User talk:Musictheory2math

Goldbach's conjecture

Badali lemma: Let $\mathbb{P}$ be the set prime numbers and $S$ is a set that has been made as below: put a point on the beginning of each member of $\Bbb{P}$ like $0.2$ or $0.19$ then $S=\{0.2,0.3,0.5,0.7,...\}$ is dense in the interval $(0.1,1)$ of real numbers.

$\,$This lemma is a base for finding formula of prime numbers, because for each member of $S$ like $a$ with its special and fixed location into $(0.1,1)$ and a small enough neighborhood like $(a-\epsilon ,a+\epsilon )$, but $a$ is in a special relation with members of $(a-\epsilon ,a+\epsilon )$ but there exists a special order on $S$ into $(0.1,1)$ and of course formula of prime numbers has whole properties related to prime numbers simultaneous. There is a musical note on the natural numbers that can be discovered by the formula of prime numbers. Musictheory2math (talk) 16:29, 25 March 2017 (CET)

True, $S$ is dense in the interval $(0.1,1)$; this fact follows easily from well-known results on Distribution of prime numbers. But I doubt that this is "This lemma is a base for finding formula of prime numbers". Boris Tsirelson (talk) 22:10, 16 March 2017 (CET)

Dear Professor Boris Tsirelson , in principle finding formula of prime numbers is very lengthy. and I am not sure be able for it but please give me few time about two month for expression my theories. Musictheory2math (talk) 16:29, 25 March 2017 (CET)

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

\[ X = (10a,10b) \cup (100a,100b) \cup (1000a,1000b) \cup \dots \, ; \]

all $ p_n $ belong to its complement

\[ Y = (0,\infty) \setminus X = (0,10a] \cup [10b,100a] \cup [100b,1000a] \cup \dots \]

Using the relation $ \frac{p_{n+1}}{p_n} \to 1 $ we take $N$ such that $ \frac{p_{n+1}}{p_n} < \frac b a $ for all $n>N$. Now, all numbers $p_n$ for $n>N$ must belong to a single interval $ [10^{k-1} b, 10^k a] $, since it cannot happen that $ p_n \le 10^k a $ and $ p_{n+1} \ge 10^k b $ (and $n>N$). We get a contradiction: $ p_n \to \infty $ but $ p_n \le 10^k a $.

And again, please sign your messages (on talk pages) with four tildas: ~~~~. Boris Tsirelson (talk) 20:57, 18 March 2017 (CET)

According to this revision and this revision, $ \frac{p_{n+1}}{p_n} < \frac b a $ implies $p_{n+i}\lt (\frac b a)^{n+i} p_n$ so $(\frac b a)^{n+i} p_n\to \infty$ where $i\to \infty$ because $\frac b a\gt 1$ so your proof is wrong! Alireza Badali 20:18, 15 September 2017 (CEST)

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. Musictheory2math (talk) 16:29, 25 March 2017 (CET)

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.) Boris Tsirelson (talk) 11:16, 19 March 2017 (CET)

Now I want state philosophy of This lemma is a base for finding formula of prime numbers: However we loose the induction axiom for finite sets (Induction axiom is unable for discovering formula of prime numbers.) but I thought that if change space from natural numbers with cardinal $\aleph_0$ to a bounded set with cardinal $\aleph_1$ in the real numbers then we can use other features like axioms and important theorems in the real numbers for working on prime numbers and I think it is a better and easier way. Musictheory2math (talk) 16:29, 25 March 2017 (CET)

I see. Well, we are free to use the whole strength of mathematics (including analysis) in the number theory; and in fact, analysis is widely used, as you may see in the article "Distribution of prime numbers".

But you still do not put four tildas at the end of each your message; please do. Boris Tsirelson (talk) 11:16, 19 March 2017 (CET)

Importance of density in the Badali lemma is similar to definition of irrational numbers from rational numbers.

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.

Assume $S_1=\{a/10^n\, |\, a\in S$ for $n=0,1,2,3,...\}$ & $L=\{(a,b)\,|\,a,b \in S_1$ & $0.01 \le a+b \lt 0.1$ & $\exists m \in \Bbb N,\, a \times 10^m,\,b \times 10^m$ are prime numbers & $a\times 10^m\neq 2\neq b\times 10^m\}$

Theorem: $S_1$ is dense in the interval $(0,1)$ and $S_1\times S_1$ is dense in the $(0,1)\times (0,1)$.

