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(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)
 
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== A new version of Goldbach's conjecture ==
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== <big>$\mathscr B$</big> $theory$ (algebraic topological analytical number theory) ==
  
'''Nanas lemma''': If $\mathbb{P}$ is the set of prime numbers and $S$ is a set that it 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.'''$($'''Nanas Dadile is my parrot and I like it so much.'''$)$'''
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Logarithm function as an inverse of the function $f:\Bbb N\to\Bbb R,\,f(n)=a^n,\,a\in\Bbb R$ has prime numbers properties because in usual definition of prime numbers multiplication operation is a point meantime we have $a^n=a\times a\times ...a,$ $(n$ times$),$ hence prime number theorem or its extensions or some other forms is applied in $B$ theory for solving problems on prime numbers exclusively and not all natural numbers.
  
'''This lemma is a base for finding formula of prime numbers.'''
 
  
There is a musical note on the natural numbers that it can be discovered by the formula of prime numbers.Alireza Badali 22:21, 8 May 2017 (CEST)
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'''Algebraic structures & topology with homotopy groups & prime number theorem and its extensions or other forms or corollaries with limitation concept'''
  
: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 lemma 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)
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Alireza Badali 00:49, 25 June 2018 (CEST)
  
::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.Alireza Badali 22:21, 8 May 2017 (CEST)
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=== [https://en.wikipedia.org/wiki/Goldbach%27s_conjecture Goldbach's conjecture] ===
  
You mean, how to prove that $S$ is dense in $(0.1,1)$, right? Well, on the page "[[Distribution of prime numbers]]", in Section 6 "The difference between prime numbers", we have $ d_n \ll p_n^\delta $, where $p_n$ is the $n$-th prime number, and $ d_n = p_{n+1}-p_n $ is the difference between adjacent prime numbers; this relation holds for all $ \delta > \frac{7}{12} $; in particular, taking $ \delta = 1 $ we get $ d_n \ll p_n $, that is, $ \frac{d_n}{p_n} \to 0 $ (as $ n \to \infty $), or equivalently, $ \frac{p_{n+1}}{p_n} \to 1 $. Now, your set $S$ consists of numbers $ s_n = 10^{-k} p_n $ for all $k$ and $n$ such that $ 10^{k-1} < p_n < 10^k $. Assume that $S$ is not dense in $(0.1,1).$ Take $a$ and $b$ such that $ 0.1 < a < b < 1 $ and $ s_n \notin (a,b) $ for all $n$; that is, no $p_n$ belongs to the set
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'''Lemma''': For each subinterval $(a,b)$ of $[0.1,1),\,\exists m\in \Bbb N$ that $\forall k\in \Bbb N$ with $k\ge m$ then $\exists t\in (a,b)$ that $t\cdot 10^k\in \Bbb P$.
\[
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:[https://math.stackexchange.com/questions/2482941/a-simple-question-about-density-in-the-interval-0-1-1/2483079#2483079 Proof] given by [https://math.stackexchange.com/users/149178/adayah @Adayah] from stackexchange site: Without loss of generality (by passing to a smaller subinterval) we can assume that $(a, b) = \left( \frac{s}{10^r}, \frac{t}{10^r} \right)$, where $s, t, r$ are positive integers and $s < t$. Let $\alpha = \frac{t}{s}$.  
X = (10a,10b) \cup (100a,100b) \cup (1000a,1000b) \cup \dots \, ;
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:The statement is now equivalent to saying that there is $m \in \mathbb{N}$ such that for every $k \geqslant m$ there is a prime $p$ with $10^{k-r} \cdot s < p < 10^{k-r} \cdot t$.
\]
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:We will prove a stronger statement: there is $m \in \mathbb{N}$ such that for every $n \geqslant m$ there is a prime $p$ such that $n < p < \alpha \cdot n$. By taking a little smaller $\alpha$ we can relax the restriction to $n < p \leqslant \alpha \cdot n$.
all $ p_n $ belong to its complement
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:Now comes the prime number theorem: $$\lim_{n \to \infty} \frac{\pi(n)}{\frac{n}{\log n}} = 1$$
\[
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:where $\pi(n) = \# \{ p \leqslant n : p$ is prime$\}.$ By the above we have $$\frac{\pi(\alpha n)}{\pi(n)} \sim \frac{\frac{\alpha n}{\log(\alpha n)}}{\frac{n}{\log(n)}} = \alpha \cdot \frac{\log n}{\log(\alpha n)} \xrightarrow{n \to \infty} \alpha$$
Y = (0,\infty) \setminus X = (0,10a] \cup [10b,100a] \cup [100b,1000a] \cup \dots
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:hence $\displaystyle \lim_{n \to \infty} \frac{\pi(\alpha n)}{\pi(n)} = \alpha$. So there is $m \in \mathbb{N}$ such that $\pi(\alpha n) > \pi(n)$ whenever $n \geqslant m$, which means there is a prime $p$ such that $n < p \leqslant \alpha \cdot n$, and that is what we wanted♦
\]
 
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)
 
  
:''''''I have special thanks to Professor [[User:Boris_Tsirelson|Boris Tsirelson]] for this beauty proof; Yours Sincerely, Alireza Badali Sarebangholi''''''
 
  
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Now we can define function $f:\{(c,d)\mid (c,d)\subseteq [0.01,0.1)\}\to\Bbb N$ that $f((c,d))$ is the least $n\in\Bbb N$ that $\exists t\in(c,d),\,\exists k\in\Bbb N$ that $p_n=t\cdot 10^{k+1}$ that $p_n$ is $n$_th prime and $\forall m\ge f((c,d))\,\,\exists u\in (c,d)$ that $u\cdot 10^{m+1}\in\Bbb P$
  
'''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.Alireza Badali 22:21, 8 May 2017 (CEST)
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and $g:(0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\to\Bbb N,$ is a function by $\forall\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))$ $g(\epsilon)=max(\{f((c,d))\mid d-c=\epsilon,$ $(c,d)\subseteq [0.01,0.1)\})$.
  
:Ah, yes, I see, this follows easily from the fact that $S$ is dense. Sounds good. Though, decimal digits are of little interest in the number theory. (I think so; but I am not an expert in the number theory.) [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 11:16, 19 March 2017 (CET)
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'''Guess''' $1$: $g$ isn't an injective function.
  
Dear Professor Boris Tsirelson, your help is very valuable to me and I think we can make a good paper together of course if you would like.Alireza Badali 22:21, 8 May 2017 (CEST)
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'''Question''' $1$: Assuming guess $1$, let $[a,a]:=\{a\}$ and $\forall n\in\Bbb N,\, h_n$ is the least subinterval of $[0.01,0.1)$ like $[a,b]$ in terms of size of $b-a$ such that $\{\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\mid g(\epsilon)=n\}\subsetneq h_n$ and obviously $g(a)=n=g(b)$ now the question is $\forall n,m\in\Bbb N$ that $m\neq n$ is $h_n\cap h_m=\emptyset$?
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:[https://math.stackexchange.com/questions/2518063/a-medium-question-about-a-set-related-to-prime-numbers/2526481#2526481 Guidance] given by [https://math.stackexchange.com/users/276986/reuns @reuns] from stackexchange site:
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:* For $n \in \mathbb{N}$ then $r(n) = 10^{-\lceil \log_{10}(n) \rceil} n$, ie. $r(19) = 0.19$. We look at the image by $r$ of the primes $\mathbb{P}$.
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:* Let $F((c,d)) = \min \{ p \in \mathbb{P}, r(p) \in (c,d)\}$ and $f((c,d)) = \pi(F(c,d))= \min \{ n, r(p_n) \in (c,d)\}$  ($\pi$ is the prime counting function)
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:* If you set $g(\epsilon) = \max_a \{ f((a,a+\epsilon))\}$ then try seing how $g(\epsilon)$ is constant on some intervals defined in term of the prime gap $g(p) = -p+\min \{ q \in \mathbb{P}, q > p\}$  and things like $ \max \{  g(p), p > 10^i, p+g(p) < 10^{i+1}\}$
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:Another guidance: The affirmative answer is given by [[Liouville theorems|Liouville's theorem on approximation of algebraic numbers]].
  
:Thank you for the compliment and the invitation, but no, I do not. Till now we did not write here anything really new in mathematics. Rather, simple exercises. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 18:50, 27 March 2017 (CEST)
 
  
::But do not you think this way about prime numbers is new and for the first time.Alireza Badali 22:21, 8 May 2017 (CEST)
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Suppose $r:\Bbb N\to (0,1)$ is a function given by $r(n)$ is obtained by putting a point at the beginning of $n$ instance $r(34880)=0.34880$ and similarly consider $\forall k\in\Bbb N,\, w_k:\Bbb N\to (0,1)$ is a function given by $\forall n\in\Bbb N,$ $w_k(n)=10^{1-k}\cdot r(n)$ and let $S=\bigcup _{k\in\Bbb N}w_k(\Bbb P)$.
  
