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(Goldbach's conjecture)
(Probably I should change integral domain for the Gauss circle problem)
 
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== Goldbach's conjecture ==
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<small>Whole my previous notes is visible in the [https://www.encyclopediaofmath.org/index.php?title=User_talk:Musictheory2math&diff=prev&oldid=43134 revision as of 18:42, 13 April 2018] Alireza Badali 21:52, 13 April 2018 (CEST)</small>
  
'''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.
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== <big>$\mathscr B$</big> $theory$ (algebraic topological analytical number theory) ==
  
$\,$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. [[User:Musictheory2math|Musictheory2math]] ([[User talk:Musictheory2math|talk]]) 16:29, 25 March 2017 (CET)
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Logarithm function as an inverse of the function $f:\Bbb N\to\Bbb R,\,f(n)=a^n,\,a\in\Bbb R$ has prime numbers properties because in usual definition of prime numbers multiplication operation is a point meantime we have $a^n=a\times a\times ...a,$ $(n$ times$),$ hence prime number theorem or its extensions or some other forms is applied in $B$ theory for solving problems on prime numbers exclusively and not all natural numbers.
  
: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". [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk: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. [[User:Musictheory2math|Musictheory2math]] ([[User talk:Musictheory2math|talk]]) 16:29, 25 March 2017 (CET)
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'''Algebraic structures on the positive numbers & topology with homotopy groups & prime number theorem and its extensions or other forms or corollaries'''
  
: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
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Alireza Badali 00:49, 25 June 2018 (CEST)
\[
 
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: <nowiki>~~~~</nowiki>. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk: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. [[User:Musictheory2math|Musictheory2math]] ([[User talk:Musictheory2math|talk]]) 16:29, 25 March 2017 (CET)
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=== [https://en.wikipedia.org/wiki/Goldbach%27s_conjecture Goldbach's conjecture] ===
  
: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.) [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 11:16, 19 March 2017 (CET)
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'''Lemma''': For each subinterval $(a,b)$ of $[0.1,1),\,\exists m\in \Bbb N$ that $\forall k\in \Bbb N$ with $k\ge m$ then $\exists t\in (a,b)$ that $t\cdot 10^k\in \Bbb P$.
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:[https://math.stackexchange.com/questions/2482941/a-simple-question-about-density-in-the-interval-0-1-1/2483079#2483079 Proof] given by [https://math.stackexchange.com/users/149178/adayah @Adayah] from stackexchange site: 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 $\alpha = \frac{t}{s}$.
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: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 $10^{k-r} \cdot s < p < 10^{k-r} \cdot t$.
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: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 $n < p \leqslant \alpha \cdot n$.
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:Now comes the prime number theorem: $$\lim_{n \to \infty} \frac{\pi(n)}{\frac{n}{\log n}} = 1$$
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:where $\pi(n) = \# \{ p \leqslant n : p$ is prime$\}.$ By the above we have $$\frac{\pi(\alpha n)}{\pi(n)} \sim \frac{\frac{\alpha n}{\log(\alpha n)}}{\frac{n}{\log(n)}} = \alpha \cdot \frac{\log n}{\log(\alpha n)} \xrightarrow{n \to \infty} \alpha$$
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: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♦
  
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. [[User:Musictheory2math|Musictheory2math]] ([[User talk: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".
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Now we can define function $f:\{(c,d)\mid (c,d)\subseteq [0.01,0.1)\}\to\Bbb N$ 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 $u\cdot 10^{m+1}\in\Bbb P$
:But you still do not put four tildas at the end of each your message; please do. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 11:16, 19 March 2017 (CET)
 
  
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and $g:(0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\to\Bbb N,$ is a function by $\forall\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))$ $g(\epsilon)=max(\{f((c,d))\mid d-c=\epsilon,$ $(c,d)\subseteq [0.01,0.1)\})$.
  
Importance of density in the Main theorem is similar to definition of irrational numbers from rational numbers.
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'''Guess''' $1$: $g$ isn't an injective function.
  
'''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.
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'''Question''' $1$: Assuming guess $1$, 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 $\{\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\mid g(\epsilon)=n\}\subsetneq h_n$ and obviously $g(a)=n=g(b)$ now the question is $\forall n,m\in\Bbb N$ that $m\neq n$ is $h_n\cap h_m=\emptyset$?
 
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:[https://math.stackexchange.com/questions/2518063/a-medium-question-about-a-set-related-to-prime-numbers/2526481#2526481 Guidance] given by [https://math.stackexchange.com/users/276986/reuns @reuns] from stackexchange site:
 
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:* 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 $\mathbb{P}$.
Assume $S_1=\{10^{-n}\cdot a\, |\, a\in S$ for $n=0,1,2,3,...\}$ & $L=\{(a,b)\,|\,a,b \in S_1,\, 0.01 \le a+b \lt 0.1,$ $\exists m \in \Bbb N,$ $a \cdot 10^m,\,b \cdot 10^m\in\Bbb P\setminus\{2\}\}$
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:* 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)
 
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:* 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}\}$
 
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:Another guidance: The affirmative answer is given by [[Liouville theorems|Liouville's theorem on approximation of algebraic numbers]].
'''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.) 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:<sub>$$\alpha = \frac{t}{\sqrt {p^2+q^2}}$$</sub>
 
  
  
'''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 $p=r$ & $q=s$.
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Suppose $r:\Bbb N\to (0,1)$ is a function given by $r(n)$ is obtained by putting a point at the beginning of $n$ instance $r(34880)=0.34880$ and similarly consider $\forall k\in\Bbb N,\, w_k:\Bbb N\to (0,1)$ is a function given by $\forall n\in\Bbb N,$ $w_k(n)=10^{1-k}\cdot r(n)$ and let $S=\bigcup _{k\in\Bbb N}w_k(\Bbb P)$.
  
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'''Theorem''' $1$: $r(\Bbb P)$ is dense in the interval $[0.1,1]$. (proof using lemma above)
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:Regarding to expression form of Goldbach's conjecture, by using this theorem, I wanted enmesh prime numbers properties (prime number theorem should be used for proving this theorem and there is no way except using prime number theorem to prove this density because there is no deference between a prime $p$ and its image $r(p)$ other than a sign or a mark as a point for instance $59$ & $0.59$.) towards Goldbach hence I planned this method.
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:A corollary: 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.
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::Question $2$: Which mathematical concept at $[0.1,1)$ could be in accordance with the prime gap at natural numbers?
  
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 $f_2 (x)$.
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'''Question''' $3$: What is equivalent to the ''prime number theorem'' in $[0.1,1)$?
  
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) $
 
  
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'''Let''' $A_n=\{p_{1n},p_{2n},p_{3n},...,p_{mn}\}$ is all primes with $n$ digits, now since $\forall i=1,2,3,...,m-1,\,r(p_{in})\lt r(p_{(i+1)n})$ and $\lim_{m\to\infty}\frac{\pi(10^{m+1})-\pi(10^m)}{\pi(10^m)}=9$ I offer (probably via group theory & prime number theorem can be solved.): '''Guess''' $2$: $$\lim_{n\to\infty}\frac{\prod_{i=1}^mr(p_{in})}{\prod_{p\in\Bbb P,\,p\lt p_{1n}}r(p)}\sim({5\over9})^9\,.$$
  
'''Guess''' $1$: $f_3 (L_1) $ is dense in the $S^1 \cap \{ (x,y)\,|\, 0 \le y$ & $2^{-0.5} \le x \}$.
 
  
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'''Theorem''' $2$: $S$ is dense in the interval $[0,1]$ and $S\times S$ is dense in the $[0,1]\times [0,1]$.
  
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 $y=\pm x,\,y=0,\,x=0$
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'''An algorithm''' that makes new cyclic groups on $\Bbb N$:
  
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Let $\Bbb N$ be that group and at first write integers as a sequence with starting from $0$ and let identity element $e=1$ be corresponding with $0$ and two generators $m$ & $n$ be corresponding with $1$ & $-1$ so we have $\Bbb N=\langle m\rangle=\langle n\rangle$ for instance: $$0,1,2,-1,-2,3,4,-3,-4,5,6,-5,-6,7,8,-7,-8,9,10,-9,-10,11,12,-11,-12,...$$ $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,...$$
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then regarding to the sequence find an even 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 $km+(2-k)$ otherwise equations won't match with definitions of members inverse, and make a table of products 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 too and of course if integers corresponding with two members don't have same signs then product will be a piecewise-defined function for example $12\star _u 15=6$ or $(4\times 3)\star _u (4\times 4-1)=6$ because $(-5)+8=3$ & $-5\to 12,\,\, 8\to 15,\,\, 3\to 6,$ that implies $(4n)\star _u (4m-1)=4m-4n+2$ where $4m-1\gt 4n$ of course it is better at first members inverse be defined for example since $(-9)+9=0$ & $0\to 1,\,\, -9\to 20,\,\, 9\to 18$ so $20\star _u 18=1$, that shows $(4m)\star _u (4m-2)=1$, and with a little bit addition and multiplication all equations will be obtained simply that for this example is:
  
'''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
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$\begin{cases} m\star _u 1=m\\ (4m)\star _u (4m-2)=1=(4m+1)\star _u (4m-1)\\ (4m-2)\star _u (4n-2)=4m+4n-5\\ (4m-2)\star _u (4n-1)=4m+4n-2\\ (4m-2)\star _u (4n)=\begin{cases} 4m-4n-1 & 4m-2\gt 4n\\ 4n-4m+1 & 4n\gt 4m-2\\ 3 & m=n+1\end{cases}\\ (4m-2)\star _u (4n+1)=\begin{cases} 4m-4n-2 & 4m-2\gt 4n+1\\ 4n-4m+4 & 4n+1\gt 4m-2\end{cases}\\ (4m-1)\star _u (4n-1)=4m+4n-1\\ (4m-1)\star _u (4n)=\begin{cases} 4m-4n+2 & 4m-1\gt 4n\\ 4n-4m & 4n\gt 4m-1\\ 2 & m=n\end{cases}\\ (4m-1)\star _u (4n+1)=\begin{cases} 4m-4n-1 & 4m-1\gt 4n+1\\ 4n-4m+1 & 4n+1\gt 4m-1\\ 3 & m=n+1\end{cases}\\ (4m)\star _u (4n)=4m+4n-3\\ (4m)\star _u (4n+1)=4m+4n\\ (4m+1)\star _u  (4n+1)=4m+4n+1\\ \Bbb N=\langle 2\rangle=\langle 4\rangle\end{cases}$
  
  
'''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.
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Problem $1$: By using matrices rewrite operation of every group on $\Bbb N$.
  
  
Let $F= \Bbb Q$ so what are Galois group of polynomials $x^4+b^2x^2+b^2$ and $(1+a^2)x^2 +a^4$.
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Assume $\forall m,n\in\Bbb N$: $\begin{cases} n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\end{cases}$
  
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and $p_n\star _1p_m=p_{n\star m}$ that $p_n$ is $n$_th prime with $e=p_1=2$, obviously $(\Bbb N,\star)$ & $(\Bbb P,\star _1)$ are groups and $\langle 2\rangle =\langle 3\rangle =(\Bbb N,\star)\simeq (\Bbb Z,+)\simeq (\Bbb P,\star _1)=\langle 3\rangle=\langle 5\rangle$.
  
'''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 $(x_1,y_1)=(x_2,y_2)$.
 
  
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'''Theorem''' $3$: $(S,\star _S)$ is a group as: $\forall p,q\in\Bbb P,\,\forall m,n\in\Bbb N,\,\forall w_m(p),w_n(q)\in S,$
  
Let $Y=\{(a,b)\,|\, (a,b)\in (S_1\times S_1)\setminus L,\, 0.01\le a+b\lt 0.1\}$ & $\forall i\in \Bbb N,$ $E_i=\{(a,b)\,|$ $(a,b)\in S_1\times S_1,$ $a+b=r_1(2i)\}$ & $O_i=\{(a,b)\,|\, (a,b)\in S_1\times S_1,\, a+b=r_1(2i-1)\}$.
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$\begin{cases} e=0.2\\ \\(w_m(p))^{-1}=w_{m^{-1}}(p^{-1}) & m\star m^{-1}=1,\, p\star _1 p^{-1}=2\\ \\w_m(p)\star _S w_n(q)=w_{m\star n} (p\star _1 q)\end{cases}$
  
 +
hence $\langle 0.02,0.3\rangle=(S,\star _S)\simeq\Bbb Z\oplus\Bbb Z$.
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:of course using algorithm above to generate cyclic groups on $\Bbb N$, we can impose another group structure on $\Bbb N$ and consequently on $\Bbb P$ but eventually $S$ with an operation analogous above operation $\star _S$ will be an Abelian group.
  
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 $\{(0.1,0),(0.05,0.05)\}$ & $\{(0.01,0),(0.005,0.005)\}$.
 
  
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'''Theorem''' $4$: $(S\times S,\star _{S\times S})$ is a group as: $\forall m_1,n_1,m_2,n_2\in\Bbb N,\,\forall p_1,p_2,q_1,q_2\in\Bbb P,$ $\forall (w_{m_1}(p_1),w_{m_2}(p_2)),(w_{n_1}(q_1),w_{n_2}(q_2))\in S\times S,$
  
'''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$(O_i \setminus L)\in \Bbb N$.
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$\begin{cases} e=(0.2,0.2)\\ \\(w_{m_1}(p_1),w_{m_2}(p_2))^{-1}=(w_{m_1^{-1}}(p_1^{-1}),w_{m_2^{-1}}(p_2^{-1}))\\ \text{such that}\quad m_1\star m_1^{-1}=1=m_2\star m_2^{-1},\, p_1\star _1p_1^{-1}=2=p_2\star _1p_2^{-1}\\ \\(w_{m_1}(p_1),w_{m_2}(p_2))\star _{S\times S} (w_{n_1}(q_1),w_{n_2}(q_2))=(w_{m_1\star n_1} (p_1\star _1 q_1),w_{m_2\star n_2}(p_2\star _1 q_2))\end{cases}$
: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.
 
  
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hence $\langle (0.02,0.2),(0.2,0.02),(0.3,0.2),(0.2,0.3)\rangle=(S\times S,\star _{S\times S})\simeq\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z$.
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:of course using algorithm above to generate cyclic groups on $\Bbb N$, we can impose another group structure on $\Bbb N$ and consequently on $\Bbb P$ but eventually $S\times S$ with an operation analogous above operation $\star _{S\times S}$ will be an Abelian group.
  
'''Guess''' $3$: $\forall i\in \Bbb N,$ cardinal$(E_i)=\aleph_0 =$cardinal$(O_i \cap L)$
 
  
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I want make some topologies having '''prime numbers properties''' presentable in the collection of '''open sets''', in principle when we image a prime $p$ to real numbers as $w_k(p)$ indeed we accompany prime numbers properties among real numbers which regarding to the expression form of prime number theorem for this aim we should use an important mathematical technique as logarithm function into some planned topologies: '''question''' $4$: 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. ($M=\Bbb R,\,B=[0,1],\,A=S$ or $M=\Bbb R^2,$ $B=[0,1]\times[0,1],$ $A=S\times S$)
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:Perhaps this technique is useful: an extension of prime number theorem: $\forall n\in\Bbb N,$ and for each subinterval $(a,b)$ of $[0.1,1),$ that $a\neq b,$ assume:
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:$\begin{cases} U_{(a,b)}:=\{n\in\Bbb N\mid a\le r(n)\le b\},\\ \\V_{(a,b)}:=\{p\in\Bbb P\mid a\le r(p)\le b\},\\ \\U_{(a,b),n}:=\{m\in U_{(a,b)}\mid m\le n\},\\ \\V_{(a,b),n}:=\{p\in V_{(a,b)}\mid p\le n\},\\ \\w_{(a,b),n}:={\#V_{(a,b),n}\over\#U_{(a,b),n}}\cdot\log n,\\ \\w_{(a,b)}:=\lim _{n\to\infty} w_{(a,b),n}\\ \\z_{(a,b),n}:={\#V_{(a,b),n}\over\#U_{(a,b),n}}\cdot\log{(\#U_{(a,b),n})}\\ \\z_{(a,b)}:=\lim_{n\to\infty}z_{(a,b),n}\end{cases}$
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::Guess $3$: $\forall (a,b)\subset [0.1,1),\,w_{(a,b)}={10\over9}\cdot(b-a)$.
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:::[https://math.stackexchange.com/questions/2683513/an-extension-of-prime-number-theorem/2683561#2683561 Answer] given by [https://math.stackexchange.com/users/82961/peter $@$Peter] from stackexchange site: Imagine a very large number $N$ and consider the range $[10^N,10^{N+1}]$. The natural logarithms of $10^N$ and $10^{N+1}$ only differ by $\ln(10)\approx 2.3$ Hence the reciprocals of the logarithms of all primes in this range virtually coincicde. Because of the approximation $$\int_a^b \frac{1}{\ln(x)}dx$$ for the number of primes in the range $[a,b]$ the number of primes is approximately the length of the interval divided by $\frac{1}{\ln(10^N)}$, so is approximately equally distributed. Hence your conjecture is true.
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:::Benfords law seems to contradict this result , but this only applies to sequences producing primes as the Mersenne primes and not if the primes are chosen randomly in the range above.
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::Guess $4$: $\forall (a,b)\subset [0.1,1),\,z_{(a,b)}={10\over9}\cdot(b-a)$.
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:::Question $5$: What does mean $\forall a\in[0.1,1),\,\forall b\in(0.1,1),\,a\lt b,\,\lim_{b\to a}z_{(a,b)}=0$?
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::Guess $5$: $\forall(a,b),(c,d)\subset[0.1,1),\,\lim_{n\to\infty}{\#V_{(a,b),n}\over\#V_{(c,d),n}}={b-a\over d-c}=\lim_{n\to\infty}{\#U_{(a,b),n}\over\#U_{(c,d),n}}$.
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:::<small>Comment given by [https://math.stackexchange.com/users/403583/dzoooks $@$Dzoooks] from stackexchange site: It shouldn't be that hard to get estimates from $V_{(a,b),n}=\{p\leq n : 10^ka\lt p\lt 10^k\text{ for some }k\}=\sqcup_{k\geq1}\{p\in[0,n]\cap(10^ka,10^kb)\},$ where the union is disjoint from $10^kb\lt10^k\leq10^{k+1}a$. Then $\#V_{(a,b),n}$ can be summed with the PNT. You'll see that a $(b-c)$ comes out of the sum..maybe</small>
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:::<small>and the PNT gives $\#\{p\in[0,n]\cap(10^ka,10^k)\}\sim\frac{(b-a)10^k}{\log b-\log a},$ for large $n$ and $k$. Factor these out of the sum, and it looks like your limit is actually $\frac{b-a}{\log b-\log a}\cdot\frac{\log d-\log c}{d-c}$.</small>
  
'''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 $[0,1]\times [0,1]$.
 
  
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'''Theorem''' $5$: Let $t_n:\Bbb N\to\Bbb N\setminus\{n\in\Bbb N: 10\mid n\}$ is a surjective strictly monotonically increasing sequence now $\{t_n\}_{n\in\Bbb N}$ is a cyclic group with: $\begin{cases} e=1\\ t_n^{-1}=t_{n^{-1}}\quad\text{that}\quad n\star n^{-1}=1\\ t_n\star _tt_m=t_{n\star m}\end{cases}$
  
'''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 $\{(x,y)\,|\, x+y=r_1(2i-1),$ $0\le y\le x\}$.
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that $(\{t_n\}_{n\in\Bbb N},\star _t)=\langle 2\rangle=\langle 3\rangle$ and let $E:=\bigcup _{k\in\Bbb N} w_k(\Bbb N\setminus\{n\in\Bbb N: 10\mid n\})$ so $(E,\star _E)$ is an Abelian group with $\forall m,n\in\Bbb N,$ $\forall a,b\in\Bbb N\setminus\{n\in\Bbb N: 10\mid n\}$: $\,\,\begin{cases} e=0.1\\ w_n(a)^{-1}=w_{n^{-1}}(a^{-1})\quad\text{that}\quad n\star n^{-1}=1,\, a\star _ta^{-1}=1\\ w_n(a)\star _Ew_m(b)=w_{n\star m}(a\star _tb)\end{cases}$
  
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that $\langle 0.01,0.2\rangle=E\simeq\Bbb Z\oplus\Bbb Z$ ♦
  
'''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.
 
  
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'''now''' assume $(S\times S)\oplus E$ is external direct sum of the groups $S\times S$ and $E$ with $e=(0.2,0.2,0.1)$ and $\langle (0.02,0.2,0.1),(0.2,0.02,0.1),(0.3,0.2,0.1),(0.2,0.3,0.1),(0.2,0.2,0.01),(0.2,0.2,0.2)\rangle=$ $(S\times S)\oplus E\simeq\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z$.
  
Now I want find a relation between $L_1$ & $W:=((S_1\times S_1) \cap A) \setminus L$.
 
  
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<small><s>'''Theorem''' $6$: $(S,\lt _1)$ is a total ordered set with order relation $\lt_1$ as: $\forall i,n,k\in\Bbb N$ if $p_n$ be $n$-th prime number, relation $\lt _1$ is defined with: $w_i(p_n)\lt _1w_i(p_{n+k})\lt _1w_{i+1}(p_n)$ or $$0.2\lt _10.3\lt _10.5\lt _10.7\lt _10.11\lt _10.13\lt _10.17\lt _1...0.02\lt _10.03\lt _10.05\lt _10.07\lt _10.011\lt _1$$ $$0.013\lt _10.017\lt _1...0.002\lt _10.003\lt _10.005\lt _10.007\lt _10.0011\lt _10.0013\lt _10.0017\lt _1...$$ and $(E,\lt_2)$ is another total ordered set with order relation $\lt _2$ as: $\forall i,n,k\in\Bbb N$ that $10\nmid n,\, 10\nmid n+k,$ $w_i(n)\lt _2w_i(n+k)\lt _2w_{i+1}(n)$ or $$0.1\lt _2 0.2\lt _2 0.3\lt _2 ...0.9\lt _2 0.11\lt _2 0.12\lt _2 ...0.19\lt _2 0.21\lt _2 ...0.01\lt _2 0.02\lt _2 0.03\lt _2 ...0.09$$ $$\lt _2 0.011\lt _2 0.012\lt _2 ...0.019\lt _2 0.021\lt _2 ...0.001\lt _2 0.002\lt _2 0.003\lt _2 ...0.009\lt _2 0.0011\lt _2 ...$$ now $M:=S\times S\times E$ is a total ordered set with order relation $\lt _3$ as: $\forall (a,b,t),(c,d,u)\in S\times S\times E,$ $(a,b,t)\lt _3(c,d,u)$ iff $\,\,\begin{cases} t\lt _2u & or\\ t=u,\,\, a+b\lt _2c+d & or\\ t=u,\,\, a+b=c+d,\,\, b\lt _1 d\end{cases}$ ♦</s></small>
  
'''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 $\{ p\cdot 10 \,|\, p \in \Bbb P \} \subset K$.
 
  
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<small><s>'''Theorem''' $6$: Assume $([0,1],\lt_1)$ is a total ordered set with: $\forall a,b\in[0,1],\,a\lt_1b$ iff $a\gt b$ then  $S$ & $E$ are total ordered sets with restriction of order relation $\lt _1$ on $S$ & $E$, now $M:=S\times S\times E$ is a total ordered set with order relation $\lt _2$ as: $\forall (a,b,t),(c,d,u)\in S\times S\times E,$ $(a,b,t)\lt _2(c,d,u)$ iff $\,\,\begin{cases} t\lt _1u & or\\ t=u,\,\, a+b\lt _1c+d & or\\ t=u,\,\, a+b=c+d,\,\, b\lt _1d\end{cases}$ ♦ </s></small>
  
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.
 
  
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'''now''' assume $M$ is a topological space ('''Hausdorff space''') induced by order relation $\lt _?$.
  
'''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 $r(m)$ & $r(n)$.
 
  
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'''Question''' $6$: Is $S$ a topological group under topology induced by order relation $\lt_?$ and is $(S\times S)\oplus E$ a topological group under topology of $M$?
  
'''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 $S_1$!
 
  
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'''A new version of Goldbach's conjecture''': For each even natural number $t$ greater than $4$ and $\forall c,m\in\Bbb N\cup\{0\}$ that $10^c\mid t,\, 10^{1+c}\nmid t$, $A_m=\{(a,b)\mid a,b\in S,\, 10^{-1-m}\le a+b\lt 10^{-m}\}$ and if $u$ is the number of digits in $t$ then $\exists (a,b)\in A_c$ such that $t=10^{c+u}\cdot (a+b),\, 10^{c+u}\cdot a,10^{c+u}\cdot b\in\Bbb P\setminus\{2\},\, (a,b,10^{-c-u}\cdot t)\in M$.
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:Using homotopy groups Goldbach's conjecture will be proved.
  
'''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 $t\cdot 10^{i-m}\neq r+s$.
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Alireza Badali 08:27, 31 March 2018 (CEST)
: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.
 
