Namespaces
Variants
Actions

Difference between revisions of "User talk:Musictheory2math"

From Encyclopedia of Mathematics
Jump to: navigation, search
(this is the last revision, so please don't insert any question because I won't be able to respond, and I leave here forever, I hope my theories be helpful. best wishes, Alireza Badali Sarebangholi)
 
(230 intermediate revisions by 2 users not shown)
Line 1: Line 1:
== Goldbach's conjecture ==
+
== <big>$\mathscr B$</big> $theory$ (algebraic topological analytical number theory) ==
  
'''Main theorem''': Let $\mathbb{P}$ be the set prime numbers and $S$ is a set that has been made as below: put a point on 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.
+
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.
  
$\,$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)
 
  
: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)
+
'''Algebraic structures & topology with homotopy groups & prime number theorem and its extensions or other forms or corollaries with limitation concept'''
  
::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 few time about two month for expression my theories. [[User:Musictheory2math|Musictheory2math]] ([[User talk:Musictheory2math|talk]]) 16:29, 25 March 2017 (CET)
+
Alireza Badali 00:49, 25 June 2018 (CEST)
  
: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
+
=== [https://en.wikipedia.org/wiki/Goldbach%27s_conjecture Goldbach's conjecture] ===
\[
 
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)
 
  
::$ \frac{p_{n+1}}{p_n} < \frac b a $ implies $p_{n+i}\lt (\frac b a)^{n+i} p_n$ so $(\frac b a)^{n+i} p_n\to \infty$ where $i\to \infty$ because $\frac b a\gt 1$ so your proof has problem! Alireza Badali 20:18, 15 September 2017 (CEST)
+
'''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 $t\cdot 10^k\in \Bbb P$.
 +
:[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 $\alpha = \frac{t}{s}$.
 +
: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♦
  
  
'''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)
+
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 $u\cdot 10^{m+1}\in\Bbb P$
  
: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)
+
and $g:(0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\to\Bbb N,$ is a function by $\forall\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))$ $g(\epsilon)=max(\{f((c,d))\mid d-c=\epsilon,$ $(c,d)\subseteq [0.01,0.1)\})$.
  
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)
+
'''Guess''' $1$: $g$ isn't an injective function.
  
: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".
+
'''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 $h_n\cap h_m=\emptyset$?
: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)
+
:[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:
 +
:* 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 theorems|Liouville's theorem on approximation of algebraic numbers]].
  
  
Importance of density in the Main theorem is similar to definition of irrational numbers from rational 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 $S=\bigcup _{k\in\Bbb N}w_k(\Bbb P)$.
  
'''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.
+
'''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.
 +
:comment given by [https://mathoverflow.net/users/3402/gerhard-paseman $@$GerhardPaseman] from stackexchange site: There are elementary methods to show your specified set is dense. Indeed, simple sieving methods and estimates known to Euler for the sum of the reciprocals of primes give a weak but for your result a sufficient upper bound on the number of primes less than $n$ (like ${n\over\log\log n}$).
 +
: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 $[0.1,1)$?
  
Assume $S_1=\{a/10^n\, |\, a\in S$ for $n=0,1,2,3,...\}$ & $L=\{(a,b)\,|\,a,b \in S_1$ & $0.01 \le a+b \lt 0.1$ & $\exists m \in \Bbb N,\, a \times 10^m,\,b \times 10^m$ are prime numbers & $a\times 10^m\neq 2\neq b\times 10^m\}$
 
  
 +
'''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''': $S_1$ is dense in the interval $(0,1)$ and $S_1\times S_1$ is dense in the $(0,1)\times (0,1)$.
 
  
 +
'''Theorem''' $2$: $S$ is dense in the interval $[0,1]$ and $S\times S$ 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 $\varphi : L \to \Bbb N,$ $\varphi ((p,q)) = \begin{cases} m+1 & n=m \\m+2 &  n=m+1 \end{cases}$
 
  
 +
'''An algorithm''' that makes new cyclic groups on $\Bbb N$:
  
'''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 $\{(x,x)\,|\,0.005 \le x \lt 0.05 \}$.
+
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 a rotation number of the form $4t,\,t\in\Bbb N$ 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:
  
 +
$\begin{cases} m\star _u 1=m\\ (4m)\star _u (4m-2)=1=(4m+1)\star _u (4m-1)\\ (4m-2)\star _u (4n-2)=4m+4n-5\\ (4m-2)\star _u (4n-1)=4m+4n-2\\ (4m-2)\star _u (4n)=\begin{cases} 4m-4n-1 & 4m-2\gt 4n\\ 4n-4m+1 & 4n\gt 4m-2\\ 3 & m=n+1\end{cases}\\ (4m-2)\star _u (4n+1)=\begin{cases} 4m-4n-2 & 4m-2\gt 4n+1\\ 4n-4m+4 & 4n+1\gt 4m-2\end{cases}\\ (4m-1)\star _u (4n-1)=4m+4n-1\\ (4m-1)\star _u (4n)=\begin{cases} 4m-4n+2 & 4m-1\gt 4n\\ 4n-4m & 4n\gt 4m-1\\ 2 & m=n\end{cases}\\ (4m-1)\star _u (4n+1)=\begin{cases} 4m-4n-1 & 4m-1\gt 4n+1\\ 4n-4m+1 & 4n+1\gt 4m-1\\ 3 & m=n+1\end{cases}\\ (4m)\star _u (4n)=4m+4n-3\\ (4m)\star _u (4n+1)=4m+4n\\ (4m+1)\star _u  (4n+1)=4m+4n+1\\ \Bbb N=\langle 2\rangle=\langle 4\rangle\end{cases}$
  
'''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 $A:=\{(x,y)\,|\, 0 \lt y\lt x,$ $0.01 \le x+y \lt 0.1\}$ & $V:=\{(a+b)\times 10^m \,|$ $(a,b) \in ((S_1 \times S_1) \cap A) \setminus L,$ there is a least member like $m$ in $\Bbb N$ such that $(a+b)\times 10^m \in \Bbb N \}$ & $r:\Bbb N \to (0,1)$ is a mapping given by $r(n)$ is obtained as put a point on the beginning of $n$ like $r(34880)=0.34880$ and similarly consider $\forall k\in\Bbb N,$ $r_k: \Bbb \N \to (0,1)$ by $r_k(n)=10^{-k}\times r(n)$.
 
  
 +
Problem $1$: By using matrices rewrite operation of every group on $\Bbb N$.
  
'''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 $0.0t_1t_2t_3...t_k=a+b$ & $10^{k+1} \times a,10^{k+1} \times b$ are prime numbers.
 
  
:Conjecture $1$ is an equivalent to Goldbach's conjecture, this conjecture has two solutions $1)$ Homotopy groups $\pi _{n \gt 1} (X)$ (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.
+
Assume $\forall m,n\in\Bbb N$: $\begin{cases} n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\end{cases}$
  
:'''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!
+
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 $\langle 2\rangle =\langle 3\rangle =(\Bbb N,\star)\simeq (\Bbb Z,+)\simeq (\Bbb P,\star _1)=\langle 3\rangle=\langle 5\rangle$.
  
::'''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: $(a^2+b^2)^{0.5}\lt 0.0t_1t_2t_3...t_k\le \sqrt 2 \times (a^2+b^2)^{0.5}$ 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>
 
  
 +
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 these techniques are useful:
 +
:$\forall n\in\Bbb N,$ 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 $3$: $\forall (a,b)\subset [0.1,1),\,w_{(a,b)}={10\over9}\cdot(b-a)$.
 +
:::[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.
 +
:::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.
 +
::::Let $e:\Bbb N\to\Bbb N,$ is a function that $\forall n\in\Bbb N$ gives the number of digits in $n$ instance $e(1320)=4$, and let $\forall n\in\Bbb N,$ $\forall k\in\Bbb N\cup\{0\},$ and for each subinterval $(a,b)$ of $[0.1,1),$ that $a\neq b,$ $\begin{cases} A_{k,(a,b)}:=\{n\mid\exists t_1\in\Bbb N,\,\exists t_2\in (a,b),\, t_2\cdot 10^{t_1}\in\Bbb N,\, 10\nmid t_2\cdot10^{t_1},\, n=t_2\cdot 10^{k+t_1}\},\\ \\B_{k,(a,b)}:=\{p\mid\exists t_1\in\Bbb N,\,\exists t_2\in (a,b),\, p=t_2\cdot 10^{t_1}\in\Bbb P,\,\exists n_1,n_2\in A_{k,(a,b)},\, n_1\le p\le n_2,\,e(n_1)=e(n_2)\},\\ \\A_{k,(a,b),n}:=\{m\in A_{k,(a,b)}\mid m\le n\},\\ \\B_{k,(a,b),n}:=\{m\in B_{k,(a,b)}\mid m\le n\},\\ \\c_{k,(a,b),n}:=(\#A_{k,(a,b),n})^{-1}\cdot\#B_{k,(a,b),n}\cdot\log n,\\ \\c_{k,(a,b)}:=\lim _{n\to\infty} c_{k,(a,b),n}\end{cases}$.
 +
:::::Guess $4$: $\forall k\in\Bbb N\cup\{0\},\,\forall (a,b)\subset [0.1,1),\,c_{k,(a,b)}=10^{-k}\cdot (b-a)$. (and we knew $\sum _{k\in\Bbb N\cup\{0\}}10^{-k}={10\over9}$)
 +
::Guess $5$: $\forall (a,b)\subset [0.1,1),\,z_{(a,b)}={10\over9}\cdot(b-a)$.
 +
:::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 $6$: $\forall(a,b),(c,d)\subset[0.1,1),\,\lim_{n\to\infty}{\#V_{(a,b),n}\over\#V_{(c,d),n}}={b-a\over d-c}=\lim_{n\to\infty}{\#U_{(a,b),n}\over\#U_{(c,d),n}}$.
 +
:::<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>
 +
:::<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>
  
'''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 $p=r$ & $q=s$.
 
  
 +
Using homotopy groups Goldbach's conjecture will be proved.
  
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 $f_2 (x)$.
+
Alireza Badali 08:27, 31 March 2018 (CEST)
  
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) $
+
==== Goldbach by $\Bbb N$ ====
  
 +
Let $\lt_1$ be a total order relation (not well ordering) on $\Bbb N$ as: $\forall m,n\in\Bbb N,\,m\lt_1n$ iff
  
'''Guess''' $1$: $f_3 (L_1) $ is dense in the $S^1 \cap \{ (x,y)\,|\, 0 \le y$ & $2^{-0.5} \le x \}$.
+
$\begin{cases} r(m)\lt r(n),\,m=m_1\times10^s,\,n=n_1\times10^s,\,10\nmid m_1,\,10\nmid n_1,\,m_1,n_1\in\Bbb N,\,s\in\Bbb N\cup\{0\} & \text{ or}\\ \\m=m_1\times10^s,\,n=n_1\times10^t,\,s\lt t,\,10\nmid m_1,\,10\nmid n_1,\,m_1,n_1,t\in\Bbb N,\,s\in\Bbb N\cup\{0\}\end{cases}$
  
 +
then assume $\mathfrak T$ is a topology on $\Bbb N$ induced by $\lt_1$ ($(\Bbb N,\mathfrak T)$ is a Hausdorff space).
  