:It is not enough to say that this ''way'' is new. The question is, does this way give new interesting ''results''? [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 21:03, 30 March 2017 (CEST)
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'''Theorem''' $1$: $r(\Bbb P)$ is dense in the interval $[0.1,1]$. (proof using lemma above)
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:Regarding to expression form of Goldbach's conjecture, by using this theorem, I wanted enmesh prime numbers properties (prime number theorem should be used for proving this theorem and there is no way except using prime number theorem to prove this density'''(?)''' because there is no deference between a prime $p$ and its image $r(p)$ other than a sign or a mark as a point for instance $59$ & $0.59$.) towards Goldbach hence I planned this method.
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: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 ${n\over\log\log n}$).
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:A corollary: For each natural number like $a=a_1a_2a_3...a_k$ that $a_j$ is $j$_th digit for $j=1,2,3,...,k$, there is a natural number like $b=b_1b_2b_3...b_r$ such that the number $c=a_1a_2a_3...a_kb_1b_2b_3...b_r$ is a prime number.
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::Question $2$: Which mathematical concept at $[0.1,1)$ could be in accordance with the prime gap at natural numbers?
  
::Dear Professor Boris Tsirelson, I thank you so much for your valuable help to me and I owe you because I was unable for proving the Nanas lemma but you proved it seemly and guided me honestly that in principle you gave me a new hope to continue.
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'''Question''' $3$: What is equivalent to the ''prime number theorem'' in $[0.1,1)$?
  
  
'''Goldbach's conjecture''' is one of the oldest and best-known unsolved problems in number theory and all of mathematics. It states: Every even integer greater than $2$ can be expressed as the sum of two primes.
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'''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''' $2$: $$\lim_{n\to\infty}\frac{\prod_{i=1}^mr(p_{in})}{\prod_{p\in\Bbb P,\,p\lt p_{1n}}r(p)}\sim({5\over9})^9\,.$$
  
Assume $S_1$={ $a/10^n$ | $a\in{S}$ for $n$=$0,1,2,3,...$ } & $L=\{(a,b)\,|\,a,b \in S_1$ & $0.01 \lt a+b \le 0.1$ & $a \times 10^m,\,b \times 10^m$ are prime numbers & $m \in \Bbb{N}\}$
 
  
'''Theorem''': $S_1$ is dense in the interval $(0,1)$.
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'''Theorem''' $2$: $S$ is dense in the interval $[0,1]$ and $S\times S$ is dense in the $[0,1]\times [0,1]$.
  
If $p,q$ are prime numbers and $n$ is the number of digits of $p+q$ and $m=$max(number of digits of $p$, number of digits of $q$), let $\varphi : L \to \Bbb N,$ $\varphi ((p,q)) = \begin{cases} m+1 & n=m \\m+2 &  n=m+1 \end{cases}$
 
  
'''Theorem''': For each $p,q$ belong to prime numbers and $\alpha \in \Bbb R$ that $0 \le \alpha,$ now if $\alpha = q/p$  then $L \cap \{(x,y)\,|\,y=\alpha x \}=\{10^{-\varphi ((p,q))}(p,q)\}$ and if $\alpha \neq q/p$ then $L \cap \{(x,y)\,|\,y=\alpha x \}=\emptyset $ and if $\alpha = 1$ then $L \cap \{(x,x)\}$ is dense in the $(0,1) \times (0,1) \cap \{(x,x)\,|\,0.005 \lt x \le 0.05 \}$.
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'''An algorithm''' that makes new cyclic groups on $\Bbb N$:
  
'''Definition''': Assume $L_1=\{(a,b)\,|\,(a,b) \in L$ & $b \lt a \}$ and $F$ be a curve contain all of points of $L_1$ such that for each $\alpha \in \Bbb R$ that $0 \lt \alpha \lt 1 ,$ $\{(x,y)\,|\, (x,y) \in F \} \cap \{(x,y)\,|\, y= \alpha x \}$ has exactly one point and let $E=(0.007,0.005)$ be a base for homotopy groups!
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Let $\Bbb N$ be that group and at first write integers as a sequence with starting from $0$ and let identity element $e=1$ be corresponding with $0$ and two generators $m$ & $n$ be corresponding with $1$ & $-1$ so we have $\Bbb N=\langle m\rangle=\langle n\rangle$ for instance: $$0,1,2,-1,-2,3,4,-3,-4,5,6,-5,-6,7,8,-7,-8,9,10,-9,-10,11,12,-11,-12,...$$ $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,...$$
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then regarding to the sequence find 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:
  
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$\begin{cases} m\star _u 1=m\\ (4m)\star _u (4m-2)=1=(4m+1)\star _u (4m-1)\\ (4m-2)\star _u (4n-2)=4m+4n-5\\ (4m-2)\star _u (4n-1)=4m+4n-2\\ (4m-2)\star _u (4n)=\begin{cases} 4m-4n-1 & 4m-2\gt 4n\\ 4n-4m+1 & 4n\gt 4m-2\\ 3 & m=n+1\end{cases}\\ (4m-2)\star _u (4n+1)=\begin{cases} 4m-4n-2 & 4m-2\gt 4n+1\\ 4n-4m+4 & 4n+1\gt 4m-2\end{cases}\\ (4m-1)\star _u (4n-1)=4m+4n-1\\ (4m-1)\star _u (4n)=\begin{cases} 4m-4n+2 & 4m-1\gt 4n\\ 4n-4m & 4n\gt 4m-1\\ 2 & m=n\end{cases}\\ (4m-1)\star _u (4n+1)=\begin{cases} 4m-4n-1 & 4m-1\gt 4n+1\\ 4n-4m+1 & 4n+1\gt 4m-1\\ 3 & m=n+1\end{cases}\\ (4m)\star _u (4n)=4m+4n-3\\ (4m)\star _u (4n+1)=4m+4n\\ (4m+1)\star _u  (4n+1)=4m+4n+1\\ \Bbb N=\langle 2\rangle=\langle 4\rangle\end{cases}$
  
'''A conjecture''': For each natural number like $t=t_1t_2t_3...t_k$, now $1)$ if $t_1=5 ,6, 7, 8, 9$ and $\forall m \in \Bbb N $ $t \neq 5 \times 10^{m-1}$ then there are two points in the $L \cap \{(x,y)\,|\,x+y=2 \times 0.00t_1t_2t_3...t_k\}$ like $(a,b),\,(b,a)$ such that $0.00t_1t_2t_3...t_k=(a+b)/2$ & $10^{k+2} \times a,\,10^{k+2} \times b$ are prime numbers and if $t$ is a prime number then $a=b=0.00t_1t_2t_3...t_k$ and $2)$ if $t_1=1, 2, 3, 4$ or $\exists k\in \Bbb N$ $t=5\times 10^{k-1}$ then there are two points in the $L \cap \{(x,y)\,|\,x+y=2 \times 0.0t_1t_2t_3...t_k\}$ like $(a,b),\,(b,a)$ such that $0.0t_1t_2t_3...t_k=(a+b)/2$ & $10^{k+1} \times a,\,10^{k+1} \times b$ are prime numbers and if $t$ is a prime number then $a=b=0.0t_1t_2t_3...t_k$.
 
  
This conjecture is ''a new version of Goldbach's conjecture''. This conjecture has two solutions $1)$ Homotopy groups $\pi _{n \gt 1} (X)$ and $2)$ Algebraic methods, but in either case we must attend to two spheres $S^2$.
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Problem $1$: By using matrices rewrite operation of every group on $\Bbb N$.
  
'''Assuming''' correctness the new version of Goldbach's conjecture, it is a way for finding formula of prime numbers, because it explains behavior of prime numbers on $(0,1) \times (0,1)$!
 
  
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Assume $\forall m,n\in\Bbb N$: $\begin{cases} n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\end{cases}$
  
'''Theorem''': $\forall p,q,r,s$ belong to prime numbers & $q \lt p$, $(p,q)$ is located at the direct line $(0,0),10^{-\varphi ((p,q))}(p,q)$ and if $(r,s)$ is belong to this line then $p=r$ & $q=s$.
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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 $\langle 2\rangle =\langle 3\rangle =(\Bbb N,\star)\simeq (\Bbb Z,+)\simeq (\Bbb P,\star _1)=\langle 3\rangle=\langle 5\rangle$.
  
Let $A=\{(x,y)\,|\,0.01 \lt x+y \le 0.1$ & $0 \lt x$ & $0 \lt y $ & $y \lt x \}$ and $S^2_2$ 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 $x$ make 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 $x$ make a direct line that be tangent on $S^2_1$ and then in this junction point make a direct line perpendicular at $S^2_2$ till cut $S^2_2$ in curve $f_2 (x)$.
 
