\, \forall i\in\Bbb N,\, 2i-1$ can be written as utmost $2i-1$ summation to form of $a\cdot 10^m+b\cdot 10^n$ that $m\neq n,\, a,b\in S_1,\, a\cdot 10^m,b\cdot 10^n$ are prime numbers and or to form of $2+b\cdot 10^n$ that $b\in S_1,\, b\cdot 10^n$ is a prime number.

**Guess** $3$: $\forall i\in \Bbb N,$ cardinal$(E_i)=\aleph_0 =$cardinal$(O_i \cap L)$

**Guess** $4$: $L_1$ is dense in $\{(x,y)\,|\, 0\le y\le x,\, 0.01\le x+y\le 0.1\}$, and $M:=\{(a,b)\,|\, a,b\in S_1,\,\exists m\in\Bbb N,$ $a\cdot 10^m,b\cdot 10^m\in\Bbb P\setminus\{2\}\}$ is dense in the $[0,1]\times [0,1]$.

**Guess** $5$: $\forall i\in \Bbb N,\, E_i\cap \{(x,y)\,|\, y\le x\}$ is dense in the $\{(x,y)\,|\, x+y=r_1(2i),\, 0\le y\le x\}$ and $O_i\cap L\cap \{(x,y)\,|\, y\le x\}$ is dense in the $\{(x,y)\,|\, x+y=r_1(2i-1),$ $0\le y\le x\}$.

**A rectangle**: Suppose $B$ is a rectangle with vertices $(0.105,-0.005),(0.05,0.05),(0.005,0.005),$ $(0.06,-0.05)$ that one of usages of this rectangle is for writing each even natural number as minus of two prime numbers but during calculations we must use $\{(x,0)\,|\, 0.01\le x\lt 0.1\}$, but the question is however this rectangle as topological isn't equivalent to the plane $\Bbb R^2$ and each point in this rectangle is corresponding to a infinite set with cardinal $\aleph_0$ in $\Bbb R^2$ but which concept on the plane $\Bbb R^2$ is corresponding to the density concept on this rectangle.

Now I want find a relation between $L_1$ & $W:=((S_1\times S_1) \cap A) \setminus L$.

**Theorem**: Let $K =\{2k \,|\, k \in \Bbb N \}$ so $r(K)$ is dense in the interval $(0.1,1)$ of real numbers. Proof from the Main theorem and this $r(p)=r(p\cdot 10)$ that $p$ is a prime number then $p\cdot 10$ is an even number and $\{ p\cdot 10 \,|\, p \in \Bbb P \} \subset K$.

Main theorem as a result of prime number theorem is a fundamental concept in number theory also multiplication operation is a base in normal definition of prime numbers so logarithm function as an inverse of $f(a)=a^n$ has some or whole prime numbers properties that has been used in prime number theorem and consequently in the Main theorem. But I want offer a new theory with researching on logarithmic functions that it can be a useful discussion in number theory.

**Now** a new definition of prime numbers based on mapping $r$ is necessary, presently I have an idea consider $\forall k\in \Bbb N,$ the sequence $b_k:\Bbb N \to \{1,2,3,4,5,6,7,8,9\},\,b_k(n)$ is the last digit in $k^n,$ so if $k=k_1k_2k_3...k_r$ then $b_k(1)=k_1$ and if $k^n=t_1t_2t_3...t_s$ so $b_k(n)=t_1,$ but for primes $k,$ it is a special different pattern than composite numbers and of course I want find some properties on $r$ for example $r(m \cdot n)$ when last digit is $1,2$ or $3$ or $4,5,6,7,8,9$, of course for $3$ penultimate digit (and probably two to last digit) is important and in addition is there any way for assessment location $r(m\cdot n)$ from $r(m)$ & $r(n)$.

**Our weakness is from basic concepts**, I want obtain a cognition of $(S_1\times S_1)\cap A$ and $L_1$ from point $(0.02,0.03)$ like theorem $2$ from $(0,0)$, but this time it is an equivalent to a new definition of $S_1$, because intersection of two direct lines contain points $\{(0.1,0),(0.01,0)\}$ & $\{(0.05,0.05),(0.005,0.005)\}$ is the point $(0,0)$ but however two direct lines contain points $\{(0.1,0),(0.05,0.05)\}$ & $\{(0.01,0),(0.005,0.005)\}$ are parallel so imposition of point $(0.02,0.03)$ as a criterion only can be equilibrated by concept of the set $S_1$!

**Hypothesis** $1$: $\forall m,t\in\Bbb N$ if $t=t_1t_2t_3...t_k\cdot 10^{m-1}$ for $t_k=2,4,6,8$ or $t=t_1t_2t_3...t_k\cdot 10^m$ for $t_k=1,3,5,7,9$ then if $\forall p,q\in \Bbb P$ that $p,q\lt t$ implies $t\neq p+q$ then $\exists M\subseteq \Bbb N$ that cardinal$(M)=\aleph_0$ so $\forall i\in M$ if $\forall r,s\in\Bbb P$ that $r,s\lt t\cdot 10^{i-m}$ then $t\cdot 10^{i-m}\neq r+s$.

- I believe the Goldbach's conjecture is truth so I think with proof by contradiction we can prove the Goldbach's conjecture.
- I think with assuming this hypothesis a contradiction will be obtained and consequently the Goldbach's conjecture will be proved.

From Dirichlet's theorem on arithmetic progressions that says: for any two positive coprime integers $a$ and $d$, there are infinitely many primes of the form $a+nd$, where $n$ is a non-negative integer, and from existence an one-to-one correspondence between the sets $A_1$ & $A_2$ in theorem $2$, I think there is a fixed in the following sets, $\forall p\in\Bbb P$ if $0.01\le r_1(p)\le 0.05$ the set $\{(a,b)\in L_1\,|\, a=r_1(p)\},$ and if $0.05\lt r_1(p)\lt 0.1$ the set $\{(a,b)\in L_1\,|\, a=r_1(p)\}\cup\{(a,b)\in L_1\,|\, a=r_2(p)\}$.

**Guess** $6$: $\forall t,r,s\in\Bbb N$ that $t=t_1t_2t_3...t_k$ is even and $r,s$ are odd and $s\le r$ and $t=r+s$ then $(10^{-k-1}r,10^{-k-1}s)\in A\cup \{(x,x)\,|$ $0.005\le x\lt0.05\}$.

If $\{\mathrm I_{\theta}\}_{\theta\in\mathcal I}$ is a partition for $U:=\{(p,q)\,|\, p,q\in\Bbb P,$ $q\le p\}$ then $\{J_{\theta}\}_{\theta\in\mathcal I}$ is a partition for $L_2:=L_1\cup \{(a,a)\in L\}$ such that $\forall\theta\in\mathcal I,$ $J_{\theta}=\{(r_s(p),r_t(q))\in L_2\,|\, r_t(q)\le r_s(p),$ $(p,q)\in\mathrm I_{\theta},$ $s,t\in\Bbb N\}\cup\{(r_t(q),r_s(p))\in L_2\,|\, r_s(p)\le r_t(q),$ $(p,q)\in\mathrm I_{\theta},$ $s,t\in\Bbb N\}$.

**In the following** I need to some partitions for $L_2$ or the same cognitions to $L_2,$ however we should be aware to the details of trapezoid shape with vertices $\{(0.1,0),(0.01,0),(0.05,0.05),(0.005,0.005)\}$, the problem is which partition is a good cognition to $L_2$

**Guess** $7$: $\forall a,d\in\Bbb N$ that gcd$(a,d)=1$ then cardinal$(\{(r_s(n),r_t(a+nd))\in L_1\,|\, n=0,1,2,3,...,$ $r,s\in\Bbb N\}\cup\{(r_t(a+nd),r_s(n))\in L_1\,|\, n=0,1,2,3,...,\, r,s\in\Bbb N\})=\aleph_0$

**Theorem** $4$: $\forall t\in\Bbb N\,\forall p\in\Bbb P\,1)\,$if $0.01\le r_1(p)\le 0.05$ then cardinal$(\{(r_1(p),u)\in L_2\})\in\Bbb N,\,2)$ if $0.05\lt r_1(p)\lt 0.1$ then cardinal$(\{(r_1(p),u)\in L_2\}\cup\{(r_2(p),u)\in L_2\})\in\Bbb N,\, 3)$ if $r_1(p)\lt 0.05$ then cardinal$(\{(u,r_t(p))\in L_2\})\in\Bbb N$ & cardinal$(\{(u,v)\in L_2\,|\, v\in\bigcup _{m\in\Bbb N} \{r_m(p)\}\})=\aleph_0$

- Proof: Be aware to number of digits in prime numbers corresponding to coordinates of each member in $L_2$ and Dirichlet theorem on arithmetic progressions.