  
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=== [https://en.wikipedia.org/wiki/Polignac%27s_conjecture Polignac's conjecture] ===
  
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 $\{(a,b)\in L_1\,|\, a=r_1(p)\}\cup\{(a,b)\in L_1\,|\, a=r_2(p)\}$.
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In previous chapter above I used an important technique by theorem $1$ for presentment of prime numbers properties as density in discussion that using prime number theorem it became applicable, anyway, but now I want perform another method for Twin prime conjecture (Polignac) in principle prime numbers properties are ubiquitous in own natural numbers.
  
  
'''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 $(10^{-k-1}r,10^{-k-1}s)\in A\cup \{(x,x)\,|$ $0.005\le x\lt0.05\}$.
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'''Theorem''' $1$: $(\Bbb N,\star _T)$ is a group with: $\forall m,n\in\Bbb N,$
  
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$\begin{cases} (12m-10)\star_T(12m-9)=1=(12m-8) \star_T(12m-5)=(12m-7) \star_T(12m-4)=\\ (12m-6) \star_T(12m-1)=(12m-3) \star_T(12m)=(12m-2) \star_T(12m+1)\\ (12m-10) \star_T(12n-10)=12m+12n-19\\ (12m-10) \star_T(12n-9)=\begin{cases} 12m-12n+1 & 12m-10\gt 12n-9\\ 12n-12m-2 & 12n-9\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-8)=12m+12n-15\\ (12m-10) \star_T(12n-7)=12m+12n-20\\ (12m-10) \star_T(12n-6)=12m+12n-11\\ (12m-10) \star_T(12n-5)=\begin{cases} 12m-12n-3 & 12m-10\gt 12n-5\\ 12n-12m+8 & 12n-5\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-4)=\begin{cases} 12m-12n-6 & 12m-10\gt 12n-4\\ 12n-12m+3 & 12n-4\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-3)=12m+12n-18\\ (12m-10) \star_T(12n-2)=\begin{cases} 12m-12n-10 & 12m-10\gt 12n-2\\ 12n-12m+11 & 12n-2\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-1)=\begin{cases} 12m-12n-7 & 12m-10\gt 12n-1\\ 12n-12m+12 & 12n-1\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n)=\begin{cases} 12m-12n-8 & 12m-10\gt 12n\\ 12n-12m+7 & 12n\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n+1)=12m+12n-10\\ (12m-9) \star_T(12n-9)=12m+12n-16\\ (12m-9) \star_T(12n-8)=\begin{cases} 12m-12n & 12m-9\gt 12n-8\\ 12n-12m+5 & 12n-8\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-7)=\begin{cases} 12m-12n-1 & 12m-9\gt 12n-7\\ 12n-12m+2 & 12n-7\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-6)=\begin{cases} 12m-12n-4 & 12m-9\gt 12n-6\\ 12n-12m+9 & 12n-6\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-5)=12m+12n-12\\ (12m-9) \star_T(12n-4)=12m+12n-17\\ (12m-9) \star_T(12n-3)=\begin{cases} 12m-12n-5 & 12m-9\gt 12n-3\\ 12n-12m+4 & 12n-3\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-2)=12m+12n-9\\ (12m-9) \star_T(12n-1)=12m+12n-14\\ (12m-9) \star_T(12n)=12m+12n-13\\ (12m-9)\star_T(12n+1)=\begin{cases} 12m-12n-9 & 12m-9\gt 12n+1\\ 12n-12m+6 & 12n+1\gt 12m-9\end{cases}\\ (12m-8) \star_T(12n-8)=12m+12n-11\\ (12m-8) \star_T(12n-7)=12m+12n-18\\ (12m-8) \star_T(12n-6)=12m+12n-7\\ (12m-8) \star_T(12n-5)=\begin{cases} 12m-12n+1 & 12m-8\gt 12n-5\\ 12n-12m-2 & 12n-5\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n-4)=\begin{cases} 12m-12n+2 & 12m-8\gt 12n-4\\ 12n-12m-1 & 12n-4\gt 12m-8\\ 2 & m=n\end{cases}\\ (12m-8) \star_T(12n-3)=12m+12n-10\\ (12m-8) \star_T(12n-2)=\begin{cases} 12m-12n-8 & 12m-8\gt 12n-2\\ 12n-12m+7 & 12n-2\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n-1)=\begin{cases} 12m-12n-3 & 12m-8\gt 12n-1\\ 12n-12m+8 & 12n-1\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n)=\begin{cases} 12m-12n-6 & 12m-8\gt 12n\\ 12n-12m+3 & 12n\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n+1)=12m+12n-8\\ (12m-7) \star_T(12n-7)=12m+12n-15\\ (12m-7) \star_T(12n-6)=12m+12n-10\\ (12m-7) \star_T(12n-5)=\begin{cases} 12m-12n-6 & 12m-7\gt 12n-5\\ 12n-12m+3 & 12n-5\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n-4)=\begin{cases} 12m-12n+1 & 12m-7\gt 12n-4\\ 12n-12m-2 & 12n-4\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n-3)=12m+12n-11\\ (12m-7) \star_T(12n-2)=\begin{cases} 12m-12n-7 & 12m-7\gt 12n-2\\ 12n-12m+12 & 12n-2\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n-1)=\begin{cases} 12m-12n-8 & 12m-7\gt 12n-1\\ 12n-12m+7 & 12n-1\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n)=\begin{cases} 12m-12n-3 & 12m-7\gt 12n\\ 12n-12m+8 & 12n\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n+1)=12m+12n-7\\ (12m-6) \star_T(12n-6)=12m+12n-3\\ (12m-6) \star_T(12n-5)=\begin{cases} 12m-12n+5 & 12m-6\gt 12n-5\\ 12n-12m & 12n-5\gt 12m-6\\ 5 & m=n\end{cases}\\ (12m-6) \star_T(12n-4)=\begin{cases} 12m-12n+4 & 12m-6\gt 12n-4\\ 12n-12m-5 & 12n-4\gt 12m-6\\ 4 & m=n\end{cases}\\ (12m-6) \star_T(12n-3)=12m+12n-8\\ (12m-6) \star_T(12n-2)=\begin{cases} 12m-12n-6 & 12m-6\gt 12n-2\\ 12n-12m+3 & 12n-2\gt 12m-6\end{cases}\\ (12m-6) \star_T(12n-1)=\begin{cases} 12m-12n+1 & 12m-6\gt 12n-1\\ 12n-12m-2 & 12n-1\gt 12m-6\end{cases}\\ (12m-6) \star_T(12n)=\begin{cases} 12m-12n+2 & 12m-6\gt 12n\\ 12n-12m-1 & 12n\gt 12m-6\\ 2 & m=n\end{cases}\\ (12m-6) \star_T(12n+1)=12m+12n-6\\ (12m-5) \star_T(12n-5)=12m+12n-14\\ (12m-5) \star_T(12n-4)=12m+12n-13\\ (12m-5) \star_T(12n-3)=\begin{cases} 12m-12n-1 & 12m-5\gt 12n-3\\ 12n-12m+2 & 12n-3\gt 12m-5\end{cases}\\ (12m-5) \star_T(12n-2)=12m+12n-5\\ (12m-5) \star_T(12n-1)=12m+12n-4\\ (12m-5) \star_T(12n)=12m+12n-9\\ (12m-5) \star_T(12n+1)=\begin{cases} 12m-12n-5 & 12m-5\gt 12n+1\\ 12n-12m+4 & 12n+1\gt 12m-5\end{cases}\\ (12m-4) \star_T(12n-4)=12m+12n-12\\ (12m-4) \star_T(12n-3)=\begin{cases} 12m-12n & 12m-4\gt 12n-3\\ 12n-12m+5 & 12n-3\gt 12m-4\end{cases}\\ (12m-4) \star_T(12n-2)=12m+12n-4\\ (12m-4) \star_T(12n-1)=12m+12n-9\\ (12m-4) \star_T(12n)=12m+12n-14\\ (12m-4) \star_T(12n+1)=\begin{cases} 12m-12n-4 & 12m-4\gt 12n+1\\ 12n-12m+9 & 12n+1\gt 12m-4\end{cases}\\ (12m-3) \star_T(12n-3)=12m+12n-7\\ (12m-3) \star_T(12n-2)=\begin{cases} 12m-12n-3 & 12m-3\gt 12n-2\\ 12n-12m+8 & 12n-2\gt 12m-3\end{cases}\\ (12m-3) \star_T(12n-1)=\begin{cases} 12m-12n-6 & 12m-3\gt 12n-1\\ 12n-12m+3 & 12n-1\gt 12m-3\end{cases}\\ (12m-3) \star_T(12n)=\begin{cases} 12m-12n+1 & 12m-3\gt 12n\\ 12n-12m-2 & 12n\gt 12m-3\end{cases}\\ (12m-3) \star_T(12n+1)=12m+12n-3\\ (12m-2) \star_T(12n-2)=12m+12n-2\\ (12m-2) \star_T(12n-1)=12m+12n-1\\ (12m-2) \star_T(12n)=12m+12n\\ (12m-2) \star_T(12n+1)=\begin{cases} 12m-12n-2 & 12m-2\gt 12n+1\\ 12n-12m+1 & 12n+1\gt 12m-2\end{cases}\\ (12m-1) \star_T(12n-1)=12m+12n\\ (12m-1) \star_T(12n)=12m+12n-5\\ (12m-1) \star_T(12n+1)=\begin{cases} 12m-12n-1 & 12m-1\gt 12n+1\\ 12n-12m+2 & 12n+1\gt 12m-1\end{cases}\\ (12m) \star_T(12n)=12m+12n-4\\ (12m) \star_T(12n+1)=\begin{cases} 12m-12n & 12m\gt 12n+1\\ 12n-12m+5 & 12n+1\gt 12m\end{cases}\\ (12m+1) \star_T(12n+1)=12m+12n+1\end{cases}$
  
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 $\forall\theta\in\mathcal I,$ $J_{\theta}=\{(r_s(p),r_t(q))\in L_2\,|\, r_t(q)\le r_s(p),$ $(p,q)\in\mathrm I_{\theta},$ $s,t\in\Bbb N\}\cup\{(r_t(q),r_s(p))\in L_2\,|\, r_s(p)\le r_t(q),$ $(p,q)\in\mathrm I_{\theta},$ $s,t\in\Bbb N\}$.
+
that $\forall k\in\Bbb N,\,\langle 2\rangle =\langle 3\rangle =\langle (2k+1)\star _T (2k+3)\rangle=(\Bbb N,\star _T)\simeq (\Bbb Z,+)$ and $\langle (2k)\star _T(2k+2)\rangle\neq\Bbb N$ and each prime in $\langle 5\rangle$ is to form of $5+12k$ or $13+12k$, $k\in\Bbb N\cup\{0\}$ and each prime in $\langle 7\rangle$ is to form of $7+12k$ or $13+12k$, $k\in\Bbb N\cup\{0\}$ and $\langle 5\rangle\cap\langle 7\rangle=\langle 13\rangle$ and $\Bbb N=\langle 5\rangle\oplus\langle 7\rangle$ but there isn't any proper subgroup including all primes of the form $11+12k,$ $k\in\Bbb N\cup\{0\}$ (probably I have to make another better).
 +
:Proof:
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$$0,-1,1,-3,-2,-5,3,2,-4,6,5,4,-6,-7,7,-9,-8,-11,9,8,-10,12,11,10,-12,-13,13,-15,$$ $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,$$ $$-14,-17,15,14,-16,18,17,16,-18,-19,19,-21,-20,-23,21,20,-22,24,23,22,-24,...$$ $$29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,...$$
  
'''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 $L_2$
 
  
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'''Guess''' $1$: For each group on $\Bbb N$ like $(\Bbb N,\star)$ generated from algorithm above, if $p_i$ be $i$_th prime number and $x_i$ be $i$_th composite number then $\exists m\in\Bbb N,\,\forall n\in\Bbb N$ that $n\ge m$ we have: $2\star3\star5\star7...\star p_n=\prod_{i=1}^{n}p_i\gt\prod _{i=1}^{n}x_i=4\star6\star8\star9...\star x_n$
  
'''Guess''' $7$: $\forall a,d\in\Bbb N$ that gcd$(a,d)=1$ then cardinal$(\{(r_s(n),r_t(a+nd))\in L_1\,|\, n=0,1,2,3,...,$ $r,s\in\Bbb N\}\cup\{(r_t(a+nd),r_s(n))\in L_1\,|\, n=0,1,2,3,...,\, r,s\in\Bbb N\})=\aleph_0$
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'''Guess''' $2$: For each group on $\Bbb N$ like $(\Bbb N,\star)$ generated from algorithm above, we have: $\lim_{n\to\infty}\prod _{n=1}^{\infty}p_n,\lim_{n\to\infty}\prod _{n=1}^{\infty}x_n\in\Bbb N,\,\,(\lim_{n\to\infty}\prod _{n=1}^{\infty}p_n)\star(\lim_{n\to\infty}\prod _{n=1}^{\infty}x_n)=1$.
  
  
'''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$(\{(u,v)\in L_2\,|\, v\in\bigcup _{m\in\Bbb N} \{r_m(p)\}\})=\aleph_0$
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now let the group $G$ be external direct sum of three copies of the group $(\Bbb N,\star _T)$, hence $G=\Bbb N\oplus\Bbb N\oplus\Bbb N$.
: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$(\{(u,v)\in L_1\,|\, u-v=c\})=\aleph_0$
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'''Theorem''' $2$: $(\Bbb N\times\Bbb N\times\Bbb N,\lt _T)$ is a well ordering set with order relation $\lt _T$ as: $\forall (m_1,n_1,t_1),(m_2,n_2,t_2)\in\Bbb N\times\Bbb N\times\Bbb N,$ $(m_1,n_1,t_1)\lt _T(m_2,n_2,t_2)$ iff $\begin{cases} t_1\lt t_2 & or\\ t_1=t_2,\, m_1-n_1\lt m_2-n_2 & or\\ t_1=t_2,\, m_1-n_1=m_2-n_2,\, n_1\lt n_2\end{cases}$
  
  
'''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 $p,q\lt 2^tm\})=\aleph_0$
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and suppose $M=\Bbb N\times\Bbb N\times\Bbb N$ is a topological space ('''Hausdorff space''') induced by order relation $\lt _T$.
  
  
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.):
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'''Question''' $1$: Is $G$ a topological group with topology of $M$?
  
  
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.
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'''Now''' regarding to the group $(\Bbb N,\star_T)$, I am planning an algebraic form of prime number theorem towards twin prime conjecture:
:[https://math.stackexchange.com/questions/2352522/density-can-be-transferred-by-a-homeomorphism Proof] given by [https://math.stackexchange.com/users/337888/daniel-schepler 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 [https://math.stackexchange.com/questions/2352522/density-can-be-transferred-by-a-homeomorphism proof] given by [https://math.stackexchange.com/users/446262/jos%C3%A9-carlos-santos 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 $[0.1,1]$
 
  
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Recall the statement of the prime number theorem: Let $x$ be a positive real number, and let $\pi(x)$ denote the number of primes that are less than or equal to $x$. Then the ratio $\pi(x)\cdot{\log x\over x}$ can be made arbitrarily close to $1$ by taking $x$ sufficiently large.
  
Assume $\forall m,n\in\Bbb N$
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Question $2$: Suppose $\pi_1(x)$ is all prime numbers of the form $4k+1$ and less than $x$ and $\pi_2(x)$ is all prime numbers of the form $4k+3$ and less than $x$. Do $\lim_{x\to\infty}\pi_1(x)\cdot{\log x\over x}=0.5=\lim_{x\to\infty}\pi_2(x)\cdot{\log x\over x}\ ?$
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:[https://math.stackexchange.com/questions/2769471/another-extension-of-prime-number-theorem/2769494#2769494 Answer] given by [https://math.stackexchange.com/users/174927/milo-brandt $@$Milo Brandt] from stackexchange site: Basically, for any $k$, the primes are equally distributed across the congruence classes $\langle n\rangle$ mod $k$ where $n$ and $k$ are coprime.
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:This result is known as the prime number theorem for arithmetic progressions. [Wikipedia](https://en.wikipedia.org/wiki/Prime_number_theorem#Prime_number_theorem_for_arithmetic_progressions) discusses it with a number of references and one can find a proof of it by Ivan Soprounov [here](http://academic.csuohio.edu/soprunov_i/pdf/primes.pdf), which makes use of the Dirichlet theorem on arithmetic progressions (which just says that $\pi_1$ and $\pi_2$ are unbounded) to prove this stronger result.
  
$\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 & $\forall (p_n,p_m),(p_s,p_t)\in\Bbb P\times\Bbb P,$ $(p_n,p_m)\star _2(p_s,p_t)=(p_n\star _1p_s,p_m\star _1p_t)$.
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Question $3$: For each neutral infinite subset $A$ of $\Bbb N$, does exist a cyclic group like $(\Bbb N,\star)$ such that $A$ is a maximal subgroup of $\Bbb N$?
  
Obviously $(\Bbb N,\star)$ & $(\Bbb P,\star _1)$ & $(\Bbb P\times\Bbb P,\star _2)$ are groups and $\langle 2\rangle =(\Bbb N,\star)\cong (\Bbb Z,+)\cong (\Bbb P,\star _1)=\langle 3\rangle$ & $\langle (3,2),(2,3)\rangle =(\Bbb P\times\Bbb P,\star _2)\cong (\Bbb Z\times\Bbb Z,+)$ & $\pi _n(\Bbb P\times\Bbb P)\cong\pi _n(\Bbb Z)\times\pi _n(\Bbb Z)$.
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Question $4$: If $(\Bbb N,\star_1)$ is a cyclic group and $n\in\Bbb N$ and $A=\{a_i\mid i\in\Bbb N\}$ is a non-trivial subgroup of $\Bbb N$ then does exist another cyclic group $(\Bbb N,\star_2)$ such that $\prod _{i=1}^{\infty}a_i=a_1\star_2a_2\star_2a_3\star_2...=n$?
  
In $(\Bbb N,\star)$ we have: $(2k+1)^n=2kn+1=(2k)^{-n},\quad (2k)^n=2kn=(2k+1)^{-n}$
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Question $5$: If $(\Bbb N,\star)$ is a cyclic group and $n\in\Bbb N$ then does exist a non-trivial subset $A=\{a_i\mid i\in\Bbb N\}$ of $\Bbb P$ with $\#(\Bbb P\setminus A)=\aleph_0$ and $\prod _{i=1}^{\infty}a_i=a_1\star a_2\star a_3\star...=n$?
  
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Question $6$: If $(\Bbb N,\star_1)$ and $(\Bbb N,\star_2)$ are cyclic groups and $A=\{a_i\mid i\in\Bbb N\}$ is a non-trivial subgroup of $(\Bbb N,\star_1)$ and $B=A\cap\Bbb P$ then does $\prod_{i=1}^{\infty}a_i=a_1\star_2a_2\star_2a_3\star_2...\in\Bbb N$?
  
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}
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'''Theorem''' $3$: $U:=\{{r(p)-r(q)\over r(s)-r(t)}\mid p,q,s,t\in\Bbb P,\,s\neq t\}$ is dense in $\Bbb R$.
\forall m,n,u,v\in\Bbb N,\, {m\over n}\star _{Q_1} 1={m\over n}\\
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:Proof given by [https://math.stackexchange.com/users/28111/noah-schweber $@$NoahSchweber] from stackexchange site: for any real number $x$ we can by the density of the image of $r$ in $[0.1,1]$ find primes $p,q,s,t$ such that $r(p)−r(q)$ is very close to $x\over n$ and $r(s)−r(t)$ is very close to $1\over n$ for some large integer $n$.
({m\over n})^{-1}={m^{-1}\over n^{-1}}\qquad m\star m^{-1}=1=n\star n^{-1}\\
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:'''Question''' $7$: Does $T:=U\cap\Bbb P$ have infinitely many primes?
{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}
+
:'''Question''' $8$: Is the set $V:=\{p+2q\mid p,q\in\Bbb P,\,p+2q\in T\}$ infinite?
{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 <math>{44 \over 12}</math> that gets <math>{11\over 3}</math>.
 
:'''Question''' $2$: does <math>(Q_1,\star _{Q_1})\cong (\Bbb Q,+)</math>?
 
:'''Problem''' $1$: Suppose <math>N_1=\{(m,n)\,|\, m,n\in\Bbb N,\, \gcd(m,n)=1\}</math>, make a group on <math>N_1</math> correspondence to <math>(Q_1,\star _{Q_1})</math> in terms of members production and isomorph to <math>(Q_1,\star _{Q_1})</math>.
 
  
 +
Alireza Badali 12:34, 28 April 2018 (CEST)
  
According to [https://en.wikipedia.org/wiki/Pairing_function#Cantor_pairing_function pairing function] let $Q_p=\{(s,t)\,|\, s,t\in\Bbb N\cup\{0\}\}$ and $Q_p$ is a cyclic group as:
+
== Some dissimilar conjectures ==
$\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}$
 
  
 +
'''Algebraic analytical number theory'''
  
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: $a_n\star _aa_m=a_{n\star m},\, e=a_1,\, G=\langle a_2\rangle$
+
Alireza Badali 16:51, 4 July 2018 (CEST)
  
 +
=== [https://en.wikipedia.org/wiki/Collatz_conjecture Collatz conjecture] ===
  
'''Algorithm''' of making of new cyclic groups on $\Bbb N$:
+
The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined as follows: start with any positive integer $n$. Then each term is obtained from the previous term as follows: if the previous term is even, the next term is one half the previous term. Otherwise, the next term is $3$ times the previous term plus $1$. The conjecture is that no matter what value of $n$, the sequence will always reach $1$. The conjecture is named after German mathematician [https://en.wikipedia.org/wiki/Lothar_Collatz Lothar Collatz], who introduced the idea in $1937$, two years after receiving his doctorate. It is also known as the $3n + 1$ conjecture.
  
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}
+
'''Theorem''' $1$: If $(\Bbb N,\star_{\Bbb N})$ is a cyclic group with $e_{\Bbb N}=1$ & $\langle m_1\rangle=\langle m_2\rangle=(\Bbb N,\star_{\Bbb N})$ and $f:\Bbb N\to\Bbb N$ is a bijection such that $f(1)=1$ then $(\Bbb N,\star _f)$ is a cyclic group with: $e_f=1$ & $\langle f(m_1)\rangle=\langle f(m_2)\rangle=(\Bbb N,\star_f)$ & $\forall m,n\in\Bbb N,$ $f(m)\star _ff(n)=f(m\star_{\Bbb N}n)$ & $(f(n))^{-1}=f(n^{-1})$ that $n\star_{\Bbb N}n^{-1}=1$.
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}
+
I want make a group in accordance with [https://www.jasondavies.com/collatz-graph/ Collatz graph] but [https://math.stackexchange.com/users/334732/robert-frost $@$RobertFrost] from stackexchange site advised me in addition, it needs to be a torsion group because then it can be used to show convergence, meantime I like apply lines in the Euclidean plane $\Bbb R^2$ too.
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.
 