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 $S^2_2 \setminus \{(0,0,2r_2)\}$.
+
'''Theorem''' $1$: $\Bbb P$ is dense in the interval $(1,10)$.
  
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 $y=\pm x,\,y=0,\,x=0$
 
  
 +
on the other hand $\Bbb N$ is a cyclic group by:
  
'''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
+
$\begin{cases} \forall m,n\in\Bbb N\\ e=1\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\\\langle 2\rangle =\langle 3\rangle =(\Bbb N,\star)\end{cases}$
  
  
'''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.
+
Question $1$: Is $\Bbb N$ a topological group?
  
  
Let $F= \Bbb Q$ so what are Galois group of polynomials $x^4+b^2x^2+b^2$ and $(1+a^2)x^2 +a^4$.
+
'''Goldbach's conjecture''': $\forall n\in\Bbb N,\,\exists p,q\in\Bbb P\setminus\{2\}$ such that $2n+3=p\star q$.
  
  
'''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 $(a,b)=(c,d)$.
+
'''Guess''' $1$: the set $P_1:=\{{p+1\over2}\mid p\in\Bbb P\}$ is dense in $\Bbb N$.
: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 $(x_1,y_1)=(x_2,y_2)$.
+
:is this content related to the prime gap?
 +
:probably ''prime number theorem'' isn't enough for proving this guess.
  
  
Let $Y=\{(a,b)\,|\, (a,b)\in (S_1\times S_1)\setminus L,\, 0.01\le a+b\lt 0.1\}$ & $\forall i\in \Bbb N,$ $E_i=\{(a,b)\,|$ $(a,b)\in S_1\times S_1,$ $a+b=r_1(2i)\}$ & $O_i=\{(a,b)\,|\, (a,b)\in S_1\times S_1,\, a+b=r_1(2i-1)\}$.
+
'''Question''' $2$: Is $\Bbb N$ metrizable?
 +
:[https://math.stackexchange.com/questions/2947518/is-this-hausdorff-space-bbb-n-metrizable-and-bounded/2947575#2947575 Answer] given by [https://math.stackexchange.com/users/15500/arthur $@$Arthur] from stackexchange site: Let $v(n)$ be the number of trailing zeroes of $n$ (i.e. the largest natural number such that $10^{v(n)}\mid n$). Then the function $n\mapsto r(n)+v(n)$ maps $\Bbb N$ to a subset of $\Bbb Q$, and using the standard ordering on $\Bbb Q$ this function respects the ordering. So $(\Bbb N,<_1)$ is order isomorphic to a subset of $(\Bbb Q,<)$. hence $d:\Bbb N\times\Bbb N\to\Bbb R,\,\forall m,n\in\Bbb N,$ $d(m,n)=\vert r(m)+v(m)-r(n)-v(n)\vert$ is distance between $m,n$.
 +
:comment given by [https://math.stackexchange.com/users/254665/danielwainfleet $@$DanielWainfleet]: If $(X,d)$ is a connected metric space and $X$ has at least $2$ points then $X$ is uncountable. Because if $a,b\in X$ with $a\neq b$ then for every $r\in(0,1)$ we have $\emptyset\neq\{c\in X\mid d(a,c)=r\cdot d(a,b)\}$. Otherwise for some $r\in(0,1)$ the open sets $\{c\in X\mid d(a,c)\lt r\cdot d(a,b)\}$,$\{c\in X\mid d(a,c)\gt r\cdot d(a,b)\}$ are disjoint and non-empty, and their union is  $X$. the $\lt_1$-order-topology on $\Bbb N$ is metrizable and therefore is not connected.
  
  
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 $\{(0.1,0),(0.05,0.05)\}$ & $\{(0.01,0),(0.005,0.005)\}$.
+
Ordered sets $(\Bbb N=\{n\times10^m\mid m\in\Bbb N\cup\{0\},\,n\in\Bbb N,\,10\nmid n\},\lt_1)$ & $(A:=\{m+r(n)\mid m\in\Bbb N\cup\{0\},\,n\in\Bbb N,\,10\nmid n\},\lt)$ have the same order type with bijective $f:\Bbb N\to A,\,f(n\times10^m)=m+r(n),$ $n\times10^m\lt_1u\times10^v$ iff $m+r(n)\lt v+r(u)$
  
 +
Alireza Badali 21:20, 17 September 2018 (CEST)
  
'''Theorem''' $3$: $1)\, \forall x\in [0.01,0.1)\setminus  r_1 (\Bbb N),\,\, \{(u,v)\,|\, u+v=x\}\cap (S_1\times S_1)=\emptyset ,\,\,$ $2)\, \forall i\in \Bbb N,$ $E_i\subsetneq L,\, O_i\cap L\neq \emptyset \neq O_i\cap Y\neq O_i,\;\;\;\;\;\; Y=(\bigcup _{i\in \Bbb N} O_i )\setminus L,$ $L=(\bigcup _{i\in \Bbb N} E_i)\cup (\bigcup _{i\in \Bbb N} (O_i\cap L)),\,\,$ $3)\, \forall i\in \Bbb N,$ cardinal$(O_i \setminus L)\in \Bbb N$.
+
==== Goldbach by odd numbers ====
: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 $a\times 10^m+b\times 10^n$ that $m\neq n,\, a,b\in S_1,\, a\times 10^m,b\times 10^n$ are prime numbers and or to form of $2+b\times 10^n$ that $b\in S_1,\, b\times 10^n$ is a prime number.
 
  
 +
Let $Z_1:=\{\pm(2n-1)\mid n\in\Bbb N\}\cup\{0\}$ and $\lt_1$ be a total order relation (not well ordering) on $Z_1$ with: $\begin{cases} \forall m,n\in\Bbb N\\ 2n-1\lt_12m-1 & \text{iff}\quad r(2n-1)\lt r(2m-1),\\ -2n+1\lt_1-2m+1 & \text{iff}\quad r(2n-1)\gt r(2m-1),\\ -2n+1\lt_10\lt_12m-1\end{cases}$
  
'''Guess''' $3$: $\forall i\in \Bbb N,$ cardinal$(E_i)=\aleph_0 =$cardinal$(O_i \cap L)$
+
then assume $\mathfrak T$ is a topology on $Z_1$ induced by $\lt_1$ ($(Z_1,\mathfrak T)$ is a Hausdorff space).
  
 +
'''Guess''' $1$: $P_1=\Bbb P\setminus\{2\}$ is dense in $N_1:=\{n\in Z_1\mid n\gt0\}$.
 +
:The topology induced by $\lt_1$ has prime numbers properties because we should apply ''prime number theorem (distribution of prime numbers)'' to prove this density  or in principle there exists an especial two sided relation between ''prime number theorem'' and this density.
 +
:and hence $Z_1\setminus\{0\}$ is a separable space under subspace topology.
  
'''Guess''' $4$: $L_1$ is dense in $\{(x,y)\,|\, 0\le y\le x,\, 0.01\le x+y\le 0.1\}$.
 
  
 +
$(\Bbb N,\star_1)$ is a cyclic group with: $\begin{cases} \forall m,n\in\Bbb N\\  m\star_11=m\\ (2n)\star_1(2n+1)=1\\ (2m)\star_1(2n)=2m+2n\\ (2m-1)\star_1(2n-1)=2m+2n-3\\ (2m)\star_1(2n-1)=\begin{cases} 2m-2n+2 & 2m\gt2n-1\\ 2n-2m-1 & 2n-1\gt2m\end{cases}\\ \Bbb N=\langle2\rangle=\langle3\rangle\end{cases}$
  
'''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 $\{(x,y)\,|\, x+y=r_1(2i-1),$ $0\le y\le x\}$.
+
hence we can consider following cyclic group $(N_1,\star_2)$ with: $\begin{cases} \forall m,n\in\Bbb N\\ 1\star_2(2m-1)=2m-1\\ (4n-1)\star_2(4n+1)=1\\ (2m-1)\star_2(2n-1)=\begin{cases} 2m+2n-1 & \text{m,n are even}\\ 2m+2n-3 & \text{m,n are odd}\\ 2m-2n-1 & m\gt n,\,m\text{ is odd},\,n\text{ is even}\\ 2n-2m+1 & m\lt n,\,m\text{ is odd},\,n\text{ is even}\end{cases}\\ N_1=\langle3\rangle=\langle5\rangle\end{cases}$
  
 +
and finally regarding sequences below we have the cyclic group $(Z_1,\star)$: $$1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49,…$$ $$0,1,-1,3,-3,5,-5,7,-7,9,-9,11,-11,13,-13,15,-15,17,-17,19,-19,21,-21,23,-23,…$$ $\begin{cases} \forall m,n\in\Bbb N,\quad e=0\\ (2m-1)\star(-2m+1)=0\\ (4m-3)\star(4n-3)=4m+4n-5\\ (4m-3)\star(-4n+3)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (4m-3)\star(4n-1)=4m+4n-3\\ (4m-3)\star(-4n+1)=\begin{cases} 4m-4n-1 & m\le n\\ 4m-4n-3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+3)=-4m-4n+5\\ (-4m+3)\star(4n-1)=\begin{cases} 4n-4m+1 & m\le n\\ 4n-4m+3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+1)=-4m-4n+3\\ (4m-1)\star(4n-1)=4m+4n-1\\ (4m-1)\star(-4n+1)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (-4m+1)\star(-4n+1)=-4m-4n+1\\ Z_1=\langle1\rangle=\langle-1\rangle\end{cases}$
  
'''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.
+
Question $1$: Is $Z_1$ a topological group?
  
  
Now I want find a relation between $L_1$ & $W:=((S_1\times S_1) \cap A) \setminus L$.
+
'''Goldbach’s conjecture''': $\forall n\in\Bbb N,\,\exists p,q\in P_1$ such that $2n+5=p\star q$.
  