  
Let $f_3: L_1 \to S^1,$ $f_3 ((a,b)) = (a^2+b^2)^{-0.5}(a,b)$ and $f_4:L_1 \to S^2,$ $f_4 ((a,b)) = (a^4+a^2b^2+b^2)^{-0.5}(a^2,ab,b) $
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I want make some topologies having '''prime numbers properties''' presentable in the collection of '''open sets''', in principle when we image a prime $p$ to real numbers as $w_k(p)$ indeed we accompany prime numbers properties among real numbers which regarding to the expression form of prime number theorem for this aim we should use an important mathematical technique as logarithm function into some planned topologies: '''question''' $4$: Let $M$ be a topological space and $A,B$ are subsets of $M$ with $A\subset B$ and $A$ is dense in $B,$ since $A$ is dense in $B,$ is there some way in which a topology on $B$ may be induced other than the subspace topology? I am also interested in specialisations, for example if $M$ is Hausdorff or Euclidean. ($M=\Bbb R,\,B=[0,1],\,A=S$ or $M=\Bbb R^2,$ $B=[0,1]\times[0,1],$ $A=S\times S$)
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:Perhaps these techniques are useful:
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:$\forall n\in\Bbb N,$ and for each subinterval $(a,b)$ of $[0.1,1),$ that $a\neq b,$ assume:
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:$\begin{cases} U_{(a,b)}:=\{n\in\Bbb N\mid a\le r(n)\le b\},\\ \\V_{(a,b)}:=\{p\in\Bbb P\mid a\le r(p)\le b\},\\ \\U_{(a,b),n}:=\{m\in U_{(a,b)}\mid m\le n\},\\ \\V_{(a,b),n}:=\{p\in V_{(a,b)}\mid p\le n\},\\ \\w_{(a,b),n}:={\#V_{(a,b),n}\over\#U_{(a,b),n}}\cdot\log n,\\ \\w_{(a,b)}:=\lim _{n\to\infty} w_{(a,b),n},\\ \\z_{(a,b),n}:={\#V_{(a,b),n}\over\#U_{(a,b),n}}\cdot\log{(\#U_{(a,b),n})},\\ \\z_{(a,b)}:=\lim_{n\to\infty}z_{(a,b),n}\end{cases}$
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::Guess $3$: $\forall (a,b)\subset [0.1,1),\,w_{(a,b)}={10\over9}\cdot(b-a)$.
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:::[https://math.stackexchange.com/questions/2683513/an-extension-of-prime-number-theorem/2683561#2683561 Answer] given by [https://math.stackexchange.com/users/82961/peter $@$Peter] from stackexchange site: Imagine a very large number $N$ and consider the range $[10^N,10^{N+1}]$. The natural logarithms of $10^N$ and $10^{N+1}$ only differ by $\ln(10)\approx 2.3$ Hence the reciprocals of the logarithms of all primes in this range virtually coincicde. Because of the approximation $$\int_a^b \frac{1}{\ln(x)}dx$$ for the number of primes in the range $[a,b]$ the number of primes is approximately the length of the interval divided by $\frac{1}{\ln(10^N)}$, so is approximately equally distributed. Hence your conjecture is true.
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:::Benfords law seems to contradict this result , but this only applies to sequences producing primes as the Mersenne primes and not if the primes are chosen randomly in the range above.
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::::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>
  
It is clear $L_1 \cup \{(x,x)\,|\, 0.005 \le x \le 0.05 \} \subseteq \bar {L_1}$ and maybe $\bar {L_1} \setminus \{(x,y)\,|\, y=x$ or $y=0\} =F$.
 
  
'''Guess''' $1$: $f_3 (L_1) $ is dense in the $S^1 \cap \{ (x,y)\,|\, 0 \le y$ & $2^{-0.5} \le x \}$.
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Using homotopy groups Goldbach's conjecture will be proved.
  
 +
Alireza Badali 08:27, 31 March 2018 (CEST)
  
Let $U: S^2_2 \setminus \{(0,0,2r_2)\} \to \{(x,y,0)\,|\, x,y \in \Bbb R \}$ such that make 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 $S^2_2 \setminus \{(0,0,2r_2)\}$.
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==== Goldbach by $\Bbb N$ ====
  
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 vector space $(\Bbb R^2,\Bbb Q,+,.)$ and now we must attend to subgroups of $G$ particularly $y=\pm x,\,y=0,\,x=0$
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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
  
'''Theorem''': Let $\Bbb P$ is the set prime numbers and $K_3 =\{p+q+r\,|\, p,q,r \in \Bbb P \}$ and put a point in the beginning each member of $K_3$ in order to be obtained $N$ so $N$ is dense in the interval $(0.1,1)$ of real numbers. Proof by Goldbach's weak conjecture.
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$\begin{cases} r(m)\lt r(n),\,m=m_1\times10^s,\,n=n_1\times10^s,\,10\nmid m_1,\,10\nmid n_1,\,m_1,n_1\in\Bbb N,\,s\in\Bbb N\cup\{0\} & \text{ or}\\ \\m=m_1\times10^s,\,n=n_1\times10^t,\,s\lt t,\,10\nmid m_1,\,10\nmid n_1,\,m_1,n_1,t\in\Bbb N,\,s\in\Bbb N\cup\{0\}\end{cases}$
  
'''Guess''' $2$: If $\Bbb P$ is the set prime numbers and $K_2 =\{p+q\,|\,p,q \in \Bbb P \}$ and put a point in the beginning each member of $K_2$ in order to be obtained $M$ so $M$ is dense in the interval $(0.1,1)$ of real numbers.
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then assume $\mathfrak T$ is a topology on $\Bbb N$ induced by $\lt_1$ ($(\Bbb N,\mathfrak T)$ is a Hausdorff space).
  
Alireza Badali 22:21, 8 May 2017 (CEST)
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'''Theorem''' $1$: $\Bbb P$ is dense in the interval $(1,10)$.
  
== Discussion about ''A new version of ...'' ==
 
  
''1)'' Let $F= \Bbb Q$, so what are Galois group of polynomials $x^4+b^2x^2+b^2$ and $(1+a^2)x^2 +a^4$.
+
on the other hand $\Bbb N$ is a cyclic group by:
  
Alireza Badali 18:37, 24 June 2017 (CEST)
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$\begin{cases} \forall m,n\in\Bbb N\\ e=1\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\\\langle 2\rangle =\langle 3\rangle =(\Bbb N,\star)\end{cases}$
  
== From ''Nanas Lemma'' ==
 
  
Assume $H$ is a mapping from $(0.1,1)$ on $(0.1,1)$ given by $H(x)=1/(10x)$. And let $T=H(S)$, $T$ is a interesting set for its members because of, a member of $S$ like $0.a_1a_2a_3...a_n$ that $a_j$ is $j$-th digit for $j=1,2,3, ... ,n$ is basically different with ${a_1.a_2a_3a_4...a_n}^{-1}$ in $T$.
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Question $1$: Is $\Bbb N$ a topological group?
  
Theorem: $T$ is dense in the $(0.1,1)$.
 
  
Theorem: $S×S$ is dense in the $(0.1,1)×(0.1,1)$. Similar theorems are right for $S×T$ & $T×T$.Alireza Badali 13:49, 17 May 2017 (CEST)
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'''Goldbach's conjecture''': $\forall n\in\Bbb N,\,\exists p,q\in\Bbb P\setminus\{2\}$ such that $2n+3=p\star q$.
  
:"Theorem: T=H(P) that P is the set of prime numbers is dense in the (0.1 , 1)." — I guess you mean H(S), not H(P). Well, this is just a special case of a simple topological fact (no number theory needed): A is dense if and only if H(A) is dense (just because H is a [[homeomorphism]]).
 
:"Theorem: C=S×S is dense in the (0.1 , 1)×(0.1 , 1) similar theorems is right for C=S×T and C=T×S and C=T×T."  — This is also a special case of a simple topological fact: $A\times B$ is dense if and only if $A$ and $B$ are dense. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 18:53, 25 March 2017 (CET)
 
  
Let $D=\mathbb{Q} \cap (0.1,1)$
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'''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.
  
Theorem: $D$ and $S$ are homeomorph by the Euclidean topology.
 
  
For each member of $D$ like $w=0.a_1a_2a_3...a_ka_{k+1}a_{k+2}...a_{n-1}a_na_{k+1}a_{k+2}...a_{n-1}a_n...$ that $a_{k+1}a_{k+2}...a_{n-1}a_n$ repeats and $k=0,1,2,3,...,n$ , assume $t=a_1a_2a_3...a_ka_{k+1}...a_n00...00$ is a natural number such that $k$ up to $0$ is inserted on the beginning of $t$ , now by the induction axiom and theorem 1 , there is the least number in the natural numbers like $b_1b_2...b_r$ such that the number $a_1a_2a_3...a_ka_{k+1}...a_{n}00...00b_1b_2...b_r$ is a prime number and so $0.a_1a_2a_3...a_ka_{k+1}...a_{n}00...00b_1b_2...b_r\in{S}$.♥ But there is a big problem, where is the rule of this homeomorphism.
+
'''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.
  
Theorem: $D$ and $T$ are homeomorph by the Euclidean topology.
 
  
Let $T_1$={ $a/10^n$ | $a\in{T}$ for $n=0,1,2,3,...$ }.
+
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 $m+r(n)\lt v+r(u)$
  
Theorem: $T_1$ is dense in the interval $(0,1)$.
+
Alireza Badali 21:20, 17 September 2018 (CEST)
  
Assume $T_1$ is the dual of $S_1$ so the combine of both $T_1$ and $S_1$ make us so stronger.
+
==== Goldbach by odd numbers ====
  
Let $W$={ (z+a)$ | $a\in{S_1 \cup T_1}$ for $z=0,1,2,3,...$ } & $G=\mathbb{Q} \setminus W$
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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: $\begin{cases} \forall m,n\in\Bbb N\\ 2n-1\lt_12m-1 & \text{iff}\quad r(2n-1)\lt r(2m-1),\\ -2n+1\lt_1-2m+1 & \text{iff}\quad r(2n-1)\gt r(2m-1),\\ -2n+1\lt_10\lt_12m-1\end{cases}$
  
Theorem: $W$ and also $G$ are dense in the $\mathbb{Q}$ and also $\mathbb{R}$.
+
then assume $\mathfrak T$ is a topology on $Z_1$ induced by $\lt_1$ ($(Z_1,\mathfrak T)$ is a Hausdorff space).
  