  
 +
<small>Question $1$: What is function of this sequence on to natural numbers? $1,2,4,3,6,5,10,7,14,8,16,9,18,11,22,12,24,13,26,15,30,17,34,19,38,20,40,21,42,23,46,25,50,...$ such that we begin from $1$ and then write $2$ then $2\times2$ then $3$ then $2\times3$ then ... but if $n$ is even and previously we have written $0.5n$ and then $n$ then ignore $n$ and continue and write $n+1$ and then $2n+2$ and so on for example we have $1,2,4,3,6,5,10$ so after $10$ we should write $7,14,...$ because previously we have written $3,6$.
 +
:[https://math.stackexchange.com/questions/2779491/what-is-function-of-this-sequence-on-to-natural-numbers/2779815#2779815 Answer] given by [https://math.stackexchange.com/users/16397/r-e-s $@$r.e.s] from stackexchange site: Following is a definition of your sequence without using recursion.
 +
:Let $S=(S_0,S_1,S_2,\ldots)$ be the increasing sequence of positive integers that are expressible as either $2^e$ or as $o_1\cdot 2^{o_2}$, where $e$ is an even nonnegative integer, $o_1>1$ is an odd positive integer and $o_2$ is an odd positive integer. Thus $$S=(1, 4, 6, 10, 14, 16, 18, 22, 24, 26, 30, 34, 38, 40,42,\ldots).$$ Let $\bar{S}$ be the complement of $S$ with respect to the positive integers; i.e., $$\bar{S}=(2, 3, 5, 7, 8, 9, 11, 12, 13, 15, 17, 19, 20, 21, 23, 25,\ldots).$$ Your sequence is then $T=(T_0,T_1,T_2,\ldots)$, where
 +
$$T_n:=\begin{cases}S_{n\over 2}&\text{ if $n$ is even}\\
 +
\bar{S}_{n-1\over 2}&\text{ if $n$ is odd.}
 +
\end{cases}
 +
$$
 +
:Thus $T=(1, 2, 4, 3, 6, 5, 10, 7, 14, 8, 16, 9, 18, 11, 22, 12, 24, 13, 26, 15, 30, 17, 34, 19, 38, 20, \ldots).$
  
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}
+
:References:
0+(-7)=-7 & 1\star _{K_4}17=17\\
+
:Sequences $S,\bar{S},T$ are OEIS [A171945](http://oeis.org/A171945), [A053661](http://oeis.org/A053661), [A034701](http://oeis.org/A034701) respectively. These are all discussed in ["The vile, dopey, evil and odious game players"](https://www.sciencedirect.com/science/article/pii/S0012365X11001427).
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}
+
:Sage code:
m\star _{K_4}1=m\\
+
    def is_in_S(n): return ( (n.valuation(2) % 2 == 0) and (n.is_power_of(2)) ) or ( (n.valuation(2) % 2 == 1) and not(n.is_power_of(2)) )
(6m-4) \star _{K_4}(6m-1)=1=(6m-3) \star _{K_4}(6m-2)=(6m) \star _{K_4}(6m+1)\\
+
    S = [n for n in [1..50] if is_in_S(n)]
(6m-4) \star _{K_4}(6n-4)=6m+6n-9\\
+
    S_ = [n for n in [1..50] if not is_in_S(n)]
(6m-4) \star _{K_4}(6n-3)=6m+6n-6\\
+
    T = []
(6m-4) \star _{K_4}(6n-2)=\begin{cases}
+
    for i in range(max(len(S),len(S_))):
6m-6n-3 & 6m-4\gt 6n-2\\
+
        if i % 2 == 0: T += [S[i/2]]
6n-6m+5 & 6n-2\gt 6m-4\end{cases}\\
+
        else: T += [S_[(i-1)/2]]
(6m-4) \star _{K_4}(6n-1)=\begin{cases}
+
    print S
6m-6n & 6m-4\gt 6n-1\\
+
    print S_
6n-6m+1 & 6n-1\gt 6m-4\end{cases}\\
+
    print T
(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!
+
    [1, 4, 6, 10, 14, 16, 18, 22, 24, 26, 30, 34, 38, 40, 42, 46, 50]
:'''Hypothesis''' $3$: With [https://en.wikipedia.org/wiki/Matrix_(mathematics) 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!
+
    [2, 3, 5, 7, 8, 9, 11, 12, 13, 15, 17, 19, 20, 21, 23, 25, 27, 28, 29, 31, 32, 33, 35, 36, 37, 39, 41, 43, 44, 45, 47, 48, 49]
 +
    [1, 2, 4, 3, 6, 5, 10, 7, 14, 8, 16, 9, 18, 11, 22, 12, 24, 13, 26, 15, 30, 17, 34, 19, 38, 20, 40, 21, 42, 23, 46, 25, 50]</small>
  
  
By using [[User_talk:Musictheory2math#Polignac.27s_conjecture|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 $u\cdot 10^{m+1}\in\Bbb P$
+
'''Theorem''' $2$: If $(\Bbb N,\star_1)$ & $(\Bbb N,\star_2)$ are cyclic groups with generators respectively $u_1$ & $v_1$ and $u_2$ & $v_2$ then $C_1=\{(m,2m)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_1}=(1,2)\\ \\\forall m,n\in\Bbb N,\,(m,2m)\star_{C_1}(n,2n)=(m\star_1n,2(m\star_1n))\\ (m,2m)^{-1}=(m^{-1},2\times m^{-1})\qquad\text{that}\quad m\star_1m^{-1}=1\\ \\C_1=\langle(u_1,2u_1)\rangle=\langle(v_1,2v_1)\rangle\end{cases}$ and $C_2=\{(3m-1,2m-1)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_2}=(2,1)\\ \\\forall m,n\in\Bbb N,\,(3m-1,2m-1)\star_{C_2}(3n-1,2n-1)=(3(m\star_2n)-1,2(m\star_2n)-1)\\ (3m-1,2m-1)^{-1}=(3\times m^{-1}-1,2\times m^{-1}-1)\qquad\text{that}\quad m\star_2 m^{-1}=1\\ \\C_2=\langle(3u_2-1,2u_2-1)\rangle=\langle(3v_2-1,2v_2-1)\rangle\end{cases}$•
 +
:And let $C:=C_1\oplus C_2$ be external direct sum of the groups $C_1$ & $C_2$. '''Problem''' $1$: What are maximal subgroups of $C$?
  
  
and $g:(0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\to\Bbb N,$ is a function by $\forall\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))$ $g(\epsilon)=max(\{f((c,d))\,|\, d-c=\epsilon,$ $(c,d)\subseteq [0.01,0.1)\})$.
+
'''Theorem''' $3$: If $(\Bbb N,\star)$ is a cyclic group with generators $u,v$ and identity element $e=1$ and $f:\Bbb N\to\Bbb R$ is an injection then $(f(\Bbb N),\star_f)$ is a cyclic group with generators $f(u),f(v)$ and identity element $e_f=f(1)$ and operation law: $\forall m,n\in\Bbb N,$ $f(m)\star_ff(n)=f(m\star n)$ and inverse law: $\forall n\in\Bbb N,$ $(f(n))^{-1}=f(n^{-1})$ that $n\star n^{-1}=1$.
  
'''Guess''' $9$: $g$ isn't an injective function.
 
  
 +
'''Suppose''' $\forall m,n\in\Bbb N,\qquad$ $\begin{cases} m\star 1=m\\ (4m)\star (4m-2)=1=(4m+1)\star (4m-1)\\ (4m-2)\star (4n-2)=4m+4n-5\\ (4m-2)\star (4n-1)=4m+4n-2\\ (4m-2)\star (4n)=\begin{cases} 4m-4n-1 & 4m-2\gt 4n\\ 4n-4m+1 & 4n\gt 4m-2\\ 3 & m=n+1\end{cases}\\ (4m-2)\star (4n+1)=\begin{cases} 4m-4n-2 & 4m-2\gt 4n+1\\ 4n-4m+4 & 4n+1\gt 4m-2\end{cases}\\ (4m-1)\star (4n-1)=4m+4n-1\\ (4m-1)\star (4n)=\begin{cases} 4m-4n+2 & 4m-1\gt 4n\\ 4n-4m & 4n\gt 4m-1\\ 2 & m=n\end{cases}\\ (4m-1)\star (4n+1)=\begin{cases} 4m-4n-1 & 4m-1\gt 4n+1\\ 4n-4m+1 & 4n+1\gt 4m-1\\ 3 & m=n+1\end{cases}\\ (4m)\star (4n)=4m+4n-3\\ (4m)\star (4n+1)=4m+4n\\ (4m+1)\star  (4n+1)=4m+4n+1\\ \Bbb N=\langle 2\rangle=\langle 4\rangle\end{cases}$
  
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 $h_n\cap h_m=\emptyset$?
+
and let $C_1=\{(m,2m)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_1}=(1,2)\\ \\\forall m,n\in\Bbb N,\,(m,2m)\star_{C_1}(n,2n)=(m\star n,2(m\star n))\\ (m,2m)^{-1}=(m^{-1},2\times m^{-1})\qquad\text{that}\quad m\star m^{-1}=1\\ \\C_1=\langle(2,4)\rangle=\langle(4,8)\rangle\end{cases}$
:[https://math.stackexchange.com/questions/2518063/a-medium-question-about-a-set-related-to-prime-numbers/2526481#2526481 Guidance] given by [https://math.stackexchange.com/users/276986/reuns @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)
+
and $C_2=\{(3m-1,2m-1)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_2}=(2,1)\\ \\\forall m,n\in\Bbb N,\, (3m-1,2m-1)\star_{C_2}(3n-1,2n-1)=(3(m\star n)-1,2(m\star n)-1)\\ (3m-1,2m-1)^{-1}=(3\times m^{-1}-1,2\times m^{-1}-1)\qquad\text{that}\quad m\star m^{-1}=1\\ \\C_2=\langle(5,3)\rangle=\langle(11,7)\rangle\end{cases}$.
  
:* 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}\}$
+
and let $C:=C_1\oplus C_2$ be external direct sum of the groups $C_1$ & $C_2$, '''Question''' $2$: What are maximal subgroups of $C$?
  
 +
Alireza Badali 10:02, 12 May 2018 (CEST)
  
'''Now I want define a group on $L$ like $(L,\star _L)$:'''
+
=== [https://en.wikipedia.org/wiki/Erdős–Straus_conjecture Erdős–Straus conjecture] ===
  
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 $\langle 5\rangle =(P_1,\star _3)\cong (\Bbb Z,+)$.
+
'''Theorem''': If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ and generators $a,b$ then $E=\{({1\over x},{1\over y},{1\over z},{-4\over n+1},n)\mid x,y,z,n\in\Bbb N\}$ is an Abelian group with: $\forall x,y,z,n,x_1,y_1,z_1,n_1\in\Bbb N$ $\begin{cases} e_E=(1,1,1,-2,1)=({1\over 1},{1\over 1},{1\over 1},{-4\over 1+1},1)\\ \\({1\over x},{1\over y},{1\over z},{-4\over n+1},n)^{-1}=({1\over x^{-1}},{1\over y^{-1}},{1\over z^{-1}},\frac{-4}{n^{-1}+1},n^{-1})\quad\text{that}\\ x\star x^{-1}=1=y\star y^{-1}=z\star z^{-1}=n\star n^{-1}\\ \\({1\over x},{1\over y},{1\over z},\frac{-4}{n+1},n)\star_E({1\over x_1},{1\over y_1},{1\over z_1},\frac{-4}{n_1+1},n_1)=(\frac{1}{x\star x_1},\frac{1}{y\star y_1},\frac{1}{z\star z_1},\frac{-4}{n\star {n_1}+1},n\star n_1)\\ \\E=\langle({1\over a},1,1,-2,1),(1,{1\over a},1,-2,1),(1,1,{1\over a},-2,1),(1,1,1,\frac{-4}{a+1},1),(1,1,1,-2,a)\rangle=\\ \langle({1\over b},1,1,-2,1),(1,{1\over b},1,-2,1),(1,1,{1\over b},-2,1),(1,1,1,\frac{-4}{b+1},1),(1,1,1,-2,b)\rangle\end{cases}$
  
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 $\exists k\in\Bbb N,$ $f(a)=a\cdot 10^k\in\Bbb P\setminus\{2\}$.
 
  
Question $4$: Let $\forall (a,b),(c,d)\in L,$ $(a,b)\star _L (c,d)=10^{-\varphi ((f(a)\star _3f(c),f(b)\star _3f(d)))}\cdot (f(a)\star _3f(c),f(b)\star _3f(d))$ and $e=(0.03,0.03)$ so is $(L,\star _L)$ an Abelian group? and if yes, is $(L,\star _L)$ non finitely generated Abelian group and does $(L,\star _L)\cong (\Bbb Q,+)$ or $(L,\star _L)\cong (\Bbb Q_{\gt 0},\cdot)$?
+
Let $(\Bbb N,\star)$ is a cyclic group with: $\begin{cases} n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\\\Bbb N=\langle 2\rangle =\langle 3\rangle \end{cases}$
:Of course we could write each another group structure on $\Bbb N$ and consequently on $P_1$.
+
:Question: Is $E_0=\{({1\over x},{1\over y},{1\over z},\frac{-4}{n+1},n)\mid x,y,z,n\in\Bbb N,\, {1\over x}+{1\over y}+{1\over z}-{4\over n+1}=0\}$ a subgroup of $E$?
  
 +
Alireza Badali 17:34, 25 May 2018 (CEST)
  
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 $f(b)),\,a_1=max(f(a),f(b)),$ $b_1=min(f(a),f(b)),\, (a,b)\in D\}$.
+
=== [https://en.wikipedia.org/wiki/Landau%27s_problems Landaus forth problem] ===
  
'''Guess''' $10$: $\{2n-1\,|\, n\in\Bbb N\}\setminus N_1$ is a finite set.
+
Friedlander–Iwaniec theorem: there are infinitely many prime numbers of the form $a^2+b^4$.
: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.
+
:I want use this theorem for [https://en.wikipedia.org/wiki/Landau%27s_problems Landaus forth problem] but prime numbers properties have been applied for Friedlander–Iwaniec theorem hence no need to prime number theorem or its other forms or extensions.
: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 $(\Bbb Q_{\gt 0},\cdot),\, ...$
 
  
 +
'''Theorem''': If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ and generators $u,v$ then $F=\{(a^2,b^4)\mid a,b\in\Bbb N\}$ is a group with: $\forall a,b,c,d\in\Bbb N\,$ $\begin{cases} e_F=(1,1)\\ (a^2,b^4)\star_F(c^2,d^4)=((a\star c)^2,(b\star d)^4)\\ (a^2,b^4)^{-1}=((a^{-1})^2,(b^{-1})^4)\qquad\text{that}\quad a\star a^{-1}=1=b\star b^{-1}\\ F=\langle (1,u^4),(u^2,1)\rangle=\langle (1,v^4),(v^2,1)\rangle\end{cases}$
  
'''Question''' $5$: $\forall k\in\Bbb N$ 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 $v_{2n+3}=v_{s_1}\star _3v_{s_2}$.
+
'''now''' let $H=\langle\{(a^2,b^4)\mid a,b\in\Bbb N,\,b\neq 1\}\rangle$ and $G=F/H$ is quotient group of $F$ by $H$. ($G$ is a group including prime numbers properties only of the form $1+n^2$.)
: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 $6$: Is there any subgroup of $P_1$ like $H$ such that $\forall v_s\in H,\, s\notin P_1$?
+
and also $L=\{1+n^2\mid n\in\Bbb N\}$ is a cyclic group with: $\forall m,n\in\Bbb N$ $\begin{cases} e_L=2=1+1^2\\ (1+n^2)\star_L(1+m^2)=1+(n\star m)^2\\ (1+n^2)^{-1}=1+(n^{-1})^2\quad\text{that}\;n\star n^{-1}=1\\ L=\langle 1+u^2\rangle=\langle 1+v^2\rangle\end{cases}$
:I want connect subgroups of $P_1$ with Goldbach's conjecture.
 
  
 +
but on the other hand we have: $L\simeq G$ hence we can apply $L$ instead $G$ of course since we are working on natural numbers generally we could consider from the beginning the group $L$ without involvement with the group $G$ anyhow.
 +
:Question $1$: For each neutral cyclic group on $\Bbb N$ then what are maximal subgroups of $L$?
  
Question $7$: 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 [http://www.genealogy.ams.org/id.php?id=90761 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 <math>k</math> repeats <math>k-1</math> times.
 
::Answer given by Professor Daniel Lazard: Your sequence is clearly a [https://en.wikipedia.org/wiki/Primitive_recursive_function 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 <math>k(k-1)/2 \ge n</math>. This may be rewritten, using the [https://en.wikipedia.org/wiki/Ceil_function ceil function], <math>f(n)=\left \lceil \frac{1+\sqrt{1+8n}}{2}\right\rceil. </math> IMO, none of these formulas is useful, as they hide the main properties of the sequence.
 
::Problem $2$: Find a bijection <math>g:\Bbb N\times(\Bbb N\setminus\{1\})\to\Bbb N\times\Bbb N</math> as <math>g(n,f(n))</math> 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: $(1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(2,3),(3,2),(4,1),(1,5),(5,1),(1,6),(2,5),(3,4),(4,3),$ $(5,2),(6,1),(1,7),(3,5),(5,3),(7,1),...$
 
::::<s>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!</s>
 
  
 +
'''Guess''' $1$: For each cyclic group structure on $\Bbb N$ like $(\Bbb N,\star)$ then for each non-trivial subgroup of $\Bbb N$ like $T$ we have $T\cap\Bbb P\neq\emptyset$.
 +
:I think this guess must be proved via prime number theorem.
  
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 $(\Bbb N,\star),\,\forall n\in\Bbb N,\,\exists p,q\in\Bbb P\setminus\{2\}\quad 2n+3=p\star q$.
+
'''For''' each neutral cyclic group on $\Bbb N$ if $L\cap\Bbb P=\{1+n_1^2,1+n_2^2,...,1+n_k^2\},\,k\in\Bbb N$ and if $A=\bigcap _{i=1}^k\langle 1+n_i^2\rangle$ so $\exists m\in\Bbb N$ that $A=\langle 1+m^2\rangle$ & $m\neq n_i$ for $i=1,2,3,...,k$ (intelligibly $k\gt1$) so we have: $A\cap\Bbb P=\emptyset$.
:This conjecture is an equivalent to Goldbach's conjecture.
+
:Question $2$: Is $A$ only unique greatest subgroup of $L$ such that $A\cap\Bbb P=\emptyset$?
: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 $8$: $\forall m\in\Bbb N$ 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 $\forall n\in\Bbb N,\,\langle n\rangle\cong\langle f(n)\rangle$ & $\Bbb N/ \langle n\rangle\cong\Bbb Z / \langle f(n)\rangle$.
 
  
 +
Alireza Badali 16:49, 28 May 2018 (CEST)
  
And $(\Bbb N,\star)$ can be defined with: $\begin{cases} 2\star 3=1,\quad 2^n=2n,\quad 3^n=2n+1\\ 2^n\star 2^m=2^{n+m},\quad 3^n\star 3^m=3^{n+m}\\ 2^n\star 3^m=\begin{cases} 2^{n-m} & 2^n\gt 3^m\\ 3^{m-n} & 3^m\gt 2^n\end{cases}\end{cases}$
+
=== [https://en.wikipedia.org/wiki/Lemoine%27s_conjecture Lemoine's conjecture] ===
: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''': If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ & generators $u,v$ then $L=\{(p_{n_1},p_{n_2},p_{n_3},-2n-5)\mid n,n_1,n_2,n_3\in\Bbb N,\,p_{n_i}$ is $n_i$_th prime for $i=1,2,3\}$ is an Abelian group with: $\forall n_1,n_2,n_3,n,m_1,m_2,m_3,m\in\Bbb N$ $\begin{cases} e_L=(2,2,2,-7)=(2,2,2,-2\times 1-5)\\ \\(p_{n_1},p_{n_2},p_{n_3},-2n-5)\star_L(p_{m_1},p_{m_2},p_{m_3},-2m-5)=(p_{n_1\star m_1},p_{n_2\star m_2},p_{n_3\star m_3},-2\times(n\star m)-5)\\ \\(p_{n_1},p_{n_2},p_{n_3},-2n-5)^{-1}=(p_{n_1^{-1}},p_{n^{-1}_2},p_{n_3^{-1}},-2\times n^{-1}-5)\quad\text{that}\\ n_1\star n_1^{-1}=1=n_2\star n_2^{-1}=n_3\star n_3^{-1}=n\star n^{-1}\\ \\L=\langle(p_u,2,2,-7),(2,p_u,2,-7),(2,2,p_u,-7),(2,2,2,-2u-5)\rangle=\\\langle(p_v,2,2,-7),(2,p_v,2,-7),(2,2,p_v,-7),(2,2,2,-2v-5)\rangle\end{cases}$
  
  
'''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 $2n+3=p_1\star p_2\star p_3...\star p_k$.
+
'''Theorem''': $\forall n\in\Bbb N,\,\exists (p_{m_1},p_{m_2},p_{m_3},-2n-5)\in(L,\star_L)$ such that $p_{m_1}+p_{m_2}+p_{m_3}-2n-5=0$.
: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.
+
:Proof using Goldbach's weak conjecture.
: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 $3$: 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 [https://en.wikipedia.org/wiki/Linear_equation linear] equation $2n=p+q$ or $2n+3=p\star q$ with [https://en.wikipedia.org/wiki/Degree_of_a_polynomial 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 [https://en.wikipedia.org/wiki/Christian_Goldbach Christian Goldbach]'s conjecture will impress mathematics certainly!
 
  
 +
'''Question''': Is $L_0=\{(p_{m_1},p_{m_2},p_{m_2},-2n-5)\mid\forall m_1,m_2\in\Bbb N,\,\exists n\in\Bbb N,$ such that $p_{m_1}+2p_{m_2}-2n-5=0\}$ a subgroup of $L$?
  
'''Problem''' $4$: Rewrite operation of the group $(\Bbb N,\star)$ by matrices.
+
Alireza Badali 19:30, 3 June 2018 (CEST)
  
 +
=== Primes with beatty sequences ===
  
'''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 $2n+3$.
+
How can we understand $\infty$? we humans only can think on natural numbers and other issues are only theorizing, algebraic theories can be some features for this aim.
:This conjecture was made compatible with $(\Bbb N,\star)$ and is an equivalent to Goldbach's conjecture.
 
  
  
Let $L_3=\{(a,b)\,|\, a,b\in S_1,\, a,b\in\{(x,y)\,|\, 0\lt x\lt 0.1,\, 0\lt y\lt 0.1\}\setminus\{(x,y)\,|\, x\lt 0.01,\, y\lt 0.01\},$ $\exists k\in\Bbb N,\, a\cdot 10^k,b\cdot 10^k\in\Bbb P\setminus\{2\}\}$.
+
[http://oeis.org/A184774 Conjecture]: If $r$ is an irrational number and $1\lt r\lt 2$, then there are infinitely many primes in the set $L=\{\text{floor}(n\cdot r)\mid n\in\Bbb N\}$.
  
'''Question''' $9$: Let $\forall (a,b),(c,d)\in L_3,\, m=1+max($number of digits in $f(a)\star _3f(c),$ number of digits in $f(b)\star _3f(d)),$ $(a,b)\star _{L_3} (c,d)=10^{-m}\cdot (f(a)\star _3f(c),f(b)\star _3f(d))$ and $e=(0.03,0.03)$ so is $(L_3,\star _{L_3})$ an Abelian group? and if yes, is $(L_3,\star _{L_3})$ non finitely generated Abelian group and does $(L_3,\star _{L_3})\cong (\Bbb Q,+)$ or $(L_3,\star _{L_3})\cong (\Bbb Q_{\gt 0},\cdot)$?
 
  
 +
'''Theorem''' $1$: If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ & generators $u,v$ and $r\in[1,2]\setminus\Bbb Q$ then $L=\{\lfloor n\cdot r\rfloor\mid n\in\Bbb N\}$ is another cyclic group with: $\forall m,n\in\Bbb N$ $\begin{cases} e_L=1\\ \lfloor n\cdot r\rfloor\star_L\lfloor m\cdot r\rfloor=\lfloor (n\star m)\cdot r\rfloor\\ (\lfloor n\cdot r\rfloor)^{-1}=\lfloor n^{-1}\cdot r\rfloor\qquad\text{that}\quad n\star n^{-1}=1\\ L=\langle\lfloor u\cdot r\rfloor\rangle=\langle\lfloor v\cdot r\rfloor\rangle\end{cases}$.
 +
:Guess $1$: $\prod_{n=1}^{\infty}\lfloor n\cdot r\rfloor=\lfloor 1\cdot r\rfloor\star\lfloor 2\cdot r\rfloor\star\lfloor 3\cdot r\rfloor\star...\in\Bbb N$.
  
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 $J_n=\{(s_1,s_2,s_3,...,s_n)\,|\,\forall i=1,2,3,...,n,\, s_i\in S_1,\,\exists k\in\Bbb N,\, s_i\cdot 10^k\in\Bbb P\}$ & $I^n=[0,1]^n$ that of course $J_n$ is dense in the $I^n,$ now I guess that under Euclidean topology $\exists n\in\Bbb N,\,\forall k\ge n$ $dim(closure(J_k))=1$.
 
 
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 $t\cdot 10^{k+1}\in \Bbb P$.
 
:[https://math.stackexchange.com/questions/2482941/a-simple-question-about-density-in-the-interval-0-1-1/2483079#2483079 Proof] given by [https://math.stackexchange.com/users/149178/adayah @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 [[User_talk:Musictheory2math#Goldbach.27s_conjecture|Main theorem]] of Goldbach's conjecture is a result of theorem $1$.
 
  
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The conjecture generalized: if $r$ is a positive irrational number and $h$ is a real number, then each of the sets $\{\text{floor}(n\cdot r+h)\mid n\in\Bbb N\}$, $\{\text{round}(n\cdot r+h)\mid n\in\Bbb N\}$, and $\{\text{ceiling}(n\cdot r+h)\mid n\in\Bbb N\}$ contains infinitely many primes.
  
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)
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'''Theorem''' $2$: If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ & generators $u,v$ & $r$ is a positive irrational number & $h\in\Bbb R$ then $G=\{n\cdot r+h\mid n\in\Bbb N\}$ is another cyclic group with: $\forall m,n\in\Bbb N$ $\begin{cases} e_G=\lfloor r+h\rfloor\\ \lfloor n\cdot r+h\rfloor\star_G\lfloor m\cdot r+h\rfloor=\lfloor (n\star m)\cdot r+h\rfloor\\ (\lfloor n\cdot r+h\rfloor)^{-1}=\lfloor n^{-1}\cdot r+h\rfloor\qquad\text{that}\quad n\star n^{-1}=1\\ L=\langle\lfloor u\cdot r+h\rfloor\rangle=\langle\lfloor v\cdot r+h\rfloor\rangle\end{cases}$.
 +
:Guess $2$: $\prod_{n=k}^{\infty}\lfloor n\cdot r+h\rfloor=\lfloor k\cdot r+h\rfloor\star\lfloor (k+1)\cdot r+h\rfloor\star\lfloor (k+2)\cdot r+h\rfloor\star...\in\Bbb N$ in which $\lfloor k\cdot r+h\rfloor\in\Bbb N$ & $\lfloor (k-1)\cdot r+h\rfloor\lt1$.
  
== Landau's forth problem ==
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Alireza Badali 19:09, 7 June 2018 (CEST)
  
Landau's forth problem: Are there infinitely many primes $p$ such that $p−1$ is a perfect square? In other words: Are there infinitely
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== Conjectures depending on the new definitions of primes ==
many primes of the form $n^2 + 1$?
 