 +
Alireza Badali 19:52, 22 September 2018 (CEST)
  
'''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 $r(p)=r(p\times 10)$ that $p$ is a prime number then $p\times 10$ is an even number and $\{ p\times 10 \,|\, p \in \Bbb P \} \subset K$.
+
===== $Z_1$ a UFD =====
  
 +
$(\Bbb N,\star_1,\circ_1)$ is an integral domain (possibly a UFD) with: $\begin{cases} \forall m,n\in\Bbb N\\  m\star_11=m\\ (2n)\star_1(2n+1)=1\\ (2m)\star_1(2n)=2m+2n\\ (2m-1)\star_1(2n-1)=2m+2n-3\\ (2m)\star_1(2n-1)=\begin{cases} 2m-2n+2 & 2m\gt2n-1\\ 2n-2m-1 & 2n-1\gt2m\end{cases}\\ 1\circ_1m=1,\quad2\circ_1m=m,\quad(3\circ_1m)\star_1m=1\\ (2m)\circ_1(2n)=2mn\\ (2m+1)\circ_1(2n+1)=2mn\\ (2m)\circ_1(2n+1)=2mn+1\end{cases}$
  
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.
+
hence $(N_1,\star_2,\circ_2)$ is an integral domain (possibly a UFD) with: $\begin{cases} \forall m,n\in\Bbb N,\,\forall v\in N_1\\ 1\star_2(2m-1)=2m-1\\ (4n-1)\star_2(4n+1)=1\\ (2m-1)\star_2(2n-1)=\begin{cases} 2m+2n-1 & \text{m,n are even}\\ 2m+2n-3 & \text{m,n are odd}\\ 2m-2n-1 & m\gt n,\,m\text{ is odd},\,n\text{ is even}\\ 2n-2m+1 & m\lt n,\,m\text{ is odd},\,n\text{ is even}\end{cases}\\ 1\circ_2v=1,\quad3\circ_2v=v,\quad(5\circ_2v)\star_2v=1\\ (8m-5) \circ_2(8n-5)=16mn-8m-8n+3\\ (8m-5) \circ_2(8n-3)=16mn-8m-8n+5\\ (8m-5) \circ_2(8n-1)=16mn-8n-1\\ (8m-5) \circ_2(8n+1)=16mn-8n+1\\ (8m-3) \circ_2(8n-3)=16mn-8m-8n+3\\ (8m-3) \circ_2(8n-1)=16mn-8n+1\\ (8m-3) \circ_2(8n+1)=16mn-8n-1\\ (8m-1) \circ_2(8n-1)=16mn-1\\ (8m-1) \circ_2(8n+1)=16mn+1\\ (8m+1) \circ_2(8n+1)=16mn-1\end{cases}$
  
 +
from this table, $m\in\Bbb N$: $$4m-2,4m-1,4m,4m+1$$ $$8m-5,8m-3,8m-1,8m+1$$ for instance $(8m-3) \circ_2(8n+1)=t(t^{-1}(8m-3)\circ_1t^{-1}(8n+1))=t((4m-1) \circ_1(4n+1))=$ $t((2(2m-1)+1)\circ_1(2(2n)+1))=t(2(2m-1)(2n))=t(4(2mn-n))=8(2mn-n)-1=16mn-8n-1$
  
'''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 $r(m \times n)$ 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 $r(m\times n)$ from $r(m)$ & $r(n)$.
+
and finally $(Z_1,\star,\circ)$ is an integral domain (possibly a UFD) with: $\begin{cases} \forall m,n\in\Bbb N,\,\forall v\in Z_1,\quad e=0\\ (2m-1)\star(-2m+1)=0\\ (4m-3)\star(4n-3)=4m+4n-5\\ (4m-3)\star(-4n+3)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (4m-3)\star(4n-1)=4m+4n-3\\ (4m-3)\star(-4n+1)=\begin{cases} 4m-4n-1 & m\le n\\ 4m-4n-3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+3)=-4m-4n+5\\ (-4m+3)\star(4n-1)=\begin{cases} 4n-4m+1 & m\le n\\ 4n-4m+3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+1)=-4m-4n+3\\ (4m-1)\star(4n-1)=4m+4n-1\\ (4m-1)\star(-4n+1)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (-4m+1)\star(-4n+1)=-4m-4n+1\\ 0\circ v=0,\quad1\circ v=v,\quad((-1)\circ v)\star v=0\\ (4m-3)\circ(4n-3)=8mn-4m-4n+1\\ (4m-3)\circ(-4n+3)=-8mn+4m+4n-1\\ (4m-3)\circ(4n-1)=8mn-4n-1\\ (4m-3)\circ(-4n+1)=-8mn+4n+1\\ (-4m+3)\circ(-4n+3)=8mn-4m-4n+1\\ (-4m+3)\circ(4n-1)=-8mn+4n+1\\ (-4m+3)\circ(-4n+1)=8mn-4n-1\\ (4m-1)\circ(4n-1)=8mn-1\\ (4m-1)\circ(-4n+1)=-8mn+1\\ (-4m+1)\circ(-4n+1)=8mn-1\end{cases}$
  
 +
from this table: $$8m-5,8m-3,8m-1,8m+1$$ $$4m-3,-4m+3,4m-1,-4m+1$$
 +
:without ring theory we have no appropriate calculations.
  
'''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 prime numbers!
 
  
 +
Guess $1$: $\forall n\in\Bbb N,\,n$ is a prime iff $2n-1$ is an irreducible element in $(Z_1,\star,\circ)$ and we have: $\begin{cases} \forall m,n,r,s\in\Bbb N,\\ m\pm n=r\qquad\text{iff}\quad(2m-1)\star(\pm(2n-1))=2r-1\\ m\cdot n=s\qquad\text{iff}\quad(2m-1)\circ(2n-1)=2s-1\quad\text{iff}\quad2s-1=(2m-1)\star(2m-1)\star...(2m-1)\,(n\text{ times})\end{cases}$.
  
'''Hypothesis''' $1$: $\forall m,t\in\Bbb N$ if $t=t_1t_2t_3...t_k\times 10^{m-1}$ for $t_k=2,4,6,8$ or $t=t_1t_2t_3...t_k\times 10^m$ 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 $r,s\lt t\times 10^{i-m}$ then $t\times 10^{i-m}\neq r+s$.
+
Irreducible elements in $(Z_1,\star,\circ)$ except $3$ are of the form $4k-3,\,k\in\Bbb N$.
: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.
 
  
 +
Guess $2$: $Y:=\{2p-1\mid p\in\Bbb P\setminus\{2\}\}$ is dense in $N_1$.
 +
:The topology induced by $\lt_1$ has prime numbers properties because we should apply ''prime number theorem (distribution of prime numbers)'' to prove this density or in principle there exists an especial two sided relation between ''prime number theorem'' and this density and in $Z_1$ there is no even number.
 +
:and hence $Z_1\setminus\{0\}$ is a separable space under subspace topology.
  
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 $\{(a,b)\in L_1\,|\, a=r_1(p)\}\cup\{(a,b)\in L_1\,|\, a=r_2(p)\}$.
 
  
 +
'''Goldbach's conjecture''': $\forall n\in\Bbb N,\,\exists r,s\in\Bbb N$, such that $4n+7=(4r-3)\star(4s-3),$ & $4r-3,4s-3$ are irreducible elements greater than $3$ in $(Z_1,\star,\circ)$.
 +
:meantime $2r-1,2s-1\in\Bbb P$ & $4n+7$ is of the form $4k-1,\,k\in\Bbb N$.
  
'''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 $(10^{-k-1}r,10^{-k-1}s)\in A\cup \{(x,x)\,|$ $0.005\le x\lt0.05\}$.
 
  
 +
Problem $1$: in order to define an infinite field based on $Z_1$, make a division algorithm like this [https://en.wikipedia.org/wiki/Division_algorithm one] in which given two elements $s,t\in Z_1$, computes their quotient and/or remainder, the result of division.
 +
:comment given by [https://mathoverflow.net/users/41291/%E1%83%9B%E1%83%90%E1%83%9B%E1%83%A3%E1%83%99%E1%83%90-%E1%83%AF%E1%83%98%E1%83%91%E1%83%9A%E1%83%90%E1%83%AB%E1%83%94 $@$მამუკაჯიბლაძე] from stackexchange site: since this isomorphic $f:\Bbb Z\to Z_1,$ $f(n)=2n-\operatorname{sign}(n)$ with inverse $g:Z_1\to\Bbb Z$ given by $g(n)=\frac{n+\operatorname{sign(n)}}2$ then $Z_1$ is an Euclidean domain, i.e. does admit an Euclidean function.
 +
:and I would define some infinite sentences by using this field.
 +
:and I want explain density by a function.
  
If $\{\mathrm I_{\theta}\}_{\theta\in\mathcal I}$ is a partition for $U:=\{(p,q)\,|\, p,q\in\Bbb P,$ $q\le p\}$ then $\{J_{\theta}\}_{\theta\in\mathcal I}$ is a partition for $L_2:=L_1\cup \{(a,a)\in L\}$ such that $\forall\theta\in\mathcal I,$ $J_{\theta}=\{(r_s(p),r_t(q))\in L_2\,|\, r_t(q)\le r_s(p),$ $(p,q)\in\mathrm I_{\theta},$ $s,t\in\Bbb N\}\cup\{(r_t(q),r_s(p))\in L_2\,|\, r_s(p)\le r_t(q),$ $(p,q)\in\mathrm I_{\theta},$ $s,t\in\Bbb N\}$.
 
  
'''In the following''' I need to some partitions for $L_2$ or the same cognitions to $L_2,$ however we should be aware to the details of trapezoid shape with vertices $\{(0.1,0),(0.01,0),(0.05,0.05),(0.005,0.005)\}$, the problem is which partition is a good cognition to $L_2$
+
Guess $3$: for each interval $(s,t)$ that $s,t\in N_1,\,\exists n,u,v\in\Bbb N,$ such that $4n+7=(4u-3)\star(4v-3)$ & $4u-3,4v-3$ are irreducible elements greater than $3$ in $(Z_1,\star,\circ)$.
  
 +
Alireza Badali 12:19, 24 September 2018 (CEST)
  
'''Guess''' $7$: $\forall a,d\in\Bbb N$ that gcd$(a,d)=1$ then cardinal$(\{(r_s(n),r_t(a+nd))\in L_1\,|\, n=0,1,2,3,...,$ $r,s\in\Bbb N\}\cup\{(r_t(a+nd),r_s(n))\in L_1\,|\, n=0,1,2,3,...,\, r,s\in\Bbb N\})=\aleph_0$
+
====== Widget theory ======
  
 +
Definition: a widget w is an element of $[0.1,1)\setminus r(\Bbb N)$ but without decimal point instance $30141592653058979320003846264...$.
  
'''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$(\{(u,v)\in L_2\,|\, v\in\bigcup _{m\in\Bbb N} \{r_m(p)\}\})=\aleph_0$
+
'''Conjecture''': $N_1$ is dense in the $W$ the set of all widgets.
: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$(\{(u,v)\in L_1\,|\, u-v=c\})=\aleph_0$
+
$\forall w_1,w_2\in W,\,w_1\lt_W w_2$ iff $a_1\lt a_2$ in which $a_i,\,i=1,2$ is corresponding to $w_i$.
  
 +
Alireza Badali 14:02, 18 October 2018 (CEST)
  
'''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 $\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ cardinal$(\{t\in\Bbb N\,|\, 2^tm\neq p+q$ for $\forall p,q\in\Bbb P$ that $p,q\lt 2^tm\})=\aleph_0$
+
===== Quotients from $Z_1$ =====
  
 +
Question $1$: does exist any best known UFD isomorphic to $(Z_1,\star,\circ)$? does exist any best known topological space homeomorphic to $(Z_1,\mathfrak T)$?
  