**Guess** $8$: $\forall c\in (\bigcup _{k=2}^{\infty} r_k(\Bbb N))\cup\{0.01\},\,$cardinal$(\{(u,v)\in L_1\,|\, u-v=c\})=\aleph_0$

**Hypothesis** $2$: $\forall k,m\in\Bbb N,\,$that gcd$(2,m)=1$ and $\forall p,q\in\Bbb P$ with $p,q\lt 2^km$ that $2^km\neq p+q$ then $\qquad\qquad$ cardinal$(\{t\in\Bbb N\,|\, 2^tm\neq p+q$ for $\forall p,q\in\Bbb P$ that $p,q\lt 2^tm\})=\aleph_0$

Now by using Chen's theorem that says: Every sufficiently large even number can be written as the sum of either two primes, or a prime and a semiprime(the product of two primes) and its extension by Tomohiro Yamada that says: Every even number greater than ${\displaystyle e^{e^{36}}\approx 1.7\cdot 10^{1872344071119348}}$ is the sum of a prime and a product of at most two primes, and Chen's theorem $II$ that is a result on the twin prime conjecture, It states that if $h$ is a positive even integer, there are infinitely many primes $p$ such that $p+h$ is either prime or the product of two primes and Ying Chun Cai's theorem that says: there exists a natural number $N$ such that every even integer $n$ larger than $N$ is a sum of a prime less than or equal to $n^{0.95}$ and a number with at most two prime factors, I want make some partitions for $L_2$ (in principle I want write Gaussian integers with prime coordinates in several equations.):

Theorem: Let $(X,T_1),\,(Y,T_2)$ be topological spaces and $H$ be a homeomorphism from $X \to Y$. If $C$ is a dense subset of $X$, $H(C)$ is dense in the $Y$ necessarily.

- Proof given by Daniel Schepler from stackexchange.com: Since density is a property which only depends on the topology, this is true, namely, suppose $U$ is a nonempty open subset of $Y$, then since $H$ is bijective, we can rewrite $U \cap H(C) = H(H^{-1}(U) \cap C)$, however, $H^{-1}(U)$ is open by continuity of $H$, and nonempty since $H$ is surjective, therefore, $H^{-1}(U) \cap C$ is nonempty since $C$ is dense; and therefore, $U \cap H(C) = H(H^{-1}(U) \cap C)$ is also nonempty. (So, this proves only using that $H$ is continuous and bijective, it is actually possible to refine the proof to work only assuming that $H$ is continuous and surjective - in that case, $U \cap H(C) \supseteq H(H^{-1}(U) \cap C)$.)
- Another proof given by José Carlos Santos from stackexchange.com: Let $y\in Y$; for proof of every neighborhood $N$ of $y$, $N\cap H(C)\neq\emptyset$, take $x\in X$ such that $f(x)=y$, then $f^{-1}(N)$ is a neighborhood of $x$ and therefore $f^{-1}(N)\cap C\neq\emptyset$. So, $N\cap H(C)\neq\emptyset$.

**Theorem**: $r(\{2p\,|\, p\in\Bbb P\})$ and $r(\{5p\,|\, p\in\Bbb P\})$ are dense in the interval $[0.1,1]$.

- Proof: Under Euclidean topology, mapping $f_1:\{(x,x)\,|\, x\in [0.005,0.05)\}\to\{(x,0)\,|\, x\in [0.01,0.1)\}$ by $f_1((a,a))=(2a,0)$ & $f_2:\{(x,0)\,|\, x\in [0.01,0.1)\}\to \{(x,x)\,|\, x\in [0.005,0.05)\}$ by $f_2((a,0))=(0.5a,0.5a)$ are homeomorphism and also $\forall r_1(p)\in [0.01,0.02]$ we have $0.5r_1(p)=r_2(5p)$ & $\forall r_1(p)\in (0.02,0.1)$ we have $0.5r_1(p)=r_1(5p)$ of course by another topology we should transfer density to $[0.1,1]$.
- Question $1$: According to gcd$(2,5)=1$ & $2\cdot 5=10$ & $(5-2)-2=1$ so $\forall q\in\Bbb P,$ is $r(\{pq\,|\, p\in\Bbb P\})$ dense in the $[0.1,1]$

Assume $\forall m,n\in\Bbb N$

$\begin{cases} n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\end{cases}$

and $p_n\star _1p_m=p_{n\star m}$ that $p_n$ is $n$_th prime & $\forall (p_n,p_m),(p_s,p_t)\in\Bbb P\times\Bbb P,$ $(p_n,p_m)\star _2(p_s,p_t)=(p_n\star _1p_s,p_m\star _1p_t)$.

Obviously $(\Bbb N,\star)$ & $(\Bbb P,\star _1)$ & $(\Bbb P\times\Bbb P,\star _2)$ are groups and $\langle 2\rangle =(\Bbb N,\star)\cong (\Bbb Z,+)\cong (\Bbb P,\star _1)=\langle 3\rangle$ & $\langle (3,2),(2,3)\rangle =(\Bbb P\times\Bbb P,\star _2)\cong (\Bbb Z\times\Bbb Z,+)$ & $\pi _n(\Bbb P\times\Bbb P)\cong\pi _n(\Bbb Z)\times\pi _n(\Bbb Z)$.

In $(\Bbb N,\star)$ we have: $(2k+1)^n=2kn+1=(2k)^{-n},\quad (2k)^n=2kn=(2k+1)^{-n}$

and let $Q_1=\{{m\over n}\,|\, m,n\in\Bbb N\}$, obviously $(Q_1,\star _{Q_1})$ is an Abelian group with:

$\begin{cases} \forall m,n,u,v\in\Bbb N,\, {m\over n}\star _{Q_1} 1={m\over n}\\ ({m\over n})^{-1}={m^{-1}\over n^{-1}}\qquad m\star m^{-1}=1=n\star n^{-1}\\ {m\over n}\star _{Q_1} {u\over v}={m_1\over n_1}\star _{Q_1} {u_1\over v_1}={{m_1\star u_1}\over {n_1\star v_1}}\quad\text{if}\,\,\begin{cases} {m\over n}={m_1\over n_1},\,\, {u\over v}={u_1\over v_1},\,\, {mu\over nv}={m_1u_1\over n_1v_1}\\ \text{gcd}(m_1,n_1)=1=\text{gcd}(m_1,v_1)=\text{gcd}(u_1,n_1)=\text{gcd}(u_1,v_1)\end{cases}\end{cases}$

- I mean is first simplification of fractions and then calculation like \({44 \over 12}\) that gets \({11\over 3}\).
**Question**$2$: does \((Q_1,\star _{Q_1})\cong (\Bbb Q,+)\)?**Problem**$1$: Suppose \(N_1=\{(m,n)\,|\, m,n\in\Bbb N,\, \gcd(m,n)=1\}\), make a group on \(N_1\) correspondence to \((Q_1,\star _{Q_1})\) in terms of members production and isomorph to \((Q_1,\star _{Q_1})\).

According to pairing function let $Q_p=\{(s,t)\,|\, s,t\in\Bbb N\cup\{0\}\}$ and $Q_p$ is a cyclic group as:
$\begin{cases}
\forall (s,t),(u,v)\in Q_p,\quad e=(0,0),\quad (Q_p,\star _p)=\langle (1,0)\rangle\\
(s,t)^{-1}=\pi ^{-1} [(0.5(s+t)(s+t+1)+t+1)^{-1}]\\
(s,t)\star _p (u,v)=\pi ^{-1} [(0.5(s+t)(s+t+1)+t+1)\star (0.5(u+v)(u+v+1)+v+1)]\end{cases}$

Of course each sequence $\{a_n\}$ that $\forall n,m\in\Bbb N,\,n\neq m$ then $a_n\neq a_m$ is a cyclic group as: $a_n\star _aa_m=a_{n\star m},\, e=a_1,\, G=\langle a_2\rangle$

**Algorithm** of making of new cyclic groups on $\Bbb N$:

Let $(\Bbb N,\star)$ be that group and we have $\Bbb N=\langle 2\rangle$ & $e=1$ and at first write integers as a sequence with starting from $0$ for instance: $$0,1,2,-1,-2,3,4,-3,-4,5,6,-5,-6,7,8,-7,-8,9,10,-9,-10,11,12,-11,-12,...$$ $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,...$$ then regarding to the sequence find a rotation number that for this sequence is $4$ and hence equations should be written with module $4$, then consider $4m-2,4m-1,4m,4m+1$ that the last should be $km+1$ and initial be $km+(k-2)$ otherwise equations won't match with members inverse definitions, and make a table of productions of those $k$ elements but during writing equations pay attention if an equation is right for given numbers it will be right generally for other numbers for example $12\star 15=6$ or $(4\times 3)\star (4\times 4-1)=6$ because $(-5)+8=3$ & $-5\to 12,\,\, 8\to 15,\,\, 3\to 6,$ that implies $(4n)\star (4m-1)=4m-4n+2$ if $4m-1\gt 4n$ of course it's better at first members inverse be defined for example since $(-9)+9=0$ & $0\to 1,\,\, -9\to 20,\,\, 9\to 18$ so $20\star 18=1$, that implies $(4m)\star (4m-2)=1$, and with a little addition and subtraction all equations will be obtained simply that for this example is:

$\begin{cases} 0+8=8 & 1\star _{\Bbb N} 15=15\\ 1+8=9 & 2\star _{\Bbb N} 15=18\\ 2+8=10 & 3\star _{\Bbb N} 15=19\\ (-1)+8=7 & 4\star _{\Bbb N} 15=14\\ (-2)+8=6 & 5\star _{\Bbb N} 15=11\\ 3+8=11 & 6\star _{\Bbb N} 15=22\\ 4+8=12 & 7\star _{\Bbb N} 15=23\\ (-3)+8=5 & 8\star _{\Bbb N} 15=10\\ (-4)+8=4 & 9\star _{\Bbb N} 15=7\\ 5+8=13 & 10\star _{\Bbb N} 15=26\\ 6+8=14 & 11\star _{\Bbb N} 15=27\\ (-5)+8=3 & 12\star _{\Bbb N} 15=6\\ (-6)+8=2 & 13\star _{\Bbb N} 15=3\\ 7+8=15 & 14\star _{\Bbb N} 15=30\\ 8+8=16 & 15\star _{\Bbb N} 15=31\\ (-7)+8=1 & 16\star _{\Bbb N} 15=2\\ (-8)+8=0 & 17\star _{\Bbb N} 15=1\end{cases}$

that its group is:

$\begin{cases} m\star _{\Bbb N} 1=m\\ (4m)\star _{\Bbb N} (4m-2)=1=(4m+1)\star _{\Bbb N} (4m-1)\\ (4m)\star _{\Bbb N} (4m+2)=3=(4m+1)\star _{\Bbb N} (4m+3)\\ (4m-2)\star _{\Bbb N} (4n-2)=4m+4n-5\\ (4m-2)\star _{\Bbb N} (4n-1)=4m+4n-2\\ (4m-2)\star _{\Bbb N} (4n)=\begin{cases} 4m-4n-1 & 4m-2\gt 4n\\ 4n-4m+1 & 4n\gt 4m-2\end{cases}\\ (4m-2)\star _{\Bbb N} (4n+1)=\begin{cases} 4m-4n-2 & 4m-2\gt 4n+1\\ 4n-4m+4 & 4n+1\gt 4m-2\end{cases}\\ (4m-1)\star _{\Bbb N} (4n-1)=4m+4n-1\\ (4m-1)\star _{\Bbb N} (4n)=\begin{cases} 4m-4n+2 & 4m-1\gt 4n\\ 4n-4m & 4n\gt 4m-1\\ 2 & m=n\end{cases}\\ (4m-1)\star _{\Bbb N} (4n+1)=\begin{cases} 4m-4n-1 & 4m-1\gt 4n+1\\ 4n-4m+1 & 4n+1\gt 4m-1\end{cases}\\ (4m)\star _{\Bbb N} (4n)=4m+4n-3\\ (4m)\star _{\Bbb N} (4n+1)=4m+4n\\ (4m+1)\star _{\Bbb N} (4n+1)=4m+4n+1\end{cases}$

that this group $(\Bbb N,\star _{\Bbb N})$ is helpful for twin prime conjecture.

and the Klein four-group $(\Bbb Z _2\times\Bbb Z _2,+)$ is a fundamental concept in the group theory that its usage is for propositions rejection so I made below group somehow similar to that group in terms of members production for proof of the Goldbach's conjecture with proof by contradiction that is: $$0,1,2,-2,-1,3,-3,4,5,-5,-4,6,-6,7,8,-8,-7,9,-9,10,11,-11,-10,12,-12,...$$ $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,...$$ and $e=1$ and $(\Bbb N,\star _{K_4})=\langle 2\rangle$ so we have:

$\begin{cases} 0+(-7)=-7 & 1\star _{K_4}17=17\\ 1+(-7)=-6 & 2\star _{K_4}17=13\\ 2+(-7)=-5 & 3\star _{K_4}17=10\\ (-2)+(-7)=-9 & 4\star _{K_4}17=19\\ (-1)+(-7)=-8 & 5\star _{K_4}17=16\\ 3+(-7)=-4 & 6\star _{K_4}17=11\\ (-3)+(-7)=-10 & 7\star _{K_4}17=23\\ 4+(-7)=-3 & 8\star _{K_4}17=7\\ 5+(-7)=-2 & 9\star _{K_4}17=4\\ (-5)+(-7)=-12 & 10\star _{K_4}17=25\\ (-4)+(-7)=-11 & 11\star _{K_4}17=22\\ 6+(-7)=-1 & 12\star _{K_4}17=5\\ (-6)+(-7)=-13 & 13\star _{K_4}17=29\\ 7+(-7)=0 & 14\star _{K_4}17=1\\ 8+(-7)=1 & 15\star _{K_4}17=2\\ (-8)+(-7)=-15 & 16\star _{K_4}17=31\\ (-7)+(-7)=-14 & 17\star _{K_4}17=28\end{cases}$

that its group is:

$\begin{cases} m\star _{K_4}1=m\\ (6m-4) \star _{K_4}(6m-1)=1=(6m-3) \star _{K_4}(6m-2)=(6m) \star _{K_4}(6m+1)\\ (6m-4) \star _{K_4}(6n-4)=6m+6n-9\\ (6m-4) \star _{K_4}(6n-3)=6m+6n-6\\ (6m-4) \star _{K_4}(6n-2)=\begin{cases} 6m-6n-3 & 6m-4\gt 6n-2\\ 6n-6m+5 & 6n-2\gt 6m-4\end{cases}\\ (6m-4) \star _{K_4}(6n-1)=\begin{cases} 6m-6n & 6m-4\gt 6n-1\\ 6n-6m+1 & 6n-1\gt 6m-4\end{cases}\\ (6m-4) \star _{K_4}(6n)=6m+6n-4\\ (6m-4) \star _{K_4}(6n+1)=\begin{cases} 6m-6n-4 & 6m-4\gt 6n+1\\ 6n-6m+4 & 6n+1\gt 6m-4\end{cases}\\ (6m-3) \star _{K_4}(6n-3)=6m+6n-4\\ (6m-3) \star _{K_4}(6n-2)=\begin{cases} 6m-6n & 6m-3\gt 6n-2\\ 6n-6m+1 & 6n-2\gt 6m-3\end{cases}\\ (6m-3) \star _{K_4}(6n-1)=\begin{cases} 6m-6n+2 & 6m-3\gt 6n-1\\ 6n-6m-2 & 6n-1\gt 6m-3\\ 2 & m=n\end{cases}\\ (6m-3) \star _{K_4}(6n)=6m+6n-3\\ (6m-3) \star _{K_4}(6n+1)=\begin{cases} 6m-6n-3 & 6m-3\gt 6n+1\\ 6n-6m+5 & 6n+1\gt 6m-3\end{cases}\\ (6m-2) \star _{K_4}(6n-2)=6m+6n-1\\ (6m-2) \star _{K_4}(6n-1)=6m+6n-5\\ (6m-2) \star _{K_4}(6n)=\begin{cases} 6m-6n-2 & 6m-2\gt 6n\\ 6n-6m+2 & 6n\gt 6m-2\end{cases}\\ (6m-2) \star _{K_4}(6n+1)=6m+6n-2\\ (6m-1) \star _{K_4}(6n-1)=6m+6n-8\\ (6m-1) \star _{K_4}(6n)=\begin{cases} 6m-6n-1 & 6m-1\gt 6n\\ 6n-6m+3 & 6n\gt 6m-1\end{cases}\\ (6m-1) \star _{K_4}(6n+1)=6m+6n-1\\ (6m) \star _{K_4}(6n)=6m+6n\\ (6m) \star _{K_4}(6n+1)=\begin{cases} 6m-6n & 6m\gt 6n+1\\ 6n-6m+1 & 6n+1\gt 6m\end{cases}\\ (6m+1) \star _{K_4}(6n+1)=6m+6n+1\end{cases}$

Each one group structure on $\Bbb N$ will show a new outlook to the numbers, $\Bbb N$ or $\Bbb Z$ (and consequently $\Bbb R)$ is an initial and basic truth but when another group on $\Bbb N$ is made indeed another copy of truth will be inserted and yield will be a better knowledge at numbers!

**Hypothesis**$3$: With matrix theory each cyclic group on $\Bbb N$ like $(\Bbb N,•)$ that is made regarding above algorithm, but binary operation $•$ can be rewritten by matrices!

By using theorem $1$ of Polignac's conjecture we can define function $f:\{(c,d)\,|\, (c,d)\subseteq [0.01,0.1)\}\to\Bbb N$ that $f((c,d))$ is the least $n\in\Bbb N$ that $\exists t\in(c,d),\,\exists k\in\Bbb N$ that $p_n=t\cdot 10^{k+1}$ that $p_n$ is $n$_th prime and $\forall m\ge f((c,d))\,\,\exists u\in (c,d)$ that $u\cdot 10^{m+1}\in\Bbb P$

and $g:(0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\to\Bbb N,$ is a function by $\forall\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))$ $g(\epsilon)=max(\{f((c,d))\,|\, d-c=\epsilon,$ $(c,d)\subseteq [0.01,0.1)\})$.

**Guess** $9$: $g$ isn't an injective function.

Question $3$: Assuming guess $9$ let $[a,a]:=\{a\}$ and $\forall n\in\Bbb N,\, h_n$ is the least subinterval of $[0.01,0.1)$ like $[a,b]$ in terms of size of $b-a$ such that $\{\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\,|\, g(\epsilon)=n\}\subsetneq h_n$ and obviously $g(a)=n=g(b)$ now the question is $\forall n,m\in\Bbb N$ that $m\neq n$ is $h_n\cap h_m=\emptyset$?

- Guidance given by @reuns from stackexchange.com:
- For $n \in \mathbb{N}$ then $r(n) = 10^{-\lceil \log_{10}(n) \rceil} n$, ie. $r(19) = 0.19$. We look at the image by $r$ of the primes $\mathbb{P}$.

- Let $F((c,d)) = \min \{ p \in \mathbb{P}, r(p) \in (c,d)\}$ and $f((c,d)) = \pi(F(c,d))= \min \{ n, r(p_n) \in (c,d)\}$ ($\pi$ is the prime counting function)

- If you set $g(\epsilon) = \max_a \{ f((a,a+\epsilon))\}$ then try seing how $g(\epsilon)$ is constant on some intervals defined in term of the prime gap $g(p) = -p+\min \{ q \in \mathbb{P}, q > p\}$ and things like $ \max \{ g(p), p > 10^i, p+g(p) < 10^{i+1}\}$

**Now I want define a group on $L$ like $(L,\star _L)$:**

Let $P_1=\{v_n\,|\,\forall n\in\Bbb N,\, v_n$ is $(n+1)$_th prime$\}$ and $\forall n,m\in\Bbb N,\, v_n\star _3 v_m=v_{n\star m}$ & $e=3$, obviously $(P_1,\star _3)$ is a group and $\langle 5\rangle =(P_1,\star _3)\cong (\Bbb Z,+)$.