  
 +
'''Algebraic analytical number theory'''
  
In analytic number theory the Friedlander–Iwaniec theorem states that there are infinitely many prime numbers of the form ${\displaystyle a^{2}+b^{4}}$.
 
  
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'''A problem''': For each cyclic group on $\Bbb N$ like $(\Bbb N,\star)$ find a new definition of prime numbers matching with the operation $\star$ in the group $(\Bbb N,\star)$.
  
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$(H)=\aleph_0$
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$\Bbb N$ is a cyclic group by: $\begin{cases} \forall m,n\in\Bbb N\\ n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\\ (\Bbb N,\star)=\langle2\rangle=\langle3\rangle\simeq(\Bbb Z,+)\end{cases}$
  
 +
in the group $(\Bbb Z,+)$ an element $p\gt 1$ is a prime iff don't exist $m,n\in\Bbb Z$ such that $p=m\times n$ & $m,n\gt1$ for instance since $12=4\times3=3+3+3+3$ then $12$ isn't a prime but $13$ is a prime, now inherently must exists an equivalent definition for prime numbers in the $(\Bbb N,\star)$.
  
'''A question''': cardinal$(H_1)\in\Bbb N$ or cardinal$(H_1)=\aleph_0$?
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prime number isn't an algebraic concept so we can not define primes by using isomorphism (and via algebraic equations primes can be defined) but since Gaussian integers contain all numbers of the form $m+ni,$ $m,n\in\Bbb N$ hence by using algebraic concepts we can solve some problems in number theory.
:This question is the same Landau's forth problem.
+
:Question: what is definition of prime numbers in the $(\Bbb N,\star)$?
  
  Alireza Badali 20:47, 21 September 2017 (CEST)
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  Alireza Badali 00:49, 25 June 2018 (CEST)
  
== Grimm's conjecture ==
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=== [https://en.wikipedia.org/wiki/Gaussian_moat Gaussian moat problem] ===
  
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 $1 ≤ i ≤ k$.
 
  
 +
Alireza Badali 18:13, 20 June 2018 (CEST)
  
Let $C=\{(x,y)\,|\, 0.01\le x\lt 0.1,\, 0.01\le y\lt 0.1\}$
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=== [https://en.wikipedia.org/wiki/Grimm%27s_conjecture Grimm's conjecture] ===
  
  
'''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)
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  Alireza Badali 18:13, 20 June 2018 (CEST)
  
== Lemoine's conjecture ==
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=== [https://en.wikipedia.org/wiki/Oppermann%27s_conjecture Oppermann'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.
 
  
 +
Alireza Badali 18:13, 20 June 2018 (CEST)
  
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 $a+b=r_1(m)-c$
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=== [https://en.wikipedia.org/wiki/Legendre%27s_conjecture Legendre's conjecture] ===
: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)
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  Alireza Badali 18:13, 20 June 2018 (CEST)
  
== About analytic number theory ==
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== Conjectures depending on the ring 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).
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'''Algebraic analytical number theory'''
  
An unproven yet, new and modern definition of prime numbers and equivalent to properties of [https://en.wikipedia.org/wiki/Riemann_zeta_function Riemann zeta function]: There is no [https://en.wikipedia.org/wiki/Polynomial polynomial] as a formula for prime numbers, because each formula or relation for prime numbers only there can be contain all prime numbers together and simultaneous like Riemann zeta function!
 
  
 +
'''An algorithm''' which makes new integral domains on $\Bbb N$: Let $(\Bbb N,\star,\circ)$ be that integral domain then identity element $i$ will be corresponding with $1$ and multiplication of natural numbers will be obtained from multiplication of integers corresponding with natural numbers and of course each natural number like $m$ multiplied by a natural number corresponding with $-1$ will be $-m$ such that $m\star(-m)=1$ & $1\circ m=1$.
  
In analytic number theory in Riemann zeta function there is an important technique as <big>$\frac{1}{n}$</big>, 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, 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 $S$.
 
  
''Below contents hasn't been reviewed and checked:''
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for instance $(\Bbb N,\star,\circ)$ is an integral domain with: $\begin{cases} \forall m,n\in\Bbb N\\ n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\\1\circ m=1\\ 2\circ m=m\\ 3\circ m=-m\qquad\text{that}\quad m\star (-m)=1\\ (2n)\circ(2m)=2mn\\ (2n+1)\circ(2m+1)=2mn\\ (2n)\circ(2m+1)=2mn+1\end{cases}$
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:Question $1$: Is $(\Bbb N,\star,\circ)$ an ''unique factorization domain'' or the same UFD? what are irreducible elements in $(\Bbb N,\star,\circ)$?
  
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 [https://mathoverflow.net/users/30186/wojowu @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.
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'''Question''' $2$: How can we make a UFD on $\Bbb N$?
  
  
Let $\mu _2:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _2(z)=\sum _{n=1}^{\infty}\frac{1}{(r(n))^z}$$
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Question $3$: Under usual total order on $\Bbb N$, do there exist any integral domain $(\Bbb N,\star,\circ)$ and an Euclidean valuation $v:\Bbb N\setminus\{1\}\to\Bbb N$ such that $(\Bbb N,\star,\circ,v)$ is an Euclidean domain? no.
  
Let $\mu _3:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _3(z)=\sum _{p\in\Bbb P}\frac{1}{(r(p))^z}$$
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'''Guess''' $1$: For each integral domain $(\Bbb N,\star,\circ)$ there exists a total order on $\Bbb N$ and an Euclidean valuation $v:\Bbb N\setminus\{1\}\to\Bbb N$ such that $(\Bbb N,\star,\circ,v)$ is an Euclidean domain.
  
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: $$\forall z\in\Bbb C\qquad\mu _5(z)=\sum _{p\in\Bbb P}\frac{1}{(r_j(p))^z}$$
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Professor [https://en.wikipedia.org/wiki/Jeffrey_Lagarias Jeffrey Clark Lagarias] advised me that you can apply group structure on $\Bbb N\cup\{0\}$ instead only $\Bbb N$ and now I see his plan is useful on the field theory, now suppose we apply two algorithms above on $\Bbb N\cup\{0\}$ hence we will have identity element for the group $(\Bbb N,\star)$ of the first algorithm is $0$ corresponding with $0$.
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:'''Problem''' $1$: If $(\Bbb N\cup\{0\},\star,\circ)$ is a UFD then what are irreducible elements in $(\Bbb N\cup\{0\},\star,\circ)$ and is $(\Bbb Q^{\ge0},\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb N\cup\{0\},\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\ {m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={-m\over n}\qquad m\star(-m)=0\end{cases}$
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::Algebraic theories on positive numbers help us to solve some open problems depending on the positive numbers.
  
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: $$\forall z\in\Bbb C\qquad\mu _7(z)=\lim _{j\to\infty}\sum _{p\in\Bbb P}\frac{1}{(r_j(p))^z}$$
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Question $4$: Is $(\Bbb N\cup\{0\},\star,\circ)$ a UFD by: $\begin{cases} \forall m,n\in\Bbb N\\ e=0\\ (2m-1)\star(2m)=0\\ (2m)\star(2n)=2m+2n\\ (2m-1)\star(2n-1)=2m+2n-1\\ (2m)\star(2n-1)=\begin{cases} 2m-2n & 2m\gt 2n-1\\ 2n-2m-1 & 2n-1\gt 2m\end{cases}\\i=1\\ 0\circ m=0\\ 2\circ m=-m\quad m\star(-m)=0\\ (2m)\circ(2n)=2mn-1\\ (2m-1)\circ(2n-1)=2mn-1\\ (2m)\circ(2n-1)=2mn\end{cases}$
  
let $\mu _8:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _8(z)=\sum _{j\in\Bbb N}\frac{1}{(r_j(j))^z}$$
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and what are irreducible elements in $(\Bbb N\cup\{0\},\star,\circ)$ and also is $(\Bbb Q^{\ge0},\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb N\cup\{0\},\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={-m\over n}\qquad m\star(-m)=0\end{cases}$
  
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: $$\forall z\in\Bbb C\qquad\mu _{10}(z)=\sum _{n\in\Bbb N}\frac{1}{(r_{a_n}(n))^z}$$
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<small>'''Conjecture''' $1$: Let $x$ be a positive real number, and let $\pi(x)$ denote the number of primes that are less than or equal to $x$ then $$\lim_{x\to\infty}\frac{x-\pi(x)}{\pi(e^u)}=1,\quad u=\sqrt{2\log(x\log x-x)}\,.$$</small>
 +
:<small>Answer given by [https://mathoverflow.net/users/37555/jan-christoph-schlage-puchta $@$Jan-ChristophSchlage-Puchta] from stackexchange site: The conjecture is obviously wrong. The numerator is at least $x/2$, the denominator is at most $e^u$, and $u\lt2\sqrt\log x$, so the limit is $\infty$.</small>
 +
::<small>'''Problem''' $2$: Find a function $f:\Bbb R\to\Bbb R$ such that $\lim_{x\to\infty}\frac{x-\pi(x)}{\pi(f(x))}=1$.</small>
 +
:::<small>Prime number theorem and its extensions or algebraic forms or corollaries allow us via infinity concept reach to some results.</small>
 +
:::<small>Prime numbers properties are stock in whole natural numbers including $\infty$ and not in any finite subset of $\Bbb N$ hence we can know them only in $\infty$, which [https://en.wikipedia.org/wiki/Prime_number_theorem prime number theorem] prepares it, but what does mean a cognition of prime numbers I think according to the [[distribution of prime numbers]], a cognition means only in $\infty$, this function $f$ can be such a cognition but only in $\infty$ because we have: $$\lim_{x\to\infty}\frac{x-\pi(x)}{\pi(f(x))}=1=\lim_{x\to\infty}\frac{f(x)-\pi(f(x))}{\pi(f(f(x)))}=\lim_{x\to\infty}\frac{f(f(x))-\pi(f(f(x)))}{\pi(f(f(f(x))))}=...$$ and I guess $f$ is to form of <big>$e^{g(x)}$</big> in which $g:\Bbb R\to\Bbb R$ is a radical logarithmic function or probably as a radical logarithmic series.</small>
 +
::::<small>'''Conjecture''' $2$: Let $h:\Bbb R\to\Bbb R,\,h(x)=\frac{f(x)}{(\log x-1)\log(f(x))}$ then $\lim_{x\to\infty}{\pi(x)\over h(x)}=1$.</small>
 +
:::::<small>Answer given by [https://mathoverflow.net/users/30186/wojowu $@$Wojowu] from stackexchange site: Since $x−\pi(x)\sim x$, you want $\pi(f(x))\sim x$, and $f(x)=x\log x$ works, and let $u=\log(x\log x)$.</small>
 +
::::::<small>'''Problem''' $3$: Based on ''prime number theorem'' very large prime numbers are equivalent to the numbers of the form $n\cdot\log n,\,n\in\Bbb N$ hence I think a test could be made to check correctness of some conjectures or problems relating to the prime numbers, and maybe some functions such as $h$ prepares it!</small>
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:::::::<small>'''Question''' $5$: If $p_n$ is $n$_th prime number then does $$\lim_{n\to\infty}\frac{p_n}{e^{\sqrt{2\log n}}\over (\log n-1)\sqrt{2\log n}}=1\,?$$</small>
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::::::::<small>Answer given by [https://mathoverflow.net/users/2926/todd-trimble $@$ToddTrimble] from stackexchange site: The numerator is asymptotically greater than $n$, and the denominator is asymptotically less.</small>
  
Let $a_n:\Bbb N\to\Bbb N$ is a sequence & $\mu _{11}:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\qquad\mu _{11}(z)=\sum _{n\in\Bbb N}\frac{1}{(r_{a_n}(p_n))^z}$$
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Alireza Badali 16:26, 26 June 2018 (CEST)
  
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: $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\,\mu _{12}(z)=\sum _{n\in\Bbb N}\frac{1}{(a_n)^z}$$
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=== [https://en.wikipedia.org/wiki/Many-worlds_interpretation Parallel universes] ===
  
Let $f_1:\Bbb C\to\Bbb C$ & $\mu _{13}:\Bbb C\to\Bbb C$ are functions as: $$\forall z\in\Bbb C\qquad\mu _{13}(z)=\sum _{n=1}^{\infty} (-1)^n\cdot (r(n))^{f_1(z)}$$
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'''An algorithm''' that makes new cyclic groups on $\Bbb Z$: Let $(\Bbb Z,\star)$ be that group and at first write integers as a sequence with starting from $0$ and then write integers with a fixed sequence below it, and let identity element $e=0$ be corresponding with $0$ and two generators $m$ & $n$ be corresponding with $1$ & $-1$, so we have $(\Bbb Z,\star)=\langle m\rangle=\langle n\rangle$ for instance: $$0,1,2,-2,-1,3,4,-4,-3,5,6,-6,-5,7,8,-8,-7,9,10,-10,-9,11,12,-12,-11,13,14,-14,-13,...$$ $$0,1,-1,2,-2,3,-3,4,-4,5,-5,6,-6,7,-7,8,-8,9,-9,10,-10,11,-11,12,-12,13,-13,14,-14,...$$ then regarding the sequence above find an even rotation number that for this sequence is $4$ (or $2k$) and hence equations should be written with module $2$ (or $k$) then consider $2m-1,2m,-2m+1,-2m$ (that general form is: $km,km-1,km-2,...,km-(k-1),-km,-km+1,-km+2,...,-km+(k-1)$) and make a table of products of those $4$ (or $2k$) elements but during writing equations pay attention if an equation is right for given numbers it will be right generally for other numbers too and of course if integers corresponding with two numbers don't have same signs then product will be a piecewise-defined function for example $7\star(-10)=2$ $=(2\times4-1)\star(-2\times5)$ because $7+(-9)=-2,\,7\to7,\,-9\to-10,\,-2\to2$ that implies $(2m-1)\star(-2n)=2n-2m$ where $2n\gt 2m-1$, of course it is better at first members inverse be defined for example since $7+(-7)=0,\,7\to7,\,-7\to-8$ so $7\star(-8)=0$ that shows $(2m-1)\star(-2m)=0$ and with a little bit addition and multiplication all equations will be obtained simply that for this example is:
  
Let $f_2:\Bbb C\to\Bbb C$ & $\mu _{14}:\Bbb C\to\Bbb C$ are functions as: $$\forall z\in\Bbb C\qquad\mu _{14}(z)=\sum _{p_n\text{is}\, n\text{_th prime number}} (-1)^n\cdot (r(p_n))^{f_2(z)}$$
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$\begin{cases} \forall t\in\Bbb Z,\quad t\star0=t\\ \forall m,n\in\Bbb N\\ (2m-1)\star(-2m)=0=(-2m+1)\star(2m)\\ (2m-1)\star(2n-1)=2m+2n-2\\ (2m-1)\star(2n)=\begin{cases} 2m-2n-1 & 2m-1\gt2n\\ 2m-2n-2 & 2n\gt 2m-1\end{cases}\\ (2m-1)\star(-2n+1)=2m+2n-1\\ (2m-1)\star(-2n)=\begin{cases} 2n-2m+1 & 2m-1\gt2n\\ 2n-2m & 2n\gt2m-1\end{cases}\\ (2m)\star(2n)=2m+2n\\ (2m)\star(-2n+1)=\begin{cases} 2m-2n+1 & 2n-1\gt2m\\ 2m-2n & 2m\gt2n-1\end{cases}\\ (2m)\star(-2n)=-2m-2n\\ (-2m+1)\star(-2n+1)=-2m-2n+1\\ (-2m+1)\star(-2n)=\begin{cases} 2m-2n+1 & 2m-1\gt2n\\ 2m-2n & 2n\gt2m-1\\ 1 & m=n\end{cases}\\ (-2m)\star(-2n)=2m+2n-2\\ \Bbb Z=\langle1\rangle=\langle-2\rangle\end{cases}$
  
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.
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'''An algorithm''' which makes new integral domains on $\Bbb Z$: Let $(\Bbb Z,\star,\circ)$ be that integral domain then identity element $i$ will be corresponding with $1$ and multiplication of integers will be obtained from multiplication of corresponding integers such that if $t:\Bbb Z\to\Bbb Z$ is a bijection that images top row on to bottom row respectively for instance in example above is seen $t(2)=-1$ & $t(-18)=18$ then we can write laws by using $t$ such as $(-2m+1)\circ(-2n)=$ $t(t^{-1}(-2m+1)\times t^{-1}(-2n))=t((2m)\times(-2n+1))=t(-2\times(2mn-m))=$ $2\times(2mn-m)=4mn-2m$ and of course each integer like $m$ multiplied by an integer corresponding with $-1$ will be $n$ such that $m\star n=0$ & $0\circ m=0$ for instance $(\Bbb Z,\star,\circ)$ is an integral domain with:
  
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$\begin{cases} \forall t\in\Bbb Z,\quad t\star0=t\\ \forall m,n\in\Bbb N\\ (2m-1)\star(-2m)=0=(-2m+1)\star(2m)\\ (2m-1)\star(2n-1)=2m+2n-2\\ (2m-1)\star(2n)=\begin{cases} 2m-2n-1 & 2m-1\gt2n\\ 2m-2n-2 & 2n\gt 2m-1\end{cases}\\ (2m-1)\star(-2n+1)=2m+2n-1\\ (2m-1)\star(-2n)=\begin{cases} 2n-2m+1 & 2m-1\gt2n\\ 2n-2m & 2n\gt2m-1\end{cases}\\ (2m)\star(2n)=2m+2n\\ (2m)\star(-2n+1)=\begin{cases} 2m-2n+1 & 2n-1\gt2m\\ 2m-2n & 2m\gt2n-1\end{cases}\\ (2m)\star(-2n)=-2m-2n\\ (-2m+1)\star(-2n+1)=-2m-2n+1\\ (-2m+1)\star(-2n)=\begin{cases} 2m-2n+1 & 2m-1\gt2n\\ 2m-2n & 2n\gt2m-1\\ 1 & m=n\end{cases}\\ (-2m)\star(-2n)=2m+2n-2\\ i=t(1)=1,\quad0\circ m=0,\quad m\star(t(-1)\circ m)=m\star(-2\circ m)=0\\ (2m-1)\circ(2n-1)=4mn-2m-2n+1\\ (2m-1)\circ(2n)=4mn-2n\\ (2m-1)\circ(-2n+1)=-4mn+2n+1\\ (2m-1)\circ(-2n)=-4mn+2m+2n-2\\ (2m)\circ(2n)=-4mn+1\\ (2m)\circ(-2n+1)=4mn\\ (2m)\circ(-2n)=-4mn+2m+1\\ (-2m+1)\circ(-2n+1)=-4mn+1\\ (-2m+1)\circ(-2n)=4mn-2m\\ (-2m)\circ(-2n)=4mn-2m-2n+1\end{cases}$
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:Question $1$: Is $(\Bbb Z,\star,\circ)$ a UFD? what are irreducible elements in $(\Bbb Z,\star,\circ)$? is $(\Bbb Q,\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$ •
  
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: $$\forall z\in\Bbb C\qquad\mu _{16}(z)=\sum _{j\in\Bbb N}\sum _{p\in\Bbb P\cap (10^{j-1},10^j)} (r_j(p))^z$$
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'''Problem''' $1$: If $(\Bbb Z,\star,\circ)$ is a UFD then what are irreducible elements in $(\Bbb Z,\star,\circ)$ and is $(\Bbb Q,\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\ {m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$
  
  
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 <math>p:\Bbb R\to\Bbb R</math> include such points <math>(n,p_n)</math> that <math>p_n</math> is <math>n</math>_th prime and by taking in mind that prime number theorem follows from normal definition of primes in terms of factorization to primes because logarithm function is inverse of the function <math>f(a)=a^n</math> and primes formulas are as logarithmic functions or other [https://en.wikipedia.org/wiki/Transcendental_function transcendental functions] but ideally there exists an unique formula as an infinite series that generates primes simply: (I had asked a question that an user from stackexchange.com grammatical corrected it.)
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Question $2$: Under usual total order on $\Bbb Z$, do there exist any integral domain $(\Bbb Z,\star,\circ)$ and an Euclidean valuation $v:\Bbb Z\setminus\{0\}\to\Bbb N$ such that $(\Bbb Z,\star,\circ,v)$ is an Euclidean domain? no.
  
Let $$\omega _1=\color{red}{0}.\color{teal}{p_1}\color{purple}{p_2}\color{teal}{p_3}\dots =\color{red}{0}.\color{blue}{2}\color{fuchsia}{3}\color{blue}{5}\color{fuchsia}{7}\color{blue}{11}\color{fuchsia}{13}\color{blue}{17}\color{fuchsia}{19}\color{blue}{23}\color{fuchsia}{29}\color{blue}{31}\color{fuchsia}{37}\dots \color{blue}{7717}\color{fuchsia}{7723}\dots$$
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'''Guess''' $1$: For each integral domain $(\Bbb Z,\star,\circ)$ there exists a total order on $\Bbb Z$ and an Euclidean valuation $v:\Bbb Z\setminus\{0\}\to\Bbb N$ such that $(\Bbb Z,\star,\circ,v)$ is an Euclidean domain.
  
(i.e. the decimal part of $\omega _1$ is obtained by concatenating the prime numbers) and
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Alireza Badali 20:32, 9 July 2018 (CEST)
  
$$\omega _2=\color{red}{0}.\color{blue}{2}0\color{fuchsia}{3}0\color{blue}{5}0\color{fuchsia}{7}0\color{blue}{11}00\color{fuchsia}{13}00\color{blue}{17}00\color{fuchsia}{19}00\color{blue}{23}00\color{fuchsia}{29}00\color{blue}{31}00\color{fuchsia}{37}00\dots\color{blue}{7717}0000\dots$$
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=== [https://en.wikipedia.org/wiki/Gauss_circle_problem Gauss circle problem] ===
  
(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 $10$).
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<small>Given this sequence: $$0,1,-1,2,-2,3,-3,4,-4,5,-5,6,-6,7,-7,8,-8,9,-9,10,-10,11,-11,12,-12,...$$ $$0,1,2,3,-3,-1,-2,4,5,6,-6,-4,-5,7,8,9,-9,-7,-8,10,11,12,-12,-10,-11,...$$ we have this integral domain $(\Bbb Z,\star,\circ)$ as:
  