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 $L_2$:
+
Alireza Badali 18:02, 7 October 2018 (CEST)
  
 +
=== [https://en.wikipedia.org/wiki/Polignac%27s_conjecture Polignac's conjecture] ===
  
'''Theorem''': $r(\{2p\,|\, p\in\Bbb P\})$ and $r(\{5p\,|\, p\in\Bbb P\})$ are dense in the interval $[0.1,1]$.
+
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.
: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: According to gcd$(2,5)=1$ & $2\times 5=10$ & $(5-2)-2=1$ so $\forall q\in\Bbb P,$ is $r(\{pq\,|\, p\in\Bbb P\})$ dense in the $[0.1,1]$
 
  
Alireza Badali 22:21, 8 May 2017 (CEST)
 
  
== Polignac's conjecture ==
+
'''Theorem''' $1$: $(\Bbb N,\star _T)$ is a group with: $\forall m,n\in\Bbb N,$
  
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.)
+
$\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}$
  
Assuming Polignac's conjecture there isn't any rhythm for prime numbers and so there isn't any formula for prime numbers!
+
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:
 +
$$0,-1,1,-3,-2,-5,3,2,-4,6,5,4,-6,-7,7,-9,-8,-11,9,8,-10,12,11,10,-12,-13,13,-15,$$ $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,$$ $$-14,-17,15,14,-16,18,17,16,-18,-19,19,-21,-20,-23,21,20,-22,24,23,22,-24,...$$ $$29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,...$$
  
  
Let $B=\{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1\}$ & $C=(S_1\times S_1)\cap \{(x,x)\,|\, 0.01\le x\lt 0.1\}$ & $T=\{(a,b)\,|\, a,b \in S_1,\, 0.01 \lt b \lt a\lt 0.1,\, \exists m \in \Bbb N,\, a \times 10^m, b \times 10^m$ or $a\times 10^{m-1}, b\times 10^m$ are consecutive prime numbers$\}$ & $\forall n \in \Bbb N,\, J_n :=\{(a,b) \,|\, (a,b) \in T,\, \exists k \in \Bbb N, a-b=r_k (2n)\}$.
+
'''Question''' $1$: 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 do $\exists m\in\Bbb N,\,\forall n\in\Bbb N$ that $n\ge m$ we have: $2\star3\star5\star7...\star p_n=\prod_{i=1}^{n}p_i\gt\prod _{i=1}^{n}x_i=4\star6\star8\star9...\star x_n$?
  
:It is clear $\bigcup _{n\in \Bbb N} J_n=T$ and Polignac's conjecture is equivalent to $\forall n \in \Bbb N,$ cardinal$(J_n)=\aleph_0$.
+
'''Question''' $2$: For which group on $\Bbb N$ like $(\Bbb N,\star)$ generated from algorithm above, do we have: $\lim_{n\to\infty}\prod _{i=1}^np_i,\lim_{n\to\infty}\prod _{i=1}^nx_i\in\Bbb N,$ $(\lim_{n\to\infty}\prod _{i=1}^np_i)\star(\lim_{n\to\infty}\prod _{i=1}^nx_i)=1$?
  
  
'''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 $n\to \infty$.
+
now let the group $G$ be external direct sum of three copies of the group $(\Bbb N,\star _T)$, hence $G=\Bbb N\oplus\Bbb N\oplus\Bbb N$.
  
  
'''Guess''' $2$: $\exists N_1 \subseteq \Bbb N,\, \forall n\in N_1,$ cardinal$(J_n)=\aleph_0=$cardinal$(N_1)$
+
'''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 $\begin{cases} t_1\lt t_2 & or\\ t_1=t_2,\, m_1-n_1\lt m_2-n_2 & or\\ t_1=t_2,\, m_1-n_1=m_2-n_2,\, n_1\lt n_2\end{cases}$
  
  
It is clear $\exists \epsilon,\epsilon_1,\epsilon_2 \in \Bbb R,\epsilon\gt 0, \epsilon_2\gt \epsilon_1\gt 0$ 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 $a\times 10^m$,$b\times 10^m$ are consecutive prime numbers, but $a\times 10^m$,$b\times 10^m$ 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 $a\times 10^{m-1},b\times 10^m$ are consecutive prime numbers but $a\times 10^{m-1}$,$b\times 10^m$ 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 $a\times 10^{m-1}$ or $a\times 10^m$ & $b\times 10^m$ are consecutive prime numbers but $a\times 10^m$ or $a\times 10^{m-1}$ & $b\times 10^m$ aren't large natural numbers.
+
and suppose $M=\Bbb N\times\Bbb N\times\Bbb N$ is a topological space ('''Hausdorff space''') induced by order relation $\lt _T$.
  
  
Theorem: $\forall c\in r_1(\Bbb P)$ cardinal$(T\cap \{(x,c)\,|\, x\in \Bbb R \})=1=$ cardinal$(T\cap \{(c,y)\,|\, y\in \Bbb R\})$
+
'''Question''' $3$: Is $G$ a topological group with topology of $M$?
  
'''Guess''' $3$: $\forall k\in \Bbb N,$ $\forall c\in r_k (\Bbb N),\, c\lt 0.09$ then cardinal$(T\cap \{(x,y)\,|\, x-y=c\}) \in \Bbb N \cup \{0\}$.
 
  
 +
'''Now''' regarding to the group $(\Bbb N,\star_T)$, I am planning an algebraic form of prime number theorem towards twin prime conjecture:
  
'''Guess''' $4$: 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 $t\times 10^{k+1}\in \Bbb P$.
 
  
 +
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.
  
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$(T\setminus \{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,\, x-y\lt \epsilon_2\})=\aleph_0$.
+
Question $4$: 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 $\lim_{x\to\infty}\pi_1(x)\cdot{\log x\over x}=0.5=\lim_{x\to\infty}\pi_2(x)\cdot{\log x\over x}\ ?$
:Proof: Be aware to number of digits in prime numbers corresponding to coordinates of each member in $T$.
+
:[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.
:'''Guess''' $5$: cardinal$(\{(a,b)\,|\, (a,b)\in T,\, 0.01\lt b\lt a\lt 0.1,$ $\epsilon_1\lt a-b\lt \epsilon_2\})\in \Bbb N\cup \{0\}$
+
:This result is known as the prime number theorem for arithmetic progressions. [https://en.wikipedia.org/wiki/Prime_number_theorem#Prime_number_theorem_for_arithmetic_progressions Wikipedia] discusses it with a number of references and one can find a proof of it by Ivan Soprounov [http://academic.csuohio.edu/soprunov_i/pdf/primes.pdf here], 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.
  
Alireza Badali 13:17, 21 August 2017 (CEST)
 
  
== Landau's forth problem ==
+
Question $5$: 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 $\Bbb N$?
  
Landau's forth problem: Are there infinitely many primes $p$ such that $p−1$ is a perfect square? In other words: Are there infinitely
+
Question $6$: 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 $\prod _{i=1}^{\infty}a_i=a_1\star_2a_2\star_2a_3\star_2...=n$?
many primes of the form $n^2 + 1$?
 
  
 +
Question $7$: 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 $\prod _{i=1}^{\infty}a_i=a_1\star a_2\star a_3\star...=n$?
  
In analytic number theory the Friedlander–Iwaniec theorem states that there are infinitely many prime numbers of the form ${\displaystyle a^{2}+b^{4}}$.
+
Question $8$: 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 $\prod_{i=1}^{\infty}a_i=a_1\star_2a_2\star_2a_3\star_2...\in\Bbb N$?
  
  
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 $a\times 10^t\in\{n^2\,|\, n\in\Bbb N\},$ $b\times 10^t\in\{n^4\,|\, n\in\Bbb N\},\, (a+b)\times 10^t\in\Bbb P\}$ and $H_1=\{(a,b)\in H\,|\, b\in\bigcup_{k\in\Bbb N} r_k(\{1\})\}$
+
'''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 [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 $n$.
 +
:'''Question''' $9$: Does $T:=U\cap\Bbb P$ have infinitely many primes?
 +
:'''Question''' $10$: Is the set $V:=\{p+2q\mid p,q\in\Bbb P,\,p+2q\in T\}$ infinite?
  
:Friedlander-Iwaniec theorem is the same cardinal$(H)=\aleph_0$
 
  
 +
'''Guess''' $1$: $\forall m,n\in\Bbb N,$ assume $p_n,q_n\in\Bbb P$ such that there is no prime number in these intervals $(p_n,m\times10^n),(m\times10^n,q_n)$ then $\lim_{n\to\infty}{q_n\over p_n}=1$.
 +
:[https://mathoverflow.net/questions/310201/a-question-relating-to-the-prime-gaps/310223#310223 Answer] given by [https://mathoverflow.net/users/3402/gerhard-paseman $@$GerhardPaseman] from stackexchange site: It turns out that $1$) explicitly for all numbers $M$ greater than $30$, there are at least two primes in $(5M/6,6M/5)$, one bigger than $M$ and one smaller than $M$, and $2$) there is $N$ large enough that for $M$ bigger than $N$ there are more than two primes in $(M,M+M^{\alpha})$, where you can pick $\alpha$ a real number larger than $0.525$. So it is like the values $p_n$ and $q_n$ will be at most a little more than $\sqrt{q_n}$ apart, which means the limit of the ratio as $n$ increases will be $1$. You can also try this with estimates from Chebyshev (before PNT) to reach the same conclusion, but it will be less obvious.
 +
::but could we replace another natural number rather than $10$ as $m\times d^n$? Yes, you can. The answer is essentially the same: the ratio as $n$ grows will eventually tend to $1$.
 +
:::'''Problem''' $1$: For each infinite strictly increasing subsequence of $\Bbb N$ like $\{a_n\}$ assume $\{p_n\}$ & $\{q_n\}$ are infinite strictly increasing subsequences of $\Bbb P$ such that $\forall n\in\Bbb N,\,p_n$ is largest prime less than $a_n$ & $q_n$ is smallest prime greater than $a_n$ then discuss on the limit below: $$\lim_{n\to\infty}{q_n\over p_n}\,.$$
  
'''A question''': cardinal$(H_1)\in\Bbb N$ or cardinal$(H_1)=\aleph_0$?
 
:This question is the same Landau's forth problem.
 
  
Alireza Badali 20:47, 21 September 2017 (CEST)
+
There is an especial (not necessarily unique), infinite and proper subsequence in prime numbers that gives the map of all prime numbers.
  
== Grimm's conjecture ==
+
Alireza Badali 12:34, 28 April 2018 (CEST)
  
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.
+
== Some dissimilar conjectures ==
  
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 $1 ≤ i ≤ k$.
+
'''Algebraic analytical number theory'''
  
 +
Alireza Badali 16:51, 4 July 2018 (CEST)
  
Let $C=\{(x,y)\,|\, 0.01\le x\lt 0.1,\, 0.01\le y\lt 0.1\}$
+
=== [https://en.wikipedia.org/wiki/Collatz_conjecture 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 [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.
  
'''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 $r_1(p_i)\times r_1(t_i)=r_2(n+i)$ or $r_3(n+i)$.
 
:This conjecture is an equivalent to Grimm's conjecture.
 
  
Alireza Badali 23:30, 20 September 2017 (CEST)
+
'''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 $n\star_{\Bbb N}n^{-1}=1$.
  
== Lemoine's conjecture ==
 
  
In number theory, Lemoine's conjecture, named after Émile Lemoine, also known as Levy's conjecture, after Hyman Levy, states that all odd integers greater than $5$ can be represented as the sum of an odd prime number and an even semiprime.
+
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.
  
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.
 