Let $\forall a\in\bigcup _{j\in\Bbb N} r_j(\Bbb P\setminus\{2\}),\, f:\bigcup _{j\in\Bbb N} r_j(\Bbb P\setminus\{2\})\to\Bbb P\setminus\{2\}$ be a function that $\exists k\in\Bbb N,$ $f(a)=a\cdot 10^k\in\Bbb P\setminus\{2\}$.

**Problem** $2$: Let $e=(0.03,0.03)$ and $\forall (a,b),(c,d)\in L,$ $$(a,b)\star _L (c,d)=10^{-\varphi ((f(a)\star _3f(c),f(b)\star _3f(d)))}\cdot (10^{-t_1(\alpha)}\cdot [f(a)\star _3f(c)],10^{-t_2(\alpha)}\cdot [f(b)\star _3f(d)])$$ that $\alpha=((a,b),(c,d)),$ where $t_1,t_2:L\times L\to\Bbb N\cup\{0\}$ are functions such that $\forall\alpha\in L\times L$ at least one of $t_1(\alpha),t_2(\alpha)$ should be $0$ so find functions $t_1,t_2$ such that $(L,\star _L)$ be a group (an Abelian group), though I guess $(L,\star _L)$ is non finitely generated Abelian group and even possibly isomorph to $(\Bbb Q,+)$ or $(\Bbb Q_{\gt 0},\cdot)$ or $A\oplus B$ that $A,B\in\{(\Bbb Q,+),(\Bbb Q_{\gt 0},\cdot)\}$.

- Of course we could write each another group structure on $\Bbb N$ and consequently on $P_1$ regarding above algorithm.
- And if under Euclidean topology (or another appropriate topology) this group $(L,\star _L)$ be a topological group it is very well, otherwise we can replace another group structure on $\Bbb N$ and then on $P_1$ and consequently on $L$.

Suppose $B=(S_1\times S_1)\cap\{(x,y)\,|\, 0.01\le x+y\lt 0.1\}$ & $D=B\setminus (L\cup\{(a,b)\in B\,|\, 2\neq f(a)=10^k\cdot a,$ $2=f(b)=10^t\cdot b,\, t\gt k,\, t,k\in\Bbb N\}\cup\{(a,b)\in B\,|\, 2=f(a)=10^k\cdot a,$ $2\neq f(b)=10^t\cdot b,\, t\lt k,$ $t,k\in\Bbb N\})$ & $N_2=\{f(a)+f(b)\,|\, (a,b)\in L\}$ & $N_1=\{a_1+10^t\cdot b_1\,|\, t=max($number of digits in $f(a),$ number of digits in $f(b))-min($number of digits in $f(a),$ number of digits in $f(b)),\,a_1=max(f(a),f(b)),$ $b_1=min(f(a),f(b)),\, (a,b)\in D\}$.

**Guess** $10$: $\{2n-1\,|\, n\in\Bbb N\}\setminus N_1$ is a finite set.

- In other words, $\{2n-1\,|\, n\in\Bbb N\}\setminus (\{p+10^k\cdot q\,|\, p,q\in\Bbb P,\, p\gt 2,\, k\in\Bbb N\}\cup\{p+2\,|\, p\in\Bbb P,\, p\gt 2\})$ is a finite set.
- Of course the Goldbach's conjecture is the same $N_2=\{2n\,|\, n\in\Bbb N\}$.

Obviously $B$ is dense in the $\{(x,y)\,|\, 0\le x,\, 0\le y,\, 0.01\le x+y\le 0.1\}$ and now I want define a group on $B$ like $(B,\star _B)$, though I guess this group will be non finitely generated Abelian group and even may be isomorph to $(\Bbb Q,+)$ or $(\Bbb Q_{\gt 0},\cdot),\, ...$

**Question** $4$: $\forall k\in\Bbb N$ are there $q_1,q_2\in P_1$ with $q_1,q_2\lt 4k+1$ such that $p_{4k+1}=p_{q_1}\star _3p_{q_2}$ and $p_{q_1},p_{q_2}\in P_1?$

**Conjecture** $3$: Under $(P_1,\star_3),\,\forall n\in\Bbb N,\,\exists s_1,s_2\in P_1$ such that $v_{2n+3}=v_{s_1}\star _3v_{s_2}$.

- According to equivalency of these three $(\Bbb Z,+)$ & $(\Bbb N,⋆)$ & $(P_1,⋆_3)$ from aspect of being group, this conjecture is an equivalent to Goldbach's conjecture.
- We can pay attention to subgroups as $\langle v_{s_1}⋆_3v_{s_2}\rangle$ as a solution for Goldbach also we can use quotient groups $P_1/ \langle v_s\rangle$.
- There exists an one-to-one correspondence between equations in $(\Bbb Z,+)$ & $(\Bbb N,⋆)$ & $(P_1,⋆_3)$ & $(\Bbb N,⋆_{K_4})$ and prime numbers properties exist in those groups although other sequences may have the same structures but prime numbers structures are specific because the own prime numbers are specific.
- There doesn't exist another way to define another cyclic group structure on $\Bbb P$ or $P_1$ for using strength of finite groups or infinite groups unless we knew the formula of prime numbers and on the other hand we can't know the formula before than knowing Goldbach and Polignac conjecture.

Question $5$: Is there any subgroup of $P_1$ like $H$ such that $\forall v_s\in H,\, s\notin P_1$?

- I want connect subgroups of $P_1$ with Goldbach's conjecture.

Question $6$: What is function of this sequence in $\Bbb N\times\Bbb N$: $(1,1),(1,2),(2,1),(1,3),(2,2),(3,1),(1,4),(2,3),(3,2),(4,1),...,(1,2k-2),(2,2k-3),...$ $,(k-1,k),(k,k-1),...,(2k-3,2),(2k-2,1),(1,2k-1),(2,2k-2),...,(k,k),...$ $,(2k-2,2),(2k-1,1),...$

- Answer given by Professor Daniel Lazard:
- $\begin{cases} (a_1,b_1)=(1,1)\\ (a_{n+1},b_{n+1})=\begin{cases} (1,a_n+1) & b_n=1\\ (a_n+1,b_n-1) & \text{else}\end{cases}\end{cases}$
- Question: What is function of this sequence: $2,3,3,4,4,4,5,5,5,5,6,6,6,6,6,...,k,k,k,...,k,...$ that \(k\) repeats \(k-1\) times.
- Answer given by Professor Daniel Lazard: Your sequence is clearly a primitive recursive function. On the other hand, I guess that, from your point of view, the best answer for your question is: $f(n)=k,$ where $k$ is the smallest integer such that \(k(k-1)/2 \ge n\). This may be rewritten, using the ceil function, \(f(n)=\left \lceil \frac{1+\sqrt{1+8n}}{2}\right\rceil. \) IMO, none of these formulas is useful, as they hide the main properties of the sequence.
- Problem $3$: Find a bijection \(g:\Bbb N\times(\Bbb N\setminus\{1\})\to\Bbb N\times\Bbb N\) as \(g(n,f(n))\) equal to function of the sequence: $(1,1),(1,2),(2,1),(1,3),(2,2),(3,1),(1,4),(2,3),(3,2),(4,1),...,(1,2k-2),$ $(2,2k-3),...,(k-1,k),(k,k-1),...,(2k-3,2),(2k-2,1),(1,2k-1),(2,2k-2),$ $...,(k-1,k+1),(k,k),(k+1,k-1),... ,(2k-2,2),(2k-1,1),...$
- Question: To define an Abelian group structure on $\Bbb N$ that is not a finitely generated Abelian group and is isomorph to $(\Bbb Q,+)$, I need to know what is function of a subsequence of $(1,1),(1,2),(2,1),(1,3),(2,2),(3,1),(1,4),(2,3),(3,2),(4,1),...,(1,2k-2),$ $(2,2k-3),...,(k-1,k),(k,k-1),...,(2k-3,2),(2k-2,1),(1,2k-1),(2,2k-2),$ $...,(k-1,k+1),(k,k),(k+1,k-1),... ,(2k-2,2),(2k-1,1),...$ such that $(u,v)$ is belong to this subsequence iff gcd$(u,v)=1$ as: $(1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(2,3),(3,2),(4,1),(1,5),(5,1),(1,6),(2,5),(3,4),(4,3),$ $(5,2),(6,1),(1,7),(3,5),(5,3),(7,1),...$
~~Unlikely hypothesis: Using homotopy theory if a group on $\Bbb N$ isomorph to $(\Bbb Q,+)$ be defined then conjecture $1$ will be proved simpler and easier!~~

- Question: To define an Abelian group structure on $\Bbb N$ that is not a finitely generated Abelian group and is isomorph to $(\Bbb Q,+)$, I need to know what is function of a subsequence of $(1,1),(1,2),(2,1),(1,3),(2,2),(3,1),(1,4),(2,3),(3,2),(4,1),...,(1,2k-2),$ $(2,2k-3),...,(k-1,k),(k,k-1),...,(2k-3,2),(2k-2,1),(1,2k-1),(2,2k-2),$ $...,(k-1,k+1),(k,k),(k+1,k-1),... ,(2k-2,2),(2k-1,1),...$ such that $(u,v)$ is belong to this subsequence iff gcd$(u,v)=1$ as: $(1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(2,3),(3,2),(4,1),(1,5),(5,1),(1,6),(2,5),(3,4),(4,3),$ $(5,2),(6,1),(1,7),(3,5),(5,3),(7,1),...$