Questions:
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$\begin{cases} \forall t\in\Bbb  Z,\quad t\star0=t,\quad\forall m,n\in\Bbb N,\\ (6m-5)\star(6m-4)=0=(6m-3)\star(-6m+3)=(-6m+5)\star(-6m+4)=(6m-2)\star(6m-1)=\\ =(6m)\star(-6m)=(-6m+2)\star(-6m+1)\\ (6m-5)\star(6n-5)=6m+6n-9\\ (6m-5)\star(6n-4)=\begin{cases} 6n-6m+2 & 6m-5\gt6n-4\\ 6m-6n+1 & 6n-4\gt6m-5\end{cases}\\ (6m-5)\star(6n-3)=-6m-6n+11\\ (6m-5)\star(6n-2)=6m+6n-6\\ (6m-5)\star(6n-1)=\begin{cases} 6n-6m+5 & 6m-5\gt6n-1\\ 6m-6n-2 & 6n-1\gt6m-5\end{cases}\\ (6m-5)\star(6n)=-6m-6n+8\\ (6m-5)\star(-6n+5)=6m+6n-8\\ (6m-5)\star(-6n+4)=\begin{cases} 6m-6n-2 & 6m-5\gt6n-4\\ 6m-6n-3 & 6n-4\gt6m-5\end{cases}\\ (6m-5)\star(-6n+3)=\begin{cases} 6m-6n & 6m-5\gt6n-3\\ 6n-6m+2 & 6n-3\gt6m-5\end{cases}\\ (6m-5)\star(-6n+2)=6m+6n-5\\ (6m-5)\star(-6n+1)=\begin{cases} 6m-6n-5 & 6m-5\gt6n-1\\ 6m-6n-6 & 6n-1\gt6m-5\end{cases}\\ (6m-5)\star(-6n)=\begin{cases} 6m-6n-3 & 6m-5\gt6n\\ 6n-6m+5 & 6n\gt6m-5\end{cases}\\ (6m-4)\star(6n-4)=-6m-6n+9\\ (6m-4)\star(6n-3)=\begin{cases} 6n-6m & 6m-4\gt6n-3\\ 6n-6m+1 & 6n-3\gt6m-4\end{cases}\\ (6m-4)\star(6n-2)=\begin{cases} 6n-6m+4 & 6m-4\gt6n-2\\ 6m-6n-1 & 6n-2\gt6m-4\end{cases}\\ (6m-4)\star(6n-1)=-6m-6n+6\\ (6m-4)\star(6n)=\begin{cases} 6n-6m+3 & 6m-4\gt6n\\ 6n-6m+4 & 6n\gt6m-4\end{cases}\\ (6m-4)\star(-6n+5)=\begin{cases} 6m-6n-1 & 6m-4\gt6n-5\\ 6n-6m+3 & 6n-5\gt6m-4\\ 3 & m=n\end{cases}\\ (6m-4)\star(-6n+4)=6m+6n-7\\ (6m-4)\star(-6n+3)=-6m-6n+10\\ (6m-4)\star(-6n+2)=\begin{cases} 6m-6n-4 & 6m-4\gt6n-2\\ 6n-6m+6 & 6n-2\gt6m-4\end{cases}\\ (6m-4)\star(-6n+1)=6m+6n-4\\ (6m-4)\star(-6n)=-6m-6n+7\\ (6m-3)\star(6n-3)=6m+6n-8\\ (6m-3)\star(6n-2)=-6m-6n+8\\ (6m-3)\star(6n-1)=\begin{cases} 6m-6n-2 & 6m-3\gt6n-1\\ 6m-6n-3 & 6n-1\gt6m-3\end{cases}\\ (6m-3)\star(6n)=6m+6n-5\\ (6m-3)\star(-6n+5)=6m+6n-6\\ (6m-3)\star(-6n+4)=\begin{cases} 6m-6n & 6m-3\gt6n-4\\ 6n-6m+2 & 6n-4\gt6m-3\\ 2 & m=n\end{cases}\\ (6m-3)\star(-6n+3)=\begin{cases} 6n-6m+2 & 6m-3\gt6n-3\\ 6m-6n+1 & 6n-3\gt6m-3\end{cases}\\ (6m-3)\star(-6n+2)=6m+6n-3\\ (6m-3)\star(-6n+1)=\begin{cases} 6m-6n-3 & 6m-3\gt6n-1\\ 6n-6m+5 & 6n-1\gt6m-3\end{cases}\\ (6m-3)\star(-6n)=\begin{cases} 6n-6m+5 & 6m-3\gt6n\\ 6m-6n-2 & 6n\gt6m-3\end{cases}\\ (6m-2)\star(6n-2)=6m+6n-3\\ (6m-2)\star(6n-1)=\begin{cases} 6n-6m+2 & 6m-2\gt6n-1\\ 6m-6n+1 & 6n-1\gt6m-2\end{cases}\\ (6m-2)\star(6n)=-6m-6n+5\\ (6m-2)\star(-6n+5)=6m+6n-5\\ (6m-2)\star(-6n+4)=\begin{cases} 6m-6n+1 & 6m-2\gt6n-4\\ 6m-6n & 6n-4\gt6m-2\end{cases}\\ (6m-2)\star(-6n+3)=\begin{cases} 6m-6n+3 & 6m-2\gt6n-3\\ 6n-6m-1 & 6n-3\gt6m-2\end{cases}\\ (6m-2)\star(-6n+2)=6m+6n-2\\ (6m-2)\star(-6n+1)=\begin{cases} 6m-6n-2 & 6m-2\gt6n-1\\ 6m-6n-3 & 6n-1\gt6m-2\end{cases}\\ (6m-2)\star(-6n)=\begin{cases} 6m-6n & 6m-2\gt6n\\ 6n-6m+2 & 6n\gt6m-2\end{cases}\\ (6m-1)\star(6n-1)=-6m-6n+3\\ (6m-1)\star(6n)=\begin{cases} 6n-6m & 6m-1\gt6n\\ 6n-6m+1 & 6n\gt6m-1\end{cases}\\ (6m-1)\star(-6n+5)=\begin{cases} 6m-6n+2 & 6m-1\gt6n-5\\ 6n-6m & 6n-5\gt6m-1\end{cases}\\ (6m-1)\star(-6n+4)=6m+6n-4\\ (6m-1)\star(-6n+3)=-6m-6n+7\\ (6m-1)\star(-6n+2)=\begin{cases} 6m-6n-1 & 6m-1\gt6n-2\\ 6n-6m+3 & 6n-2\gt6m-1\\ 3 & m=n\end{cases}\\ (6m-1)\star(-6n+1)=6m+6n-1\\ (6m-1)\star(-6n)=-6m-6n+4\\ (6m)\star(6n)=6m+6n-2\\ (6m)\star(-6n+5)=6m+6n-3\\ (6m)\star(-6n+4)=\begin{cases} 6m-6n+3 & 6m\gt6n-4\\ 6n-6m-1 & 6n-4\gt6m\end{cases}\\ (6m)\star(-6n+3)=\begin{cases} 6n-6m-1 & 6m\gt6n-3\\ 6m-6n+4 & 6n-3\gt6m\end{cases}\\ (6m)\star(-6n+2)=6m+6n\\ (6m)\star(-6n+1)=\begin{cases} 6m-6n & 6m\gt6n-1\\ 6m-6n-1 & 6n-1\gt6m\\ 2 & m=n\end{cases}\\ (6m)\star(-6n)=\begin{cases} 6n-6m+2 & 6m\gt6n\\ 6m-6n+1 & 6n\gt6m\end{cases}\\ (-6m+5)\star(-6n+5)=-6m-6n+8\\ (-6m+5)\star(-6n+4)=\begin{cases} 6n-6m+2 & 6m-5\gt6n-4\\ 6m-6n+1 & 6n-4\gt6m-5\end{cases}\\ (-6m+5)\star(-6n+3)=\begin{cases} 6m-6n+1 & 6m-5\gt6n-3\\ 6m-6n & 6n-3\gt6m-5\\ 1 & m=n\end{cases}\\ (-6m+5)\star(-6n+2)=-6m-6n+5\\ (-6m+5)\star(-6n+1)=\begin{cases} 6n-6m+5 & 6m-5\gt6n-1\\ 6m-6n-2 & 6n-1\gt6m-5\end{cases}\\ (-6m+5)\star(-6n)=\begin{cases} 6m-6n-2 & 6m-5\gt6n\\ 6m-6n-3 & 6n\gt6m-5\end{cases}\\ (-6m+4)\star(-6n+4)=-6m-6n+7\\ (-6m+4)\star(-6n+3)=-6m-6n+6\\ (-6m+4)\star(-6n+2)=\begin{cases} 6n-6m+4 & 6m-4\gt6n-2\\ 6m-6n-1 & 6n-2\gt6m-4\end{cases}\\ (-6m+4)\star(-6n+1)=-6m-6n+4\\ (-6m+4)\star(-6n)=-6m-6n+3\\ (-6m+3)\star(-6n+3)=6m+6n-7\\ (-6m+3)\star(-6n+2)=\begin{cases} 6n-6m+3 & 6m-3\gt6n-2\\ 6n-6m+4 & 6n-2\gt6m-3\end{cases}\\ (-6m+3)\star(-6n+1)=-6m-6n+3\\ (-6m+3)\star(-6n)=6m+6n-4\\ (-6m+2)\star(-6n+2)=-6m-6n+2\\ (-6m+2)\star(-6n+1)=\begin{cases} 6n-6m+2 & 6m-2\gt6n-1\\ 6m-6n+1 & 6n-1\gt6m-2\end{cases}\\ (-6m+2)\star(-6n)=\begin{cases} 6m-6n+1 & 6m-2\gt6n\\ 6m-6n & 6n\gt6m-2\\ 1 & m=n\end{cases}\\ (-6m+1)\star(-6n+1)=-6m-6n+1\\ (-6m+1)\star(-6n)=-6m-6n\\ (-6m)\star(-6n)=6m+6n-1\\ t\circ1=t,\quad t\circ0=0,\quad t\star(2\circ t)=0\end{cases}$
  
$1.$ 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?
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$\begin{cases} (6m-5)\circ(6n-5)=36mn-30m-30n+25\\ (6m-5)\circ(6n-4)=36mn-30m-30n+26\\ (6m-5)\circ(6n-3)=36mn-24m-30n+21\\ (6m-5)\circ(6n-2)=36mn-12m-30n+10\\ (6m-5)\circ(6n-1)=36mn-12m-30n+11\\ (6m-5)\circ(6n)=36mn-6m-30n+6\\ (6m-5)\circ(-6n+5)=-36mn+18m+30n-13\\ (6m-5)\circ(-6n+4)=-36mn+18m+30n-14\\ (6m-5)\circ(-6n+3)=-36mn+24m+30n-21\\ (6m-5)\circ(-6n+2)=-36mn+30n+2\\ (6m-5)\circ(-6n+1)=-36mn+30n+1\\ (6m-5)\circ(-6n)=-36mn+6m+30n-6\\ (6m-4)\circ(6n-4)=36mn-30m-30n+25\\ (6m-4)\circ(6n-3)=-36mn+24m+30n-21\\ (6m-4)\circ(6n-2)=36mn-12m-30n+11\\ (6m-4)\circ(6n-1)=36mn-12m-30n+10\\ (6m-4)\circ(6n)=-36mn+6m+30n-6\\ (6m-4)\circ(-6n+5)=-36mn+18m+30n-14\\ (6m-4)\circ(-6n+4)=-36mn+18m+30n-13\\ (6m-4)\circ(-6n+3)=36mn-24m-30n+21\\ (6m-4)\circ(-6n+2)=-36mn+30n+1\\ (6m-4)\circ(-6n+1)=-36mn+30n+2\\ (6m-4)\circ(-6n)=36mn-6m-30n+6\\ (6m-3)\circ(6n-3)=36mn-24m-24n+16\\ (6m-3)\circ(6n-2)=36mn-12m-24n+9\\ (6m-3)\circ(6n-1)=-36mn+12m+24n-9\\ (6m-3)\circ(6n)=36mn-6m-24n+4\\ (6m-3)\circ(-6n+5)=-36mn+18m+24n-10\\ (6m-3)\circ(-6n+4)=-36mn+18m+24n-11\\ (6m-3)\circ(-6n+3)=36mn-24m-24n+17\\ (6m-3)\circ(-6n+2)=-36mn+24n+2\\ (6m-3)\circ(-6n+1)=-36mn+24n+1\\ (6m-3)\circ(-6n)=36mn-6m-24n+5\\ (6m-2)\circ(6n-2)=36mn-12m-12n+4\\ (6m-2)\circ(6n-1)=36mn-12m-12n+5\\ (6m-2)\circ(6n)=36mn-6m-12n+3\\ (6m-2)\circ(-6n+5)=-36mn+18m+12n-4\\ (6m-2)\circ(-6n+4)=-36mn+18m+12n-5\\ (6m-2)\circ(-6n+3)=-36mn+24m+12n-9\\ (6m-2)\circ(-6n+2)=-36mn+12n+2\\ (6m-2)\circ(-6n+1)=-36mn+12n+1\\ (6m-2)\circ(-6n)=-36mn+6m+12n-3\\ (6m-1)\circ(6n-1)=36mn-12m-12n+4\\ (6m-1)\circ(6n)=-36mn+6m+12n-3\\ (6m-1)\circ(-6n+5)=-36mn+18m+12n-5\\ (6m-1)\circ(-6n+4)=-36mn+18m+12n-4\\ (6m-1)\circ(-6n+3)=36mn-24m-12n+9\\ (6m-1)\circ(-6n+2)=-36mn+12n+1\\ (6m-1)\circ(-6n+1)=-36mn+12n+2\\ (6m-1)\circ(-6n)=36mn-6m-12n+3\\ (6m)\circ(6n)=36mn-6m-6n+1\\ (6m)\circ(-6n+5)=-36mn+18m+6n-1\\ (6m)\circ(-6n+4)=-36mn+18m+6n-2\\ (6m)\circ(-6n+3)=36mn-24m-6n+5\\ (6m)\circ(-6n+2)=-36mn+6n+2\\ (6m)\circ(-6n+1)=-36mn+6n+1\\ (6m)\circ(-6n)=36mn-6m-6n+2\\ (-6m+5)\circ(-6n+5)=-36mn+18m+18n-7\\ (-6m+5)\circ(-6n+4)=-36mn+18m+18n-8\\ (-6m+5)\circ(-6n+3)=-36mn+24m+18n-11\\ (-6m+5)\circ(-6n+2)=-36mn+18n+2\\ (-6m+5)\circ(-6n+1)=-36mn+18n+1\\ (-6m+5)\circ(-6n)=-36mn+6m+18n-2\\ (-6m+4)\circ(-6n+4)=-36mn+18m+18n-7\\ (-6m+4)\circ(-6n+3)=-36mn+24m+18n-10\\ (-6m+4)\circ(-6n+2)=-36mn+18n+1\\ (-6m+4)\circ(-6n+1)=-36mn+18n+2\\ (-6m+4)\circ(-6n)=-36mn+6m+18n-1\\ (-6m+3)\circ(-6n+3)=36mn-24m-24n+16\\ (-6m+3)\circ(-6n+2)=-36mn+24n+1\\ (-6m+3)\circ(-6n+1)=-36mn+24n+2\\ (-6m+3)\circ(-6n)=36mn-6m-24n+4\\ (-6m+2)\circ(-6n+2)=-36mn+2\\ (-6m+2)\circ(-6n+1)=-36mn+1\\ (-6m+2)\circ(-6n)=-36mn+6m+1\\ (-6m+1)\circ(-6n+1)=-36mn+2\\ (-6m+1)\circ(-6n)=-36mn+6m+2\\ (-6m)\circ(-6n)=36mn-6m-6n+1\end{cases}$
  
$2.$ 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?
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but some calculations are such as: $(6m-4)\circ(6n-3)=t(t^{-1}(6m-4)\times t^{-1}(6n-3))=t((-6m+5)\times(6n-4))=$ $t(-6(6mn-4m-5n+4)+4)=-6(6mn-4m-5n+4)+3=-36mn+24m+30n-21,$ $(6m-4)\circ(-6n+1)=$ $t(t^{-1}(6m-4)\times t^{-1}(-6n+1))=t((-6m+5)\times(-6n))=t(6(6mn-5n))=-6(6mn-5n)+2=-36mn+30n+2,$ $(-6m+1)\circ(-6n+1)=t(t^{-1}(-6m+1)\times t^{-1}(-6n+1))=t((-6m)\times(-6n))=t(6(6mn))=-6(6mn)+2=$ $-36mn+2$.</small>
  
Alireza Badali 20:44, 12 October 2017 (CEST)
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Question $1$: Is $(\Bbb Z,\star,\circ)$ a UFD? what are irreducible elements in $(\Bbb Z,\star,\circ)$? is $(\Bbb Q,\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$
  
== 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 $(mn)^i=m^i n^i$. [[User:Passer By|Passer By]] ([[User talk:Passer By|talk]]) 13:14, 3 December 2017 (CET)
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'''Problem''' $1$: Reinterpret (possibly via matrices) ''Gauss circle problem'' under the field $(\Bbb Q,\star_1,\circ_1)$ (not in whole $\Bbb R$).
: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 theorems|Liouville's theorem on approximation of algebraic numbers]]. [[User:Passer By|Passer By]] ([[User talk:Passer By|talk]]) 19:05, 4 December 2017 (CET)
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Alireza Badali 00:49, 25 June 2018 (CEST)
:Thank you. Alireza Badali 16:00, 6 December 2017 (CET)
 
  
==Question==
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== Comments ==
  
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 [https://en.wikipedia.org/wiki/Formula_for_primes], [http://mathworld.wolfram.com/PrimeFormulas.html], [https://www.quora.com/What-is-the-formula-of-prime-numbers], [https://primes.utm.edu/notes/faq/p_n.html], [https://math.stackexchange.com/questions/1257/is-there-a-known-mathematical-equation-to-find-the-nth-prime], [https://thatsmaths.com/2016/06/09/prime-generating-formulae/] etc. [[User:Passer By|Passer By]] ([[User talk:Passer By|talk]]) 19:24, 4 December 2017 (CET)
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Please just insert your comment here! Alireza Badali 20:47, 15 April 2018 (CEST)
: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. [[User:Passer By|Passer By]] ([[User talk: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. [[User:Passer By|Passer By]] ([[User talk: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)
 
::::::Not mathematics at all. Bye. [[User:Passer By|Passer By]] ([[User talk:Passer By|talk]]) 15:30, 7 December 2017 (CET)
 
:::::::Okay, thank you for your attention to my page. Alireza Badali 20:13, 7 December 2017 (CET)
 

Latest revision as of 19:51, 12 August 2018

Whole my previous notes is visible in the revision as of 18:42, 13 April 2018 Alireza Badali 21:52, 13 April 2018 (CEST)

$\mathscr B$ $theory$ (algebraic topological analytical number theory)

Logarithm function as an inverse of the function $f:\Bbb N\to\Bbb R,\,f(n)=a^n,\,a\in\Bbb R$ has prime numbers properties because in usual definition of prime numbers multiplication operation is a point meantime we have $a^n=a\times a\times ...a,$ $(n$ times$),$ hence prime number theorem or its extensions or some other forms is applied in $B$ theory for solving problems on prime numbers exclusively and not all natural numbers.


Algebraic structures on the positive numbers & topology with homotopy groups & prime number theorem and its extensions or other forms or corollaries

Alireza Badali 00:49, 25 June 2018 (CEST)

Goldbach's conjecture

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

Proof given by @Adayah from stackexchange site: 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♦


Now we can define function $f:\{(c,d)\mid (c,d)\subseteq [0.01,0.1)\}\to\Bbb N$ 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 $\forall\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))$ $g(\epsilon)=max(\{f((c,d))\mid d-c=\epsilon,$ $(c,d)\subseteq [0.01,0.1)\})$.

Guess $1$: $g$ isn't an injective function.

Question $1$: Assuming guess $1$, 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 $\{\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\mid g(\epsilon)=n\}\subsetneq h_n$ 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 site:
  • 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}\}$
Another guidance: The affirmative answer is given by Liouville's theorem on approximation of algebraic numbers.


Suppose $r:\Bbb N\to (0,1)$ is a function given by $r(n)$ is obtained by putting a point at the beginning of $n$ instance $r(34880)=0.34880$ and similarly consider $\forall k\in\Bbb N,\, w_k:\Bbb N\to (0,1)$ is a function given by $\forall n\in\Bbb N,$ $w_k(n)=10^{1-k}\cdot r(n)$ and let $S=\bigcup _{k\in\Bbb N}w_k(\Bbb P)$.

Theorem $1$: $r(\Bbb P)$ is dense in the interval $[0.1,1]$. (proof using lemma above)

Regarding to expression form of Goldbach's conjecture, by using this theorem, I wanted enmesh prime numbers properties (prime number theorem should be used for proving this theorem and there is no way except using prime number theorem to prove this density because there is no deference between a prime $p$ and its image $r(p)$ other than a sign or a mark as a point for instance $59$ & $0.59$.) towards Goldbach hence I planned this method.
A corollary: 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.
Question $2$: Which mathematical concept at $[0.1,1)$ could be in accordance with the prime gap at natural numbers?

Question $3$: What is equivalent to the prime number theorem in $[0.1,1)$?


Let $A_n=\{p_{1n},p_{2n},p_{3n},...,p_{mn}\}$ is all primes with $n$ digits, now since $\forall i=1,2,3,...,m-1,\,r(p_{in})\lt r(p_{(i+1)n})$ and $\lim_{m\to\infty}\frac{\pi(10^{m+1})-\pi(10^m)}{\pi(10^m)}=9$ I offer (probably via group theory & prime number theorem can be solved.): Guess $2$: $$\lim_{n\to\infty}\frac{\prod_{i=1}^mr(p_{in})}{\prod_{p\in\Bbb P,\,p\lt p_{1n}}r(p)}\sim({5\over9})^9\,.$$


Theorem $2$: $S$ is dense in the interval $[0,1]$ and $S\times S$ is dense in the $[0,1]\times [0,1]$.


An algorithm that makes new cyclic groups on $\Bbb N$:

Let $\Bbb N$ be that group and at first write integers as a sequence with starting from $0$ and let identity element $e=1$ be corresponding with $0$ and two generators $m$ & $n$ be corresponding with $1$ & $-1$ so we have $\Bbb N=\langle m\rangle=\langle n\rangle$ 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 an even 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 $km+(2-k)$ otherwise equations won't match with definitions of members inverse, and make a table of products 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 too and of course if integers corresponding with two members don't have same signs then product will be a piecewise-defined function for example $12\star _u 15=6$ or $(4\times 3)\star _u (4\times 4-1)=6$ because $(-5)+8=3$ & $-5\to 12,\,\, 8\to 15,\,\, 3\to 6,$ that implies $(4n)\star _u (4m-1)=4m-4n+2$ where $4m-1\gt 4n$ of course it is better at first members inverse be defined for example since $(-9)+9=0$ & $0\to 1,\,\, -9\to 20,\,\, 9\to 18$ so $20\star _u 18=1$, that shows $(4m)\star _u (4m-2)=1$, and with a little bit addition and multiplication all equations will be obtained simply that for this example is:

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


Problem $1$: By using matrices rewrite operation of every group on $\Bbb N$.


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 with $e=p_1=2$, obviously $(\Bbb N,\star)$ & $(\Bbb P,\star _1)$ are groups and $\langle 2\rangle =\langle 3\rangle =(\Bbb N,\star)\simeq (\Bbb Z,+)\simeq (\Bbb P,\star _1)=\langle 3\rangle=\langle 5\rangle$.


Theorem $3$: $(S,\star _S)$ is a group as: $\forall p,q\in\Bbb P,\,\forall m,n\in\Bbb N,\,\forall w_m(p),w_n(q)\in S,$

$\begin{cases} e=0.2\\ \\(w_m(p))^{-1}=w_{m^{-1}}(p^{-1}) & m\star m^{-1}=1,\, p\star _1 p^{-1}=2\\ \\w_m(p)\star _S w_n(q)=w_{m\star n} (p\star _1 q)\end{cases}$

hence $\langle 0.02,0.3\rangle=(S,\star _S)\simeq\Bbb Z\oplus\Bbb Z$.

of course using algorithm above to generate cyclic groups on $\Bbb N$, we can impose another group structure on $\Bbb N$ and consequently on $\Bbb P$ but eventually $S$ with an operation analogous above operation $\star _S$ will be an Abelian group.


Theorem $4$: $(S\times S,\star _{S\times S})$ is a group as: $\forall m_1,n_1,m_2,n_2\in\Bbb N,\,\forall p_1,p_2,q_1,q_2\in\Bbb P,$ $\forall (w_{m_1}(p_1),w_{m_2}(p_2)),(w_{n_1}(q_1),w_{n_2}(q_2))\in S\times S,$

$\begin{cases} e=(0.2,0.2)\\ \\(w_{m_1}(p_1),w_{m_2}(p_2))^{-1}=(w_{m_1^{-1}}(p_1^{-1}),w_{m_2^{-1}}(p_2^{-1}))\\ \text{such that}\quad m_1\star m_1^{-1}=1=m_2\star m_2^{-1},\, p_1\star _1p_1^{-1}=2=p_2\star _1p_2^{-1}\\ \\(w_{m_1}(p_1),w_{m_2}(p_2))\star _{S\times S} (w_{n_1}(q_1),w_{n_2}(q_2))=(w_{m_1\star n_1} (p_1\star _1 q_1),w_{m_2\star n_2}(p_2\star _1 q_2))\end{cases}$

hence $\langle (0.02,0.2),(0.2,0.02),(0.3,0.2),(0.2,0.3)\rangle=(S\times S,\star _{S\times S})\simeq\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z$.

of course using algorithm above to generate cyclic groups on $\Bbb N$, we can impose another group structure on $\Bbb N$ and consequently on $\Bbb P$ but eventually $S\times S$ with an operation analogous above operation $\star _{S\times S}$ will be an Abelian group.


I want make some topologies having prime numbers properties presentable in the collection of open sets, in principle when we image a prime $p$ to real numbers as $w_k(p)$ indeed we accompany prime numbers properties among real numbers which regarding to the expression form of prime number theorem for this aim we should use an important mathematical technique as logarithm function into some planned topologies: question $4$: 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. ($M=\Bbb R,\,B=[0,1],\,A=S$ or $M=\Bbb R^2,$ $B=[0,1]\times[0,1],$ $A=S\times S$)

Perhaps this technique is useful: an extension of prime number theorem: $\forall n\in\Bbb N,$ and for each subinterval $(a,b)$ of $[0.1,1),$ that $a\neq b,$ assume:
$\begin{cases} U_{(a,b)}:=\{n\in\Bbb N\mid a\le r(n)\le b\},\\ \\V_{(a,b)}:=\{p\in\Bbb P\mid a\le r(p)\le b\},\\ \\U_{(a,b),n}:=\{m\in U_{(a,b)}\mid m\le n\},\\ \\V_{(a,b),n}:=\{p\in V_{(a,b)}\mid p\le n\},\\ \\w_{(a,b),n}:={\#V_{(a,b),n}\over\#U_{(a,b),n}}\cdot\log n,\\ \\w_{(a,b)}:=\lim _{n\to\infty} w_{(a,b),n}\\ \\z_{(a,b),n}:={\#V_{(a,b),n}\over\#U_{(a,b),n}}\cdot\log{(\#U_{(a,b),n})}\\ \\z_{(a,b)}:=\lim_{n\to\infty}z_{(a,b),n}\end{cases}$
Guess $3$: $\forall (a,b)\subset [0.1,1),\,w_{(a,b)}={10\over9}\cdot(b-a)$.
Answer given by $@$Peter from stackexchange site: Imagine a very large number $N$ and consider the range $[10^N,10^{N+1}]$. The natural logarithms of $10^N$ and $10^{N+1}$ only differ by $\ln(10)\approx 2.3$ Hence the reciprocals of the logarithms of all primes in this range virtually coincicde. Because of the approximation $$\int_a^b \frac{1}{\ln(x)}dx$$ for the number of primes in the range $[a,b]$ the number of primes is approximately the length of the interval divided by $\frac{1}{\ln(10^N)}$, so is approximately equally distributed. Hence your conjecture is true.
Benfords law seems to contradict this result , but this only applies to sequences producing primes as the Mersenne primes and not if the primes are chosen randomly in the range above.
Guess $4$: $\forall (a,b)\subset [0.1,1),\,z_{(a,b)}={10\over9}\cdot(b-a)$.
Question $5$: What does mean $\forall a\in[0.1,1),\,\forall b\in(0.1,1),\,a\lt b,\,\lim_{b\to a}z_{(a,b)}=0$?
Guess $5$: $\forall(a,b),(c,d)\subset[0.1,1),\,\lim_{n\to\infty}{\#V_{(a,b),n}\over\#V_{(c,d),n}}={b-a\over d-c}=\lim_{n\to\infty}{\#U_{(a,b),n}\over\#U_{(c,d),n}}$.
Comment given by $@$Dzoooks from stackexchange site: It shouldn't be that hard to get estimates from $V_{(a,b),n}=\{p\leq n : 10^ka\lt p\lt 10^k\text{ for some }k\}=\sqcup_{k\geq1}\{p\in[0,n]\cap(10^ka,10^kb)\},$ where the union is disjoint from $10^kb\lt10^k\leq10^{k+1}a$. Then $\#V_{(a,b),n}$ can be summed with the PNT. You'll see that a $(b-c)$ comes out of the sum..maybe
and the PNT gives $\#\{p\in[0,n]\cap(10^ka,10^k)\}\sim\frac{(b-a)10^k}{\log b-\log a},$ for large $n$ and $k$. Factor these out of the sum, and it looks like your limit is actually $\frac{b-a}{\log b-\log a}\cdot\frac{\log d-\log c}{d-c}$.