  
 +
<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 $T=(1, 2, 4, 3, 6, 5, 10, 7, 14, 8, 16, 9, 18, 11, 22, 12, 24, 13, 26, 15, 30, 17, 34, 19, 38, 20, \ldots).$
  
Theorem: $\forall n\in\Bbb N$ that $m=2n+5,\, \exists (a,b)\in\{(x,y)\,|\, 0\lt y\le x,\, x+y\lt 0.1,\, x,y\in S_1,\, \exists t\in\Bbb N,$ $x\times 10^t,y\times 10^t\in\Bbb P\},\,\exists k\in\Bbb N,\, \exists c\in r_k(\Bbb P)$ such that $a+b=r_1(m)-c$
+
----------------------------
:In principle $(r_1(m),-c)\in \{(x,y)\,|\, -x\lt y\lt 0,\, 0.01\le x\lt 0.1\}$
 
:This theorem is an equivalent to Goldbach's weak conjecture.
 
  
 +
:References:
 +
:Sequences $S,\bar{S},T$ are OEIS [http://oeis.org/A171945 A171945], [http://oeis.org/A053661 A053661], [http://oeis.org/A034701 A034701] respectively. These are all discussed in ''[https://www.sciencedirect.com/science/article/pii/S0012365X11001427 The vile, dopey, evil and odious game players]''.
  
'''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)
+
: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
  
== Some new zeta functions ==
+
    [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]</small>
  
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).
 
  
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 $S$.
+
<small>'''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 $C$?</small>
  
  
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: $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\,\mu _1(z)=\lim_{n\to\infty}\frac{1}{n^2}\sum _{a\in A_n} a^{-f(z)}$$
+
'''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 $n\star n^{-1}=1$.
  
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 $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\, f(z)=\sqrt 2 +\frac{z}{N+|z|}$$ 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.
 
  
 +
<small>'''Suppose''' $\forall m,n\in\Bbb N,\qquad$ $\begin{cases} m\star 1=m\\ (4m)\star (4m-2)=1=(4m+1)\star (4m-1)\\ (4m-2)\star (4n-2)=4m+4n-5\\ (4m-2)\star (4n-1)=4m+4n-2\\ (4m-2)\star (4n)=\begin{cases} 4m-4n-1 & 4m-2\gt 4n\\ 4n-4m+1 & 4n\gt 4m-2\\ 3 & m=n+1\end{cases}\\ (4m-2)\star (4n+1)=\begin{cases} 4m-4n-2 & 4m-2\gt 4n+1\\ 4n-4m+4 & 4n+1\gt 4m-2\end{cases}\\ (4m-1)\star (4n-1)=4m+4n-1\\ (4m-1)\star (4n)=\begin{cases} 4m-4n+2 & 4m-1\gt 4n\\ 4n-4m & 4n\gt 4m-1\\ 2 & m=n\end{cases}\\ (4m-1)\star (4n+1)=\begin{cases} 4m-4n-1 & 4m-1\gt 4n+1\\ 4n-4m+1 & 4n+1\gt 4m-1\\ 3 & m=n+1\end{cases}\\ (4m)\star (4n)=4m+4n-3\\ (4m)\star (4n+1)=4m+4n\\ (4m+1)\star  (4n+1)=4m+4n+1\\ \Bbb N=\langle 2\rangle=\langle 4\rangle\end{cases}$
  
Let $\mu _2:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\,\mu _2(z)=\sum _{n=1}^{\infty}\frac{1}{(r(n))^z}$$
+
and let $C_1=\{(m,2m)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_1}=(1,2)\\ \\\forall m,n\in\Bbb N,\,(m,2m)\star_{C_1}(n,2n)=(m\star n,2(m\star n))\\ (m,2m)^{-1}=(m^{-1},2\times m^{-1})\qquad\text{that}\quad m\star m^{-1}=1\\ \\C_1=\langle(2,4)\rangle=\langle(4,8)\rangle\end{cases}$
  
Let $\mu _3:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\,\mu _3(z)=\sum _{p\in\Bbb P}\frac{1}{(r(p))^z}$$
+
and $C_2=\{(3m-1,2m-1)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_2}=(2,1)\\ \\\forall m,n\in\Bbb N,\, (3m-1,2m-1)\star_{C_2}(3n-1,2n-1)=(3(m\star n)-1,2(m\star n)-1)\\ (3m-1,2m-1)^{-1}=(3\times m^{-1}-1,2\times m^{-1}-1)\qquad\text{that}\quad m\star m^{-1}=1\\ \\C_2=\langle(5,3)\rangle=\langle(11,7)\rangle\end{cases}$.
  
Let $\forall j\in\Bbb N,\,\mu _4:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\,\mu _4(z)=\sum _{n=1}^{\infty}\frac{1}{(r_j(n))^z}$$
+
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 $C$?</small>
  
Let $\forall j\in\Bbb N,\,\mu _5:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\,\mu _5(z)=\sum _{p\in\Bbb P}\frac{1}{(r_j(p))^z}$$
 
  
Let $\mu _6:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\,\mu _6(z)=\lim j\to\infty\sum _{n=1}^{\infty}\frac{1}{(r_j(n))^z}$$
+
<small>'''Question''' $3$: If $(\Bbb N,\star)$ is a cyclic group with generators $u,v$ & identity element $1$ then could $(\Bbb N,\star_1)$ be another cyclic group with: $\begin{cases} \forall m,n\in\Bbb N,\\ e=1\\ m\star_1n=(2m)\star(2n) & \text{if } m,n\text{ aren't of the form } 6k+4,\,k\in\Bbb N\\ (6m+4)\star_1n=(2m+1)\star(2n) & \text{if } n\text{ isn't of the form } 6k+4,\,k\in\Bbb N\\ (6m+4)\star_1(6n+4)=(2m+1)\star(2n+1)\\ n^{-1}=k & \text{if } n\text{ isn't of the form } 6t+4,\,t\in\Bbb N,\,k\star(2n)^{-1}=1\\ (6m+4)^{-1}=k & k\star(2m+1)^{-1}=1\\ \Bbb N=\langle u_1\rangle=\langle v_1\rangle & \begin{cases} u_1=\begin{cases} 2k+1 & \text{if } u\text{ is of the form } 6k+4,\,k\in\Bbb N\\ 2u & \text{otherwise}\end{cases}\\ v_1=\begin{cases} 2k+1 & \text{if } v\text{ is of the form } 6k+4,\,k\in\Bbb N\\ 2v & \text{otherwise}\end{cases}\end{cases}\end{cases}$ ? maybe.</small>
  
Let $\mu _7:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\,\mu _7(z)=\lim j\to\infty\sum _{p\in\Bbb P}\frac{1}{(r_j(p))^z}$$
 
  
let $\mu _8:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\,\mu _8(z)=\sum _{j\in\Bbb N}\frac{1}{(r_j(j))^z}$$
+
Question $4$: Has any relation on the Collatz tree been discovered other than its definitions? In order to define a group structure on the [https://www.jasondavies.com/collatz-graph/ Collatz tree] I need such relations but other than its [https://en.wikipedia.org/wiki/Collatz_conjecture definitions], please introduce them (ideally suited a relation could be equivalent to its definition) if exist. if such a relation there was then via a group on $\Bbb N$, we could define a group on the Collatz tree.
  
Let $p_j$ is $j$_th prime number & $\mu _9:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\,\mu _9(z)=\sum _{j\in\Bbb N}\frac{1}{(r_j(p_j))^z}$$
+
Alireza Badali 10:02, 12 May 2018 (CEST)
  
Let $a_n:\Bbb N\to\Bbb N$ is a sequence & $\mu _{10}:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\,\mu _{10}(z)=\sum _{n\in\Bbb N}\frac{1}{(r_{a_n}(n))^z}$$
+
=== [https://en.wikipedia.org/wiki/Erdős–Straus_conjecture Erdős–Straus conjecture] ===
  
Let $a_n:\Bbb N\to\Bbb N$ is a sequence & $\mu _{11}:\Bbb C\to\Bbb C$ is a function as: $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\,\mu _{11}(z)=\sum _{n\in\Bbb N}\frac{1}{(r_{a_n}(p_n))^z}$$
+
'''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 $\omega =0.p_1p_2p_3...=0.23571113171923293137...$ 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: $$\forall z\in\Bbb C\,\,\,\,\,\,\,\,\,\,\mu _{12}(z)=\sum _{n\in\Bbb N}\frac{1}{(r_{a_n}(p_n))^z}$$
 
  
Now some theorems on these functions about density should be presented.
+
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 20:44, 12 October 2017 (CEST)
+
  Alireza Badali 17:34, 25 May 2018 (CEST)
 +
 
 +
=== [https://en.wikipedia.org/wiki/Landau%27s_problems 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 [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.
 +
 
 +
 
 +
'''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)
 +
 
 +
=== [https://en.wikipedia.org/wiki/Lemoine%27s_conjecture 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.
 +
 
 +
 
 +
[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 $L=\{\text{floor}(n\cdot r)\mid n\in\Bbb N\}$.
 +
 
 +
 
 +
'''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)
 +
 
 +
=== [https://en.wikipedia.org/wiki/Gaussian_moat Gaussian moat problem] ===
 +
 
 +
 
 +
 
 +
Alireza Badali 18:13, 20 June 2018 (CEST)
 +
 
 +
=== [https://en.wikipedia.org/wiki/Grimm%27s_conjecture Grimm's conjecture] ===
 +
 
 +
 
 +
 
 +
Alireza Badali 18:13, 20 June 2018 (CEST)
 +
 
 +
=== [https://en.wikipedia.org/wiki/Oppermann%27s_conjecture Oppermann's conjecture] ===
 +
 
 +
 
 +
 
 +
Alireza Badali 18:13, 20 June 2018 (CEST)
 +
 
 +
=== [https://en.wikipedia.org/wiki/Legendre%27s_conjecture 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 [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 $(\Bbb N\cup\{0\},\star)$ 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: $\begin{cases} \forall m,n,u,v\in\Bbb N\cup\{0\},\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\ {m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={-m\over n}\qquad m\star(-m)=0\end{cases}$•
 +
::in addition under a total order relation, an unique & specific division algorithm like this [https://en.wikipedia.org/wiki/Division_algorithm one] in accordance with $(\Bbb N\cup\{0\},\star,\circ)$ is needed which given two natural numbers $m$ and $n$, computes their quotient and/or remainder, the result of division.
 +
 
 +
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: $\begin{cases} \forall m,n,u,v\in\Bbb N\cup\{0\},\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={-m\over n}\qquad m\star(-m)=0\end{cases}$
 +
:in addition an unique & specific division algorithm like this [https://en.wikipedia.org/wiki/Division_algorithm one] in accordance with $(\Bbb N\cup\{0\},\star,\circ)$ is needed which given two natural numbers $m$ and $n$, computes their quotient and/or remainder, the result of division•
 +
 
 +
 
 +
<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>
 +
:::::::<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>
 +
::::::::<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>
 +
 
 +
Alireza Badali 16:26, 26 June 2018 (CEST)
 +
 
 +
=== [https://en.wikipedia.org/wiki/Many-worlds_interpretation 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 a rotation number of the form $4t,\,t\in\Bbb N$ that for this sequence is $4$ (or $4t$) and hence equations should be written with module $2$ (or $2t$) 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 $4t$) 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: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$ •
 +
::in addition an unique & specific division algorithm like this [https://en.wikipedia.org/wiki/Division_algorithm one] in accordance with $(\Bbb Z,\star,\circ)$ is needed which given two integers $m$ and $n$, computes their quotient and/or remainder, the result of division•
 +
 