If $p,q$ are prime numbers and $n$ is the number of digits in $p+q$ and $m=$max(number of digits in $p$, number of digits in $q$), let $\varphi : L \to \Bbb N,$ $\varphi ((p,q)) = \begin{cases} m+1 & n=m \\m+2 & n=m+1 \end{cases}$

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 $\{(x,x)\,|\,0.005 \le x \lt 0.05 \}$.

Definition: Assume $L_1=\{(a,b)\,|\,(a,b) \in L$ & $b \lt a \}$ of course members in $L$ & $L_1$ are corresponding to prime numbers as multiplication and sum and minus and let $E=(0.007,0.005)$ (and also $5$ points to form of $(0.007+\epsilon _1,0.005-\epsilon _2)$ that $\epsilon _2 \approx 2\epsilon _1$) is a base for homotopy groups! and let $A:=\{(x,y)\,|\, 0 \lt y\lt x,$ $0.01 \le x+y \lt 0.1\}$ & $V:=\{(a+b)\times 10^m \,|$ $(a,b) \in ((S_1 \times S_1) \cap A) \setminus L,$ there is a least member like $m$ in $\Bbb N$ such that $(a+b)\times 10^m \in \Bbb N \}$ & $r:\Bbb N \to (0,1)$ is a mapping given by $r(n)$ is obtained as put a point on the beginning of $n$ like $r(34880)=0.34880$ and similarly consider $\forall k \in \Bbb \N \cup \{0\} ,\, r_k: \Bbb \N \to (0,1)$ by $r_k(n)=10^{-k}\times r(n)$.

Conjecture $1$: For each even natural number like $t=t_1t_2t_3...t_k$, then $\exists (a,b),(b,a)\in L \cap$ $\{(x,y)\,|$ $x+y=0.0t_1t_2t_3...t_k\}$ such that $0.0t_1t_2t_3...t_k=a+b$ & $10^{k+1} \times a,10^{k+1} \times b$ are prime numbers.

Conjecture $1$ is an equivalent to Goldbach's conjecture, this conjecture has two solutions $1)$ Homotopy groups $\pi _{n \gt 1} (X)$ (by using cognition $L_1$ from homotopy groups this conjecture is solved of course we must attend to two spheres because $S^2$ minus the tallest point in north pole as topological and algebraic is an equivalent with plane $\Bbb R^2$ (except $\infty$) and also every mapping is made between these two spheres easily if these spheres aren't concentric.) and $2)$ Algebraic methods.

Assuming conjecture $1$, it guides us to finding formula of prime numbers at $(0,1) \times (0,1),$ in natural numbers based on each natural number is equal to half of an even number so in natural numbers main role is with even numbers but when we change space from $\Bbb N$ to $r(\Bbb N)$ then main role will be with $r( \{2k-1\, | \, k \in \Bbb N \} )$, because $r(\{2k-1\,|\,k\in \Bbb N\})\subset r(\{2k\,|\,k\in \Bbb N\})$ or in principle $r(\Bbb N)=r(\{2k\,|\,k\in \Bbb N\})$ for example $0.400=0.40=0.4$ or $0.500=0.50=0.5$ but however a smaller proper subset of $r( \{2k-1\, | \, k \in \Bbb N \} \cup \{2\} )$ namely $S$ is helpful, but for finding formula of prime numbers we need to all power of Badali lemma not only what such that is stated in above conjecture namely for example we must attend to the set $V$ too!

Conjecture $2$: For each even natural number like $t=t_1t_2t_3...t_k,$ $\exists x\in \{\alpha (a^2+b^2)^{0.5}\,|$ $(a,b) \in L,$ $\alpha \in (1,\sqrt 2] \} \cap r_1(\{2k\,|\, k\in \Bbb N \} )$ such that $t=10^{k+1} x$.