Theorem $5$: Let $t_n:\Bbb N\to\Bbb N\setminus\{n\in\Bbb N: 10\mid n\}$ is a surjective strictly monotonically increasing sequence now $\{t_n\}_{n\in\Bbb N}$ is a cyclic group with: $\begin{cases} e=1\\ t_n^{-1}=t_{n^{-1}}\quad\text{that}\quad n\star n^{-1}=1\\ t_n\star _tt_m=t_{n\star m}\end{cases}$

that $(\{t_n\}_{n\in\Bbb N},\star _t)=\langle 2\rangle=\langle 3\rangle$ and let $E:=\bigcup _{k\in\Bbb N} w_k(\Bbb N\setminus\{n\in\Bbb N: 10\mid n\})$ so $(E,\star _E)$ is an Abelian group with $\forall m,n\in\Bbb N,$ $\forall a,b\in\Bbb N\setminus\{n\in\Bbb N: 10\mid n\}$: $\,\,\begin{cases} e=0.1\\ w_n(a)^{-1}=w_{n^{-1}}(a^{-1})\quad\text{that}\quad n\star n^{-1}=1,\, a\star _ta^{-1}=1\\ w_n(a)\star _Ew_m(b)=w_{n\star m}(a\star _tb)\end{cases}$

that $\langle 0.01,0.2\rangle=E\simeq\Bbb Z\oplus\Bbb Z$ ♦


now assume $(S\times S)\oplus E$ is external direct sum of the groups $S\times S$ and $E$ with $e=(0.2,0.2,0.1)$ and $\langle (0.02,0.2,0.1),(0.2,0.02,0.1),(0.3,0.2,0.1),(0.2,0.3,0.1),(0.2,0.2,0.01),(0.2,0.2,0.2)\rangle=$ $(S\times S)\oplus E\simeq\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z$.


Theorem $6$: $(S,\lt _1)$ is a total ordered set with order relation $\lt_1$ as: $\forall i,n,k\in\Bbb N$ if $p_n$ be $n$-th prime number, relation $\lt _1$ is defined with: $w_i(p_n)\lt _1w_i(p_{n+k})\lt _1w_{i+1}(p_n)$ or $$0.2\lt _10.3\lt _10.5\lt _10.7\lt _10.11\lt _10.13\lt _10.17\lt _1...0.02\lt _10.03\lt _10.05\lt _10.07\lt _10.011\lt _1$$ $$0.013\lt _10.017\lt _1...0.002\lt _10.003\lt _10.005\lt _10.007\lt _10.0011\lt _10.0013\lt _10.0017\lt _1...$$ and $(E,\lt_2)$ is another total ordered set with order relation $\lt _2$ as: $\forall i,n,k\in\Bbb N$ that $10\nmid n,\, 10\nmid n+k,$ $w_i(n)\lt _2w_i(n+k)\lt _2w_{i+1}(n)$ or $$0.1\lt _2 0.2\lt _2 0.3\lt _2 ...0.9\lt _2 0.11\lt _2 0.12\lt _2 ...0.19\lt _2 0.21\lt _2 ...0.01\lt _2 0.02\lt _2 0.03\lt _2 ...0.09$$ $$\lt _2 0.011\lt _2 0.012\lt _2 ...0.019\lt _2 0.021\lt _2 ...0.001\lt _2 0.002\lt _2 0.003\lt _2 ...0.009\lt _2 0.0011\lt _2 ...$$ now $M:=S\times S\times E$ is a total ordered set with order relation $\lt _3$ as: $\forall (a,b,t),(c,d,u)\in S\times S\times E,$ $(a,b,t)\lt _3(c,d,u)$ iff $\,\,\begin{cases} t\lt _2u & or\\ t=u,\,\, a+b\lt _2c+d & or\\ t=u,\,\, a+b=c+d,\,\, b\lt _1 d\end{cases}$ ♦


Theorem $6$: Assume $([0,1],\lt_1)$ is a total ordered set with: $\forall a,b\in[0,1],\,a\lt_1b$ iff $a\gt b$ then $S$ & $E$ are total ordered sets with restriction of order relation $\lt _1$ on $S$ & $E$, now $M:=S\times S\times E$ is a total ordered set with order relation $\lt _2$ as: $\forall (a,b,t),(c,d,u)\in S\times S\times E,$ $(a,b,t)\lt _2(c,d,u)$ iff $\,\,\begin{cases} t\lt _1u & or\\ t=u,\,\, a+b\lt _1c+d & or\\ t=u,\,\, a+b=c+d,\,\, b\lt _1d\end{cases}$ ♦


now assume $M$ is a topological space (Hausdorff space) induced by order relation $\lt _?$.


Question $6$: Is $S$ a topological group under topology induced by order relation $\lt_?$ and is $(S\times S)\oplus E$ a topological group under topology of $M$?


A new version of Goldbach's conjecture: For each even natural number $t$ greater than $4$ and $\forall c,m\in\Bbb N\cup\{0\}$ that $10^c\mid t,\, 10^{1+c}\nmid t$, $A_m=\{(a,b)\mid a,b\in S,\, 10^{-1-m}\le a+b\lt 10^{-m}\}$ and if $u$ is the number of digits in $t$ then $\exists (a,b)\in A_c$ such that $t=10^{c+u}\cdot (a+b),\, 10^{c+u}\cdot a,10^{c+u}\cdot b\in\Bbb P\setminus\{2\},\, (a,b,10^{-c-u}\cdot t)\in M$.

Using homotopy groups Goldbach's conjecture will be proved.
Alireza Badali 08:27, 31 March 2018 (CEST)

Polignac's conjecture

In previous chapter above I used an important technique by theorem $1$ for presentment of prime numbers properties as density in discussion that using prime number theorem it became applicable, anyway, but now I want perform another method for Twin prime conjecture (Polignac) in principle prime numbers properties are ubiquitous in own natural numbers.


Theorem $1$: $(\Bbb N,\star _T)$ is a group with: $\forall m,n\in\Bbb N,$

$\begin{cases} (12m-10)\star_T(12m-9)=1=(12m-8) \star_T(12m-5)=(12m-7) \star_T(12m-4)=\\ (12m-6) \star_T(12m-1)=(12m-3) \star_T(12m)=(12m-2) \star_T(12m+1)\\ (12m-10) \star_T(12n-10)=12m+12n-19\\ (12m-10) \star_T(12n-9)=\begin{cases} 12m-12n+1 & 12m-10\gt 12n-9\\ 12n-12m-2 & 12n-9\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-8)=12m+12n-15\\ (12m-10) \star_T(12n-7)=12m+12n-20\\ (12m-10) \star_T(12n-6)=12m+12n-11\\ (12m-10) \star_T(12n-5)=\begin{cases} 12m-12n-3 & 12m-10\gt 12n-5\\ 12n-12m+8 & 12n-5\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-4)=\begin{cases} 12m-12n-6 & 12m-10\gt 12n-4\\ 12n-12m+3 & 12n-4\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-3)=12m+12n-18\\ (12m-10) \star_T(12n-2)=\begin{cases} 12m-12n-10 & 12m-10\gt 12n-2\\ 12n-12m+11 & 12n-2\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-1)=\begin{cases} 12m-12n-7 & 12m-10\gt 12n-1\\ 12n-12m+12 & 12n-1\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n)=\begin{cases} 12m-12n-8 & 12m-10\gt 12n\\ 12n-12m+7 & 12n\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n+1)=12m+12n-10\\ (12m-9) \star_T(12n-9)=12m+12n-16\\ (12m-9) \star_T(12n-8)=\begin{cases} 12m-12n & 12m-9\gt 12n-8\\ 12n-12m+5 & 12n-8\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-7)=\begin{cases} 12m-12n-1 & 12m-9\gt 12n-7\\ 12n-12m+2 & 12n-7\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-6)=\begin{cases} 12m-12n-4 & 12m-9\gt 12n-6\\ 12n-12m+9 & 12n-6\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-5)=12m+12n-12\\ (12m-9) \star_T(12n-4)=12m+12n-17\\ (12m-9) \star_T(12n-3)=\begin{cases} 12m-12n-5 & 12m-9\gt 12n-3\\ 12n-12m+4 & 12n-3\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-2)=12m+12n-9\\ (12m-9) \star_T(12n-1)=12m+12n-14\\ (12m-9) \star_T(12n)=12m+12n-13\\ (12m-9)\star_T(12n+1)=\begin{cases} 12m-12n-9 & 12m-9\gt 12n+1\\ 12n-12m+6 & 12n+1\gt 12m-9\end{cases}\\ (12m-8) \star_T(12n-8)=12m+12n-11\\ (12m-8) \star_T(12n-7)=12m+12n-18\\ (12m-8) \star_T(12n-6)=12m+12n-7\\ (12m-8) \star_T(12n-5)=\begin{cases} 12m-12n+1 & 12m-8\gt 12n-5\\ 12n-12m-2 & 12n-5\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n-4)=\begin{cases} 12m-12n+2 & 12m-8\gt 12n-4\\ 12n-12m-1 & 12n-4\gt 12m-8\\ 2 & m=n\end{cases}\\ (12m-8) \star_T(12n-3)=12m+12n-10\\ (12m-8) \star_T(12n-2)=\begin{cases} 12m-12n-8 & 12m-8\gt 12n-2\\ 12n-12m+7 & 12n-2\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n-1)=\begin{cases} 12m-12n-3 & 12m-8\gt 12n-1\\ 12n-12m+8 & 12n-1\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n)=\begin{cases} 12m-12n-6 & 12m-8\gt 12n\\ 12n-12m+3 & 12n\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n+1)=12m+12n-8\\ (12m-7) \star_T(12n-7)=12m+12n-15\\ (12m-7) \star_T(12n-6)=12m+12n-10\\ (12m-7) \star_T(12n-5)=\begin{cases} 12m-12n-6 & 12m-7\gt 12n-5\\ 12n-12m+3 & 12n-5\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n-4)=\begin{cases} 12m-12n+1 & 12m-7\gt 12n-4\\ 12n-12m-2 & 12n-4\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n-3)=12m+12n-11\\ (12m-7) \star_T(12n-2)=\begin{cases} 12m-12n-7 & 12m-7\gt 12n-2\\ 12n-12m+12 & 12n-2\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n-1)=\begin{cases} 12m-12n-8 & 12m-7\gt 12n-1\\ 12n-12m+7 & 12n-1\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n)=\begin{cases} 12m-12n-3 & 12m-7\gt 12n\\ 12n-12m+8 & 12n\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n+1)=12m+12n-7\\ (12m-6) \star_T(12n-6)=12m+12n-3\\ (12m-6) \star_T(12n-5)=\begin{cases} 12m-12n+5 & 12m-6\gt 12n-5\\ 12n-12m & 12n-5\gt 12m-6\\ 5 & m=n\end{cases}\\ (12m-6) \star_T(12n-4)=\begin{cases} 12m-12n+4 & 12m-6\gt 12n-4\\ 12n-12m-5 & 12n-4\gt 12m-6\\ 4 & m=n\end{cases}\\ (12m-6) \star_T(12n-3)=12m+12n-8\\ (12m-6) \star_T(12n-2)=\begin{cases} 12m-12n-6 & 12m-6\gt 12n-2\\ 12n-12m+3 & 12n-2\gt 12m-6\end{cases}\\ (12m-6) \star_T(12n-1)=\begin{cases} 12m-12n+1 & 12m-6\gt 12n-1\\ 12n-12m-2 & 12n-1\gt 12m-6\end{cases}\\ (12m-6) \star_T(12n)=\begin{cases} 12m-12n+2 & 12m-6\gt 12n\\ 12n-12m-1 & 12n\gt 12m-6\\ 2 & m=n\end{cases}\\ (12m-6) \star_T(12n+1)=12m+12n-6\\ (12m-5) \star_T(12n-5)=12m+12n-14\\ (12m-5) \star_T(12n-4)=12m+12n-13\\ (12m-5) \star_T(12n-3)=\begin{cases} 12m-12n-1 & 12m-5\gt 12n-3\\ 12n-12m+2 & 12n-3\gt 12m-5\end{cases}\\ (12m-5) \star_T(12n-2)=12m+12n-5\\ (12m-5) \star_T(12n-1)=12m+12n-4\\ (12m-5) \star_T(12n)=12m+12n-9\\ (12m-5) \star_T(12n+1)=\begin{cases} 12m-12n-5 & 12m-5\gt 12n+1\\ 12n-12m+4 & 12n+1\gt 12m-5\end{cases}\\ (12m-4) \star_T(12n-4)=12m+12n-12\\ (12m-4) \star_T(12n-3)=\begin{cases} 12m-12n & 12m-4\gt 12n-3\\ 12n-12m+5 & 12n-3\gt 12m-4\end{cases}\\ (12m-4) \star_T(12n-2)=12m+12n-4\\ (12m-4) \star_T(12n-1)=12m+12n-9\\ (12m-4) \star_T(12n)=12m+12n-14\\ (12m-4) \star_T(12n+1)=\begin{cases} 12m-12n-4 & 12m-4\gt 12n+1\\ 12n-12m+9 & 12n+1\gt 12m-4\end{cases}\\ (12m-3) \star_T(12n-3)=12m+12n-7\\ (12m-3) \star_T(12n-2)=\begin{cases} 12m-12n-3 & 12m-3\gt 12n-2\\ 12n-12m+8 & 12n-2\gt 12m-3\end{cases}\\ (12m-3) \star_T(12n-1)=\begin{cases} 12m-12n-6 & 12m-3\gt 12n-1\\ 12n-12m+3 & 12n-1\gt 12m-3\end{cases}\\ (12m-3) \star_T(12n)=\begin{cases} 12m-12n+1 & 12m-3\gt 12n\\ 12n-12m-2 & 12n\gt 12m-3\end{cases}\\ (12m-3) \star_T(12n+1)=12m+12n-3\\ (12m-2) \star_T(12n-2)=12m+12n-2\\ (12m-2) \star_T(12n-1)=12m+12n-1\\ (12m-2) \star_T(12n)=12m+12n\\ (12m-2) \star_T(12n+1)=\begin{cases} 12m-12n-2 & 12m-2\gt 12n+1\\ 12n-12m+1 & 12n+1\gt 12m-2\end{cases}\\ (12m-1) \star_T(12n-1)=12m+12n\\ (12m-1) \star_T(12n)=12m+12n-5\\ (12m-1) \star_T(12n+1)=\begin{cases} 12m-12n-1 & 12m-1\gt 12n+1\\ 12n-12m+2 & 12n+1\gt 12m-1\end{cases}\\ (12m) \star_T(12n)=12m+12n-4\\ (12m) \star_T(12n+1)=\begin{cases} 12m-12n & 12m\gt 12n+1\\ 12n-12m+5 & 12n+1\gt 12m\end{cases}\\ (12m+1) \star_T(12n+1)=12m+12n+1\end{cases}$

that $\forall k\in\Bbb N,\,\langle 2\rangle =\langle 3\rangle =\langle (2k+1)\star _T (2k+3)\rangle=(\Bbb N,\star _T)\simeq (\Bbb Z,+)$ and $\langle (2k)\star _T(2k+2)\rangle\neq\Bbb N$ and each prime in $\langle 5\rangle$ is to form of $5+12k$ or $13+12k$, $k\in\Bbb N\cup\{0\}$ and each prime in $\langle 7\rangle$ is to form of $7+12k$ or $13+12k$, $k\in\Bbb N\cup\{0\}$ and $\langle 5\rangle\cap\langle 7\rangle=\langle 13\rangle$ and $\Bbb N=\langle 5\rangle\oplus\langle 7\rangle$ but there isn't any proper subgroup including all primes of the form $11+12k,$ $k\in\Bbb N\cup\{0\}$ (probably I have to make another better).

Proof:

$$0,-1,1,-3,-2,-5,3,2,-4,6,5,4,-6,-7,7,-9,-8,-11,9,8,-10,12,11,10,-12,-13,13,-15,$$ $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,$$ $$-14,-17,15,14,-16,18,17,16,-18,-19,19,-21,-20,-23,21,20,-22,24,23,22,-24,...$$ $$29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,...$$


Guess $1$: For each group on $\Bbb N$ like $(\Bbb N,\star)$ generated from algorithm above, if $p_i$ be $i$_th prime number and $x_i$ be $i$_th composite number then $\exists m\in\Bbb N,\,\forall n\in\Bbb N$ that $n\ge m$ we have: $2\star3\star5\star7...\star p_n=\prod_{i=1}^{n}p_i\gt\prod _{i=1}^{n}x_i=4\star6\star8\star9...\star x_n$

Guess $2$: For each group on $\Bbb N$ like $(\Bbb N,\star)$ generated from algorithm above, we have: $\lim_{n\to\infty}\prod _{n=1}^{\infty}p_n,\lim_{n\to\infty}\prod _{n=1}^{\infty}x_n\in\Bbb N,\,\,(\lim_{n\to\infty}\prod _{n=1}^{\infty}p_n)\star(\lim_{n\to\infty}\prod _{n=1}^{\infty}x_n)=1$.


now let the group $G$ be external direct sum of three copies of the group $(\Bbb N,\star _T)$, hence $G=\Bbb N\oplus\Bbb N\oplus\Bbb N$.


Theorem $2$: $(\Bbb N\times\Bbb N\times\Bbb N,\lt _T)$ is a well ordering set with order relation $\lt _T$ as: $\forall (m_1,n_1,t_1),(m_2,n_2,t_2)\in\Bbb N\times\Bbb N\times\Bbb N,$ $(m_1,n_1,t_1)\lt _T(m_2,n_2,t_2)$ iff $\begin{cases} t_1\lt t_2 & or\\ t_1=t_2,\, m_1-n_1\lt m_2-n_2 & or\\ t_1=t_2,\, m_1-n_1=m_2-n_2,\, n_1\lt n_2\end{cases}$ ♦


and suppose $M=\Bbb N\times\Bbb N\times\Bbb N$ is a topological space (Hausdorff space) induced by order relation $\lt _T$.


Question $1$: Is $G$ a topological group with topology of $M$?


Now regarding to the group $(\Bbb N,\star_T)$, I am planning an algebraic form of prime number theorem towards twin prime conjecture:


Recall the statement of the prime number theorem: Let $x$ be a positive real number, and let $\pi(x)$ denote the number of primes that are less than or equal to $x$. Then the ratio $\pi(x)\cdot{\log x\over x}$ can be made arbitrarily close to $1$ by taking $x$ sufficiently large.

Question $2$: Suppose $\pi_1(x)$ is all prime numbers of the form $4k+1$ and less than $x$ and $\pi_2(x)$ is all prime numbers of the form $4k+3$ and less than $x$. Do $\lim_{x\to\infty}\pi_1(x)\cdot{\log x\over x}=0.5=\lim_{x\to\infty}\pi_2(x)\cdot{\log x\over x}\ ?$

Answer given by $@$Milo Brandt from stackexchange site: Basically, for any $k$, the primes are equally distributed across the congruence classes $\langle n\rangle$ mod $k$ where $n$ and $k$ are coprime.
This result is known as the prime number theorem for arithmetic progressions. [Wikipedia](https://en.wikipedia.org/wiki/Prime_number_theorem#Prime_number_theorem_for_arithmetic_progressions) discusses it with a number of references and one can find a proof of it by Ivan Soprounov [here](http://academic.csuohio.edu/soprunov_i/pdf/primes.pdf), which makes use of the Dirichlet theorem on arithmetic progressions (which just says that $\pi_1$ and $\pi_2$ are unbounded) to prove this stronger result.


Question $3$: For each neutral infinite subset $A$ of $\Bbb N$, does exist a cyclic group like $(\Bbb N,\star)$ such that $A$ is a maximal subgroup of $\Bbb N$?

Question $4$: If $(\Bbb N,\star_1)$ is a cyclic group and $n\in\Bbb N$ and $A=\{a_i\mid i\in\Bbb N\}$ is a non-trivial subgroup of $\Bbb N$ then does exist another cyclic group $(\Bbb N,\star_2)$ such that $\prod _{i=1}^{\infty}a_i=a_1\star_2a_2\star_2a_3\star_2...=n$?

Question $5$: If $(\Bbb N,\star)$ is a cyclic group and $n\in\Bbb N$ then does exist a non-trivial subset $A=\{a_i\mid i\in\Bbb N\}$ of $\Bbb P$ with $\#(\Bbb P\setminus A)=\aleph_0$ and $\prod _{i=1}^{\infty}a_i=a_1\star a_2\star a_3\star...=n$?

Question $6$: If $(\Bbb N,\star_1)$ and $(\Bbb N,\star_2)$ are cyclic groups and $A=\{a_i\mid i\in\Bbb N\}$ is a non-trivial subgroup of $(\Bbb N,\star_1)$ and $B=A\cap\Bbb P$ then does $\prod_{i=1}^{\infty}a_i=a_1\star_2a_2\star_2a_3\star_2...\in\Bbb N$?


Theorem $3$: $U:=\{{r(p)-r(q)\over r(s)-r(t)}\mid p,q,s,t\in\Bbb P,\,s\neq t\}$ is dense in $\Bbb R$.

Proof given by $@$NoahSchweber from stackexchange site: for any real number $x$ we can by the density of the image of $r$ in $[0.1,1]$ find primes $p,q,s,t$ such that $r(p)−r(q)$ is very close to $x\over n$ and $r(s)−r(t)$ is very close to $1\over n$ for some large integer $n$.
Question $7$: Does $T:=U\cap\Bbb P$ have infinitely many primes?
Question $8$: Is the set $V:=\{p+2q\mid p,q\in\Bbb P,\,p+2q\in T\}$ infinite?
Alireza Badali 12:34, 28 April 2018 (CEST)

Some dissimilar conjectures

Algebraic analytical number theory

Alireza Badali 16:51, 4 July 2018 (CEST)

Collatz conjecture

The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined as follows: start with any positive integer $n$. Then each term is obtained from the previous term as follows: if the previous term is even, the next term is one half the previous term. Otherwise, the next term is $3$ times the previous term plus $1$. The conjecture is that no matter what value of $n$, the sequence will always reach $1$. The conjecture is named after German mathematician Lothar Collatz, who introduced the idea in $1937$, two years after receiving his doctorate. It is also known as the $3n + 1$ conjecture.


Theorem $1$: If $(\Bbb N,\star_{\Bbb N})$ is a cyclic group with $e_{\Bbb N}=1$ & $\langle m_1\rangle=\langle m_2\rangle=(\Bbb N,\star_{\Bbb N})$ and $f:\Bbb N\to\Bbb N$ is a bijection such that $f(1)=1$ then $(\Bbb N,\star _f)$ is a cyclic group with: $e_f=1$ & $\langle f(m_1)\rangle=\langle f(m_2)\rangle=(\Bbb N,\star_f)$ & $\forall m,n\in\Bbb N,$ $f(m)\star _ff(n)=f(m\star_{\Bbb N}n)$ & $(f(n))^{-1}=f(n^{-1})$ that $n\star_{\Bbb N}n^{-1}=1$.


I want make a group in accordance with Collatz graph but $@$RobertFrost from stackexchange site advised me in addition, it needs to be a torsion group because then it can be used to show convergence, meantime I like apply lines in the Euclidean plane $\Bbb R^2$ too.


Question $1$: What is function of this sequence on to natural numbers? $1,2,4,3,6,5,10,7,14,8,16,9,18,11,22,12,24,13,26,15,30,17,34,19,38,20,40,21,42,23,46,25,50,...$ such that we begin from $1$ and then write $2$ then $2\times2$ then $3$ then $2\times3$ then ... but if $n$ is even and previously we have written $0.5n$ and then $n$ then ignore $n$ and continue and write $n+1$ and then $2n+2$ and so on for example we have $1,2,4,3,6,5,10$ so after $10$ we should write $7,14,...$ because previously we have written $3,6$.