 +
 
 +
'''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: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\ {m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$•
 +
:in addition under a total order relation, an unique & specific division algorithm like this [https://en.wikipedia.org/wiki/Division_algorithm one] in accordance with $(\Bbb Z,\star,\circ)$ is needed which given two integers $m$ and $n$, computes their quotient and/or remainder, the result of division•
 +
 
 +
 
 +
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)
 +
 
 +
=== [https://en.wikipedia.org/wiki/Gauss_circle_problem Gauss circle problem] ===
 +
 
 +
<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:
 +
 
 +
$\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: $(6m-4)\circ(6n-3)=t(t^{-1}(6m-4)\times t^{-1}(6n-3))=t((-6m+5)\times(6n-4))=$ $t(-6(6mn-4m-5n+4)+4)=-6(6mn-4m-5n+4)+3=-36mn+24m+30n-21,$ $(6m-4)\circ(-6n+1)=$ $t(t^{-1}(6m-4)\times t^{-1}(-6n+1))=t((-6m+5)\times(-6n))=t(6(6mn-5n))=-6(6mn-5n)+2=-36mn+30n+2,$ $(-6m+1)\circ(-6n+1)=t(t^{-1}(-6m+1)\times t^{-1}(-6n+1))=t((-6m)\times(-6n))=t(6(6mn))=-6(6mn)+2=$ $-36mn+2$.</small>
 +
 
 +
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: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$
 +
:in addition an unique & specific division algorithm like this [https://en.wikipedia.org/wiki/Division_algorithm one] in accordance with $(\Bbb Z,\star,\circ)$ is needed which given two integers $m$ and $n$, computes their quotient and/or remainder, the result of division•
 +
 
 +
 
 +
'''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)

Latest revision as of 05:22, 21 October 2018

$\mathscr B$ $theory$ (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 & topology with homotopy groups & prime number theorem and its extensions or other forms or corollaries with limitation concept

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.
comment given by $@$GerhardPaseman from stackexchange site: There are elementary methods to show your specified set is dense. Indeed, simple sieving methods and estimates known to Euler for the sum of the reciprocals of primes give a weak but for your result a sufficient upper bound on the number of primes less than $n$ (like ${n\over\log\log n}$).
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 a rotation number of the form $4t,\,t\in\Bbb N$ 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 $\langle 2\rangle =\langle 3\rangle =(\Bbb N,\star)\simeq (\Bbb Z,+)\simeq (\Bbb P,\star _1)=\langle 3\rangle=\langle 5\rangle$.


