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$\mathscr B$ $theory$ (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 $3$: $\forall (a,b)\subset [0.1,1),\,w_{(a,b)}={10\over9}\cdot(b-a)$.
Answer given by $@$Peter from stackexchange site: Imagine a very large number $N$ and consider the range $[10^N,10^{N+1}]$. The natural logarithms of $10^N$ and $10^{N+1}$ only differ by $\ln(10)\approx 2.3$ Hence the reciprocals of the logarithms of all primes in this range virtually coincicde. Because of the approximation $$\int_a^b \frac{1}{\ln(x)}dx$$ for the number of primes in the range $[a,b]$ the number of primes is approximately the length of the interval divided by $\frac{1}{\ln(10^N)}$, so is approximately equally distributed. Hence your conjecture is true.
Benfords law seems to contradict this result , but this only applies to sequences producing primes as the Mersenne primes and not if the primes are chosen randomly in the range above.
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 $6$: $\forall(a,b),(c,d)\subset[0.1,1),\,\lim_{n\to\infty}{\#V_{(a,b),n}\over\#V_{(c,d),n}}={b-a\over d-c}=\lim_{n\to\infty}{\#U_{(a,b),n}\over\#U_{(c,d),n}}$.
Comment given by $@$Dzoooks from stackexchange site: It shouldn't be that hard to get estimates from $V_{(a,b),n}=\{p\leq n : 10^ka\lt p\lt 10^k\text{ for some }k\}=\sqcup_{k\geq1}\{p\in[0,n]\cap(10^ka,10^kb)\},$ where the union is disjoint from $10^kb\lt10^k\leq10^{k+1}a$. Then $\#V_{(a,b),n}$ can be summed with the PNT. You'll see that a $(b-c)$ comes out of the sum..maybe
and the PNT gives $\#\{p\in[0,n]\cap(10^ka,10^k)\}\sim\frac{(b-a)10^k}{\log b-\log a},$ for large $n$ and $k$. Factor these out of the sum, and it looks like your limit is actually $\frac{b-a}{\log b-\log a}\cdot\frac{\log d-\log c}{d-c}$.


Using homotopy groups Goldbach's conjecture will be proved.

Alireza Badali 08:27, 31 March 2018 (CEST)

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?

Answer given by $@$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 $@$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 $(Z_1,\star)$: $$1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49,…$$ $$0,1,-1,3,-3,5,-5,7,-7,9,-9,11,-11,13,-13,15,-15,17,-17,19,-19,21,-21,23,-23,…$$ $\begin{cases} \forall m,n\in\Bbb N,\quad e=0\\ (2m-1)\star(-2m+1)=0\\ (4m-3)\star(4n-3)=4m+4n-5\\ (4m-3)\star(-4n+3)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (4m-3)\star(4n-1)=4m+4n-3\\ (4m-3)\star(-4n+1)=\begin{cases} 4m-4n-1 & m\le n\\ 4m-4n-3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+3)=-4m-4n+5\\ (-4m+3)\star(4n-1)=\begin{cases} 4n-4m+1 & m\le n\\ 4n-4m+3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+1)=-4m-4n+3\\ (4m-1)\star(4n-1)=4m+4n-1\\ (4m-1)\star(-4n+1)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (-4m+1)\star(-4n+1)=-4m-4n+1\\ Z_1=\langle1\rangle=\langle-1\rangle\end{cases}$

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, $m\in\Bbb N$: $$4m-2,4m-1,4m,4m+1$$ $$8m-5,8m-3,8m-1,8m+1$$ for instance $(8m-3) \circ_2(8n+1)=t(t^{-1}(8m-3)\circ_1t^{-1}(8n+1))=t((4m-1) \circ_1(4n+1))=$ $t((2(2m-1)+1)\circ_1(2(2n)+1))=t(2(2m-1)(2n))=t(4(2mn-n))=8(2mn-n)-1=16mn-8n-1$