Answer given by $@$r.e.s from stackexchange site: Following is a definition of your sequence without using recursion.
Let $S=(S_0,S_1,S_2,\ldots)$ be the increasing sequence of positive integers that are expressible as either $2^e$ or as $o_1\cdot 2^{o_2}$, where $e$ is an even nonnegative integer, $o_1>1$ is an odd positive integer and $o_2$ is an odd positive integer. Thus $$S=(1, 4, 6, 10, 14, 16, 18, 22, 24, 26, 30, 34, 38, 40,42,\ldots).$$ Let $\bar{S}$ be the complement of $S$ with respect to the positive integers; i.e., $$\bar{S}=(2, 3, 5, 7, 8, 9, 11, 12, 13, 15, 17, 19, 20, 21, 23, 25,\ldots).$$ Your sequence is then $T=(T_0,T_1,T_2,\ldots)$, where

$$T_n:=\begin{cases}S_{n\over 2}&\text{ if $n$ is even}\\ \bar{S}_{n-1\over 2}&\text{ if $n$ is odd.} \end{cases} $$

Thus $T=(1, 2, 4, 3, 6, 5, 10, 7, 14, 8, 16, 9, 18, 11, 22, 12, 24, 13, 26, 15, 30, 17, 34, 19, 38, 20, \ldots).$

References:
Sequences $S,\bar{S},T$ are OEIS [A171945](http://oeis.org/A171945), [A053661](http://oeis.org/A053661), [A034701](http://oeis.org/A034701) respectively. These are all discussed in ["The vile, dopey, evil and odious game players"](https://www.sciencedirect.com/science/article/pii/S0012365X11001427).

Sage code:
   def is_in_S(n): return ( (n.valuation(2) % 2 == 0) and (n.is_power_of(2))  ) or ( (n.valuation(2) % 2 == 1) and not(n.is_power_of(2))  )
   S = [n for n in [1..50] if is_in_S(n)]
   S_ = [n for n in [1..50] if not is_in_S(n)]
   T = []
   for i in range(max(len(S),len(S_))):
       if i % 2 == 0: T += [S[i/2]]
       else: T += [S_[(i-1)/2]]
   print S
   print S_
   print T
   [1, 4, 6, 10, 14, 16, 18, 22, 24, 26, 30, 34, 38, 40, 42, 46, 50]
   [2, 3, 5, 7, 8, 9, 11, 12, 13, 15, 17, 19, 20, 21, 23, 25, 27, 28, 29, 31, 32, 33, 35, 36, 37, 39, 41, 43, 44, 45, 47, 48, 49]
   [1, 2, 4, 3, 6, 5, 10, 7, 14, 8, 16, 9, 18, 11, 22, 12, 24, 13, 26, 15, 30, 17, 34, 19, 38, 20, 40, 21, 42, 23, 46, 25, 50]


Theorem $2$: If $(\Bbb N,\star_1)$ & $(\Bbb N,\star_2)$ are cyclic groups with generators respectively $u_1$ & $v_1$ and $u_2$ & $v_2$ then $C_1=\{(m,2m)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_1}=(1,2)\\ \\\forall m,n\in\Bbb N,\,(m,2m)\star_{C_1}(n,2n)=(m\star_1n,2(m\star_1n))\\ (m,2m)^{-1}=(m^{-1},2\times m^{-1})\qquad\text{that}\quad m\star_1m^{-1}=1\\ \\C_1=\langle(u_1,2u_1)\rangle=\langle(v_1,2v_1)\rangle\end{cases}$ and $C_2=\{(3m-1,2m-1)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_2}=(2,1)\\ \\\forall m,n\in\Bbb N,\,(3m-1,2m-1)\star_{C_2}(3n-1,2n-1)=(3(m\star_2n)-1,2(m\star_2n)-1)\\ (3m-1,2m-1)^{-1}=(3\times m^{-1}-1,2\times m^{-1}-1)\qquad\text{that}\quad m\star_2 m^{-1}=1\\ \\C_2=\langle(3u_2-1,2u_2-1)\rangle=\langle(3v_2-1,2v_2-1)\rangle\end{cases}$•

And let $C:=C_1\oplus C_2$ be external direct sum of the groups $C_1$ & $C_2$. Problem $1$: What are maximal subgroups of $C$?


Theorem $3$: If $(\Bbb N,\star)$ is a cyclic group with generators $u,v$ and identity element $e=1$ and $f:\Bbb N\to\Bbb R$ is an injection then $(f(\Bbb N),\star_f)$ is a cyclic group with generators $f(u),f(v)$ and identity element $e_f=f(1)$ and operation law: $\forall m,n\in\Bbb N,$ $f(m)\star_ff(n)=f(m\star n)$ and inverse law: $\forall n\in\Bbb N,$ $(f(n))^{-1}=f(n^{-1})$ that $n\star n^{-1}=1$.


Suppose $\forall m,n\in\Bbb N,\qquad$ $\begin{cases} m\star 1=m\\ (4m)\star (4m-2)=1=(4m+1)\star (4m-1)\\ (4m-2)\star (4n-2)=4m+4n-5\\ (4m-2)\star (4n-1)=4m+4n-2\\ (4m-2)\star (4n)=\begin{cases} 4m-4n-1 & 4m-2\gt 4n\\ 4n-4m+1 & 4n\gt 4m-2\\ 3 & m=n+1\end{cases}\\ (4m-2)\star (4n+1)=\begin{cases} 4m-4n-2 & 4m-2\gt 4n+1\\ 4n-4m+4 & 4n+1\gt 4m-2\end{cases}\\ (4m-1)\star (4n-1)=4m+4n-1\\ (4m-1)\star (4n)=\begin{cases} 4m-4n+2 & 4m-1\gt 4n\\ 4n-4m & 4n\gt 4m-1\\ 2 & m=n\end{cases}\\ (4m-1)\star (4n+1)=\begin{cases} 4m-4n-1 & 4m-1\gt 4n+1\\ 4n-4m+1 & 4n+1\gt 4m-1\\ 3 & m=n+1\end{cases}\\ (4m)\star (4n)=4m+4n-3\\ (4m)\star (4n+1)=4m+4n\\ (4m+1)\star (4n+1)=4m+4n+1\\ \Bbb N=\langle 2\rangle=\langle 4\rangle\end{cases}$

and let $C_1=\{(m,2m)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_1}=(1,2)\\ \\\forall m,n\in\Bbb N,\,(m,2m)\star_{C_1}(n,2n)=(m\star n,2(m\star n))\\ (m,2m)^{-1}=(m^{-1},2\times m^{-1})\qquad\text{that}\quad m\star m^{-1}=1\\ \\C_1=\langle(2,4)\rangle=\langle(4,8)\rangle\end{cases}$

and $C_2=\{(3m-1,2m-1)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_2}=(2,1)\\ \\\forall m,n\in\Bbb N,\, (3m-1,2m-1)\star_{C_2}(3n-1,2n-1)=(3(m\star n)-1,2(m\star n)-1)\\ (3m-1,2m-1)^{-1}=(3\times m^{-1}-1,2\times m^{-1}-1)\qquad\text{that}\quad m\star m^{-1}=1\\ \\C_2=\langle(5,3)\rangle=\langle(11,7)\rangle\end{cases}$.

and let $C:=C_1\oplus C_2$ be external direct sum of the groups $C_1$ & $C_2$, Question $2$: What are maximal subgroups of $C$?

Alireza Badali 10:02, 12 May 2018 (CEST)

Erdős–Straus conjecture

Theorem: If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ and generators $a,b$ then $E=\{({1\over x},{1\over y},{1\over z},{-4\over n+1},n)\mid x,y,z,n\in\Bbb N\}$ is an Abelian group with: $\forall x,y,z,n,x_1,y_1,z_1,n_1\in\Bbb N$ $\begin{cases} e_E=(1,1,1,-2,1)=({1\over 1},{1\over 1},{1\over 1},{-4\over 1+1},1)\\ \\({1\over x},{1\over y},{1\over z},{-4\over n+1},n)^{-1}=({1\over x^{-1}},{1\over y^{-1}},{1\over z^{-1}},\frac{-4}{n^{-1}+1},n^{-1})\quad\text{that}\\ x\star x^{-1}=1=y\star y^{-1}=z\star z^{-1}=n\star n^{-1}\\ \\({1\over x},{1\over y},{1\over z},\frac{-4}{n+1},n)\star_E({1\over x_1},{1\over y_1},{1\over z_1},\frac{-4}{n_1+1},n_1)=(\frac{1}{x\star x_1},\frac{1}{y\star y_1},\frac{1}{z\star z_1},\frac{-4}{n\star {n_1}+1},n\star n_1)\\ \\E=\langle({1\over a},1,1,-2,1),(1,{1\over a},1,-2,1),(1,1,{1\over a},-2,1),(1,1,1,\frac{-4}{a+1},1),(1,1,1,-2,a)\rangle=\\ \langle({1\over b},1,1,-2,1),(1,{1\over b},1,-2,1),(1,1,{1\over b},-2,1),(1,1,1,\frac{-4}{b+1},1),(1,1,1,-2,b)\rangle\end{cases}$•


Let $(\Bbb N,\star)$ is a cyclic group with: $\begin{cases} n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\\\Bbb N=\langle 2\rangle =\langle 3\rangle \end{cases}$

Question: Is $E_0=\{({1\over x},{1\over y},{1\over z},\frac{-4}{n+1},n)\mid x,y,z,n\in\Bbb N,\, {1\over x}+{1\over y}+{1\over z}-{4\over n+1}=0\}$ a subgroup of $E$?
Alireza Badali 17:34, 25 May 2018 (CEST)

Landaus forth problem

Friedlander–Iwaniec theorem: there are infinitely many prime numbers of the form $a^2+b^4$.

I want use this theorem for Landaus forth problem but prime numbers properties have been applied for Friedlander–Iwaniec theorem hence no need to prime number theorem or its other forms or extensions.


Theorem: If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ and generators $u,v$ then $F=\{(a^2,b^4)\mid a,b\in\Bbb N\}$ is a group with: $\forall a,b,c,d\in\Bbb N\,$ $\begin{cases} e_F=(1,1)\\ (a^2,b^4)\star_F(c^2,d^4)=((a\star c)^2,(b\star d)^4)\\ (a^2,b^4)^{-1}=((a^{-1})^2,(b^{-1})^4)\qquad\text{that}\quad a\star a^{-1}=1=b\star b^{-1}\\ F=\langle (1,u^4),(u^2,1)\rangle=\langle (1,v^4),(v^2,1)\rangle\end{cases}$


now let $H=\langle\{(a^2,b^4)\mid a,b\in\Bbb N,\,b\neq 1\}\rangle$ and $G=F/H$ is quotient group of $F$ by $H$. ($G$ is a group including prime numbers properties only of the form $1+n^2$.)

and also $L=\{1+n^2\mid n\in\Bbb N\}$ is a cyclic group with: $\forall m,n\in\Bbb N$ $\begin{cases} e_L=2=1+1^2\\ (1+n^2)\star_L(1+m^2)=1+(n\star m)^2\\ (1+n^2)^{-1}=1+(n^{-1})^2\quad\text{that}\;n\star n^{-1}=1\\ L=\langle 1+u^2\rangle=\langle 1+v^2\rangle\end{cases}$

but on the other hand we have: $L\simeq G$ hence we can apply $L$ instead $G$ of course since we are working on natural numbers generally we could consider from the beginning the group $L$ without involvement with the group $G$ anyhow.

Question $1$: For each neutral cyclic group on $\Bbb N$ then what are maximal subgroups of $L$?


Guess $1$: For each cyclic group structure on $\Bbb N$ like $(\Bbb N,\star)$ then for each non-trivial subgroup of $\Bbb N$ like $T$ we have $T\cap\Bbb P\neq\emptyset$.

I think this guess must be proved via prime number theorem.


For each neutral cyclic group on $\Bbb N$ if $L\cap\Bbb P=\{1+n_1^2,1+n_2^2,...,1+n_k^2\},\,k\in\Bbb N$ and if $A=\bigcap _{i=1}^k\langle 1+n_i^2\rangle$ so $\exists m\in\Bbb N$ that $A=\langle 1+m^2\rangle$ & $m\neq n_i$ for $i=1,2,3,...,k$ (intelligibly $k\gt1$) so we have: $A\cap\Bbb P=\emptyset$.

Question $2$: Is $A$ only unique greatest subgroup of $L$ such that $A\cap\Bbb P=\emptyset$?
Alireza Badali 16:49, 28 May 2018 (CEST)

Lemoine's conjecture

Theorem: If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ & generators $u,v$ then $L=\{(p_{n_1},p_{n_2},p_{n_3},-2n-5)\mid n,n_1,n_2,n_3\in\Bbb N,\,p_{n_i}$ is $n_i$_th prime for $i=1,2,3\}$ is an Abelian group with: $\forall n_1,n_2,n_3,n,m_1,m_2,m_3,m\in\Bbb N$ $\begin{cases} e_L=(2,2,2,-7)=(2,2,2,-2\times 1-5)\\ \\(p_{n_1},p_{n_2},p_{n_3},-2n-5)\star_L(p_{m_1},p_{m_2},p_{m_3},-2m-5)=(p_{n_1\star m_1},p_{n_2\star m_2},p_{n_3\star m_3},-2\times(n\star m)-5)\\ \\(p_{n_1},p_{n_2},p_{n_3},-2n-5)^{-1}=(p_{n_1^{-1}},p_{n^{-1}_2},p_{n_3^{-1}},-2\times n^{-1}-5)\quad\text{that}\\ n_1\star n_1^{-1}=1=n_2\star n_2^{-1}=n_3\star n_3^{-1}=n\star n^{-1}\\ \\L=\langle(p_u,2,2,-7),(2,p_u,2,-7),(2,2,p_u,-7),(2,2,2,-2u-5)\rangle=\\\langle(p_v,2,2,-7),(2,p_v,2,-7),(2,2,p_v,-7),(2,2,2,-2v-5)\rangle\end{cases}$•


Theorem: $\forall n\in\Bbb N,\,\exists (p_{m_1},p_{m_2},p_{m_3},-2n-5)\in(L,\star_L)$ such that $p_{m_1}+p_{m_2}+p_{m_3}-2n-5=0$.

Proof using Goldbach's weak conjecture.


Question: Is $L_0=\{(p_{m_1},p_{m_2},p_{m_2},-2n-5)\mid\forall m_1,m_2\in\Bbb N,\,\exists n\in\Bbb N,$ such that $p_{m_1}+2p_{m_2}-2n-5=0\}$ a subgroup of $L$?

Alireza Badali 19:30, 3 June 2018 (CEST)

Primes with beatty sequences

How can we understand $\infty$? we humans only can think on natural numbers and other issues are only theorizing, algebraic theories can be some features for this aim.


Conjecture: If $r$ is an irrational number and $1\lt r\lt 2$, then there are infinitely many primes in the set $L=\{\text{floor}(n\cdot r)\mid n\in\Bbb N\}$.


Theorem $1$: If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ & generators $u,v$ and $r\in[1,2]\setminus\Bbb Q$ then $L=\{\lfloor n\cdot r\rfloor\mid n\in\Bbb N\}$ is another cyclic group with: $\forall m,n\in\Bbb N$ $\begin{cases} e_L=1\\ \lfloor n\cdot r\rfloor\star_L\lfloor m\cdot r\rfloor=\lfloor (n\star m)\cdot r\rfloor\\ (\lfloor n\cdot r\rfloor)^{-1}=\lfloor n^{-1}\cdot r\rfloor\qquad\text{that}\quad n\star n^{-1}=1\\ L=\langle\lfloor u\cdot r\rfloor\rangle=\langle\lfloor v\cdot r\rfloor\rangle\end{cases}$.

Guess $1$: $\prod_{n=1}^{\infty}\lfloor n\cdot r\rfloor=\lfloor 1\cdot r\rfloor\star\lfloor 2\cdot r\rfloor\star\lfloor 3\cdot r\rfloor\star...\in\Bbb N$.


The conjecture generalized: if $r$ is a positive irrational number and $h$ is a real number, then each of the sets $\{\text{floor}(n\cdot r+h)\mid n\in\Bbb N\}$, $\{\text{round}(n\cdot r+h)\mid n\in\Bbb N\}$, and $\{\text{ceiling}(n\cdot r+h)\mid n\in\Bbb N\}$ contains infinitely many primes.


Theorem $2$: If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ & generators $u,v$ & $r$ is a positive irrational number & $h\in\Bbb R$ then $G=\{n\cdot r+h\mid n\in\Bbb N\}$ is another cyclic group with: $\forall m,n\in\Bbb N$ $\begin{cases} e_G=\lfloor r+h\rfloor\\ \lfloor n\cdot r+h\rfloor\star_G\lfloor m\cdot r+h\rfloor=\lfloor (n\star m)\cdot r+h\rfloor\\ (\lfloor n\cdot r+h\rfloor)^{-1}=\lfloor n^{-1}\cdot r+h\rfloor\qquad\text{that}\quad n\star n^{-1}=1\\ L=\langle\lfloor u\cdot r+h\rfloor\rangle=\langle\lfloor v\cdot r+h\rfloor\rangle\end{cases}$.

Guess $2$: $\prod_{n=k}^{\infty}\lfloor n\cdot r+h\rfloor=\lfloor k\cdot r+h\rfloor\star\lfloor (k+1)\cdot r+h\rfloor\star\lfloor (k+2)\cdot r+h\rfloor\star...\in\Bbb N$ in which $\lfloor k\cdot r+h\rfloor\in\Bbb N$ & $\lfloor (k-1)\cdot r+h\rfloor\lt1$.
Alireza Badali 19:09, 7 June 2018 (CEST)

Conjectures depending on the new definitions of primes

Algebraic analytical number theory


A problem: For each cyclic group on $\Bbb N$ like $(\Bbb N,\star)$ find a new definition of prime numbers matching with the operation $\star$ in the group $(\Bbb N,\star)$.


$\Bbb N$ is a cyclic group by: $\begin{cases} \forall m,n\in\Bbb N\\ n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\\ (\Bbb N,\star)=\langle2\rangle=\langle3\rangle\simeq(\Bbb Z,+)\end{cases}$

in the group $(\Bbb Z,+)$ an element $p\gt 1$ is a prime iff don't exist $m,n\in\Bbb Z$ such that $p=m\times n$ & $m,n\gt1$ for instance since $12=4\times3=3+3+3+3$ then $12$ isn't a prime but $13$ is a prime, now inherently must exists an equivalent definition for prime numbers in the $(\Bbb N,\star)$.

prime number isn't an algebraic concept so we can not define primes by using isomorphism (and via algebraic equations primes can be defined) but since Gaussian integers contain all numbers of the form $m+ni,$ $m,n\in\Bbb N$ hence by using algebraic concepts we can solve some problems in number theory.

Question: what is definition of prime numbers in the $(\Bbb N,\star)$?
Alireza Badali 00:49, 25 June 2018 (CEST)

Gaussian moat problem

Alireza Badali 18:13, 20 June 2018 (CEST)

Grimm's conjecture

Alireza Badali 18:13, 20 June 2018 (CEST)

Oppermann's conjecture

Alireza Badali 18:13, 20 June 2018 (CEST)

Legendre's conjecture

Alireza Badali 18:13, 20 June 2018 (CEST)

Conjectures depending on the ring theory

Algebraic analytical number theory


An algorithm which makes new integral domains on $\Bbb N$: Let $(\Bbb N,\star,\circ)$ be that integral domain then identity element $i$ will be corresponding with $1$ and multiplication of natural numbers will be obtained from multiplication of integers corresponding with natural numbers and of course each natural number like $m$ multiplied by a natural number corresponding with $-1$ will be $-m$ such that $m\star(-m)=1$ & $1\circ m=1$.


for instance $(\Bbb N,\star,\circ)$ is an integral domain with: $\begin{cases} \forall m,n\in\Bbb N\\ n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\\1\circ m=1\\ 2\circ m=m\\ 3\circ m=-m\qquad\text{that}\quad m\star (-m)=1\\ (2n)\circ(2m)=2mn\\ (2n+1)\circ(2m+1)=2mn\\ (2n)\circ(2m+1)=2mn+1\end{cases}$

Question $1$: Is $(\Bbb N,\star,\circ)$ an unique factorization domain or the same UFD? what are irreducible elements in $(\Bbb N,\star,\circ)$?


Question $2$: How can we make a UFD on $\Bbb N$?


Question $3$: Under usual total order on $\Bbb N$, do there exist any integral domain $(\Bbb N,\star,\circ)$ and an Euclidean valuation $v:\Bbb N\setminus\{1\}\to\Bbb N$ such that $(\Bbb N,\star,\circ,v)$ is an Euclidean domain? no.

Guess $1$: For each integral domain $(\Bbb N,\star,\circ)$ there exists a total order on $\Bbb N$ and an Euclidean valuation $v:\Bbb N\setminus\{1\}\to\Bbb N$ such that $(\Bbb N,\star,\circ,v)$ is an Euclidean domain.


Professor Jeffrey Clark Lagarias advised me that you can apply group structure on $\Bbb N\cup\{0\}$ instead only $\Bbb N$ and now I see his plan is useful on the field theory, now suppose we apply two algorithms above on $\Bbb N\cup\{0\}$ hence we will have identity element for the group $(\Bbb N,\star)$ of the first algorithm is $0$ corresponding with $0$.

Problem $1$: If $(\Bbb N\cup\{0\},\star,\circ)$ is a UFD then what are irreducible elements in $(\Bbb N\cup\{0\},\star,\circ)$ and is $(\Bbb Q^{\ge0},\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb N\cup\{0\},\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\ {m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={-m\over n}\qquad m\star(-m)=0\end{cases}$•
Algebraic theories on positive numbers help us to solve some open problems depending on the positive numbers.