''A'' '''conjecture''': For each member of $G$ like $g$, there are two members of $W$ like $a,b$ in the interval $(g-0.5,g+0.5)$ such that $g=(a+b)/2$.
+
'''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.
  
Of course it is enough that the conjecture just is proved for the interval $(0,1)$ namely assume $g\in{(0,1) \cap {G}}$.
 
  
 +
$(\Bbb N,\star_1)$ is a cyclic group with: $\begin{cases} \forall m,n\in\Bbb N\\  m\star_11=m\\ (2n)\star_1(2n+1)=1\\ (2m)\star_1(2n)=2m+2n\\ (2m-1)\star_1(2n-1)=2m+2n-3\\ (2m)\star_1(2n-1)=\begin{cases} 2m-2n+2 & 2m\gt2n-1\\ 2n-2m-1 & 2n-1\gt2m\end{cases}\\ \Bbb N=\langle2\rangle=\langle3\rangle\end{cases}$
  
''An important question'': Is there any proper infinite subset of $S$ such that it is dense in the interval $(0.1,1)$ of real numbers?
+
hence we can consider following cyclic group $(N_1,\star_2)$ with: $\begin{cases} \forall m,n\in\Bbb N\\ 1\star_2(2m-1)=2m-1\\ (4n-1)\star_2(4n+1)=1\\ (2m-1)\star_2(2n-1)=\begin{cases} 2m+2n-1 & \text{m,n are even}\\ 2m+2n-3 & \text{m,n are odd}\\ 2m-2n-1 & m\gt n,\,m\text{ is odd},\,n\text{ is even}\\ 2n-2m+1 & m\lt n,\,m\text{ is odd},\,n\text{ is even}\end{cases}\\ N_1=\langle3\rangle=\langle5\rangle\end{cases}$
  
Assuming $P_1$={ $p$ | $p$ is a prime number & number of digits of $p$ is a prime number } that $P_1$ is called the set of ''primer numbers'', now whether $P_1$ is dense in the $(0.1,1)$$?$ This is a suitable definition for information security and coding theory<big>'''!'''</big>
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and finally regarding sequences below we have the cyclic group $(Z_1,\star)$: $$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,…$$ $\begin{cases} \forall m,n\in\Bbb N,\quad e=0\\ (2m-1)\star(-2m+1)=0\\ (4m-3)\star(4n-3)=4m+4n-5\\ (4m-3)\star(-4n+3)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (4m-3)\star(4n-1)=4m+4n-3\\ (4m-3)\star(-4n+1)=\begin{cases} 4m-4n-1 & m\le n\\ 4m-4n-3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+3)=-4m-4n+5\\ (-4m+3)\star(4n-1)=\begin{cases} 4n-4m+1 & m\le n\\ 4n-4m+3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+1)=-4m-4n+3\\ (4m-1)\star(4n-1)=4m+4n-1\\ (4m-1)\star(-4n+1)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (-4m+1)\star(-4n+1)=-4m-4n+1\\ Z_1=\langle1\rangle=\langle-1\rangle\end{cases}$
  
Alireza Badali 22:21, 8 May 2017 (CEST)
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Question $1$: Is $Z_1$ a topological group?
  
==Other problems==
 
  
'''1)''' About set theory: So many years ago I heard of my friend that it has been proved that each set is order-able with total order, Is this right? and please say by who and when.Alireza Badali 22:21, 8 May 2017 (CEST)
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'''Goldbach’s conjecture''': $\forall n\in\Bbb N,\,\exists p,q\in P_1$ such that $2n+5=p\star q$.
  
:Yes. See [[Axiom of choice]]. There, find this: "Many postulates equivalent to the axiom of choice were subsequently discovered. Among these are: 1) The well-ordering theorem: On any set there exists a total order". [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 16:46, 8 April 2017 (CEST)
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Alireza Badali 19:52, 22 September 2018 (CEST)
  
::So is there any set with cardinal between $\aleph_0$ and $\aleph_1$ and not equal to $\aleph_0$ and $\aleph_1$, and also for $\aleph_1$ and $\aleph_2$ and so on, I think so the well-ordering theorem answers this question.Alireza Badali 22:21, 8 May 2017 (CEST)
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===== $Z_1$ a UFD =====
  
:See [[Continuum hypothesis]]. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 07:27, 9 April 2017 (CEST)
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$(\Bbb N,\star_1,\circ_1)$ is an integral domain (possibly a UFD) with: $\begin{cases} \forall m,n\in\Bbb N\\  m\star_11=m\\ (2n)\star_1(2n+1)=1\\ (2m)\star_1(2n)=2m+2n\\ (2m-1)\star_1(2n-1)=2m+2n-3\\ (2m)\star_1(2n-1)=\begin{cases} 2m-2n+2 & 2m\gt2n-1\\ 2n-2m-1 & 2n-1\gt2m\end{cases}\\ 1\circ_1m=1,\quad2\circ_1m=m,\quad(3\circ_1m)\star_1m=1\\ (2m)\circ_1(2n)=2mn\\ (2m+1)\circ_1(2n+1)=2mn\\ (2m)\circ_1(2n+1)=2mn+1\end{cases}$
:Also, did you try Wikipedia? Visit this: [https://en.wikipedia.org/wiki/Continuum_hypothesis WP:Continuum_hypothesis]. And this: [https://en.wikipedia.org/wiki/Wikipedia:WikiProject_Mathematics WP:Project Mathematics]. And [https://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics WP:Reference desk/Mathematics]. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 07:38, 9 April 2017 (CEST)
 
  
::Dear Professor Boris Tsirelson, I thank you so much and yes I should go to the Wikipedia more and I think the well-ordering theorem results each set is exactly a line. Yours Sincerely Badali
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hence $(N_1,\star_2,\circ_2)$ is an integral domain (possibly a UFD) with: $\begin{cases} \forall m,n\in\Bbb N,\,\forall v\in N_1\\ 1\star_2(2m-1)=2m-1\\ (4n-1)\star_2(4n+1)=1\\ (2m-1)\star_2(2n-1)=\begin{cases} 2m+2n-1 & \text{m,n are even}\\ 2m+2n-3 & \text{m,n are odd}\\ 2m-2n-1 & m\gt n,\,m\text{ is odd},\,n\text{ is even}\\ 2n-2m+1 & m\lt n,\,m\text{ is odd},\,n\text{ is even}\end{cases}\\ 1\circ_2v=1,\quad3\circ_2v=v,\quad(5\circ_2v)\star_2v=1\\ (8m-5) \circ_2(8n-5)=16mn-8m-8n+3\\ (8m-5) \circ_2(8n-3)=16mn-8m-8n+5\\ (8m-5) \circ_2(8n-1)=16mn-8n-1\\ (8m-5) \circ_2(8n+1)=16mn-8n+1\\ (8m-3) \circ_2(8n-3)=16mn-8m-8n+3\\ (8m-3) \circ_2(8n-1)=16mn-8n+1\\ (8m-3) \circ_2(8n+1)=16mn-8n-1\\ (8m-1) \circ_2(8n-1)=16mn-1\\ (8m-1) \circ_2(8n+1)=16mn+1\\ (8m+1) \circ_2(8n+1)=16mn-1\end{cases}$
  
'''2)''' About '''<big>?!</big>''': The set of real numbers is dense in the whole of mathematics( whole and not set ).
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from this table, $m\in\Bbb N$: $$4m-2,4m-1,4m,4m+1$$ $$8m-5,8m-3,8m-1,8m+1$$ for instance $(8m-3) \circ_2(8n+1)=t(t^{-1}(8m-3)\circ_1t^{-1}(8n+1))=t((4m-1) \circ_1(4n+1))=$ $t((2(2m-1)+1)\circ_1(2(2n)+1))=t(2(2m-1)(2n))=t(4(2mn-n))=8(2mn-n)-1=16mn-8n-1$
  
'''3)''' A wonderful definition in the mathematical philosophy (and perhaps mathematical logic too): In between the all of mathematical concepts, there are some special concepts that are not logical or are in contradiction with other concepts but in the whole of mathematics a changing occurred that made mathematics logical, I define this changing the curvature of mathematical concepts.
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and finally $(Z_1,\star,\circ)$ is an integral domain (possibly a UFD) with: $\begin{cases} \forall m,n\in\Bbb N,\,\forall v\in Z_1,\quad e=0\\ (2m-1)\star(-2m+1)=0\\ (4m-3)\star(4n-3)=4m+4n-5\\ (4m-3)\star(-4n+3)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (4m-3)\star(4n-1)=4m+4n-3\\ (4m-3)\star(-4n+1)=\begin{cases} 4m-4n-1 & m\le n\\ 4m-4n-3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+3)=-4m-4n+5\\ (-4m+3)\star(4n-1)=\begin{cases} 4n-4m+1 & m\le n\\ 4n-4m+3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+1)=-4m-4n+3\\ (4m-1)\star(4n-1)=4m+4n-1\\ (4m-1)\star(-4n+1)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (-4m+1)\star(-4n+1)=-4m-4n+1\\ 0\circ v=0,\quad1\circ v=v,\quad((-1)\circ v)\star v=0\\ (4m-3)\circ(4n-3)=8mn-4m-4n+1\\ (4m-3)\circ(-4n+3)=-8mn+4m+4n-1\\ (4m-3)\circ(4n-1)=8mn-4n-1\\ (4m-3)\circ(-4n+1)=-8mn+4n+1\\ (-4m+3)\circ(-4n+3)=8mn-4m-4n+1\\ (-4m+3)\circ(4n-1)=-8mn+4n+1\\ (-4m+3)\circ(-4n+1)=8mn-4n-1\\ (4m-1)\circ(4n-1)=8mn-1\\ (4m-1)\circ(-4n+1)=-8mn+1\\ (-4m+1)\circ(-4n+1)=8mn-1\end{cases}$
  
'''4)''' About mathematical logic: Complex set theory: It seems in different sections of mathematics we connect sets to each other by function concept that of course the own function is a set with a property too, but whether function concept has itself importance as a independent and basic axiom like Axiom of Choice or not, because in mathematics every work is occurred by function, in principle the Axiom of choice or natural numbers(or real numbers and or complex numbers) or function concept are dense in the whole of mathematics, and I want offer that a set theory is made just like complex numbers that set concept is real line and function concept is imagine line. Of course I think that the own perception is a function of own human not more and not less.
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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.
  