A conjecture equivalent to Goldbach's weak conjecture: $\forall n\in\Bbb N,\, \exists p,q,r\in\Bbb P,$ $\quad 2n+7=\begin{cases}
4\star p+8 & \text{or}\\
p\star q\star r+2\end{cases}$

**Conjecture** $4$: Under $(\Bbb N,\star),\,\forall n\in\Bbb N,\,\exists p,q\in\Bbb P\setminus\{2\}\quad 2n+3=p\star q$.

- This conjecture is an equivalent to Goldbach's conjecture.
- We have: $\langle p\star q\rangle=\{1\}\cup\{n\cdot (p+q-2)+1\,|\, n\in\Bbb N\}\cup\{n\cdot (p+q-2)\,|\, n\in\Bbb N\}$
- Question $7$: $\forall m\in\Bbb N$ do $\exists p,q\in\Bbb P,\,\exists n,k\in\Bbb N$ such that $2m+3\neq p\star q$ and $(2m+3)^n=(p\star q)^k$ or that $2n(m+1)=k(p+q-2)$?
- Clearly $f:\Bbb N\to\Bbb Z,\,\, f(1)=0,\, f(2n)=n,\, f(2n+1)=-n,$ is an isomorphism and $\forall n\in\Bbb N,\,\langle n\rangle\cong\langle f(n)\rangle$ & $\Bbb N/ \langle n\rangle\cong\Bbb Z / \langle f(n)\rangle$.

And $(\Bbb N,\star)$ can be defined with: $\begin{cases} 2\star 3=1,\quad 2^n=2n,\quad 3^n=2n+1\\ 2^n\star 2^m=2^{n+m},\quad 3^n\star 3^m=3^{n+m}\\ 2^n\star 3^m=\begin{cases} 2^{n-m} & 2^n\gt 3^m\\ 3^{m-n} & 3^m\gt 2^n\end{cases}\end{cases}$

- We have $\langle 2^n\rangle=\langle 3^n\rangle=\{1\}\cup\{2^{nk}\,|\, k\in\Bbb N\}\cup\{3^{nk}\,|\, k\in\Bbb N\}$.

Suppose $t:\Bbb N\to\Bbb N$ is a sequence with $n\mapsto t(n)$ such that $3^{t(n)}\in\Bbb P\setminus\{2\}$ so codomain of $t$ is $T:=\{1,2,3,5,6,8,9,11,14,15,18,20,21,23,26,29,30,33,35,36,39,41,44,48,50,51,53,54,...\}$.

**Theorem**: $\forall n\in\Bbb N,\,\exists k\in\Bbb N\setminus\{1\},\,\exists p_1,p_2,p_3,...,p_k\in\Bbb P\setminus\{2\}$ such that $2n+3=p_1\star p_2\star p_3...\star p_k$.

- Proof by induction axiom and that $\forall n\in\Bbb N,\,\exists m\in\Bbb N,$ such that $2n+3=3^m$ and of course each odd prime number is to form of $3^m$ as $5=3^2,7=3^3,11=3^5,13=3^6,17=3^8,19=3^9,...$ too.
- I think there exists a minimum value of $k$ that $\forall n\in\Bbb N,$ holds this equation $2n+3=p_1\star p_2\star p_3...\star p_k$, so find this $k$.

**Guess** $11$: $\forall p_1,p_2,p_3\in\Bbb P\setminus\{2\},\,\exists q_1,q_2\in\Bbb P\setminus\{2\}$ such that $p_1\star p_2\star p_3=q_1\star q_2$.

- Problem $4$: Make another definition of prime numbers embedded into $(\Bbb N,\star)$ and free of normal definition of prime numbers in $(\Bbb Z,+)$.
- Regarding to the $\Bbb N/ \langle n\rangle\cong\Bbb Z / \langle f(n)\rangle$, a new definition of prime numbers will be obtained and we will approach to this guess $10$ (or in fact to the Goldbach), but should be noted the linear equation $2n=p+q$ or $2n+3=p\star q$ with degree one won't be proved only by group theory and indeed we need to prime numbers properties that is in prime number theorem as a logarithm function that is used in proof of the Main theorem, so finally Goldbach conjecture will be proved through conjecture $1$ and for conjecture $1$ I offer homotopy groups, but Goldbach is the key of whole mathematics because with Goldbach and Polignac and some other conjectures, prime numbers will be recognized and then some new numbers except algebraic numbers with cardinal $\aleph _0$ will be discovered and consequently whole mathematics and mathematical logic will be improved widely hence this German great mathematician Christian Goldbach's conjecture will impress mathematics certainly!

**Problem** $5$: Rewrite operation of the group $(\Bbb N,\star)$ by matrices.

**Conjecture** $5$: $\forall n\in\Bbb N,\,\exists (a,b)\in\{(x,y)\,|\, x,y\in S_1,\,\exists m\in\Bbb N,\, x\cdot 10^m,y\cdot 10^m\in\Bbb P\setminus\{2\},$ $0\lt y\le x\lt 0.1,$ $0.01\le x\},$ $\exists (c,0)\in\{(-x,0)\,|\, x\in\{10^{-n}\,|\, n\in\Bbb N\}\}$ such that $c=-10^{-k}$ & $2n+3=10^k\cdot (a+b+c)$ & $10^k\cdot a,10^k\cdot b\in\Bbb P\setminus\{2\}$ & $k$ is the number of digits in $2n+3$.

- This conjecture was made compatible with group $(\Bbb N,\star)$ and is an equivalent to Goldbach's conjecture.

Let $L_3=\{(a,b)\,|\, a,b\in S_1,\, a,b\in\{(x,y)\,|\, 0\lt x\lt 0.1,\, 0\lt y\lt 0.1\}\setminus\{(x,y)\,|\, x\lt 0.01,\, y\lt 0.01\},$ $\exists k\in\Bbb N,\, a\cdot 10^k,b\cdot 10^k\in\Bbb P\setminus\{2\}\}$.

**Problem** $6$: Let $e=(0.03,0.03)$ & $\forall (a,b),(c,d)\in L_3,\, m=1+max($number of digits in $f(a)\star _3f(c),$ number of digits in $f(b)\star _3f(d)),$ $$(a,b)\star _{L_3} (c,d)=10^{-m}\cdot (10^{-u_1(\alpha)}\cdot [f(a)\star _3f(c)],10^{-u_2(\alpha)}\cdot [f(b)\star _3f(d)])$$ that $\alpha=((a,b),(c,d))$ where $u_1,u_2:L_3\times L_3\to\Bbb N\cup\{0\}$ are functions such that $\forall\alpha\in L_3\times L_3$ at least one of $u_1(\alpha),u_2(\alpha)$ should be $0$ so find functions $u_1,u_2$ such that $(L_3,\star _{L_3})$ be a group (an Abelian group), though I guess $(L_3,\star _{L_3})$ is non finitely generated Abelian group and possibly isomorph to $(\Bbb Q,+)$ or $(\Bbb Q_{\gt 0},\cdot)$ or $A\oplus B$ that $A,B\in\{(\Bbb Q,+),(\Bbb Q_{\gt 0},\cdot)\}$.

- Of course we could write each another group structure on $\Bbb N$ and consequently on $P_1$ regarding above algorithm.
- And if under Euclidean topology (or another appropriate topology) this group $(L_3,\star _{L_3})$ be a topological group it is very well, otherwise we can replace another group structure on $\Bbb N$ and then on $P_1$ and consequently on $L_3$.

Question $8$: Let $e=0.2$ & $\forall ,\, r_m(p),r_n(q)\in S_1,\, r_m(p)\star _{S_1} r_n(q)=r_{(m+1)\star (n+1)-1} (p\star _1 q)$ that $p,q\in\Bbb P,$ $m,n\in\Bbb N\cup\{0\}$ now is $(S_1,\star _{S_1})$ a group (an Abelian group) and non finitely generated Abelian group and possibly isomorph to $(\Bbb Q,+)$ or $(\Bbb Q_{\gt 0},\cdot)$?

Question $9$: Let $e=(0.2,0.2)$ & $\forall (r_{m_1}(p_1),r_{m_2}(p_2)),(r_{n_1}(q_1),r_{n_2}(q_2))\in S_1\times S_1,$ $(r_{m_1}(p_1),r_{m_2}(p_2))\star _{S_1\times S_1} (r_{n_1}(q_1),r_{n_2}(q_2))=(r_{(m_1+1)\star (n_1+1)-1} (p_1\star _1 q_1),r_{(m_2+1)\star (n_2+1)-1}(p_2\star _1 q_2)$ that $m_1,n_1,m_2,m_2\in\Bbb N\cup\{0\},\, p_1,p_2,q_1,q_2\in\Bbb P$ now is $(S_1\times S_1,\star _{S_1\times S_1})$ a group (an Abelian group) and non finitely generated Abelian group and possibly isomorph to $(\Bbb Q,+)$ or $(\Bbb Q_{\gt 0},\cdot)$ or $A\oplus B$ that $A,B\in\{(\Bbb Q,+),(\Bbb Q_{\gt 0},\cdot)\}$?

Question $10$: Let $M$ be a topological space and $A,B$ are subsets of $M$ with $A\subset B$ and $A$ is dense in $B,$ since $A$ is dense in $B,$ is there some way in which a topology on $B$ may be induced other than the subspace topology? I am also interested in specialisations, for example if $M$ is Hausdorff or Euclidean.