Question $4$: Is $(\Bbb N\cup\{0\},\star,\circ)$ a UFD by: $\begin{cases} \forall m,n\in\Bbb N\\ e=0\\ (2m-1)\star(2m)=0\\ (2m)\star(2n)=2m+2n\\ (2m-1)\star(2n-1)=2m+2n-1\\ (2m)\star(2n-1)=\begin{cases} 2m-2n & 2m\gt 2n-1\\ 2n-2m-1 & 2n-1\gt 2m\end{cases}\\i=1\\ 0\circ m=0\\ 2\circ m=-m\quad m\star(-m)=0\\ (2m)\circ(2n)=2mn-1\\ (2m-1)\circ(2n-1)=2mn-1\\ (2m)\circ(2n-1)=2mn\end{cases}$

and what are irreducible elements in $(\Bbb N\cup\{0\},\star,\circ)$ and also is $(\Bbb Q^{\ge0},\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb N\cup\{0\},\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={-m\over n}\qquad m\star(-m)=0\end{cases}$ •


Conjecture $1$: Let $x$ be a positive real number, and let $\pi(x)$ denote the number of primes that are less than or equal to $x$ then $$\lim_{x\to\infty}\frac{x-\pi(x)}{\pi(e^u)}=1,\quad u=\sqrt{2\log(x\log x-x)}\,.$$

Answer given by $@$Jan-ChristophSchlage-Puchta from stackexchange site: The conjecture is obviously wrong. The numerator is at least $x/2$, the denominator is at most $e^u$, and $u\lt2\sqrt\log x$, so the limit is $\infty$.
Problem $2$: Find a function $f:\Bbb R\to\Bbb R$ such that $\lim_{x\to\infty}\frac{x-\pi(x)}{\pi(f(x))}=1$.
Prime number theorem and its extensions or algebraic forms or corollaries allow us via infinity concept reach to some results.
Prime numbers properties are stock in whole natural numbers including $\infty$ and not in any finite subset of $\Bbb N$ hence we can know them only in $\infty$, which prime number theorem prepares it, but what does mean a cognition of prime numbers I think according to the distribution of prime numbers, a cognition means only in $\infty$, this function $f$ can be such a cognition but only in $\infty$ because we have: $$\lim_{x\to\infty}\frac{x-\pi(x)}{\pi(f(x))}=1=\lim_{x\to\infty}\frac{f(x)-\pi(f(x))}{\pi(f(f(x)))}=\lim_{x\to\infty}\frac{f(f(x))-\pi(f(f(x)))}{\pi(f(f(f(x))))}=...$$ and I guess $f$ is to form of $e^{g(x)}$ in which $g:\Bbb R\to\Bbb R$ is a radical logarithmic function or probably as a radical logarithmic series.
Conjecture $2$: Let $h:\Bbb R\to\Bbb R,\,h(x)=\frac{f(x)}{(\log x-1)\log(f(x))}$ then $\lim_{x\to\infty}{\pi(x)\over h(x)}=1$.
Answer given by $@$Wojowu from stackexchange site: Since $x−\pi(x)\sim x$, you want $\pi(f(x))\sim x$, and $f(x)=x\log x$ works, and let $u=\log(x\log x)$.
Problem $3$: Based on prime number theorem very large prime numbers are equivalent to the numbers of the form $n\cdot\log n,\,n\in\Bbb N$ hence I think a test could be made to check correctness of some conjectures or problems relating to the prime numbers, and maybe some functions such as $h$ prepares it!
Question $5$: If $p_n$ is $n$_th prime number then does $$\lim_{n\to\infty}\frac{p_n}{e^{\sqrt{2\log n}}\over (\log n-1)\sqrt{2\log n}}=1\,?$$
Answer given by $@$ToddTrimble from stackexchange site: The numerator is asymptotically greater than $n$, and the denominator is asymptotically less.
Alireza Badali 16:26, 26 June 2018 (CEST)

Parallel universes

An algorithm that makes new cyclic groups on $\Bbb Z$: Let $(\Bbb Z,\star)$ be that group and at first write integers as a sequence with starting from $0$ and then write integers with a fixed sequence below it, and let identity element $e=0$ be corresponding with $0$ and two generators $m$ & $n$ be corresponding with $1$ & $-1$, so we have $(\Bbb Z,\star)=\langle m\rangle=\langle n\rangle$ for instance: $$0,1,2,-2,-1,3,4,-4,-3,5,6,-6,-5,7,8,-8,-7,9,10,-10,-9,11,12,-12,-11,13,14,-14,-13,...$$ $$0,1,-1,2,-2,3,-3,4,-4,5,-5,6,-6,7,-7,8,-8,9,-9,10,-10,11,-11,12,-12,13,-13,14,-14,...$$ then regarding the sequence above find an even rotation number that for this sequence is $4$ (or $2k$) and hence equations should be written with module $2$ (or $k$) then consider $2m-1,2m,-2m+1,-2m$ (that general form is: $km,km-1,km-2,...,km-(k-1),-km,-km+1,-km+2,...,-km+(k-1)$) and make a table of products of those $4$ (or $2k$) elements but during writing equations pay attention if an equation is right for given numbers it will be right generally for other numbers too and of course if integers corresponding with two numbers don't have same signs then product will be a piecewise-defined function for example $7\star(-10)=2$ $=(2\times4-1)\star(-2\times5)$ because $7+(-9)=-2,\,7\to7,\,-9\to-10,\,-2\to2$ that implies $(2m-1)\star(-2n)=2n-2m$ where $2n\gt 2m-1$, of course it is better at first members inverse be defined for example since $7+(-7)=0,\,7\to7,\,-7\to-8$ so $7\star(-8)=0$ that shows $(2m-1)\star(-2m)=0$ and with a little bit addition and multiplication all equations will be obtained simply that for this example is:

$\begin{cases} \forall t\in\Bbb Z,\quad t\star0=t\\ \forall m,n\in\Bbb N\\ (2m-1)\star(-2m)=0=(-2m+1)\star(2m)\\ (2m-1)\star(2n-1)=2m+2n-2\\ (2m-1)\star(2n)=\begin{cases} 2m-2n-1 & 2m-1\gt2n\\ 2m-2n-2 & 2n\gt 2m-1\end{cases}\\ (2m-1)\star(-2n+1)=2m+2n-1\\ (2m-1)\star(-2n)=\begin{cases} 2n-2m+1 & 2m-1\gt2n\\ 2n-2m & 2n\gt2m-1\end{cases}\\ (2m)\star(2n)=2m+2n\\ (2m)\star(-2n+1)=\begin{cases} 2m-2n+1 & 2n-1\gt2m\\ 2m-2n & 2m\gt2n-1\end{cases}\\ (2m)\star(-2n)=-2m-2n\\ (-2m+1)\star(-2n+1)=-2m-2n+1\\ (-2m+1)\star(-2n)=\begin{cases} 2m-2n+1 & 2m-1\gt2n\\ 2m-2n & 2n\gt2m-1\\ 1 & m=n\end{cases}\\ (-2m)\star(-2n)=2m+2n-2\\ \Bbb Z=\langle1\rangle=\langle-2\rangle\end{cases}$


An algorithm which makes new integral domains on $\Bbb Z$: Let $(\Bbb Z,\star,\circ)$ be that integral domain then identity element $i$ will be corresponding with $1$ and multiplication of integers will be obtained from multiplication of corresponding integers such that if $t:\Bbb Z\to\Bbb Z$ is a bijection that images top row on to bottom row respectively for instance in example above is seen $t(2)=-1$ & $t(-18)=18$ then we can write laws by using $t$ such as $(-2m+1)\circ(-2n)=$ $t(t^{-1}(-2m+1)\times t^{-1}(-2n))=t((2m)\times(-2n+1))=t(-2\times(2mn-m))=$ $2\times(2mn-m)=4mn-2m$ and of course each integer like $m$ multiplied by an integer corresponding with $-1$ will be $n$ such that $m\star n=0$ & $0\circ m=0$ for instance $(\Bbb Z,\star,\circ)$ is an integral domain with:

$\begin{cases} \forall t\in\Bbb Z,\quad t\star0=t\\ \forall m,n\in\Bbb N\\ (2m-1)\star(-2m)=0=(-2m+1)\star(2m)\\ (2m-1)\star(2n-1)=2m+2n-2\\ (2m-1)\star(2n)=\begin{cases} 2m-2n-1 & 2m-1\gt2n\\ 2m-2n-2 & 2n\gt 2m-1\end{cases}\\ (2m-1)\star(-2n+1)=2m+2n-1\\ (2m-1)\star(-2n)=\begin{cases} 2n-2m+1 & 2m-1\gt2n\\ 2n-2m & 2n\gt2m-1\end{cases}\\ (2m)\star(2n)=2m+2n\\ (2m)\star(-2n+1)=\begin{cases} 2m-2n+1 & 2n-1\gt2m\\ 2m-2n & 2m\gt2n-1\end{cases}\\ (2m)\star(-2n)=-2m-2n\\ (-2m+1)\star(-2n+1)=-2m-2n+1\\ (-2m+1)\star(-2n)=\begin{cases} 2m-2n+1 & 2m-1\gt2n\\ 2m-2n & 2n\gt2m-1\\ 1 & m=n\end{cases}\\ (-2m)\star(-2n)=2m+2n-2\\ i=t(1)=1,\quad0\circ m=0,\quad m\star(t(-1)\circ m)=m\star(-2\circ m)=0\\ (2m-1)\circ(2n-1)=4mn-2m-2n+1\\ (2m-1)\circ(2n)=4mn-2n\\ (2m-1)\circ(-2n+1)=-4mn+2n+1\\ (2m-1)\circ(-2n)=-4mn+2m+2n-2\\ (2m)\circ(2n)=-4mn+1\\ (2m)\circ(-2n+1)=4mn\\ (2m)\circ(-2n)=-4mn+2m+1\\ (-2m+1)\circ(-2n+1)=-4mn+1\\ (-2m+1)\circ(-2n)=4mn-2m\\ (-2m)\circ(-2n)=4mn-2m-2n+1\end{cases}$

Question $1$: Is $(\Bbb Z,\star,\circ)$ a UFD? what are irreducible elements in $(\Bbb Z,\star,\circ)$? is $(\Bbb Q,\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$ •


Problem $1$: If $(\Bbb Z,\star,\circ)$ is a UFD then what are irreducible elements in $(\Bbb Z,\star,\circ)$ and is $(\Bbb Q,\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\ {m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$•


Question $2$: Under usual total order on $\Bbb Z$, do there exist any integral domain $(\Bbb Z,\star,\circ)$ and an Euclidean valuation $v:\Bbb Z\setminus\{0\}\to\Bbb N$ such that $(\Bbb Z,\star,\circ,v)$ is an Euclidean domain? no.

Guess $1$: For each integral domain $(\Bbb Z,\star,\circ)$ there exists a total order on $\Bbb Z$ and an Euclidean valuation $v:\Bbb Z\setminus\{0\}\to\Bbb N$ such that $(\Bbb Z,\star,\circ,v)$ is an Euclidean domain.

Alireza Badali 20:32, 9 July 2018 (CEST)

Gauss circle problem

Given this sequence: $$0,1,-1,2,-2,3,-3,4,-4,5,-5,6,-6,7,-7,8,-8,9,-9,10,-10,11,-11,12,-12,...$$ $$0,1,2,3,-3,-1,-2,4,5,6,-6,-4,-5,7,8,9,-9,-7,-8,10,11,12,-12,-10,-11,...$$ we have this integral domain $(\Bbb Z,\star,\circ)$ as:

$\begin{cases} \forall t\in\Bbb Z,\quad t\star0=t,\quad\forall m,n\in\Bbb N,\\ (6m-5)\star(6m-4)=0=(6m-3)\star(-6m+3)=(-6m+5)\star(-6m+4)=(6m-2)\star(6m-1)=\\ =(6m)\star(-6m)=(-6m+2)\star(-6m+1)\\ (6m-5)\star(6n-5)=6m+6n-9\\ (6m-5)\star(6n-4)=\begin{cases} 6n-6m+2 & 6m-5\gt6n-4\\ 6m-6n+1 & 6n-4\gt6m-5\end{cases}\\ (6m-5)\star(6n-3)=-6m-6n+11\\ (6m-5)\star(6n-2)=6m+6n-6\\ (6m-5)\star(6n-1)=\begin{cases} 6n-6m+5 & 6m-5\gt6n-1\\ 6m-6n-2 & 6n-1\gt6m-5\end{cases}\\ (6m-5)\star(6n)=-6m-6n+8\\ (6m-5)\star(-6n+5)=6m+6n-8\\ (6m-5)\star(-6n+4)=\begin{cases} 6m-6n-2 & 6m-5\gt6n-4\\ 6m-6n-3 & 6n-4\gt6m-5\end{cases}\\ (6m-5)\star(-6n+3)=\begin{cases} 6m-6n & 6m-5\gt6n-3\\ 6n-6m+2 & 6n-3\gt6m-5\end{cases}\\ (6m-5)\star(-6n+2)=6m+6n-5\\ (6m-5)\star(-6n+1)=\begin{cases} 6m-6n-5 & 6m-5\gt6n-1\\ 6m-6n-6 & 6n-1\gt6m-5\end{cases}\\ (6m-5)\star(-6n)=\begin{cases} 6m-6n-3 & 6m-5\gt6n\\ 6n-6m+5 & 6n\gt6m-5\end{cases}\\ (6m-4)\star(6n-4)=-6m-6n+9\\ (6m-4)\star(6n-3)=\begin{cases} 6n-6m & 6m-4\gt6n-3\\ 6n-6m+1 & 6n-3\gt6m-4\end{cases}\\ (6m-4)\star(6n-2)=\begin{cases} 6n-6m+4 & 6m-4\gt6n-2\\ 6m-6n-1 & 6n-2\gt6m-4\end{cases}\\ (6m-4)\star(6n-1)=-6m-6n+6\\ (6m-4)\star(6n)=\begin{cases} 6n-6m+3 & 6m-4\gt6n\\ 6n-6m+4 & 6n\gt6m-4\end{cases}\\ (6m-4)\star(-6n+5)=\begin{cases} 6m-6n-1 & 6m-4\gt6n-5\\ 6n-6m+3 & 6n-5\gt6m-4\\ 3 & m=n\end{cases}\\ (6m-4)\star(-6n+4)=6m+6n-7\\ (6m-4)\star(-6n+3)=-6m-6n+10\\ (6m-4)\star(-6n+2)=\begin{cases} 6m-6n-4 & 6m-4\gt6n-2\\ 6n-6m+6 & 6n-2\gt6m-4\end{cases}\\ (6m-4)\star(-6n+1)=6m+6n-4\\ (6m-4)\star(-6n)=-6m-6n+7\\ (6m-3)\star(6n-3)=6m+6n-8\\ (6m-3)\star(6n-2)=-6m-6n+8\\ (6m-3)\star(6n-1)=\begin{cases} 6m-6n-2 & 6m-3\gt6n-1\\ 6m-6n-3 & 6n-1\gt6m-3\end{cases}\\ (6m-3)\star(6n)=6m+6n-5\\ (6m-3)\star(-6n+5)=6m+6n-6\\ (6m-3)\star(-6n+4)=\begin{cases} 6m-6n & 6m-3\gt6n-4\\ 6n-6m+2 & 6n-4\gt6m-3\\ 2 & m=n\end{cases}\\ (6m-3)\star(-6n+3)=\begin{cases} 6n-6m+2 & 6m-3\gt6n-3\\ 6m-6n+1 & 6n-3\gt6m-3\end{cases}\\ (6m-3)\star(-6n+2)=6m+6n-3\\ (6m-3)\star(-6n+1)=\begin{cases} 6m-6n-3 & 6m-3\gt6n-1\\ 6n-6m+5 & 6n-1\gt6m-3\end{cases}\\ (6m-3)\star(-6n)=\begin{cases} 6n-6m+5 & 6m-3\gt6n\\ 6m-6n-2 & 6n\gt6m-3\end{cases}\\ (6m-2)\star(6n-2)=6m+6n-3\\ (6m-2)\star(6n-1)=\begin{cases} 6n-6m+2 & 6m-2\gt6n-1\\ 6m-6n+1 & 6n-1\gt6m-2\end{cases}\\ (6m-2)\star(6n)=-6m-6n+5\\ (6m-2)\star(-6n+5)=6m+6n-5\\ (6m-2)\star(-6n+4)=\begin{cases} 6m-6n+1 & 6m-2\gt6n-4\\ 6m-6n & 6n-4\gt6m-2\end{cases}\\ (6m-2)\star(-6n+3)=\begin{cases} 6m-6n+3 & 6m-2\gt6n-3\\ 6n-6m-1 & 6n-3\gt6m-2\end{cases}\\ (6m-2)\star(-6n+2)=6m+6n-2\\ (6m-2)\star(-6n+1)=\begin{cases} 6m-6n-2 & 6m-2\gt6n-1\\ 6m-6n-3 & 6n-1\gt6m-2\end{cases}\\ (6m-2)\star(-6n)=\begin{cases} 6m-6n & 6m-2\gt6n\\ 6n-6m+2 & 6n\gt6m-2\end{cases}\\ (6m-1)\star(6n-1)=-6m-6n+3\\ (6m-1)\star(6n)=\begin{cases} 6n-6m & 6m-1\gt6n\\ 6n-6m+1 & 6n\gt6m-1\end{cases}\\ (6m-1)\star(-6n+5)=\begin{cases} 6m-6n+2 & 6m-1\gt6n-5\\ 6n-6m & 6n-5\gt6m-1\end{cases}\\ (6m-1)\star(-6n+4)=6m+6n-4\\ (6m-1)\star(-6n+3)=-6m-6n+7\\ (6m-1)\star(-6n+2)=\begin{cases} 6m-6n-1 & 6m-1\gt6n-2\\ 6n-6m+3 & 6n-2\gt6m-1\\ 3 & m=n\end{cases}\\ (6m-1)\star(-6n+1)=6m+6n-1\\ (6m-1)\star(-6n)=-6m-6n+4\\ (6m)\star(6n)=6m+6n-2\\ (6m)\star(-6n+5)=6m+6n-3\\ (6m)\star(-6n+4)=\begin{cases} 6m-6n+3 & 6m\gt6n-4\\ 6n-6m-1 & 6n-4\gt6m\end{cases}\\ (6m)\star(-6n+3)=\begin{cases} 6n-6m-1 & 6m\gt6n-3\\ 6m-6n+4 & 6n-3\gt6m\end{cases}\\ (6m)\star(-6n+2)=6m+6n\\ (6m)\star(-6n+1)=\begin{cases} 6m-6n & 6m\gt6n-1\\ 6m-6n-1 & 6n-1\gt6m\\ 2 & m=n\end{cases}\\ (6m)\star(-6n)=\begin{cases} 6n-6m+2 & 6m\gt6n\\ 6m-6n+1 & 6n\gt6m\end{cases}\\ (-6m+5)\star(-6n+5)=-6m-6n+8\\ (-6m+5)\star(-6n+4)=\begin{cases} 6n-6m+2 & 6m-5\gt6n-4\\ 6m-6n+1 & 6n-4\gt6m-5\end{cases}\\ (-6m+5)\star(-6n+3)=\begin{cases} 6m-6n+1 & 6m-5\gt6n-3\\ 6m-6n & 6n-3\gt6m-5\\ 1 & m=n\end{cases}\\ (-6m+5)\star(-6n+2)=-6m-6n+5\\ (-6m+5)\star(-6n+1)=\begin{cases} 6n-6m+5 & 6m-5\gt6n-1\\ 6m-6n-2 & 6n-1\gt6m-5\end{cases}\\ (-6m+5)\star(-6n)=\begin{cases} 6m-6n-2 & 6m-5\gt6n\\ 6m-6n-3 & 6n\gt6m-5\end{cases}\\ (-6m+4)\star(-6n+4)=-6m-6n+7\\ (-6m+4)\star(-6n+3)=-6m-6n+6\\ (-6m+4)\star(-6n+2)=\begin{cases} 6n-6m+4 & 6m-4\gt6n-2\\ 6m-6n-1 & 6n-2\gt6m-4\end{cases}\\ (-6m+4)\star(-6n+1)=-6m-6n+4\\ (-6m+4)\star(-6n)=-6m-6n+3\\ (-6m+3)\star(-6n+3)=6m+6n-7\\ (-6m+3)\star(-6n+2)=\begin{cases} 6n-6m+3 & 6m-3\gt6n-2\\ 6n-6m+4 & 6n-2\gt6m-3\end{cases}\\ (-6m+3)\star(-6n+1)=-6m-6n+3\\ (-6m+3)\star(-6n)=6m+6n-4\\ (-6m+2)\star(-6n+2)=-6m-6n+2\\ (-6m+2)\star(-6n+1)=\begin{cases} 6n-6m+2 & 6m-2\gt6n-1\\ 6m-6n+1 & 6n-1\gt6m-2\end{cases}\\ (-6m+2)\star(-6n)=\begin{cases} 6m-6n+1 & 6m-2\gt6n\\ 6m-6n & 6n\gt6m-2\\ 1 & m=n\end{cases}\\ (-6m+1)\star(-6n+1)=-6m-6n+1\\ (-6m+1)\star(-6n)=-6m-6n\\ (-6m)\star(-6n)=6m+6n-1\\ t\circ1=t,\quad t\circ0=0,\quad t\star(2\circ t)=0\end{cases}$

$\begin{cases} (6m-5)\circ(6n-5)=36mn-30m-30n+25\\ (6m-5)\circ(6n-4)=36mn-30m-30n+26\\ (6m-5)\circ(6n-3)=36mn-24m-30n+21\\ (6m-5)\circ(6n-2)=36mn-12m-30n+10\\ (6m-5)\circ(6n-1)=36mn-12m-30n+11\\ (6m-5)\circ(6n)=36mn-6m-30n+6\\ (6m-5)\circ(-6n+5)=-36mn+18m+30n-13\\ (6m-5)\circ(-6n+4)=-36mn+18m+30n-14\\ (6m-5)\circ(-6n+3)=-36mn+24m+30n-21\\ (6m-5)\circ(-6n+2)=-36mn+30n+2\\ (6m-5)\circ(-6n+1)=-36mn+30n+1\\ (6m-5)\circ(-6n)=-36mn+6m+30n-6\\ (6m-4)\circ(6n-4)=36mn-30m-30n+25\\ (6m-4)\circ(6n-3)=-36mn+24m+30n-21\\ (6m-4)\circ(6n-2)=36mn-12m-30n+11\\ (6m-4)\circ(6n-1)=36mn-12m-30n+10\\ (6m-4)\circ(6n)=-36mn+6m+30n-6\\ (6m-4)\circ(-6n+5)=-36mn+18m+30n-14\\ (6m-4)\circ(-6n+4)=-36mn+18m+30n-13\\ (6m-4)\circ(-6n+3)=36mn-24m-30n+21\\ (6m-4)\circ(-6n+2)=-36mn+30n+1\\ (6m-4)\circ(-6n+1)=-36mn+30n+2\\ (6m-4)\circ(-6n)=36mn-6m-30n+6\\ (6m-3)\circ(6n-3)=36mn-24m-24n+16\\ (6m-3)\circ(6n-2)=36mn-12m-24n+9\\ (6m-3)\circ(6n-1)=-36mn+12m+24n-9\\ (6m-3)\circ(6n)=36mn-6m-24n+4\\ (6m-3)\circ(-6n+5)=-36mn+18m+24n-10\\ (6m-3)\circ(-6n+4)=-36mn+18m+24n-11\\ (6m-3)\circ(-6n+3)=36mn-24m-24n+17\\ (6m-3)\circ(-6n+2)=-36mn+24n+2\\ (6m-3)\circ(-6n+1)=-36mn+24n+1\\ (6m-3)\circ(-6n)=36mn-6m-24n+5\\ (6m-2)\circ(6n-2)=36mn-12m-12n+4\\ (6m-2)\circ(6n-1)=36mn-12m-12n+5\\ (6m-2)\circ(6n)=36mn-6m-12n+3\\ (6m-2)\circ(-6n+5)=-36mn+18m+12n-4\\ (6m-2)\circ(-6n+4)=-36mn+18m+12n-5\\ (6m-2)\circ(-6n+3)=-36mn+24m+12n-9\\ (6m-2)\circ(-6n+2)=-36mn+12n+2\\ (6m-2)\circ(-6n+1)=-36mn+12n+1\\ (6m-2)\circ(-6n)=-36mn+6m+12n-3\\ (6m-1)\circ(6n-1)=36mn-12m-12n+4\\ (6m-1)\circ(6n)=-36mn+6m+12n-3\\ (6m-1)\circ(-6n+5)=-36mn+18m+12n-5\\ (6m-1)\circ(-6n+4)=-36mn+18m+12n-4\\ (6m-1)\circ(-6n+3)=36mn-24m-12n+9\\ (6m-1)\circ(-6n+2)=-36mn+12n+1\\ (6m-1)\circ(-6n+1)=-36mn+12n+2\\ (6m-1)\circ(-6n)=36mn-6m-12n+3\\ (6m)\circ(6n)=36mn-6m-6n+1\\ (6m)\circ(-6n+5)=-36mn+18m+6n-1\\ (6m)\circ(-6n+4)=-36mn+18m+6n-2\\ (6m)\circ(-6n+3)=36mn-24m-6n+5\\ (6m)\circ(-6n+2)=-36mn+6n+2\\ (6m)\circ(-6n+1)=-36mn+6n+1\\ (6m)\circ(-6n)=36mn-6m-6n+2\\ (-6m+5)\circ(-6n+5)=-36mn+18m+18n-7\\ (-6m+5)\circ(-6n+4)=-36mn+18m+18n-8\\ (-6m+5)\circ(-6n+3)=-36mn+24m+18n-11\\ (-6m+5)\circ(-6n+2)=-36mn+18n+2\\ (-6m+5)\circ(-6n+1)=-36mn+18n+1\\ (-6m+5)\circ(-6n)=-36mn+6m+18n-2\\ (-6m+4)\circ(-6n+4)=-36mn+18m+18n-7\\ (-6m+4)\circ(-6n+3)=-36mn+24m+18n-10\\ (-6m+4)\circ(-6n+2)=-36mn+18n+1\\ (-6m+4)\circ(-6n+1)=-36mn+18n+2\\ (-6m+4)\circ(-6n)=-36mn+6m+18n-1\\ (-6m+3)\circ(-6n+3)=36mn-24m-24n+16\\ (-6m+3)\circ(-6n+2)=-36mn+24n+1\\ (-6m+3)\circ(-6n+1)=-36mn+24n+2\\ (-6m+3)\circ(-6n)=36mn-6m-24n+4\\ (-6m+2)\circ(-6n+2)=-36mn+2\\ (-6m+2)\circ(-6n+1)=-36mn+1\\ (-6m+2)\circ(-6n)=-36mn+6m+1\\ (-6m+1)\circ(-6n+1)=-36mn+2\\ (-6m+1)\circ(-6n)=-36mn+6m+2\\ (-6m)\circ(-6n)=36mn-6m-6n+1\end{cases}$

but some calculations are such as: $(6m-4)\circ(6n-3)=t(t^{-1}(6m-4)\times t^{-1}(6n-3))=t((-6m+5)\times(6n-4))=$ $t(-6(6mn-4m-5n+4)+4)=-6(6mn-4m-5n+4)+3=-36mn+24m+30n-21,$ $(6m-4)\circ(-6n+1)=$ $t(t^{-1}(6m-4)\times t^{-1}(-6n+1))=t((-6m+5)\times(-6n))=t(6(6mn-5n))=-6(6mn-5n)+2=-36mn+30n+2,$ $(-6m+1)\circ(-6n+1)=t(t^{-1}(-6m+1)\times t^{-1}(-6n+1))=t((-6m)\times(-6n))=t(6(6mn))=-6(6mn)+2=$ $-36mn+2$.

Question $1$: Is $(\Bbb Z,\star,\circ)$ a UFD? what are irreducible elements in $(\Bbb Z,\star,\circ)$? is $(\Bbb Q,\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$


Problem $1$: Reinterpret (possibly via matrices) Gauss circle problem under the field $(\Bbb Q,\star_1,\circ_1)$ (not in whole $\Bbb R$).

Alireza Badali 00:49, 25 June 2018 (CEST)

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Please just insert your comment here! Alireza Badali 20:47, 15 April 2018 (CEST)

How to Cite This Entry:
Musictheory2math. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Musictheory2math&oldid=42831