Conjecture $2$ is an equivalent to conjecture $1$, because $\forall t=t_1t_2t_3...t_k \in \Bbb N$ that $t$ is even, $\forall (a,b)\in \{(x,y)\,|\, x+y=0.0t_1t_2t_3...t_k,\, 0\lt y\le x \}$ we have: $(a^2+b^2)^{0.5}\lt 0.0t_1t_2t_3...t_k\le \sqrt 2 \times (a^2+b^2)^{0.5}$ so by intermediate value theorem we have $0.0t_1t_2t_3...t_k=\alpha (a^2+b^2)^{0.5}$ that $1\lt \alpha \le \sqrt 2$. But now if $a=10^{-k-1} p,b=10^{-k-1} q$ for $p,q$ belong to prime numbers we have:_{$$\alpha = \frac{t}{\sqrt {p^2+q^2}}$$}

Theorem: $\forall p,q,r,s$ belong to prime numbers & $q \lt p$, $(p,q)$ is located at the direct line contain the points $(0,0),10^{-\varphi ((p,q))}(p,q)$ and if $(r,s)$ is belong to this line then $p=r$ & $q=s$.

Let $S^2_2$ be a sphere with center $(0,0,r_2)$ and radius $r_2$ and $S^2_1$ be a sphere with center $(0.007,0.005,c)$ and radius $r_1$ such that $S^2_1$ is into the $S^2_2$ now suppose $f_1,f_2$ are two mapping from $A$ to $S^2_2$ such that $1)$ if $x \in A,$ $f_1 (x)$ is a curve on $S^2_2$ that is obtained as below: from $x$ 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) $

Guess $1$: $f_3 (L_1) $ is dense in the $S^1 \cap \{ (x,y)\,|\, 0 \le y$ & $2^{-0.5} \le x \}$.

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 $S^2_2 \setminus \{(0,0,2r_2)\}$.

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 $y=\pm x,\,y=0,\,x=0$

Theorem: Let $\Bbb P$ is the set prime numbers and $K_3 =\{p+q+r\,|\, p,q,r \in \Bbb P \}$ so $r(K_3)$ is dense in the interval $(0.1,1)$ of real numbers. Proof by Goldbach's weak conjecture.

Guess $2$: If $\Bbb P$ is the set prime numbers and $K_2 =\{p+q\,|\,p,q \in \Bbb P \}$ so $r(K_2)$ is dense in the interval $(0.1,1)$ of real numbers.

Let $F= \Bbb Q$, so what are Galois group of polynomials $x^4+b^2x^2+b^2$ and $(1+a^2)x^2 +a^4$.

Theorem $2$: If $(a,b),(c,d)\in \{(u,v)\,|\, u,v\in S_1$ & $0.01\le u+v\lt 0.1$ & $0\lt v\lt u \}$ and $(a,b),(c,d),(0,0)$ are located at a direct line then $(a,b)=(c,d)$.

Proof: Suppose $A_1=\{(x,y)\,|\, y \lt x\lt 0.01,\, x+y\ge 0.01\}$ & $A_2=\{(x,y)\,|\, y\lt x\lt 0.1,$ $x+y\ge 0.1\}$ so $\forall (x,y) \in A_2 :\,\, 0.1(x,y)\in A_1$ & $\forall (x,y) \in A_1 :\,\, 10(x,y)\in A_2$ so theorem can be proved in $A_3=\{(x,y)\,|\, 0\lt y\lt x\lt 0.1,\, x\ge 0.01\}$ instead $A$, but in $A_3$ we have: $\forall (x_1,y_1),(x_2,y_2)\in A_3\cap (S_1\times S_1)$ so $x_1=10^{-r_1}p_1,\, y_1=10^{-s_1}q_1,$ $x_2=10^{-r_2}p_2,$ $y_2=10^{-s_2}q_2$ and if ${{y_1}\over {x_1}}$=${{y_2}\over {x_2}}$ then ${{10^{-s_1}q_1}\over {10^{-r_1}p_1}}$=${{10^{-s_2}q_2}\over {10^{-r_2}p_2}}$ so $p_1=p_2,\, q_1=q_2$ so $x_1=x_2$ so $y_1=y_2$ therefore $(x_1,y_1)=(x_2,y_2)$.