- If a topology based on density be made then we will be able to import prime numbers properties into a topology!
- Indeed since $S_1$ is dense in the $[0,1]$ so each topology on $(0,1)^n$ and $\Bbb R^n,\, n\in\Bbb N$ and also on $(0,1)\times (0,1)$ has prime numbers properties and consequently $\forall x\in\Bbb R^n,\,\forall n\in\Bbb N,\, x$ has a relation with prime numbers!

- of course this wordage isn't mathematically but I will write as topological!

Theorem: Suppose $\forall n\in\Bbb N,\, W_n:=\{(z_1+s_1,z_2+s_2,z_3+s_3,...,z_n+s_n)\,|\, \forall i=1,2,3,...,n,\, z_i\in\Bbb Z,$ $s_i\in S_1\}$ then $W_n$ is dense in the $\Bbb R^n$. (under Euclidean topology)

Guess $12$: Suppose $J_n=\{(s_1,s_2,s_3,...,s_n)\,|\,\exists k\in\Bbb N,\, s_i\cdot 10^k\in\Bbb P,\,\forall i=1,2,3,...,n,\, s_i\in S_1\}$ & $I^n=[0,1]^n$ that of course $S_1^n$ is dense in the $I^n,$ now I guess that under Euclidean topology $\exists n\in\Bbb N,\,\forall k\ge n$ $dimension(closure(J_k))=1$.

$r:\Bbb P\to S,\, p\mapsto r(p)$ is a bijection, under this bijection $\Bbb P$ & $S$ have the same order type, in other words $\lt _{\Bbb P}$ is an order relation on $\Bbb P$ iff $\lt _S$ be an order relation on $S$ such that $p\lt _{\Bbb P} q$ iff $r(p)\lt _S r(q)$.

**Guess** $13$: There are no topology except probably indiscrete (trivial) or discrete topology on $\Bbb P$ neither on $S$ such that these two topological space be homeomorph.

Suppose $h:\Bbb P\to Power(\Bbb P),\, h(p)=\{q\,|\, \exists b_1b_2b_3...b_k\in\Bbb N,\,q=a_1a_2...a_nb_1b_2...b_k\in\Bbb P,$ $p=a_1a_2a_3...a_n\}$ is an injection.

Alireza Badali 22:21, 8 May 2017 (CEST)

## Polignac's conjecture

In number theory, Polignac's conjecture was made by Alphonse de Polignac in 1849 and states: For any positive even number $n$, there are infinitely many prime gaps of size $n$. In other words: There are infinitely many cases of two consecutive prime numbers with difference $n$. (Tattersall, J.J. (2005), Elementary number theory in nine chapters, Cambridge University Press, ISBN: 978-0-521-85014-8, p. 112) Although the conjecture has not yet been proven or disproven for any given value of n, in 2013 an important breakthrough was made by Zhang Yitang who proved that there are infinitely many prime gaps of size n for some value of n < 70,000,000.(Zhang, Yitang (2014). "Bounded gaps between primes". Annals of Mathematics. 179 (3): 1121–1174. MR 3171761. Zbl 1290.11128. doi:10.4007/annals.2014.179.3.7. _ Klarreich, Erica (19 May 2013). "Unheralded Mathematician Bridges the Prime Gap". Simons Science News. Retrieved 21 May 2013.) Later that year, James Maynard announced a related breakthrough which proved that there are infinitely many prime gaps of some size less than or equal to 600.(Augereau, Benjamin (15 January 2013). “An old mathematical puzzle soon to be unraveled? Phys.org. Retrieved 10 February 2013.)

Assuming Polignac's conjecture there isn't any rhythm for prime numbers and so there isn't any formula for prime numbers!

Let $B=\{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1\}$ & $C=(S_1\times S_1)\cap \{(x,x)\,|\, 0.01\le x\lt 0.1\}$ & $T=\{(a,b)\,|\, a,b \in S_1,\, 0.01 \lt b \lt a\lt 0.1,\, \exists m \in \Bbb N,\, a \cdot 10^m, b \cdot 10^m$ or $a\cdot 10^{m-1}, b\cdot 10^m$ are consecutive prime numbers$\}$ & $\forall n \in \Bbb N,\, J_n :=\{(a,b) \,|\, (a,b) \in T,\, \exists k \in \Bbb N, a-b=r_k (2n)\}$.

- Obviously $\bigcup _{n\in \Bbb N} J_n=T$ and Polignac's conjecture is equivalent to $\forall n \in \Bbb N,$ cardinal$(J_n)=\aleph_0$.

**Guess** $1$: $\forall (a,a) \in C$ there are some sequences in $T$ like $\{a_n\}$ that $a_n\to (a,a)$ where $n\to \infty$ and there are some sequences in $T$ like $b_n$ that $b_n\to (0.1,0.01)$ where $n\to \infty$.

**Guess** $2$: $\exists N_1 \subseteq \Bbb N,\, \forall n\in N_1,$ cardinal$(J_n)=\aleph_0=$cardinal$(N_1)$

Obviously $\exists \epsilon,\epsilon_1,\epsilon_2 \in \Bbb R,\epsilon\gt 0, \epsilon_2\gt \epsilon_1\gt 0$ that $\forall (a,b) \in T \cap \{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,$ $x-y\lt \epsilon\}$ that $\exists m\in \Bbb N$ that $a\cdot 10^m$,$b\cdot 10^m$ are consecutive prime numbers, but $a\cdot 10^m$,$b\cdot 10^m$ are large natural numbers, and $\forall (a,b)\in T\cap$ $\{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,$ $0.11-\epsilon_1 \lt x+y\lt 0.11+\epsilon_1,$ $x-y\gt 0.09-\epsilon_2 \}$ that $\exists m\in \Bbb N$ that $a\cdot 10^{m-1},b\cdot 10^m$ are consecutive prime numbers but $a\cdot 10^{m-1}$,$b\cdot 10^m$ are large natural numbers and $\forall (a,b)\in T\setminus (\{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,$ $x-y\lt \epsilon\}$ $\cup$ $\{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,$ $0.11-\epsilon_1 \lt x+y\lt 0.11+\epsilon_1,\, x-y\gt 0.09-\epsilon_2 \})$ that $\exists m\in \Bbb N$ that $a\cdot 10^{m-1}$ or $a\cdot 10^m$ & $b\cdot 10^m$ are consecutive prime numbers but $a\cdot 10^m$ or $a\cdot 10^{m-1}$ & $b\cdot 10^m$ aren't large natural numbers.

Theorem: $\forall c\in r_1(\Bbb P)$ cardinal$(T\cap \{(x,c)\,|\, x\in \Bbb R \})=1=$ cardinal$(T\cap \{(c,y)\,|\, y\in \Bbb R\})$

**Guess** $3$: $\forall k\in \Bbb N,$ $\forall c\in r_k (\Bbb N),\, c\lt 0.09$ then cardinal$(T\cap \{(x,y)\,|\, x-y=c\}) \in \Bbb N \cup \{0\}$.

**Theorem** $1$: For each subinterval of $[0.01,0.1)$ like $(a,b),\,\exists m\in \Bbb N$ that $\forall k\in \Bbb N$ with $k\ge m$ then $\exists t\in (a,b)$ that $t\cdot 10^{k+1}\in \Bbb P$.

- Proof given by @Adayah from stackexchange.com: Without loss of generality (by passing to a smaller subinterval) we can assume that $(a, b) = \left( \frac{s}{10^r}, \frac{t}{10^r} \right)$, where $s, t, r$ are positive integers and $s < t$. Let $\alpha = \frac{t}{s}$.

- The statement is now equivalent to saying that there is $m \in \mathbb{N}$ such that for every $k \geqslant m$ there is a prime $p$ with $10^{k-r} \cdot s < p < 10^{k-r} \cdot t$.

- We will prove a stronger statement: there is $m \in \mathbb{N}$ such that for every $n \geqslant m$ there is a prime $p$ such that $n < p < \alpha \cdot n$. By taking a little smaller $\alpha$ we can relax the restriction to $n < p \leqslant \alpha \cdot n$.

- Now comes the prime number theorem: $$\lim_{n \to \infty} \frac{\pi(n)}{\frac{n}{\log n}} = 1$$

- where $\pi(n) = \# \{ p \leqslant n : p$ is prime$\}.$ By the above we have $$\frac{\pi(\alpha n)}{\pi(n)} \sim \frac{\frac{\alpha n}{\log(\alpha n)}}{\frac{n}{\log(n)}} = \alpha \cdot \frac{\log n}{\log(\alpha n)} \xrightarrow{n \to \infty} \alpha$$

- hence $\displaystyle \lim_{n \to \infty} \frac{\pi(\alpha n)}{\pi(n)} = \alpha$. So there is $m \in \mathbb{N}$ such that $\pi(\alpha n) > \pi(n)$ whenever $n \geqslant m$, which means there is a prime $p$ such that $n < p \leqslant \alpha \cdot n$, and that is what we wanted.
- Clearly the Main theorem of Goldbach's conjecture is a result of theorem $1$.

Theorem: $\forall \epsilon_1,\epsilon_2$ that $0\lt \epsilon_1\lt \epsilon_2\lt 0.09$ then cardinal$(T\setminus \{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,\, x-y\gt \epsilon_1\})$ $=$ cardinal$(T\setminus \{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,\, x-y\lt \epsilon_2\})=\aleph_0$.