'''5)''' Most important question: The formula of prime numbers what impact will create on mindset of human.
 
  
'''6)''' About physics: The time dimension is a periodic phenomenon. Fifth dimension is thought with more speed than light.
+
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: $\begin{cases} \forall m,n,r,s\in\Bbb N,\\ m\pm n=r\qquad\text{iff}\quad(2m-1)\star(\pm(2n-1))=2r-1\\ m\cdot n=s\qquad\text{iff}\quad(2m-1)\circ(2n-1)=2s-1\quad\text{iff}\quad2s-1=(2m-1)\star(2m-1)\star...(2m-1)\,(n\text{ times})\end{cases}$.
  
'''7)''' Artificial intelligence is the ability to lying without any goofs of course only for a limited time.
+
Irreducible elements in $(Z_1,\star,\circ)$ except $3$ are of the form $4k-3,\,k\in\Bbb N$.
  
'''8)''' Each mathematical theory is a graph or a hyper graph.
+
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.
  
'''9)''' Mathematical logic is the language of Mathematics''(''There is not any difference between language of Mathematics and English literature or Persian literature or each other language, because both Mathematical and literary just state statements about some subjects and each one by own logical principles, of course language of Mathematics is very elementary till now compared with literature of a language and must be improved.'')''  and Mathematical philosophy is the way of thinking about Mathematics, always there was a special relationship between Mathematical philosophy and Mathematical logic, but nobody knows which one is first and basic, and there is no boundary between them, however, weakness of Mathematics is from ''1)'' disagreement between the Mathematical philosophy and the Mathematical logic''(''Of course whatever this disagreement is decreased, equally Mathematics grows.'')'' and ''2)'' weakness of own Mathematical logic, but music can be help to removing this problem, because of everything has harmony is a music too, In principle I want say that whatever this disagreement between expression and thinking is decreased equally Mathematics grows.
 
  
'''10)''' The Goldbach`s conjecture is a way for finding formula of prime numbers<big>!</big>
+
'''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 $4k-1,\,k\in\Bbb N$.
  
'''11)''' Whether the formula of number of topologies on a finite set is equivalent with the formula of prime numbers.
 
  
'''12)''' Whether always there is a topological space in each Mathematical proposition that embraces all of content of proposition?
+
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.
  
'''13)'''  Only algebraic theories or algebraic structures join cardinal of a set with its members and however make us so stronger but restrict the field of our act.
 
  
'''14)''' Universe is a wave toward truth! so lying is swimming in the opposite direction of the flow of water instead that universal wave helps honest man!
+
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 $(Z_1,\star,\circ)$.
  
'''15)''' Most our knowledge about algebraic structures on numbers is result of normal addition and multiplication (of course I understand why $\sqrt 2 \times {1 \over \sqrt 2}$ is $1$ in order to how being product this two number that result to be $1$ and this can be resulted from $1)$ another operation like $÷$ in definition algebraic structure is needed namely an algebraic structure with three operation, normal addition & multiplication & division as $(\Bbb R ,+,×,÷)$, or from $2)$ choice axiom ) but I want define a group on the natural numbers for this, I must obtain rule of sequence $a(n)$, $a:\Bbb N \to \Bbb Q,$ by $a(n)={n \over k(n)}$ such that $k:\Bbb N \to \Bbb Q$, $k(n)$ is the number of digits of $n$ so $a(1)=1,\,a(2)=2,\,...\,a(9)=9,\,a(10)=5,\,a(11)=5.5,\,...\,a(100)=33.\bar 3 ,\,a(101)=33.\bar 6 ,\, ... $
+
Alireza Badali 12:19, 24 September 2018 (CEST)
  
Alireza Badali 22:21, 8 May 2017 (CEST)
+
====== 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)
 +
 
 +
===== 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)
 +
 
 +
=== [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:
 +
$$0,-1,1,-3,-2,-5,3,2,-4,6,5,4,-6,-7,7,-9,-8,-11,9,8,-10,12,11,10,-12,-13,13,-15,$$ $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,$$ $$-14,-17,15,14,-16,18,17,16,-18,-19,19,-21,-20,-23,21,20,-22,24,23,22,-24,...$$ $$29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,...$$
 +
 
 +
 
 +
'''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: $$\lim_{n\to\infty}{q_n\over p_n}\,.$$
 +
 
 +
 
 +
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 $$S=(1, 4, 6, 10, 14, 16, 18, 22, 24, 26, 30, 34, 38, 40,42,\ldots).$$ Let $\bar{S}$ be the complement of $S$ with respect to the positive integers; i.e., $$\bar{S}=(2, 3, 5, 7, 8, 9, 11, 12, 13, 15, 17, 19, 20, 21, 23, 25,\ldots).$$ Your sequence is then $T=(T_0,T_1,T_2,\ldots)$, where
 +
$$T_n:=\begin{cases}S_{n\over 2}&\text{ if $n$ is even}\\
 +
\bar{S}_{n-1\over 2}&\text{ if $n$ is odd.}
 +
\end{cases}
 +
$$
 +
:Thus $T=(1, 2, 4, 3, 6, 5, 10, 7, 14, 8, 16, 9, 18, 11, 22, 12, 24, 13, 26, 15, 30, 17, 34, 19, 38, 20, \ldots).$
 +
 
 +
----------------------------
 +
 
 +
:References:
 +
:Sequences $S,\bar{S},T$ are OEIS [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)
 +
 
 +
=== [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 $(\Bbb N,\star)$.
 +
 