Let $Y=\{(a,b)\,|\, (a,b)\in (S_1\times S_1)\setminus L,\, 0.01\le a+b\lt 0.1\}$ & $\forall i\in \Bbb N,$ $E_i=\{(a,b)\,|$ $(a,b)\in S_1\times S_1,$ $a+b=r_1(2i)\}$ & $O_i=\{(a,b)\,|\, (a,b)\in S_1\times S_1,\, a+b=r_1(2i-1)\}$.

In theorem $2$ I obtained a cognition to $(S_1\times S_1)\cap A$ from $(0,0)$ but now I want do it from $\infty$, in the shape trapezoid with vertices $\{(0.1,0),(0.01,0),(0.05,0.05),(0.005,0.005)\}$, intersection of two direct lines contain points $\{(0.1,0),(0.01,0)\}$ & $\{(0.05,0.05),(0.005,0.005)\}$ is $(0,0)$ so we can describe $(S_1\times S_1)\cap A$ from $(0,0)$ but when we look at two parallel lines contain points $\{(0.1,0),(0.05,0.05)\}$ & $\{(0.01,0),(0.005,0.005)\}$ there isn't any point as a criterion for description of $Y$ or $L$ only inaccessible $\infty$ remains to description or the same these parallel lines contain points $\{(0.1,0),(0.05,0.05)\}$ & $\{(0.01,0),(0.005,0.005)\}$.

Theorem $3$: $1)\, \forall x\in [0.01,0.1)\setminus r_1 (\Bbb N),\,\, \{(u,v)\,|\, u+v=x\}\cap (S_1\times S_1)=\emptyset ,\,\,$ $2)\, \forall i\in \Bbb N,$ $E_i\subsetneq L,\, O_i\cap L\neq \emptyset \neq O_i\cap Y\neq O_i,\;\;\;\;\;\; Y=(\bigcup _{i\in \Bbb N} O_i )\setminus L,$ $L=(\bigcup _{i\in \Bbb N} E_i)\cup (\bigcup _{i\in \Bbb N} (O_i\cap L)),\,\,$ $3)\, \forall i\in \Bbb N,$ cardinal$(O_i \setminus L)\in \Bbb N$.

Proof: $1,2)\, \forall (u,v)\in \{(x,y)\,|\, 0\lt y,x,\, 0.01\le x+y\lt 0.1\}$ be aware to summation $u+v$ at the lines $x+y=c$ for $0.01\le c\lt 0.1 \,\,\,3)\, \forall i\in\Bbb N,\, 2i-1$ can be written as utmost $2i-1$ summation to form of $a\times 10^m+b\times 10^n$ that $m\neq n,\, a,b\in S_1,\, a\times 10^m,b\times 10^n$ are prime numbers and or to form of $2+b\times 10^n$ that $b\in S_1,\, b\times 10^n$ is a prime number.

Guess $3$: $\forall i\in \Bbb N,$ cardinal$(E_i)=\aleph_0 =$cardinal$(O_i \cap L)$

Guess $4$: $L_1$ is dense in $\{(x,y)\,|\, 0\le y\le x,\, 0.01\le x+y\le 0.1\}$ and $((S_1\times S_1)\cap A)) \setminus L$ isn't dense in $\{(x,y)\,|\, 0\le y\le x,\, 0.01\le x+y\le 0.1\}$.

Guess $5$: $\forall i\in \Bbb N,\, E_i\cap \{(x,y)\,|\, y\le x\}$ is dense in the $\{(x,y)\,|\, x+y=r_1(2i),\, 0\le y\le x\}$ and $O_i\cap L\cap \{(x,y)\,|\, y\le x\}$ is dense in the $\{(x,y)\,|\, x+y=r_1(2i-1),$ $0\le y\le x\}$.

A rectangle: Suppose $B$ is a rectangle with vertices $(0.105,-0.005),(0.05,0.05),(0.005,0.005),$ $(0.06,-0.05)$ that one of usages of this rectangle is for writing each even natural number as minus of two prime numbers but during calculations we must use $\{(x,0)\,|\, 0.01\le x\lt 0.1\}$, but the question is however this rectangle as topological isn't equivalent to the plane $\Bbb R^2$ and each point in this rectangle is corresponding to a infinite set with cardinal $\aleph_0$ in $\Bbb R^2$ but what concept on the plane $\Bbb R^2$ is corresponding to the density concept on this rectangle.

