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

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

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.com: Without loss of generality (by passing to a smaller subinterval) we can assume that $(a, b) = \left( \frac{s}{10^r}, \frac{t}{10^r} \right)$, where $s, t, r$ are positive integers and $s < t$. Let $\alpha = \frac{t}{s}$.
The statement is now equivalent to saying that there is $m \in \mathbb{N}$ such that for every $k \geqslant m$ there is a prime $p$ with $10^{k-r} \cdot s < p < 10^{k-r} \cdot t$.
We will prove a stronger statement: there is $m \in \mathbb{N}$ such that for every $n \geqslant m$ there is a prime $p$ such that $n < p < \alpha \cdot n$. By taking a little smaller $\alpha$ we can relax the restriction to $n < p \leqslant \alpha \cdot n$.
Now comes the prime number theorem: $$\lim_{n \to \infty} \frac{\pi(n)}{\frac{n}{\log n}} = 1$$
where $\pi(n) = \# \{ p \leqslant n : p$ is prime$\}.$ By the above we have $$\frac{\pi(\alpha n)}{\pi(n)} \sim \frac{\frac{\alpha n}{\log(\alpha n)}}{\frac{n}{\log(n)}} = \alpha \cdot \frac{\log n}{\log(\alpha n)} \xrightarrow{n \to \infty} \alpha$$
hence $\displaystyle \lim_{n \to \infty} \frac{\pi(\alpha n)}{\pi(n)} = \alpha$. So there is $m \in \mathbb{N}$ such that $\pi(\alpha n) > \pi(n)$ whenever $n \geqslant m$, which means there is a prime $p$ such that $n < p \leqslant \alpha \cdot n$, and that is what we wanted.♦


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.com:
  • For $n \in \mathbb{N}$ then $r(n) = 10^{-\lceil \log_{10}(n) \rceil} n$, ie. $r(19) = 0.19$. We look at the image by $r$ of the primes $\mathbb{P}$.
  • Let $F((c,d)) = \min \{ p \in \mathbb{P}, r(p) \in (c,d)\}$ and $f((c,d)) = \pi(F(c,d))= \min \{ n, r(p_n) \in (c,d)\}$ ($\pi$ is the prime counting function)
  • If you set $g(\epsilon) = \max_a \{ f((a,a+\epsilon))\}$ then try seing how $g(\epsilon)$ is constant on some intervals defined in term of the prime gap $g(p) = -p+\min \{ q \in \mathbb{P}, q > p\}$ and things like $ \max \{ g(p), p > 10^i, p+g(p) < 10^{i+1}\}$
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 above lemma)

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.

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

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


Assume $\forall m,n\in\Bbb N$: $\begin{cases} n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\end{cases}$

and $p_n\star _1p_m=p_{n\star m}$ that $p_n$ is $n$_th prime with $e=p_1=2$, obviously $(\Bbb N,\star)$ & $(\Bbb P,\star _1)$ are groups and $\langle 2\rangle =\langle 3\rangle =(\Bbb N,\star)\simeq (\Bbb Z,+)\simeq (\Bbb P,\star _1)=\langle 3\rangle=\langle 5\rangle$.


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

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

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

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


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

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

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

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


I want make some topologies having prime numbers properties presentable in 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 into 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 $2$: 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.

Perhaps this technique is useful: $\forall n\in\Bbb N,$ and for each subinterval $(a,b)$ of $[0.1,1),$ that $a\neq b,$
$\begin{cases} U_{(a,b)}:=\{n\in\Bbb N\mid a\le r(n)\le b\},\\ \\V_{(a,b)}:=\{p\in\Bbb P\mid a\le r(p)\le b\},\\ \\U_{(a,b),n}:=\{m\in U_{(a,b)}\mid m\le n\},\\ \\V_{(a,b),n}:=\{m\in V_{(a,b)}\mid m\le n\},\\ \\w_{(a,b),n}:=(\#U_{(a,b),n})^{-1}\cdot\#V_{(a,b),n}\cdot\log n,\\ \\w_{(a,b)}:=\lim _{n\to\infty} w_{(a,b),n}\end{cases}$
Guess $2$: $\forall (a,b)\subset [0.1,1),\,w_{(a,b)}=0.9^{-1}\cdot (b-a)$.
Answer given by $@$Peter: 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.


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

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

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


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


Theorem $6$: $(S,\lt _1)$ is a well-ordering set with order relation $\lt _1$ as: $\forall i,n,k\in\Bbb N$ if $p_n$ be $n$-th prime number, relation $\lt _1$ is defined with: $w_i(p_n)\lt _1w_i(p_{n+k})\lt _1w_{i+1}(p_n)$ or $$0.2\lt _10.3\lt _10.5\lt _10.7\lt _10.11\lt _10.13\lt _10.17\lt _1...0.02\lt _10.03\lt _10.05\lt _10.07\lt _10.011\lt _1$$ $$0.013\lt _10.017\lt _1...0.002\lt _10.003\lt _10.005\lt _10.007\lt _10.0011\lt _10.0013\lt _10.0017\lt _1...$$

and $(E,\lt _2)$ is another well ordering set with order relation $\lt _2$ as: $\forall i,n,k\in\Bbb N$ that $10\nmid n,\, 10\nmid n+k,$ $w_i(n)\lt _2w_i(n+k)\lt _2w_{i+1}(n)$ or $$0.1\lt _2 0.2\lt _2 0.3\lt _2 ...0.9\lt _2 0.11\lt _2 0.12\lt _2 ...0.19\lt _2 0.21\lt _2 ...0.01\lt _2 0.02\lt _2 0.03\lt _2 ...0.09$$ $$\lt _2 0.011\lt _2 0.012\lt _2 ...0.019\lt _2 0.021\lt _2 ...0.001\lt _2 0.002\lt _2 0.003\lt _2 ...0.009\lt _2 0.0011\lt _2 ...$$

now $M:=S\times S\times E$ is a well-ordering set with order relation $\lt _3$ as: $\forall (a,b,t),(c,d,u)\in S\times S\times E,$ $(a,b,t)\lt _3(c,d,u)$ if $\,\,\begin{cases} t\lt _2u & or\\ t=u,\,\, a+b\lt _2c+d & or\\ t=u,\,\, a+b=c+d,\,\, b\lt _1 d\end{cases}$ ♦


now assume $M$ is a topological space induced by order relation $\lt _3$.


Question $3$: Is $S$ a topological group under topology induced by order relation $\lt_1$ and is $(S\times S)\oplus E$ a topological group under topology of $M$?


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

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

Comments

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

How to Cite This Entry:
Musictheory2math. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Musictheory2math&oldid=43179