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prime number isn't an algebraic concept so we can not define primes by using isomorphism (and via algebraic equations primes can be defined) but since Gaussian integers contain all numbers of the form $m+ni,$ $m,n\in\Bbb N$ hence by using algebraic concepts we can solve some problems in number theory.
 +
:Question: what is definition of prime numbers in the $(\Bbb N,\star)$?
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 +
Alireza Badali 00:49, 25 June 2018 (CEST)
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=== [https://en.wikipedia.org/wiki/Gaussian_moat Gaussian moat problem] ===
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Alireza Badali 18:13, 20 June 2018 (CEST)
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=== [https://en.wikipedia.org/wiki/Grimm%27s_conjecture Grimm's conjecture] ===
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Alireza Badali 18:13, 20 June 2018 (CEST)
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=== [https://en.wikipedia.org/wiki/Oppermann%27s_conjecture Oppermann's conjecture] ===
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Alireza Badali 18:13, 20 June 2018 (CEST)
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=== [https://en.wikipedia.org/wiki/Legendre%27s_conjecture Legendre's conjecture] ===
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Alireza Badali 18:13, 20 June 2018 (CEST)
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== Conjectures depending on the ring theory ==
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'''Algebraic analytical number theory'''
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'''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 $m\star(-m)=1$ & $1\circ m=1$.
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for instance $(\Bbb N,\star,\circ)$ is an integral domain with: $\begin{cases} \forall m,n\in\Bbb N\\ n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\\1\circ m=1\\ 2\circ m=m\\ 3\circ m=-m\qquad\text{that}\quad m\star (-m)=1\\ (2n)\circ(2m)=2mn\\ (2n+1)\circ(2m+1)=2mn\\ (2n)\circ(2m+1)=2mn+1\end{cases}$
 +
:Question $1$: Is $(\Bbb N,\star,\circ)$ an ''unique factorization domain'' or the same UFD? what are irreducible elements in $(\Bbb N,\star,\circ)$?
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'''Question''' $2$: How can we make a UFD on $\Bbb N$?
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Question $3$: Under usual total order on $\Bbb N$, do there exist any integral domain $(\Bbb N,\star,\circ)$ and an Euclidean valuation $v:\Bbb N\setminus\{1\}\to\Bbb N$ such that $(\Bbb N,\star,\circ,v)$ is an Euclidean domain? no.
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'''Guess''' $1$: For each integral domain $(\Bbb N,\star,\circ)$ there exists a total order on $\Bbb N$ and an Euclidean valuation $v:\Bbb N\setminus\{1\}\to\Bbb N$ such that $(\Bbb N,\star,\circ,v)$ is an Euclidean domain.
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Professor [https://en.wikipedia.org/wiki/Jeffrey_Lagarias Jeffrey Clark Lagarias] advised me that you can apply group structure on $\Bbb N\cup\{0\}$ instead only $\Bbb N$ and now I see his plan is useful on the field theory, now suppose we apply two algorithms above on $\Bbb N\cup\{0\}$ hence we will have identity element for the group $(\Bbb N\cup\{0\},\star)$ of the first algorithm is $0$ corresponding with $0$.
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:'''Problem''' $1$: If $(\Bbb N\cup\{0\},\star,\circ)$ is a UFD then what are irreducible elements in $(\Bbb N\cup\{0\},\star,\circ)$ and is $(\Bbb Q^{\ge0},\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb N\cup\{0\},\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\ {m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={-m\over n}\qquad m\star(-m)=0\end{cases}$•
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::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.
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Question $4$: Is $(\Bbb N\cup\{0\},\star,\circ)$ a UFD by: $\begin{cases} \forall m,n\in\Bbb N\\ e=0\\ (2m-1)\star(2m)=0\\ (2m)\star(2n)=2m+2n\\ (2m-1)\star(2n-1)=2m+2n-1\\ (2m)\star(2n-1)=\begin{cases} 2m-2n & 2m\gt 2n-1\\ 2n-2m-1 & 2n-1\gt 2m\end{cases}\\i=1\\ 0\circ m=0\\ 2\circ m=-m\quad m\star(-m)=0\\ (2m)\circ(2n)=2mn-1\\ (2m-1)\circ(2n-1)=2mn-1\\ (2m)\circ(2n-1)=2mn\end{cases}$
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and what are irreducible elements in $(\Bbb N\cup\{0\},\star,\circ)$ and also is $(\Bbb Q^{\ge0},\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb N\cup\{0\},\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={-m\over n}\qquad m\star(-m)=0\end{cases}$
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: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•
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<small>'''Conjecture''' $1$: Let $x$ be a positive real number, and let $\pi(x)$ denote the number of primes that are less than or equal to $x$ then $$\lim_{x\to\infty}\frac{x-\pi(x)}{\pi(e^u)}=1,\quad u=\sqrt{2\log(x\log x-x)}\,.$$</small>
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:<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>
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::<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>
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:::<small>Prime number theorem and its extensions or algebraic forms or corollaries allow us via infinity concept reach to some results.</small>
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:::<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>
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::::<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>
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:::::<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>
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::::::<small>'''Problem''' $3$: Based on ''prime number theorem'' very large prime numbers are equivalent to the numbers of the form $n\cdot\log n,\,n\in\Bbb N$ hence I think a test could be made to check correctness of some conjectures or problems relating to the prime numbers, and maybe some functions such as $h$ prepares it!</small>
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:::::::<small>'''Question''' $5$: If $p_n$ is $n$_th prime number then does $$\lim_{n\to\infty}\frac{p_n}{e^{\sqrt{2\log n}}\over (\log n-1)\sqrt{2\log n}}=1\,?$$</small>
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::::::::<small>Answer given by [https://mathoverflow.net/users/2926/todd-trimble $@$ToddTrimble] from stackexchange site: The numerator is asymptotically greater than $n$, and the denominator is asymptotically less.</small>
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Alireza Badali 16:26, 26 June 2018 (CEST)
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=== [https://en.wikipedia.org/wiki/Many-worlds_interpretation Parallel universes] ===
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'''An algorithm''' that makes new cyclic groups on $\Bbb Z$: Let $(\Bbb Z,\star)$ be that group and at first write integers as a sequence with starting from $0$ and then write integers with a fixed sequence below it, and let identity element $e=0$ be corresponding with $0$ and two generators $m$ & $n$ be corresponding with $1$ & $-1$, so we have $(\Bbb Z,\star)=\langle m\rangle=\langle n\rangle$ for instance: $$0,1,2,-2,-1,3,4,-4,-3,5,6,-6,-5,7,8,-8,-7,9,10,-10,-9,11,12,-12,-11,13,14,-14,-13,...$$ $$0,1,-1,2,-2,3,-3,4,-4,5,-5,6,-6,7,-7,8,-8,9,-9,10,-10,11,-11,12,-12,13,-13,14,-14,...$$ then regarding the sequence above find 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:
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$\begin{cases} \forall t\in\Bbb Z,\quad t\star0=t\\ \forall m,n\in\Bbb N\\ (2m-1)\star(-2m)=0=(-2m+1)\star(2m)\\ (2m-1)\star(2n-1)=2m+2n-2\\ (2m-1)\star(2n)=\begin{cases} 2m-2n-1 & 2m-1\gt2n\\ 2m-2n-2 & 2n\gt 2m-1\end{cases}\\ (2m-1)\star(-2n+1)=2m+2n-1\\ (2m-1)\star(-2n)=\begin{cases} 2n-2m+1 & 2m-1\gt2n\\ 2n-2m & 2n\gt2m-1\end{cases}\\ (2m)\star(2n)=2m+2n\\ (2m)\star(-2n+1)=\begin{cases} 2m-2n+1 & 2n-1\gt2m\\ 2m-2n & 2m\gt2n-1\end{cases}\\ (2m)\star(-2n)=-2m-2n\\ (-2m+1)\star(-2n+1)=-2m-2n+1\\ (-2m+1)\star(-2n)=\begin{cases} 2m-2n+1 & 2m-1\gt2n\\ 2m-2n & 2n\gt2m-1\\ 1 & m=n\end{cases}\\ (-2m)\star(-2n)=2m+2n-2\\ \Bbb Z=\langle1\rangle=\langle-2\rangle\end{cases}$
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'''An algorithm''' which makes new integral domains on $\Bbb Z$: Let $(\Bbb Z,\star,\circ)$ be that integral domain then identity element $i$ will be corresponding with $1$ and multiplication of integers will be obtained from multiplication of corresponding integers such that if $t:\Bbb Z\to\Bbb Z$ is a bijection that images top row on to bottom row respectively for instance in example above is seen $t(2)=-1$ & $t(-18)=18$ then we can write laws by using $t$ such as $(-2m+1)\circ(-2n)=$ $t(t^{-1}(-2m+1)\times t^{-1}(-2n))=t((2m)\times(-2n+1))=$ $t(-2\times(2mn-m))=$ $2\times(2mn-m)=4mn-2m$ and of course each integer like $m$ multiplied by an integer corresponding with $-1$ will be $n$ such that $m\star n=0$ & $0\circ m=0$ for instance $(\Bbb Z,\star,\circ)$ is an integral domain with:
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$\begin{cases} \forall t\in\Bbb Z,\quad t\star0=t\\ \forall m,n\in\Bbb N\\ (2m-1)\star(-2m)=0=(-2m+1)\star(2m)\\ (2m-1)\star(2n-1)=2m+2n-2\\ (2m-1)\star(2n)=\begin{cases} 2m-2n-1 & 2m-1\gt2n\\ 2m-2n-2 & 2n\gt 2m-1\end{cases}\\ (2m-1)\star(-2n+1)=2m+2n-1\\ (2m-1)\star(-2n)=\begin{cases} 2n-2m+1 & 2m-1\gt2n\\ 2n-2m & 2n\gt2m-1\end{cases}\\ (2m)\star(2n)=2m+2n\\ (2m)\star(-2n+1)=\begin{cases} 2m-2n+1 & 2n-1\gt2m\\ 2m-2n & 2m\gt2n-1\end{cases}\\ (2m)\star(-2n)=-2m-2n\\ (-2m+1)\star(-2n+1)=-2m-2n+1\\ (-2m+1)\star(-2n)=\begin{cases} 2m-2n+1 & 2m-1\gt2n\\ 2m-2n & 2n\gt2m-1\\ 1 & m=n\end{cases}\\ (-2m)\star(-2n)=2m+2n-2\\ i=t(1)=1,\quad0\circ m=0,\quad m\star(t(-1)\circ m)=m\star(-2\circ m)=0\\ (2m-1)\circ(2n-1)=4mn-2m-2n+1\\ (2m-1)\circ(2n)=4mn-2n\\ (2m-1)\circ(-2n+1)=-4mn+2n+1\\ (2m-1)\circ(-2n)=-4mn+2m+2n-2\\ (2m)\circ(2n)=-4mn+1\\ (2m)\circ(-2n+1)=4mn\\ (2m)\circ(-2n)=-4mn+2m+1\\ (-2m+1)\circ(-2n+1)=-4mn+1\\ (-2m+1)\circ(-2n)=4mn-2m\\ (-2m)\circ(-2n)=4mn-2m-2n+1\end{cases}$
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:Question $1$: Is $(\Bbb Z,\star,\circ)$ a UFD? what are irreducible elements in $(\Bbb Z,\star,\circ)$? is $(\Bbb Q,\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$ •
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::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•
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'''Problem''' $1$: If $(\Bbb Z,\star,\circ)$ is a UFD then what are irreducible elements in $(\Bbb Z,\star,\circ)$ and is $(\Bbb Q,\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\ {m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$•
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: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•
 +
 