Now I want find a relation between $L_1$ & $W:=((S_1\times S_1) \cap A) \setminus L$.

Theorem: Let $K =\{2k \,|\, k \in \Bbb N \}$ so $r(K)$ is dense in the interval $(0.1,1)$ of real numbers. Proof by the Badali Lemma and this $r(p)=r(p\times 10)$ that $p$ is a prime number then $p\times 10$ is an even number and $\{ p\times 10 \,|\, p \in \Bbb P \} \subset K$.

Badali lemma as a result of Dirichlet theorem on arithmetic progressions and prime number theorem or some distributions of prime numbers, is a fundamental concept in number theory also multiplication operation is a base in normal definition of prime numbers so logarithm function as an inverse of $f(a)=a^n$ has some or whole prime numbers properties that has been used in prime number theorem and consequently in the Badali lemma. But I want offer a new theory with researching on logarithmic functions that it can be a useful discussion in number theory.

Now a new definition of prime numbers based on mapping $r$ is necessary, presently I have an idea consider $\forall k\in \Bbb N,$ the sequence $b_k:\Bbb N \to \{1,2,3,4,5,6,7,8,9\},\,b_k(n)$ is the last digit in $k^n,$ so if $k=k_1k_2k_3...k_r$ then $b_k(1)=k_1$ and if $k^n=t_1t_2t_3...t_s$ so $b_k(n)=t_1,$ but for primes $k,$ it is a special different pattern than composite numbers and of course I want find some properties on $r$ for example $r(m \times n)$ when last digit is $1,2$ or $3$ or $4,5,6,7,8,9$, of course for $3$ penultimate digit (and probably two to last digit) is important and in addition is there any way for assessment location $r(m\times n)$ from $r(m)$ & $r(n)$.

Our weakness is from basic concepts, I want obtain a cognition of $(S_1\times S_1)\cap A$ and $L_1$ from point $(0.02,0.03)$ like theorem $2$ from $(0,0)$, but this time it is an equivalent to a new definition of prime numbers, so we will enable bring up so many effective theorems in number theory, because intersection of two direct lines contain points $\{(0.1,0),(0.01,0)\}$ & $\{(0.05,0.05),(0.005,0.005)\}$ is the point $(0,0)$ but however two direct lines contain points $\{(0.1,0),(0.05,0.05)\}$ & $\{(0.01,0),(0.005,0.005)\}$ are parallel so imposition of point $(0.02,0.03)$ as a criterion only can be equilibrated by concept of prime numbers!

A hypothesis: $\forall m,t\in\Bbb N$ if $t=t_1t_2t_3...t_k\times 10^{m-1}$ for $t_k=2,4,6,8$ or $t=t_1t_2t_3...t_k\times 10^m$ for $t_k=1,3,5,7,9$ then if $\forall p,q\in \Bbb P$ that $p,q\lt t,$ $t\neq p+q$ then $\exists M\subseteq \Bbb N,$ cardinal$(M)=\aleph_0,$ $\forall i\in M$ if $\forall r,s\in\Bbb P$ that $r,s\lt t\times 10^{i-m}$ then $t\times 10^{i-m}\neq r+s$.

I believe the Goldbach's conjecture is truth so I think with proof by contradiction we can prove the Goldbach's conjecture.

I think with assuming this hypothesis, if we can prove density of $L_1$ in $A$ (by Dirichlet's theorem on arithmetic progressions) then a contradiction will be obtained and consequently the Goldbach's conjecture will be proved.

From Dirichlet's theorem on arithmetic progressions that says: for any two positive coprime integers $a$ and $d$, there are infinitely many primes of the form $a+nd$, where $n$ is a non-negative integer, and from existence an one-to-one correspondence between the sets $A_1$ & $A_2$ in theorem $2$, I think there is an invariant quantity in the following sets, $\forall p\in\Bbb P$ if $0.01\le r_1(p)\le 0.05$ the set $\{(a,b)\in L_1\,|\, a=r_1(p)\},$ and if $0.05\lt r_1(p)\lt 0.1$ the set $\{(a,b)\in L_1\,|\, a=r_1(p)\}\cup\{(a,b)\in L_1\,|\, a=r_2(p)\}$.

In the following I need to make a partition for $L_1$.