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 these techniques are useful:
$\forall n\in\Bbb N,$ 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 $3$: $\forall (a,b)\subset [0.1,1),\,w_{(a,b)}={10\over9}\cdot(b-a)$. :::[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 '"`UNIQ-MathJax6-QINU`"' 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. ::::Let $e:\Bbb N\to\Bbb N,$ is a function that $\forall n\in\Bbb N$ gives the number of digits in $n$ instance $e(1320)=4$, and let $\forall n\in\Bbb N,$ $\forall k\in\Bbb N\cup\{0\},$ and for each subinterval $(a,b)$ of $[0.1,1),$ that $a\neq b,$ $\begin{cases} A_{k,(a,b)}:=\{n\mid\exists t_1\in\Bbb N,\,\exists t_2\in (a,b),\, t_2\cdot 10^{t_1}\in\Bbb N,\, 10\nmid t_2\cdot10^{t_1},\, n=t_2\cdot 10^{k+t_1}\},\\ \\B_{k,(a,b)}:=\{p\mid\exists t_1\in\Bbb N,\,\exists t_2\in (a,b),\, p=t_2\cdot 10^{t_1}\in\Bbb P,\,\exists n_1,n_2\in A_{k,(a,b)},\, n_1\le p\le n_2,\,e(n_1)=e(n_2)\},\\ \\A_{k,(a,b),n}:=\{m\in A_{k,(a,b)}\mid m\le n\},\\ \\B_{k,(a,b),n}:=\{m\in B_{k,(a,b)}\mid m\le n\},\\ \\c_{k,(a,b),n}:=(\#A_{k,(a,b),n})^{-1}\cdot\#B_{k,(a,b),n}\cdot\log n,\\ \\c_{k,(a,b)}:=\lim _{n\to\infty} c_{k,(a,b),n}\end{cases}$. :::::Guess $4$: $\forall k\in\Bbb N\cup\{0\},\,\forall (a,b)\subset [0.1,1),\,c_{k,(a,b)}=10^{-k}\cdot (b-a)$. (and we knew $\sum _{k\in\Bbb N\cup\{0\}}10^{-k}={10\over9}$) ::Guess $5$: $\forall (a,b)\subset [0.1,1),\,z_{(a,b)}={10\over9}\cdot(b-a)$. :::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 $6$: $\forall(a,b),(c,d)\subset[0.1,1),\,\lim_{n\to\infty}{\#V_{(a,b),n}\over\#V_{(c,d),n}}={b-a\over d-c}=\lim_{n\to\infty}{\#U_{(a,b),n}\over\#U_{(c,d),n}}$. :::<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> :::<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> Using homotopy groups Goldbach's conjecture will be proved. Alireza Badali 08:27, 31 March 2018 (CEST) ===='"`UNIQ--h-2--QINU`"' Goldbach by $\Bbb N$ ==== Let $\lt_1$ be a total order relation (not well ordering) on $\Bbb N$ as: $\forall m,n\in\Bbb N,\,m\lt_1n$ iff $\begin{cases} r(m)\lt r(n),\,m=m_1\times10^s,\,n=n_1\times10^s,\,10\nmid m_1,\,10\nmid n_1,\,m_1,n_1\in\Bbb N,\,s\in\Bbb N\cup\{0\} & \text{ or}\\ \\m=m_1\times10^s,\,n=n_1\times10^t,\,s\lt t,\,10\nmid m_1,\,10\nmid n_1,\,m_1,n_1,t\in\Bbb N,\,s\in\Bbb N\cup\{0\}\end{cases}$ then assume $\mathfrak T$ is a topology on $\Bbb N$ induced by $\lt_1$ ($(\Bbb N,\mathfrak T)$ is a Hausdorff space). '''Theorem''' $1$: $\Bbb P$ is dense in the interval $(1,10)$. on the other hand $\Bbb N$ is a cyclic group by: $\begin{cases} \forall m,n\in\Bbb N\\ e=1\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\\\langle 2\rangle =\langle 3\rangle =(\Bbb N,\star)\end{cases}$ Question $1$: Is $\Bbb N$ a topological group? '''Goldbach's conjecture''': $\forall n\in\Bbb N,\,\exists p,q\in\Bbb P\setminus\{2\}$ such that $2n+3=p\star q$. '''Guess''' $1$: the set $P_1:=\{{p+1\over2}\mid p\in\Bbb P\}$ is dense in $\Bbb N$. :is this content related to the prime gap? :probably ''prime number theorem'' isn't enough for proving this guess. '''Question''' $2$: Is $\Bbb N$ metrizable? :[https://math.stackexchange.com/questions/2947518/is-this-hausdorff-space-bbb-n-metrizable-and-bounded/2947575#2947575 Answer] given by [https://math.stackexchange.com/users/15500/arthur $@$Arthur] from stackexchange site: Let $v(n)$ be the number of trailing zeroes of $n$ (i.e. the largest natural number such that $10^{v(n)}\mid n$). Then the function $n\mapsto r(n)+v(n)$ maps $\Bbb N$ to a subset of $\Bbb Q$, and using the standard ordering on $\Bbb Q$ this function respects the ordering. So $(\Bbb N,<_1)$ is order isomorphic to a subset of $(\Bbb Q,<)$. hence $d:\Bbb N\times\Bbb N\to\Bbb R,\,\forall m,n\in\Bbb N,$ $d(m,n)=\vert r(m)+v(m)-r(n)-v(n)\vert$ is distance between $m,n$. :comment given by [https://math.stackexchange.com/users/254665/danielwainfleet $@$DanielWainfleet]: If $(X,d)$ is a connected metric space and $X$ has at least $2$ points then $X$ is uncountable. Because if $a,b\in X$ with $a\neq b$ then for every $r\in(0,1)$ we have $\emptyset\neq\{c\in X\mid d(a,c)=r\cdot d(a,b)\}$. Otherwise for some $r\in(0,1)$ the open sets $\{c\in X\mid d(a,c)\lt r\cdot d(a,b)\}$,$\{c\in X\mid d(a,c)\gt r\cdot d(a,b)\}$ are disjoint and non-empty, and their union is $X$. the $\lt_1$-order-topology on $\Bbb N$ is metrizable and therefore is not connected. Ordered sets $(\Bbb N=\{n\times10^m\mid m\in\Bbb N\cup\{0\},\,n\in\Bbb N,\,10\nmid n\},\lt_1)$ & $(A:=\{m+r(n)\mid m\in\Bbb N\cup\{0\},\,n\in\Bbb N,\,10\nmid n\},\lt)$ have the same order type with bijective $f:\Bbb N\to A,\,f(n\times10^m)=m+r(n),$ $n\times10^m\lt_1u\times10^v$ iff $m+r(n)\lt v+r(u)$ Alireza Badali 21:20, 17 September 2018 (CEST) ==== Goldbach by odd numbers ==== Let $Z_1:=\{\pm(2n-1)\mid n\in\Bbb N\}\cup\{0\}$ and $\lt_1$ be a total order relation (not well ordering) on $Z_1$ with: $\begin{cases} \forall m,n\in\Bbb N\\ 2n-1\lt_12m-1 & \text{iff}\quad r(2n-1)\lt r(2m-1),\\ -2n+1\lt_1-2m+1 & \text{iff}\quad r(2n-1)\gt r(2m-1),\\ -2n+1\lt_10\lt_12m-1\end{cases}$ then assume $\mathfrak T$ is a topology on $Z_1$ induced by $\lt_1$ ($(Z_1,\mathfrak T)$ is a Hausdorff space). '''Guess''' $1$: $P_1=\Bbb P\setminus\{2\}$ is dense in $N_1:=\{n\in Z_1\mid n\gt0\}$. :The topology induced by $\lt_1$ has prime numbers properties because we should apply ''prime number theorem (distribution of prime numbers)'' to prove this density or in principle there exists an especial two sided relation between ''prime number theorem'' and this density. :and hence $Z_1\setminus\{0\}$ is a separable space under subspace topology. $(\Bbb N,\star_1)$ is a cyclic group with: $\begin{cases} \forall m,n\in\Bbb N\\ m\star_11=m\\ (2n)\star_1(2n+1)=1\\ (2m)\star_1(2n)=2m+2n\\ (2m-1)\star_1(2n-1)=2m+2n-3\\ (2m)\star_1(2n-1)=\begin{cases} 2m-2n+2 & 2m\gt2n-1\\ 2n-2m-1 & 2n-1\gt2m\end{cases}\\ \Bbb N=\langle2\rangle=\langle3\rangle\end{cases}$ hence we can consider following cyclic group $(N_1,\star_2)$ with: $\begin{cases} \forall m,n\in\Bbb N\\ 1\star_2(2m-1)=2m-1\\ (4n-1)\star_2(4n+1)=1\\ (2m-1)\star_2(2n-1)=\begin{cases} 2m+2n-1 & \text{m,n are even}\\ 2m+2n-3 & \text{m,n are odd}\\ 2m-2n-1 & m\gt n,\,m\text{ is odd},\,n\text{ is even}\\ 2n-2m+1 & m\lt n,\,m\text{ is odd},\,n\text{ is even}\end{cases}\\ N_1=\langle3\rangle=\langle5\rangle\end{cases}$ and finally regarding sequences below we have the cyclic group $(Z_1,\star)$: '"`UNIQ-MathJax7-QINU`"' '"`UNIQ-MathJax8-QINU`"' $\begin{cases} \forall m,n\in\Bbb N,\quad e=0\\ (2m-1)\star(-2m+1)=0\\ (4m-3)\star(4n-3)=4m+4n-5\\ (4m-3)\star(-4n+3)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (4m-3)\star(4n-1)=4m+4n-3\\ (4m-3)\star(-4n+1)=\begin{cases} 4m-4n-1 & m\le n\\ 4m-4n-3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+3)=-4m-4n+5\\ (-4m+3)\star(4n-1)=\begin{cases} 4n-4m+1 & m\le n\\ 4n-4m+3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+1)=-4m-4n+3\\ (4m-1)\star(4n-1)=4m+4n-1\\ (4m-1)\star(-4n+1)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (-4m+1)\star(-4n+1)=-4m-4n+1\\ Z_1=\langle1\rangle=\langle-1\rangle\end{cases}$ Question $1$: Is $Z_1$ a topological group? '''Goldbach’s conjecture''': $\forall n\in\Bbb N,\,\exists p,q\in P_1$ such that $2n+5=p\star q$. Alireza Badali 19:52, 22 September 2018 (CEST) ===== $Z_1$ a UFD ===== $(\Bbb N,\star_1,\circ_1)$ is an integral domain (possibly a UFD) with: $\begin{cases} \forall m,n\in\Bbb N\\ m\star_11=m\\ (2n)\star_1(2n+1)=1\\ (2m)\star_1(2n)=2m+2n\\ (2m-1)\star_1(2n-1)=2m+2n-3\\ (2m)\star_1(2n-1)=\begin{cases} 2m-2n+2 & 2m\gt2n-1\\ 2n-2m-1 & 2n-1\gt2m\end{cases}\\ 1\circ_1m=1,\quad2\circ_1m=m,\quad(3\circ_1m)\star_1m=1\\ (2m)\circ_1(2n)=2mn\\ (2m+1)\circ_1(2n+1)=2mn\\ (2m)\circ_1(2n+1)=2mn+1\end{cases}$ hence $(N_1,\star_2,\circ_2)$ is an integral domain (possibly a UFD) with: $\begin{cases} \forall m,n\in\Bbb N,\,\forall v\in N_1\\ 1\star_2(2m-1)=2m-1\\ (4n-1)\star_2(4n+1)=1\\ (2m-1)\star_2(2n-1)=\begin{cases} 2m+2n-1 & \text{m,n are even}\\ 2m+2n-3 & \text{m,n are odd}\\ 2m-2n-1 & m\gt n,\,m\text{ is odd},\,n\text{ is even}\\ 2n-2m+1 & m\lt n,\,m\text{ is odd},\,n\text{ is even}\end{cases}\\ 1\circ_2v=1,\quad3\circ_2v=v,\quad(5\circ_2v)\star_2v=1\\ (8m-5) \circ_2(8n-5)=16mn-8m-8n+3\\ (8m-5) \circ_2(8n-3)=16mn-8m-8n+5\\ (8m-5) \circ_2(8n-1)=16mn-8n-1\\ (8m-5) \circ_2(8n+1)=16mn-8n+1\\ (8m-3) \circ_2(8n-3)=16mn-8m-8n+3\\ (8m-3) \circ_2(8n-1)=16mn-8n+1\\ (8m-3) \circ_2(8n+1)=16mn-8n-1\\ (8m-1) \circ_2(8n-1)=16mn-1\\ (8m-1) \circ_2(8n+1)=16mn+1\\ (8m+1) \circ_2(8n+1)=16mn-1\end{cases}$ from this table, $m\in\Bbb N$: '"`UNIQ-MathJax9-QINU`"' '"`UNIQ-MathJax10-QINU`"' for instance $(8m-3) \circ_2(8n+1)=t(t^{-1}(8m-3)\circ_1t^{-1}(8n+1))=t((4m-1) \circ_1(4n+1))=$ $t((2(2m-1)+1)\circ_1(2(2n)+1))=t(2(2m-1)(2n))=t(4(2mn-n))=8(2mn-n)-1=16mn-8n-1$ and finally $(Z_1,\star,\circ)$ is an integral domain (possibly a UFD) with: $\begin{cases} \forall m,n\in\Bbb N,\,\forall v\in Z_1,\quad e=0\\ (2m-1)\star(-2m+1)=0\\ (4m-3)\star(4n-3)=4m+4n-5\\ (4m-3)\star(-4n+3)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (4m-3)\star(4n-1)=4m+4n-3\\ (4m-3)\star(-4n+1)=\begin{cases} 4m-4n-1 & m\le n\\ 4m-4n-3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+3)=-4m-4n+5\\ (-4m+3)\star(4n-1)=\begin{cases} 4n-4m+1 & m\le n\\ 4n-4m+3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+1)=-4m-4n+3\\ (4m-1)\star(4n-1)=4m+4n-1\\ (4m-1)\star(-4n+1)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (-4m+1)\star(-4n+1)=-4m-4n+1\\ 0\circ v=0,\quad1\circ v=v,\quad((-1)\circ v)\star v=0\\ (4m-3)\circ(4n-3)=8mn-4m-4n+1\\ (4m-3)\circ(-4n+3)=-8mn+4m+4n-1\\ (4m-3)\circ(4n-1)=8mn-4n-1\\ (4m-3)\circ(-4n+1)=-8mn+4n+1\\ (-4m+3)\circ(-4n+3)=8mn-4m-4n+1\\ (-4m+3)\circ(4n-1)=-8mn+4n+1\\ (-4m+3)\circ(-4n+1)=8mn-4n-1\\ (4m-1)\circ(4n-1)=8mn-1\\ (4m-1)\circ(-4n+1)=-8mn+1\\ (-4m+1)\circ(-4n+1)=8mn-1\end{cases}$ from this table: '"`UNIQ-MathJax11-QINU`"' '"`UNIQ-MathJax12-QINU`"' :without ring theory we have no appropriate calculations. Guess $1$: $\forall n\in\Bbb N,\,n$ is a prime iff $2n-1$ is an irreducible element in $(Z_1,\star,\circ)$ and we have: $\begin{cases} \forall m,n,r,s\in\Bbb N,\\ m\pm n=r\qquad\text{iff}\quad(2m-1)\star(\pm(2n-1))=2r-1\\ m\cdot n=s\qquad\text{iff}\quad(2m-1)\circ(2n-1)=2s-1\quad\text{iff}\quad2s-1=(2m-1)\star(2m-1)\star...(2m-1)\,(n\text{ times})\end{cases}$. Irreducible elements in $(Z_1,\star,\circ)$ except $3$ are of the form $4k-3,\,k\in\Bbb N$. Guess $2$: $Y:=\{2p-1\mid p\in\Bbb P\setminus\{2\}\}$ is dense in $N_1$. :The topology induced by $\lt_1$ has prime numbers properties because we should apply ''prime number theorem (distribution of prime numbers)'' to prove this density or in principle there exists an especial two sided relation between ''prime number theorem'' and this density and in $Z_1$ there is no even number. :and hence $Z_1\setminus\{0\}$ is a separable space under subspace topology. '''Goldbach's conjecture''': $\forall n\in\Bbb N,\,\exists r,s\in\Bbb N$, such that $4n+7=(4r-3)\star(4s-3),$ & $4r-3,4s-3$ are irreducible elements greater than $3$ in $(Z_1,\star,\circ)$. :meantime $2r-1,2s-1\in\Bbb P$ & $4n+7$ is of the form $4k-1,\,k\in\Bbb N$. Problem $1$: in order to define an infinite field based on $Z_1$, make a division algorithm like this [https://en.wikipedia.org/wiki/Division_algorithm one] in which given two elements $s,t\in Z_1$, computes their quotient and/or remainder, the result of division. :comment given by [https://mathoverflow.net/users/41291/%E1%83%9B%E1%83%90%E1%83%9B%E1%83%A3%E1%83%99%E1%83%90-%E1%83%AF%E1%83%98%E1%83%91%E1%83%9A%E1%83%90%E1%83%AB%E1%83%94 $@$მამუკაჯიბლაძე] from stackexchange site: since this isomorphic $f:\Bbb Z\to Z_1,$ $f(n)=2n-\operatorname{sign}(n)$ with inverse $g:Z_1\to\Bbb Z$ given by $g(n)=\frac{n+\operatorname{sign(n)}}2$ then $Z_1$ is an Euclidean domain, i.e. does admit an Euclidean function. :and I would define some infinite sentences by using this field. :and I want explain density by a function. Guess $3$: for each interval $(s,t)$ that $s,t\in N_1,\,\exists n,u,v\in\Bbb N,$ such that $4n+7=(4u-3)\star(4v-3)$ & $4u-3,4v-3$ are irreducible elements greater than $3$ in $(Z_1,\star,\circ)$. Alireza Badali 12:19, 24 September 2018 (CEST) ======'"`UNIQ--h-3--QINU`"' Widget theory ====== Definition: a widget w is an element of $[0.1,1)\setminus r(\Bbb N)$ but without decimal point instance $30141592653058979320003846264...