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 +
Question $2$: Under usual total order on $\Bbb Z$, do there exist any integral domain $(\Bbb Z,\star,\circ)$ and an Euclidean valuation $v:\Bbb Z\setminus\{0\}\to\Bbb N$ such that $(\Bbb Z,\star,\circ,v)$ is an Euclidean domain? no.
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'''Guess''' $1$: For each integral domain $(\Bbb Z,\star,\circ)$ there exists a total order on $\Bbb Z$ and an Euclidean valuation $v:\Bbb Z\setminus\{0\}\to\Bbb N$ such that $(\Bbb Z,\star,\circ,v)$ is an Euclidean domain.
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Alireza Badali 20:32, 9 July 2018 (CEST)
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=== [https://en.wikipedia.org/wiki/Gauss_circle_problem Gauss circle problem] ===
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<small>Given this sequence: $$0,1,-1,2,-2,3,-3,4,-4,5,-5,6,-6,7,-7,8,-8,9,-9,10,-10,11,-11,12,-12,...$$ $$0,1,2,3,-3,-1,-2,4,5,6,-6,-4,-5,7,8,9,-9,-7,-8,10,11,12,-12,-10,-11,...$$ we have this integral domain $(\Bbb Z,\star,\circ)$ as:
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$\begin{cases} \forall t\in\Bbb  Z,\quad t\star0=t,\quad\forall m,n\in\Bbb N,\\ (6m-5)\star(6m-4)=0=(6m-3)\star(-6m+3)=(-6m+5)\star(-6m+4)=(6m-2)\star(6m-1)=\\ =(6m)\star(-6m)=(-6m+2)\star(-6m+1)\\ (6m-5)\star(6n-5)=6m+6n-9\\ (6m-5)\star(6n-4)=\begin{cases} 6n-6m+2 & 6m-5\gt6n-4\\ 6m-6n+1 & 6n-4\gt6m-5\end{cases}\\ (6m-5)\star(6n-3)=-6m-6n+11\\ (6m-5)\star(6n-2)=6m+6n-6\\ (6m-5)\star(6n-1)=\begin{cases} 6n-6m+5 & 6m-5\gt6n-1\\ 6m-6n-2 & 6n-1\gt6m-5\end{cases}\\ (6m-5)\star(6n)=-6m-6n+8\\ (6m-5)\star(-6n+5)=6m+6n-8\\ (6m-5)\star(-6n+4)=\begin{cases} 6m-6n-2 & 6m-5\gt6n-4\\ 6m-6n-3 & 6n-4\gt6m-5\end{cases}\\ (6m-5)\star(-6n+3)=\begin{cases} 6m-6n & 6m-5\gt6n-3\\ 6n-6m+2 & 6n-3\gt6m-5\end{cases}\\ (6m-5)\star(-6n+2)=6m+6n-5\\ (6m-5)\star(-6n+1)=\begin{cases} 6m-6n-5 & 6m-5\gt6n-1\\ 6m-6n-6 & 6n-1\gt6m-5\end{cases}\\ (6m-5)\star(-6n)=\begin{cases} 6m-6n-3 & 6m-5\gt6n\\ 6n-6m+5 & 6n\gt6m-5\end{cases}\\ (6m-4)\star(6n-4)=-6m-6n+9\\ (6m-4)\star(6n-3)=\begin{cases} 6n-6m & 6m-4\gt6n-3\\ 6n-6m+1 & 6n-3\gt6m-4\end{cases}\\ (6m-4)\star(6n-2)=\begin{cases} 6n-6m+4 & 6m-4\gt6n-2\\ 6m-6n-1 & 6n-2\gt6m-4\end{cases}\\ (6m-4)\star(6n-1)=-6m-6n+6\\ (6m-4)\star(6n)=\begin{cases} 6n-6m+3 & 6m-4\gt6n\\ 6n-6m+4 & 6n\gt6m-4\end{cases}\\ (6m-4)\star(-6n+5)=\begin{cases} 6m-6n-1 & 6m-4\gt6n-5\\ 6n-6m+3 & 6n-5\gt6m-4\\ 3 & m=n\end{cases}\\ (6m-4)\star(-6n+4)=6m+6n-7\\ (6m-4)\star(-6n+3)=-6m-6n+10\\ (6m-4)\star(-6n+2)=\begin{cases} 6m-6n-4 & 6m-4\gt6n-2\\ 6n-6m+6 & 6n-2\gt6m-4\end{cases}\\ (6m-4)\star(-6n+1)=6m+6n-4\\ (6m-4)\star(-6n)=-6m-6n+7\\ (6m-3)\star(6n-3)=6m+6n-8\\ (6m-3)\star(6n-2)=-6m-6n+8\\ (6m-3)\star(6n-1)=\begin{cases} 6m-6n-2 & 6m-3\gt6n-1\\ 6m-6n-3 & 6n-1\gt6m-3\end{cases}\\ (6m-3)\star(6n)=6m+6n-5\\ (6m-3)\star(-6n+5)=6m+6n-6\\ (6m-3)\star(-6n+4)=\begin{cases} 6m-6n & 6m-3\gt6n-4\\ 6n-6m+2 & 6n-4\gt6m-3\\ 2 & m=n\end{cases}\\ (6m-3)\star(-6n+3)=\begin{cases} 6n-6m+2 & 6m-3\gt6n-3\\ 6m-6n+1 & 6n-3\gt6m-3\end{cases}\\ (6m-3)\star(-6n+2)=6m+6n-3\\ (6m-3)\star(-6n+1)=\begin{cases} 6m-6n-3 & 6m-3\gt6n-1\\ 6n-6m+5 & 6n-1\gt6m-3\end{cases}\\ (6m-3)\star(-6n)=\begin{cases} 6n-6m+5 & 6m-3\gt6n\\ 6m-6n-2 & 6n\gt6m-3\end{cases}\\ (6m-2)\star(6n-2)=6m+6n-3\\ (6m-2)\star(6n-1)=\begin{cases} 6n-6m+2 & 6m-2\gt6n-1\\ 6m-6n+1 & 6n-1\gt6m-2\end{cases}\\ (6m-2)\star(6n)=-6m-6n+5\\ (6m-2)\star(-6n+5)=6m+6n-5\\ (6m-2)\star(-6n+4)=\begin{cases} 6m-6n+1 & 6m-2\gt6n-4\\ 6m-6n & 6n-4\gt6m-2\end{cases}\\ (6m-2)\star(-6n+3)=\begin{cases} 6m-6n+3 & 6m-2\gt6n-3\\ 6n-6m-1 & 6n-3\gt6m-2\end{cases}\\ (6m-2)\star(-6n+2)=6m+6n-2\\ (6m-2)\star(-6n+1)=\begin{cases} 6m-6n-2 & 6m-2\gt6n-1\\ 6m-6n-3 & 6n-1\gt6m-2\end{cases}\\ (6m-2)\star(-6n)=\begin{cases} 6m-6n & 6m-2\gt6n\\ 6n-6m+2 & 6n\gt6m-2\end{cases}\\ (6m-1)\star(6n-1)=-6m-6n+3\\ (6m-1)\star(6n)=\begin{cases} 6n-6m & 6m-1\gt6n\\ 6n-6m+1 & 6n\gt6m-1\end{cases}\\ (6m-1)\star(-6n+5)=\begin{cases} 6m-6n+2 & 6m-1\gt6n-5\\ 6n-6m & 6n-5\gt6m-1\end{cases}\\ (6m-1)\star(-6n+4)=6m+6n-4\\ (6m-1)\star(-6n+3)=-6m-6n+7\\ (6m-1)\star(-6n+2)=\begin{cases} 6m-6n-1 & 6m-1\gt6n-2\\ 6n-6m+3 & 6n-2\gt6m-1\\ 3 & m=n\end{cases}\\ (6m-1)\star(-6n+1)=6m+6n-1\\ (6m-1)\star(-6n)=-6m-6n+4\\ (6m)\star(6n)=6m+6n-2\\ (6m)\star(-6n+5)=6m+6n-3\\ (6m)\star(-6n+4)=\begin{cases} 6m-6n+3 & 6m\gt6n-4\\ 6n-6m-1 & 6n-4\gt6m\end{cases}\\ (6m)\star(-6n+3)=\begin{cases} 6n-6m-1 & 6m\gt6n-3\\ 6m-6n+4 & 6n-3\gt6m\end{cases}\\ (6m)\star(-6n+2)=6m+6n\\ (6m)\star(-6n+1)=\begin{cases} 6m-6n & 6m\gt6n-1\\ 6m-6n-1 & 6n-1\gt6m\\ 2 & m=n\end{cases}\\ (6m)\star(-6n)=\begin{cases} 6n-6m+2 & 6m\gt6n\\ 6m-6n+1 & 6n\gt6m\end{cases}\\ (-6m+5)\star(-6n+5)=-6m-6n+8\\ (-6m+5)\star(-6n+4)=\begin{cases} 6n-6m+2 & 6m-5\gt6n-4\\ 6m-6n+1 & 6n-4\gt6m-5\end{cases}\\ (-6m+5)\star(-6n+3)=\begin{cases} 6m-6n+1 & 6m-5\gt6n-3\\ 6m-6n & 6n-3\gt6m-5\\ 1 & m=n\end{cases}\\ (-6m+5)\star(-6n+2)=-6m-6n+5\\ (-6m+5)\star(-6n+1)=\begin{cases} 6n-6m+5 & 6m-5\gt6n-1\\ 6m-6n-2 & 6n-1\gt6m-5\end{cases}\\ (-6m+5)\star(-6n)=\begin{cases} 6m-6n-2 & 6m-5\gt6n\\ 6m-6n-3 & 6n\gt6m-5\end{cases}\\ (-6m+4)\star(-6n+4)=-6m-6n+7\\ (-6m+4)\star(-6n+3)=-6m-6n+6\\ (-6m+4)\star(-6n+2)=\begin{cases} 6n-6m+4 & 6m-4\gt6n-2\\ 6m-6n-1 & 6n-2\gt6m-4\end{cases}\\ (-6m+4)\star(-6n+1)=-6m-6n+4\\ (-6m+4)\star(-6n)=-6m-6n+3\\ (-6m+3)\star(-6n+3)=6m+6n-7\\ (-6m+3)\star(-6n+2)=\begin{cases} 6n-6m+3 & 6m-3\gt6n-2\\ 6n-6m+4 & 6n-2\gt6m-3\end{cases}\\ (-6m+3)\star(-6n+1)=-6m-6n+3\\ (-6m+3)\star(-6n)=6m+6n-4\\ (-6m+2)\star(-6n+2)=-6m-6n+2\\ (-6m+2)\star(-6n+1)=\begin{cases} 6n-6m+2 & 6m-2\gt6n-1\\ 6m-6n+1 & 6n-1\gt6m-2\end{cases}\\ (-6m+2)\star(-6n)=\begin{cases} 6m-6n+1 & 6m-2\gt6n\\ 6m-6n & 6n\gt6m-2\\ 1 & m=n\end{cases}\\ (-6m+1)\star(-6n+1)=-6m-6n+1\\ (-6m+1)\star(-6n)=-6m-6n\\ (-6m)\star(-6n)=6m+6n-1\\ t\circ1=t,\quad t\circ0=0,\quad t\star(2\circ t)=0\end{cases}$
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$\begin{cases} (6m-5)\circ(6n-5)=36mn-30m-30n+25\\ (6m-5)\circ(6n-4)=36mn-30m-30n+26\\ (6m-5)\circ(6n-3)=36mn-24m-30n+21\\ (6m-5)\circ(6n-2)=36mn-12m-30n+10\\ (6m-5)\circ(6n-1)=36mn-12m-30n+11\\ (6m-5)\circ(6n)=36mn-6m-30n+6\\ (6m-5)\circ(-6n+5)=-36mn+18m+30n-13\\ (6m-5)\circ(-6n+4)=-36mn+18m+30n-14\\ (6m-5)\circ(-6n+3)=-36mn+24m+30n-21\\ (6m-5)\circ(-6n+2)=-36mn+30n+2\\ (6m-5)\circ(-6n+1)=-36mn+30n+1\\ (6m-5)\circ(-6n)=-36mn+6m+30n-6\\ (6m-4)\circ(6n-4)=36mn-30m-30n+25\\ (6m-4)\circ(6n-3)=-36mn+24m+30n-21\\ (6m-4)\circ(6n-2)=36mn-12m-30n+11\\ (6m-4)\circ(6n-1)=36mn-12m-30n+10\\ (6m-4)\circ(6n)=-36mn+6m+30n-6\\ (6m-4)\circ(-6n+5)=-36mn+18m+30n-14\\ (6m-4)\circ(-6n+4)=-36mn+18m+30n-13\\ (6m-4)\circ(-6n+3)=36mn-24m-30n+21\\ (6m-4)\circ(-6n+2)=-36mn+30n+1\\ (6m-4)\circ(-6n+1)=-36mn+30n+2\\ (6m-4)\circ(-6n)=36mn-6m-30n+6\\ (6m-3)\circ(6n-3)=36mn-24m-24n+16\\ (6m-3)\circ(6n-2)=36mn-12m-24n+9\\ (6m-3)\circ(6n-1)=-36mn+12m+24n-9\\ (6m-3)\circ(6n)=36mn-6m-24n+4\\ (6m-3)\circ(-6n+5)=-36mn+18m+24n-10\\ (6m-3)\circ(-6n+4)=-36mn+18m+24n-11\\ (6m-3)\circ(-6n+3)=36mn-24m-24n+17\\ (6m-3)\circ(-6n+2)=-36mn+24n+2\\ (6m-3)\circ(-6n+1)=-36mn+24n+1\\ (6m-3)\circ(-6n)=36mn-6m-24n+5\\ (6m-2)\circ(6n-2)=36mn-12m-12n+4\\ (6m-2)\circ(6n-1)=36mn-12m-12n+5\\ (6m-2)\circ(6n)=36mn-6m-12n+3\\ (6m-2)\circ(-6n+5)=-36mn+18m+12n-4\\ (6m-2)\circ(-6n+4)=-36mn+18m+12n-5\\ (6m-2)\circ(-6n+3)=-36mn+24m+12n-9\\ (6m-2)\circ(-6n+2)=-36mn+12n+2\\ (6m-2)\circ(-6n+1)=-36mn+12n+1\\ (6m-2)\circ(-6n)=-36mn+6m+12n-3\\ (6m-1)\circ(6n-1)=36mn-12m-12n+4\\ (6m-1)\circ(6n)=-36mn+6m+12n-3\\ (6m-1)\circ(-6n+5)=-36mn+18m+12n-5\\ (6m-1)\circ(-6n+4)=-36mn+18m+12n-4\\ (6m-1)\circ(-6n+3)=36mn-24m-12n+9\\ (6m-1)\circ(-6n+2)=-36mn+12n+1\\ (6m-1)\circ(-6n+1)=-36mn+12n+2\\ (6m-1)\circ(-6n)=36mn-6m-12n+3\\ (6m)\circ(6n)=36mn-6m-6n+1\\ (6m)\circ(-6n+5)=-36mn+18m+6n-1\\ (6m)\circ(-6n+4)=-36mn+18m+6n-2\\ (6m)\circ(-6n+3)=36mn-24m-6n+5\\ (6m)\circ(-6n+2)=-36mn+6n+2\\ (6m)\circ(-6n+1)=-36mn+6n+1\\ (6m)\circ(-6n)=36mn-6m-6n+2\\ (-6m+5)\circ(-6n+5)=-36mn+18m+18n-7\\ (-6m+5)\circ(-6n+4)=-36mn+18m+18n-8\\ (-6m+5)\circ(-6n+3)=-36mn+24m+18n-11\\ (-6m+5)\circ(-6n+2)=-36mn+18n+2\\ (-6m+5)\circ(-6n+1)=-36mn+18n+1\\ (-6m+5)\circ(-6n)=-36mn+6m+18n-2\\ (-6m+4)\circ(-6n+4)=-36mn+18m+18n-7\\ (-6m+4)\circ(-6n+3)=-36mn+24m+18n-10\\ (-6m+4)\circ(-6n+2)=-36mn+18n+1\\ (-6m+4)\circ(-6n+1)=-36mn+18n+2\\ (-6m+4)\circ(-6n)=-36mn+6m+18n-1\\ (-6m+3)\circ(-6n+3)=36mn-24m-24n+16\\ (-6m+3)\circ(-6n+2)=-36mn+24n+1\\ (-6m+3)\circ(-6n+1)=-36mn+24n+2\\ (-6m+3)\circ(-6n)=36mn-6m-24n+4\\ (-6m+2)\circ(-6n+2)=-36mn+2\\ (-6m+2)\circ(-6n+1)=-36mn+1\\ (-6m+2)\circ(-6n)=-36mn+6m+1\\ (-6m+1)\circ(-6n+1)=-36mn+2\\ (-6m+1)\circ(-6n)=-36mn+6m+2\\ (-6m)\circ(-6n)=36mn-6m-6n+1\end{cases}$
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but some calculations are such as: $(6m-4)\circ(6n-3)=t(t^{-1}(6m-4)\times t^{-1}(6n-3))=t((-6m+5)\times(6n-4))=$ $t(-6(6mn-4m-5n+4)+4)=-6(6mn-4m-5n+4)+3=-36mn+24m+30n-21,$ $(6m-4)\circ(-6n+1)=$ $t(t^{-1}(6m-4)\times t^{-1}(-6n+1))=t((-6m+5)\times(-6n))=t(6(6mn-5n))=-6(6mn-5n)+2=-36mn+30n+2,$ $(-6m+1)\circ(-6n+1)=t(t^{-1}(-6m+1)\times t^{-1}(-6n+1))=t((-6m)\times(-6n))=t(6(6mn))=-6(6mn)+2=$ $-36mn+2$.</small>
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Question $1$: Is $(\Bbb Z,\star,\circ)$ a UFD? what are irreducible elements in $(\Bbb Z,\star,\circ)$? is $(\Bbb Q,\star_1,\circ_1)$ a field by: $\begin{cases} \forall m,n,u,v\in\Bbb Z,\,\,n\neq0\neq v\\ e_1=0,\qquad i_1=1\\{m\over n}\star_1{u\over v}=\frac{(m\circ v)\star(u\circ n)}{n\circ v}\\ {m\over n}\circ_1{u\over v}=\frac{m\circ u}{n\circ v}\\ ({m\over n})^{-1}={n\over m}\,\qquad m\neq0\\ -({m\over n})={w\over n}\qquad\,\,\,m\star w=0\end{cases}$
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: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•
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'''Problem''' $1$: Reinterpret (possibly via matrices) ''Gauss circle problem'' under the field $(\Bbb Q,\star_1,\circ_1)$ (not in whole $\Bbb R$).
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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=41679