Guess $6$: $\forall t,r,s\in\Bbb N,\, t=t_1t_2t_3...t_k$ is even$,\, r,s$ are odd$,\, s\le r,\, t=r+s$ then $(10^{-k-1}r,10^{-k-1}s)\in A\cup \{(x,x)\,|$ $0.005\le x\lt0.05\}$.

Alireza Badali 22:21, 8 May 2017 (CEST)

Polignac's conjecture

In number theory, Polignac's conjecture was made by Alphonse de Polignac in 1849 and states: For any positive even number n, there are infinitely many prime gaps of size n. In other words: There are infinitely many cases of two consecutive prime numbers with difference n. (Tattersall, J.J. (2005), Elementary number theory in nine chapters, Cambridge University Press, ISBN 978-0-521-85014-8, p. 112) Although the conjecture has not yet been proven or disproven for any given value of n, in 2013 an important breakthrough was made by Zhang Yitang who proved that there are infinitely many prime gaps of size n for some value of n < 70,000,000.(Zhang, Yitang (2014). "Bounded gaps between primes". Annals of Mathematics. 179 (3): 1121–1174. MR 3171761. Zbl 1290.11128. doi:10.4007/annals.2014.179.3.7. _ Klarreich, Erica (19 May 2013). "Unheralded Mathematician Bridges the Prime Gap". Simons Science News. Retrieved 21 May 2013.) Later that year, James Maynard announced a related breakthrough which proved that there are infinitely many prime gaps of some size less than or equal to 600.(Augereau, Benjamin (15 January 2013). “An old mathematical puzzle soon to be unraveled? Phys.org. Retrieved 10 February 2013.)

Assuming Polignac's conjecture there isn't any rhythm for prime numbers and so there isn't any formula for prime numbers!

Let $B=\{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1\}$ & $C=(S_1\times S_1)\cap \{(x,x)\,|\, 0.01\le x\lt 0.1\}$ & $L_2=\{(a,b)\,|\, a,b \in S_1,\, 0.01 \lt b \lt a\lt 0.1,\, \exists m \in \Bbb N,\, a \times 10^m, b \times 10^m$ or $a\times 10^{m-1}, b\times 10^m$ are consecutive prime numbers$\}$ & $\forall n \in \Bbb N,\, J_n :=\{(a,b) \,|\, (a,b) \in L_2,\, \exists k \in \Bbb N, a-b=r_k (2n)\}$.

It is clear $\bigcup _{n\in \Bbb N} J_n=L_2$ and Polignac's conjecture is equivalent to $\forall n \in \Bbb N,$ cardinal$(J_n)=\aleph_0$ and graph of $L_2$ is a new definition of prime numbers.

Guess $1$: $\forall (a,a) \in C$ there are some sequences in $L_2$ like $\{a_n\}$ that $a_n\to (a,a)$ where $n\to \infty$ and there are some sequences in $L_2$ like $b_n$ that $b_n\to (0.1,0.01)$ where $n\to \infty$.

Guess $2$: $\exists N_1 \subseteq \Bbb N,\, \forall n\in N_1,$ cardinal$(J_n)=\aleph_0=$cardinal$(N_1)$

It is clear $\exists \epsilon,\epsilon_1,\epsilon_2 \in \Bbb R,\epsilon\gt 0, \epsilon_2\gt \epsilon_1\gt 0,$ $\forall (a,b) \in L_2 \cap \{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,\, x-y\lt \epsilon\}$ that $\exists m\in \Bbb N,\, a\times 10^m,b\times 10^m$ are consecutive prime numbers, but $a\times 10^m,b\times 10^m$ are large natural numbers, and $\forall (a,b)\in L_2\cap$ $\{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,$ $0.11-\epsilon_1 \lt x+y\lt 0.11+\epsilon_1,$ $x-y\gt 0.09-\epsilon_2 \}$ that $\exists m\in \Bbb N,\, a\times 10^{m-1},b\times 10^m$ are consecutive prime numbers, but $a\times 10^{m-1},b\times 10^m$ are large natural numbers and $\forall (a,b)\in L_2\setminus (\{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,$ $x-y\lt \epsilon\}\cup\{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,$ $0.11-\epsilon_1 \lt x+y\lt 0.11+\epsilon_1,\, x-y\gt 0.09-\epsilon_2 \})$ that $\exists m\in \Bbb N,\, a\times 10^{m-1}$ or $a\times 10^m$ & $b\times 10^m$ are consecutive prime numbers, but $a\times 10^m$ or $a\times 10^{m-1}$ & $b\times 10^m$ aren't large natural numbers.