- Proof: Be aware to number of digits in prime numbers corresponding to coordinates of each member in $T$.
**Guess**$4$: cardinal$(\{(a,b)\,|\, (a,b)\in T,\, 0.01\lt b\lt a\lt 0.1,$ $\epsilon_1\lt a-b\lt \epsilon_2\})\in \Bbb N\cup \{0\}$

Alireza Badali 13:17, 21 August 2017 (CEST)

## Landau's forth problem

Landau's forth problem: Are there infinitely many primes $p$ such that $p−1$ is a perfect square? In other words: Are there infinitely many primes of the form $n^2 + 1$?

In analytic number theory the Friedlander–Iwaniec theorem states that there are infinitely many prime numbers of the form ${\displaystyle a^{2}+b^{4}}$.

Suppose $H=\{(a,b)\,|\, a\in\bigcup_{k\in\Bbb N} r_k(\{n^2\,|\, n\in\Bbb N\}),\, b\in\bigcup_{k\in\Bbb N} r_k(\{n^4\,|\, n\in\Bbb N\})$ & $\exists t\in\Bbb N$ that $a\cdot 10^t\in\{n^2\,|\, n\in\Bbb N\},$ $b\cdot 10^t\in\{n^4\,|\, n\in\Bbb N\},\, (a+b)\cdot 10^t\in\Bbb P\}$ and $H_1=\{(a,b)\in H\,|\, b\in\bigcup_{k\in\Bbb N} r_k(\{1\})\}$

- Friedlander-Iwaniec theorem is the same cardinal$(H)=\aleph_0$

**A question**: cardinal$(H_1)\in\Bbb N$ or cardinal$(H_1)=\aleph_0$?

- This question is the same Landau's forth problem.

Alireza Badali 20:47, 21 September 2017 (CEST)

## Grimm's conjecture

In mathematics, and in particular number theory, Grimm's conjecture (named after Karl Albert Grimm) states that to each element of a set of consecutive composite numbers one can assign a distinct prime that divides it. It was first published in American Mathematical Monthly, 76(1969) 1126-1128.

Formal statement: Suppose $n + 1, n + 2, …, n + k$ are all composite numbers, then there are $k$ distinct primes $p_i$ such that $p_i$ divides $n+i$ for $1 ≤ i ≤ k$.

Let $C=\{(x,y)\,|\, 0.01\le x\lt 0.1,\, 0.01\le y\lt 0.1\}$

**A conjecture**: Suppose $n+1,n+2,n+3,...,n+k$ are all composite numbers then $\forall i=1,2,3,...,k$ $\exists (r_1(p_i),r_1(t_i))\in C$ that $p_i\in\Bbb P,$ $t_i\in\Bbb N$ & $\forall j=1,2,3,...,k$ that $i\neq j$ implies $p_i\neq p_j$ we have $r_1(p_i)\cdot r_1(t_i)=r_2(n+i)$ or $r_3(n+i)$.

- This conjecture is an equivalent to Grimm's conjecture.

Alireza Badali 23:30, 20 September 2017 (CEST)

## Lemoine's conjecture

In number theory, Lemoine's conjecture, named after Émile Lemoine, also known as Levy's conjecture, after Hyman Levy, states that all odd integers greater than $5$ can be represented as the sum of an odd prime number and an even semiprime.

Formal definition: To put it algebraically, $2n + 1 = p + 2q$ always has a solution in primes $p$ and $q$ (not necessarily distinct) for $n > 2$. The Lemoine conjecture is similar to but stronger than Goldbach's weak conjecture.

Theorem: $\forall n\in\Bbb N$ that $m=2n+5,\, \exists (a,b)\in\{(x,y)\,|\, 0\lt y\le x,\, x+y\lt 0.1,\, x,y\in S_1,\, \exists t\in\Bbb N,$ $x\cdot 10^t,y\cdot 10^t\in\Bbb P\},\,\exists k\in\Bbb N,\, \exists c\in r_k(\Bbb P)$ such that $a+b=r_1(m)-c$

- In principle $(r_1(m),-c)\in \{(x,y)\,|\, -x\lt y\lt 0,\, 0.01\le x\lt 0.1\}$
- This theorem is an equivalent to Goldbach's weak conjecture.

**A conjecture**: $\forall n\in\Bbb N$ that $m=2n+5,\, \exists k_1,k_2\in\Bbb N,\, \exists b\in r_{k_1}(\Bbb P)$ that $b\lt 0.05,\, \exists a\in r_{k_2}(\Bbb P)$ such that $2b=r_1(m)-a$

- In principle $(b,b)\in\{(x,x)\,|\, 0\lt x\lt 0.05\}$ & $(r_1(m),-a)\in\{(x,y)\,|\, -x\lt y\lt 0,\, 0.01\le x\lt 0.1\}$
- This conjecture is an equivalent to Lemoine's conjecture.

Alireza Badali 00:30, 27 September 2017 (CEST)

## About analytic number theory

In mathematics, analytic number theory is a branch of number theory that uses methods from mathematical analysis to solve problems about the integers. It is often said to have begun with Peter Gustav Lejeune Dirichlet's 1837 introduction of Dirichlet L-functions to give the first proof of Dirichlet's theorem on arithmetic progressions. It is well known for its results on prime numbers (involving the Prime Number Theorem and Riemann zeta function) and additive number theory (such as the Goldbach conjecture and Waring's problem).

There is no $n$-variable polynomial, $n\in\Bbb N$ as a formula for prime numbers, because each formula or relation for prime numbers only there can be include all prime numbers together and simultaneous like Riemann zeta function!

Primental (prime-transcendental) property: A continuous function $f:\Bbb R\to\Bbb R,$ has primental property if $\exists z_1,z_2,t_1,t_2\in\Bbb Z,$ $\exists m\in\Bbb N,\,\forall n\ge m,\, f(z_1+n\cdot z_2)=t_1+p_n\cdot t_2$ that $p_n$ is $n$_th prime number and each function $f$ with this property is called a primental.

Question: Is there any primental that be polynomial too?

- Answer given by @ArnaudMortier from stackexchange.com site: The answer is no, $f$ can't be a polynomial. Because the asymptotics of $p_n$ are $$p_n\sim n\log n $$ and if $f$ is a polynomial then $$f(z_1+n\cdot z_2)\sim C\cdot n^\alpha$$ where $C$ is some constant and $\alpha\in \mathbb{N}$.
- By transitivity you would have $$C\cdot n^\alpha\sim t_2 n\log n$$ which can't be true.
- There is no polynomial with integer coefficients that outputs only primes. However, there are multivariable polynomials whose integer positive values are exactly the prime numbers (mathworld.wolfram.com/Prime-GeneratingPolynomial.html)
- For the exact same reason, multivariable functions satisfy this condition: $\exists z_i,s_i∈\Bbb Z$ for $i=1,2,3,...,k$ & $\exists z,s∈\Bbb Z$ & $\exists m∈\Bbb N,\,\forall n≥m$ $f(z_1+ns_1,z_2+ns_2,z_3+ns_3,...,z_k+ns_k)=z+p_ns$, can not be multivariable polynomials.
- Incidentally, it is known that there exists a polynomial $f(t_1,…,t_n,p)$ such that $p∈\Bbb Z$ is prime if and only if there exist $t_1,...,t_n∈\Bbb Z$ such that $f(t_1,…,t_n,p)=0$. (Though actually writing out such a polynomial f would likely be painful and result in a long expression.)

- This is a philosophy: this question is useful for Riemann hypothesis, because normal definition of prime numbers & prime numbers properties & distribution of prime numbers in $\Bbb N$ & prime number theorem & the set $S$ & density of $S$ in $[0.1,1]$ & distribution of $S$ in $[0.1,1]$ &theorem $1$ of Polignac with question $3$ of Goldbach are equivalent all together! hence prime numbers properties are present in whole $\Bbb R^2$ (or $\Bbb R^n,\, n\in\Bbb N$) so they are present in each infinite subset of $\Bbb R^2$ but in curves located at $\Bbb R^2$ they are present in fraction of $y$-coordinate or ordinate by $x$-coordinate or abscissa or the same trigonometric functions, on the other hand differential of polynomials are again polynomials and we can with linear transformation move polynomials to each position in $\Bbb R^2$.

My thought line on the formula of prime numbers: According to important theorems in number theory and distribution of prime numbers I think there doesn't exist any polynomial \(p:\Bbb R\to\Bbb R\) include all such points \((n,p_n)\) that \(p_n\) is \(n\)_th prime and with that in mind prime number theorem is obtained from normal definition of prime numbers in terms of natural numbers factorization to prime numbers that it is because of logarithm function is inverse of the function \(f(a)=a^n\) and finally primes formulas are as logarithmic functions or other transcendental functions but as an imagination there exists an special formula as an infinite series that generates primes simply: (I had asked a question that an user from stackexchange.com grammatical corrected it.)

Let $$\omega _1=\color{red}{0}.\color{teal}{p_1}\color{purple}{p_2}\color{teal}{p_3}\dots =\color{red}{0}.\color{blue}{2}\color{fuchsia}{3}\color{blue}{5}\color{fuchsia}{7}\color{blue}{11}\color{fuchsia}{13}\color{blue}{17}\color{fuchsia}{19}\color{blue}{23}\color{fuchsia}{29}\color{blue}{31}\color{fuchsia}{37}\dots \color{blue}{7717}\color{fuchsia}{7723}\dots$$

(i.e. the decimal part of $\omega _1$ is obtained by concatenating the prime numbers) and $$\omega _2=\color{red}{0}.\color{blue}{2}0\color{fuchsia}{3}0\color{blue}{5}0\color{fuchsia}{7}0\color{blue}{11}00\color{fuchsia}{13}00\color{blue}{17}00\color{fuchsia}{19}00\color{blue}{23}00\color{fuchsia}{29}00\color{blue}{31}00\color{fuchsia}{37}00\dots\color{blue}{7717}0000\dots$$

(i.e. the decimal part of $\omega _2$ is obtained by concatenating the prime numbers, each of them followed by a number of copies of $0$ equal to the number of its digits in base $10$).