$. '''Conjecture''': $N_1$ is dense in the $W$ the set of all widgets. $\forall w_1,w_2\in W,\,w_1\lt_W w_2$ iff $a_1\lt a_2$ in which $a_i,\,i=1,2$ is corresponding to $w_i$. Alireza Badali 14:02, 18 October 2018 (CEST) ====='"`UNIQ--h-4--QINU`"' Quotients from $Z_1$ ===== Question $1$: does exist any best known UFD isomorphic to $(Z_1,\star,\circ)$? does exist any best known topological space homeomorphic to $(Z_1,\mathfrak T)$? Alireza Badali 18:02, 7 October 2018 (CEST) ==='"`UNIQ--h-5--QINU`"' [https://en.wikipedia.org/wiki/Polignac%27s_conjecture 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: '"`UNIQ-MathJax13-QINU`"' '"`UNIQ-MathJax14-QINU`"' '"`UNIQ-MathJax15-QINU`"' '"`UNIQ-MathJax16-QINU`"' '''Question''' $1$: 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 do $\exists m\in\Bbb N,\,\forall n\in\Bbb N$ that $n\ge m$ we have: $2\star3\star5\star7...\star p_n=\prod_{i=1}^{n}p_i\gt\prod _{i=1}^{n}x_i=4\star6\star8\star9...\star x_n$? '''Question''' $2$: For which group on $\Bbb N$ like $(\Bbb N,\star)$ generated from algorithm above, do we have: $\lim_{n\to\infty}\prod _{i=1}^np_i,\lim_{n\to\infty}\prod _{i=1}^nx_i\in\Bbb N,$ $(\lim_{n\to\infty}\prod _{i=1}^np_i)\star(\lim_{n\to\infty}\prod _{i=1}^nx_i)=1$? 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''' $3$: 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 $4$: 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 $\lim_{x\to\infty}\pi_1(x)\cdot{\log x\over x}=0.5=\lim_{x\to\infty}\pi_2(x)\cdot{\log x\over x}\ ?$ :[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. :This result is known as the prime number theorem for arithmetic progressions. [https://en.wikipedia.org/wiki/Prime_number_theorem#Prime_number_theorem_for_arithmetic_progressions Wikipedia] discusses it with a number of references and one can find a proof of it by Ivan Soprounov [http://academic.csuohio.edu/soprunov_i/pdf/primes.pdf here], 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 $5$: 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 $6$: 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 $7$: 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 $8$: 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 [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 $n$. :'''Question''' $9$: Does $T:=U\cap\Bbb P$ have infinitely many primes? :'''Question''' $10$: Is the set $V:=\{p+2q\mid p,q\in\Bbb P,\,p+2q\in T\}$ infinite? '''Guess''' $1$: $\forall m,n\in\Bbb N,$ assume $p_n,q_n\in\Bbb P$ such that there is no prime number in these intervals $(p_n,m\times10^n),(m\times10^n,q_n)$ then $\lim_{n\to\infty}{q_n\over p_n}=1$. :[https://mathoverflow.net/questions/310201/a-question-relating-to-the-prime-gaps/310223#310223 Answer] given by [https://mathoverflow.net/users/3402/gerhard-paseman $@$GerhardPaseman] from stackexchange site: It turns out that $1$) explicitly for all numbers $M$ greater than $30$, there are at least two primes in $(5M/6,6M/5)$, one bigger than $M$ and one smaller than $M$, and $2$) there is $N$ large enough that for $M$ bigger than $N$ there are more than two primes in $(M,M+M^{\alpha})$, where you can pick $\alpha$ a real number larger than $0.525$. So it is like the values $p_n$ and $q_n$ will be at most a little more than $\sqrt{q_n}$ apart, which means the limit of the ratio as $n$ increases will be $1$. You can also try this with estimates from Chebyshev (before PNT) to reach the same conclusion, but it will be less obvious. ::but could we replace another natural number rather than $10$ as $m\times d^n$? Yes, you can. The answer is essentially the same: the ratio as $n$ grows will eventually tend to $1$. :::'''Problem''' $1$: For each infinite strictly increasing subsequence of $\Bbb N$ like $\{a_n\}$ assume $\{p_n\}$ & $\{q_n\}$ are infinite strictly increasing subsequences of $\Bbb P$ such that $\forall n\in\Bbb N,\,p_n$ is largest prime less than $a_n$ & $q_n$ is smallest prime greater than $a_n$ then discuss on the limit below: '"`UNIQ-MathJax17-QINU`"' There is an especial (not necessarily unique), infinite and proper subsequence in prime numbers that gives the map of all prime numbers. Alireza Badali 12:34, 28 April 2018 (CEST) == Some dissimilar conjectures == '''Algebraic analytical number theory''' Alireza Badali 16:51, 4 July 2018 (CEST) === [https://en.wikipedia.org/wiki/Collatz_conjecture 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 [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. '''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 [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. <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 '"`UNIQ-MathJax18-QINU`"' Let $\bar{S}$ be the complement of $S$ with respect to the positive integers; i.e., '"`UNIQ-MathJax19-QINU`"' Your sequence is then $T=(T_0,T_1,T_2,\ldots)$, where '"`UNIQ-MathJax20-QINU`"' :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 [http://oeis.org/A171945 A171945], [http://oeis.org/A053661 A053661], [http://oeis.org/A034701 A034701] respectively. These are all discussed in ''[https://www.sciencedirect.com/science/article/pii/S0012365X11001427 The vile, dopey, evil and odious game players]''. -------------------------- :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]</small> <small>'''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 $C$?</small> '''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 $n\star n^{-1}=1$. <small>'''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 $C$?</small> <small>'''Question''' $3$: If $(\Bbb N,\star)$ is a cyclic group with generators $u,v$ & identity element $1$ then could $(\Bbb N,\star_1)$ be another cyclic group with: $\begin{cases} \forall m,n\in\Bbb N,\\ e=1\\ m\star_1n=(2m)\star(2n) & \text{if } m,n\text{ aren't of the form } 6k+4,\,k\in\Bbb N\\ (6m+4)\star_1n=(2m+1)\star(2n) & \text{if } n\text{ isn't of the form } 6k+4,\,k\in\Bbb N\\ (6m+4)\star_1(6n+4)=(2m+1)\star(2n+1)\\ n^{-1}=k & \text{if } n\text{ isn't of the form } 6t+4,\,t\in\Bbb N,\,k\star(2n)^{-1}=1\\ (6m+4)^{-1}=k & k\star(2m+1)^{-1}=1\\ \Bbb N=\langle u_1\rangle=\langle v_1\rangle & \begin{cases} u_1=\begin{cases} 2k+1 & \text{if } u\text{ is of the form } 6k+4,\,k\in\Bbb N\\ 2u & \text{otherwise}\end{cases}\\ v_1=\begin{cases} 2k+1 & \text{if } v\text{ is of the form } 6k+4,\,k\in\Bbb N\\ 2v & \text{otherwise}\end{cases}\end{cases}\end{cases}$ ? maybe.</small> Question $4$: Has any relation on the Collatz tree been discovered other than its definitions? In order to define a group structure on the [https://www.jasondavies.com/collatz-graph/ Collatz tree] I need such relations but other than its [https://en.wikipedia.org/wiki/Collatz_conjecture definitions], please introduce them (ideally suited a relation could be equivalent to its definition) if exist. if such a relation there was then via a group on $\Bbb N$, we could define a group on the Collatz tree. Alireza Badali 10:02, 12 May 2018 (CEST) === [https://en.wikipedia.org/wiki/Erdős–Straus_conjecture 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) ==='"`UNIQ--h-6--QINU`"' [https://en.wikipedia.org/wiki/Landau%27s_problems 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 [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. '''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) ==='"`UNIQ--h-7--QINU`"' [https://en.wikipedia.org/wiki/Lemoine%27s_conjecture 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) ==='"`UNIQ--h-8--QINU`"' 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. [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 $L=\{\text{floor}(n\cdot r)\mid n\in\Bbb N\}$. '''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) =='"`UNIQ--h-9--QINU`"' 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) ==='"`UNIQ--h-10--QINU`"' [https://en.wikipedia.org/wiki/Gaussian_moat Gaussian moat problem] === Alireza Badali 18:13, 20 June 2018 (CEST) ==='"`UNIQ--h-11--QINU`"' [https://en.wikipedia.org/wiki/Grimm%27s_conjecture Grimm's conjecture] === Alireza Badali 18:13, 20 June 2018 (CEST) ==='"`UNIQ--h-12--QINU`"' [https://en.wikipedia.org/wiki/Oppermann%27s_conjecture Oppermann's conjecture] === Alireza Badali 18:13, 20 June 2018 (CEST) ==='"`UNIQ--h-13--QINU`"' [https://en.wikipedia.org/wiki/Legendre%27s_conjecture Legendre's conjecture] === Alireza Badali 18:13, 20 June 2018 (CEST) =='"`UNIQ--h-14--QINU`"' 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 [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 $(\Bbb N\cup\{0\},\star)$ 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: $\begin{cases} \forall m,n,u,v\in\Bbb N\cup\{0\},\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\ {m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={-m\over n}\qquad m\star(-m)=0\end{cases}$• ::in addition under a total order relation, an unique & specific division algorithm like this [https://en.wikipedia.org/wiki/Division_algorithm one] in accordance with $(\Bbb N\cup\{0\},\star,\circ)$ is needed which given two natural numbers $m$ and $n$, computes their quotient and/or remainder, the result of division. 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: $\begin{cases} \forall m,n,u,v\in\Bbb N\cup\{0\},\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={-m\over n}\qquad m\star(-m)=0\end{cases}$ :in addition an unique & specific division algorithm like this [https://en.wikipedia.org/wiki/Division_algorithm one] in accordance with $(\Bbb N\cup\{0\},\star,\circ)$ is needed which given two natural numbers $m$ and $n$, computes their quotient and/or remainder, the result of division• <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 '"`UNIQ-MathJax21-QINU`"'</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: '"`UNIQ-MathJax22-QINU`"' 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> :::::::<small>'''Question''' $5$: If $p_n$ is $n$_th prime number then does '"`UNIQ-MathJax23-QINU`"'</small> ::::::::<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> Alireza Badali 16:26, 26 June 2018 (CEST) ==='"`UNIQ--h-15--QINU`"' [https://en.wikipedia.org/wiki/Many-worlds_interpretation 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: '"`UNIQ-MathJax24-QINU`"' '"`UNIQ-MathJax25-QINU`"' then regarding the sequence above find a rotation number of the form $4t,\,t\in\Bbb N$ that for this sequence is $4$ (or $4t$) and hence equations should be written with module $2$ (or $2t$) 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 $4t$) 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: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$ • ::in addition an unique & specific division algorithm like this [https://en.wikipedia.org/wiki/Division_algorithm one] in accordance with $(\Bbb Z,\star,\circ)$ is needed which given two integers $m$ and $n$, computes their quotient and/or remainder, the result of division• '''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: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\ {m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$• :in addition under a total order relation, an unique & specific division algorithm like this [https://en.wikipedia.org/wiki/Division_algorithm one] in accordance with $(\Bbb Z,\star,\circ)$ is needed which given two integers $m$ and $n$, computes their quotient and/or remainder, the result of division• 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) ==='"`UNIQ--h-16--QINU`"' [https://en.wikipedia.org/wiki/Gauss_circle_problem Gauss circle problem] === <small>Given this sequence: '"`UNIQ-MathJax26-QINU`"' '"`UNIQ-MathJax27-QINU`"' 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: $(6m-4)\circ(6n-3)=t(t^{-1}(6m-4)\times t^{-1}(6n-3))=t((-6m+5)\times(6n-4))=$ $t(-6(6mn-4m-5n+4)+4)=-6(6mn-4m-5n+4)+3=-36mn+24m+30n-21,$ $(6m-4)\circ(-6n+1)=$ $t(t^{-1}(6m-4)\times t^{-1}(-6n+1))=t((-6m+5)\times(-6n))=t(6(6mn-5n))=-6(6mn-5n)+2=-36mn+30n+2,$ $(-6m+1)\circ(-6n+1)=t(t^{-1}(-6m+1)\times t^{-1}(-6n+1))=t((-6m)\times(-6n))=t(6(6mn))=-6(6mn)+2=$ $-36mn+2$.</small> 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: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$ :in addition an unique & specific division algorithm like this [https://en.wikipedia.org/wiki/Division_algorithm one] in accordance with $(\Bbb Z,\star,\circ)$ is needed which given two integers $m$ and $n$, computes their quotient and/or remainder, the result of division• '''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)
How to Cite This Entry:
Musictheory2math. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Musictheory2math&oldid=42085