and finally $(Z_1,\star,\circ)$ is an integral domain (possibly a UFD) with: $\begin{cases} \forall m,n\in\Bbb N,\,\forall v\in Z_1,\quad e=0\\ (2m-1)\star(-2m+1)=0\\ (4m-3)\star(4n-3)=4m+4n-5\\ (4m-3)\star(-4n+3)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (4m-3)\star(4n-1)=4m+4n-3\\ (4m-3)\star(-4n+1)=\begin{cases} 4m-4n-1 & m\le n\\ 4m-4n-3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+3)=-4m-4n+5\\ (-4m+3)\star(4n-1)=\begin{cases} 4n-4m+1 & m\le n\\ 4n-4m+3 & m\gt n\end{cases}\\ (-4m+3)\star(-4n+1)=-4m-4n+3\\ (4m-1)\star(4n-1)=4m+4n-1\\ (4m-1)\star(-4n+1)=\begin{cases} 4m-4n+1 & m\lt n\\ 4m-4n-1 & m\gt n\end{cases}\\ (-4m+1)\star(-4n+1)=-4m-4n+1\\ 0\circ v=0,\quad1\circ v=v,\quad((-1)\circ v)\star v=0\\ (4m-3)\circ(4n-3)=8mn-4m-4n+1\\ (4m-3)\circ(-4n+3)=-8mn+4m+4n-1\\ (4m-3)\circ(4n-1)=8mn-4n-1\\ (4m-3)\circ(-4n+1)=-8mn+4n+1\\ (-4m+3)\circ(-4n+3)=8mn-4m-4n+1\\ (-4m+3)\circ(4n-1)=-8mn+4n+1\\ (-4m+3)\circ(-4n+1)=8mn-4n-1\\ (4m-1)\circ(4n-1)=8mn-1\\ (4m-1)\circ(-4n+1)=-8mn+1\\ (-4m+1)\circ(-4n+1)=8mn-1\end{cases}$

from this table: $$8m-5,8m-3,8m-1,8m+1$$ $$4m-3,-4m+3,4m-1,-4m+1$$

without ring theory we have no appropriate calculations.


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 one in which given two elements $s,t\in Z_1$, computes their quotient and/or remainder, the result of division.

comment given by $@$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)
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)

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 $\lim_{x\to\infty}\pi_1(x)\cdot{\log x\over x}=0.5=\lim_{x\to\infty}\pi_2(x)\cdot{\log x\over x}\ ?$

Answer given by $@$Milo Brandt from stackexchange site: Basically, for any $k$, the primes are equally distributed across the congruence classes $\langle n\rangle$ mod $k$ where $n$ and $k$ are coprime.
This result is known as the prime number theorem for arithmetic progressions. Wikipedia discusses it with a number of references and one can find a proof of it by Ivan Soprounov 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 $@$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 $\lim_{n\to\infty}{q_n\over p_n}=1$.

Answer given by $@$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)

Collatz conjecture

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


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


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


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

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

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

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

References:
Sequences $S,\bar{S},T$ are OEIS A171945, A053661, A034701 respectively. These are all discussed in 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]


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

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


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


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

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

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

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


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.


Question $4$: Has any relation on the Collatz tree been discovered other than its definitions? In order to define a group structure on the Collatz tree I need such relations but other than its 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)

Erdős–Straus conjecture

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


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

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

Landaus forth problem

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

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


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


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

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

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

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


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

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


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

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

Lemoine's conjecture

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


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

Proof using Goldbach's weak conjecture.


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

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

Primes with beatty sequences

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


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


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

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


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


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

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

Conjectures depending on the new definitions of primes

Algebraic analytical number theory


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


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

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

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

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

Gaussian moat problem

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

Grimm's conjecture

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

Oppermann's conjecture

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

Legendre's conjecture

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

Conjectures depending on the ring theory

Algebraic analytical number theory


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


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

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


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


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

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


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


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

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

Parallel universes

An algorithm that makes new cyclic groups on $\Bbb Z$: Let $(\Bbb Z,\star)$ be that group and at first write integers as a sequence with starting from $0$ and then write integers with a fixed sequence below it, and let identity element $e=0$ be corresponding with $0$ and two generators $m$ & $n$ be corresponding with $1$ & $-1$, so we have $(\Bbb Z,\star)=\langle m\rangle=\langle n\rangle$ for instance: $$0,1,2,-2,-1,3,4,-4,-3,5,6,-6,-5,7,8,-8,-7,9,10,-10,-9,11,12,-12,-11,13,14,-14,-13,...$$ $$0,1,-1,2,-2,3,-3,4,-4,5,-5,6,-6,7,-7,8,-8,9,-9,10,-10,11,-11,12,-12,13,-13,14,-14,...$$ then regarding the sequence above find 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 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 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)

Gauss circle problem

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

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

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

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

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

in addition an unique & specific division algorithm like this 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://www.encyclopediaofmath.org/index.php?title=Musictheory2math&oldid=43435