Questions:

$1.$ Is $\omega _1$ or $\omega _2$ or another some similar number transcendental, and if yes is this a contradiction to the existence of a formula for prime numbers?

$2.$ For each sequence $a_n: \mathbb N \to \mathbb N$ is there any sequence like $b_n: \mathbb N \to \mathbb N$ such that the number $\theta :=0.a_10 \dots 0a_20 \dots 0a_30 \dots 0 \dots$ obtained by concatenating the numbers $a_n$, each of them followed by a number of copies of $0$ equal to $b_n$, is a transcendental number?

**In analytic number theory** in Riemann zeta function there is an important technique as $\frac{1}{n}$, inverse of natural numbers but I want add another technique to this collection as putting a point at the beginning of natural numbers like $6484070\to 0.6484070$ so we will have a stronger theory.

*Below functions $\mu _1,\mu _2,\mu _3,...,\mu _{16}$ haven't been reviewed and checked:*

Let $S=\{0.2,0.3,0.5,0.7,0.11,...\}$ and $s_1=0.2,\, s_2=0.3,\, s_3=0.5,...$ that $s_k$ is $k$_th member in $S$, suppose $A_n=\{s_is_j\,|\, s_i,s_j\in \{s_1,s_2,s_3,...,s_n\},$ $s_i\neq s_j\}$ & $\mu _1:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _1(z)=\lim_{n\to\infty}\frac{1}{n^2}\sum _{a\in A_n} a^{-f(z)}$$

such that $f:\Bbb C\to\Bbb C$ is an injective function that $\forall n\in\Bbb N,\,\forall z\in\Bbb C,\, n^{f(z)}$ isn't a Gaussian integer, for example this function has been given by @Wojowu from stackexchange.com: $$\forall z\in\Bbb C\qquad f(z)=\sqrt 2 +\frac{z}{N+|z|}$$ for large sufficiently $N\in\Bbb N$ of course $f$ maps $\Bbb C$ injectively into a disk around $\sqrt 2$ of radius $N^{-1}$ for large sufficiently $N,$ this disk contains no solution $z$ of $n^z\in\Bbb Z [i]$, according to definition of the function $\mu _1$ I think it is in a near relation with Riemann zeta function.

Let $\mu _2:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _2(z)=\sum _{n=1}^{\infty}\frac{1}{(r(n))^z}$$

Let $\mu _3:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _3(z)=\sum _{p\in\Bbb P}\frac{1}{(r(p))^z}$$

Let $\forall j\in\Bbb N,\,\mu _4:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _4(z)=\sum _{n=1}^{\infty}\frac{1}{(r_j(n))^z}$$

Let $\forall j\in\Bbb N,\,\mu _5:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _5(z)=\sum _{p\in\Bbb P}\frac{1}{(r_j(p))^z}$$

and $\mu _6:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _6(z)=\lim _{j\to\infty}\sum _{n=1}^{\infty}\frac{1}{(r_j(n))^z}$$

Let $\mu _7:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _7(z)=\lim _{j\to\infty}\sum _{p\in\Bbb P}\frac{1}{(r_j(p))^z}$$

let $\mu _8:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _8(z)=\sum _{j\in\Bbb N}\frac{1}{(r_j(j))^z}$$

Let $p_j$ is $j$_th prime number & $\mu _9:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _9(z)=\sum _{j\in\Bbb N}\frac{1}{(r_j(p_j))^z}$$

Let $a_n:\Bbb N\to\Bbb N$ is a sequence & $\mu _{10}:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _{10}(z)=\sum _{n\in\Bbb N}\frac{1}{(r_{a_n}(n))^z}$$

Let $a_n:\Bbb N\to\Bbb N$ is a sequence & $\mu _{11}:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _{11}(z)=\sum _{n\in\Bbb N}\frac{1}{(r_{a_n}(p_n))^z}$$

Let $\omega =0.p_1p_2p_3...=0.23571113171923293137...$ that in principle in $\omega$ prime numbers has been arranged respectively, now assume $a_n:\Bbb N\to (0,1)$ is a sequence that $\sum _{n\in\Bbb N} a_n=\omega$ & $\mu _{12}:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\,\mu _{12}(z)=\sum _{n\in\Bbb N}\frac{1}{(a_n)^z}$$

Let $f_1:\Bbb C\to\Bbb C$ & $\mu _{13}:\Bbb C\to\Bbb C$ are functions as: $$\forall z\in\Bbb C\qquad\mu _{13}(z)=\sum _{n=1}^{\infty} (-1)^n\cdot (r(n))^{f_1(z)}$$

Let $f_2:\Bbb C\to\Bbb C$ & $\mu _{14}:\Bbb C\to\Bbb C$ are functions as: $$\forall z\in\Bbb C\qquad\mu _{14}(z)=\sum _{p_n\text{is}\, n\text{_th prime number}} (-1)^n\cdot (r(p_n))^{f_2(z)}$$

Find $f_1$ & $f_2$ as much as possible simple such that $\mu _{13}(i[\Bbb Q])$ is dense in the $\mu _{13}(\Bbb C)$ and $\mu _{14}(i[\Bbb Q])$ is dense in the $\mu _{14}(\Bbb C)$.

Now some theorems on these functions about density should be presented.

Let $\mu_{15}:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _{15}(z)=\sum _{j\in\Bbb N}\sum _{n\in\Bbb N\cap [10^{j-1},10^j)} (r_j(n))^z$$

Let $\mu_{16}:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _{16}(z)=\sum _{j\in\Bbb N}\sum _{p\in\Bbb P\cap (10^{j-1},10^j)} (r_j(p))^z$$

Alireza Badali 20:44, 12 October 2017 (CEST)

## Some notes

I see, you like to densely embed natural numbers into a continuum. You may also try to embed them into the unit circle on the complex plane by $n\mapsto i^n=\cos(\log n)+i\sin(\log n)$. Then $(mn)^i=m^i n^i$. Passer By (talk) 13:14, 3 December 2017 (CET)

- Very nice, thank you so much. Alireza Badali 21:33, 3 December 2017 (CET)

To your Question 3 above: the affirmative answer is given by Liouville's theorem on approximation of algebraic numbers. Passer By (talk) 19:05, 4 December 2017 (CET)

- Thank you. Alireza Badali 16:00, 6 December 2017 (CET)

## Question

You often mention "Formula of prime numbers". What do you mean? This is not a well-defined mathematical object, but a vague idea, with a lot of *non-equivalent* interpretations. See for instance [1], [2], [3], [4], [5], [6] etc. Passer By (talk) 19:24, 4 December 2017 (CET)

- Thank you and each one of these wordage about the formula of prime numbers is a theory separately one by one and probably non-equivalent or even incompatible so please don't consider their relations together. Alireza Badali 16:00, 6 December 2017 (CET)
- OK, then I do not consider these interpretations of the phrase "Formula of prime numbers" (since yours is different from them all). My question is, what is your interpretation of the phrase "Formula of prime numbers"? We cannot ask "does it exist or not" if we do not know what exactly is meant by "it". The answer is affirmative for some interpretations and negative for other interpretations. Passer By (talk) 20:08, 6 December 2017 (CET)
- Formula of prime numbers is a subsequence in $\Bbb N$ that has a special order and this order is the same formula of prime numbers, but this order isn't located on a polynomial necessarily. Alireza Badali 21:36, 6 December 2017 (CET)
- I know what is a sequence and subsequence, and I understand that prime numbers may be treated as a subsequence of the sequence of natural numbers; but I do not know what is "special order"; I also do not know what is "order located on a polynomial"; thus, I get no answer to my question. Passer By (talk) 23:52, 6 December 2017 (CET)
- Prime numbers are non repetitive and each natural number is a production of them but apparently they are indomitable (like me) and I think they are similar to transcendental numbers, order is special because prime numbers are special however natural numbers is defined by prime numbers as a production of them although another definition is given by induction axiom or another definition is in nature that all people see it in their routines, about polynomial I want say there is no polynomial as a formula for primes although there is a special polynomial like $y=a+bx^3$ or something such that $\exists n\in\Bbb N\,\,\forall m\gt n,\,\,p_m$ is around that curve and finally I want say discussion on numbers will continue forevermore and this is fate but there exists a unique dreamy logarithmic function that gives primes pleasantly and way for reaching to secrets is in homotopy theory. Alireza Badali 12:36, 7 December 2017 (CET)

- I know what is a sequence and subsequence, and I understand that prime numbers may be treated as a subsequence of the sequence of natural numbers; but I do not know what is "special order"; I also do not know what is "order located on a polynomial"; thus, I get no answer to my question. Passer By (talk) 23:52, 6 December 2017 (CET)

- Formula of prime numbers is a subsequence in $\Bbb N$ that has a special order and this order is the same formula of prime numbers, but this order isn't located on a polynomial necessarily. Alireza Badali 21:36, 6 December 2017 (CET)

- OK, then I do not consider these interpretations of the phrase "Formula of prime numbers" (since yours is different from them all). My question is, what is your interpretation of the phrase "Formula of prime numbers"? We cannot ask "does it exist or not" if we do not know what exactly is meant by "it". The answer is affirmative for some interpretations and negative for other interpretations. Passer By (talk) 20:08, 6 December 2017 (CET)

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Musictheory2math.

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