Theorem: $\forall c\in r_1(\Bbb P)$ cardinal$(L_2\cap \{(x,c)\,|\, x\in \Bbb R \})=1=$ cardinal$(L_2\cap \{(c,y)\,|\, y\in \Bbb R\})$ and $\forall k\in \Bbb N,$ $\forall c\in r_k (\Bbb N),\, c\lt 0.09$ cardinal$(L_2\cap \{(x,y)\,|\, x-y=c\}) \in \Bbb N \cup \{0\}$.

Guess $3$: For each subinterval of $[0.01,0.1)$ like $(a,b),\, \exists m\in \Bbb N,\, \forall k\in \Bbb N,\, k\ge m,\, \exists t\in (a,b),$ $t\times 10^{k+1}\in \Bbb P$.

Theorem: $\forall \epsilon_1,\epsilon_2,\, 0\lt \epsilon_1\lt \epsilon_2\lt 0.09,$ cardinal$(\{(a,b)\,|\, (a,b)\in L_2,\, 0.01\lt b\lt a\lt 0.1,$ $\epsilon_1\lt a-b\lt \epsilon_2\})\in \Bbb N\cup \{0\}$ and cardinal$(L_2\setminus \{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,\, x-y\gt \epsilon_1\})=\,\,\,\,$cardinal$(L_2\setminus \{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,\, x-y\lt \epsilon_2\})=\aleph_0$.

Proof: Be aware to number of digits in prime numbers corresponding to coordinates of each member in $L_2$.

Alireza Badali 13:17, 21 August 2017 (CEST)

Landau's forth problem

Landau's forth problem: Are there infinitely many primes $p$ such that $p−1$ is a perfect square? In other words: Are there infinitely many primes of the form $n^2 + 1$?

In analytic number theory the Friedlander–Iwaniec theorem states that there are infinitely many prime numbers of the form ${\displaystyle a^{2}+b^{4}}$.

Suppose $H=\{(a,b)\,|\, a\in\bigcup_{k\in\Bbb N} r_k(\{n^2\,|\, n\in\Bbb N\}),\, b\in\bigcup_{k\in\Bbb N} r_k(\{n^4\,|\, n\in\Bbb N\}),\,\exists t\in\Bbb N,$ $a\times 10^t\in\{n^2\,|\, n\in\Bbb N\},$ $b\times 10^t\in\{n^4\,|\, n\in\Bbb N\},\, (a+b)\times 10^t\in\Bbb P\}$ & $H_1=\{(a,b)\,|\, (a,b)\in H,$ $b\in\bigcup_{k\in\Bbb N} r_k(\{1\})\}$

Friedlander-Iwaniec theorem is the same cardinal$(H)=\aleph_0$

A question: cardinal$(H_1)\in\Bbb N$ or cardinal$(H_1)=\aleph_0$?

This question is the same Landau's forth problem.

Alireza Badali 20:47, 21 September 2017 (CEST)

Grimm's conjecture

In mathematics, and in particular number theory, Grimm's conjecture (named after Karl Albert Grimm) states that to each element of a set of consecutive composite numbers one can assign a distinct prime that divides it. It was first published in American Mathematical Monthly, 76(1969) 1126-1128.

Formal statement: Suppose $n + 1, n + 2, …, n + k$ are all composite numbers, then there are $k$ distinct primes $p_i$ such that $p_i$ divides $n+i$ for $1 ≤ i ≤ k$.

Let $C=\{(x,y)\,|\, 0.01\le x\lt 0.1,\, 0.01\le y\lt 0.1\}$

A conjecture: Suppose $n+1,n+2,n+3,...,n+k$ are all composite numbers then $\forall i=1,2,3,...,k$ $\exists (r_1(p_i),r_1(t_i))\in C$ that $p_i\in\Bbb P,$ $t_i\in\Bbb N$ & $\forall j=1,2,3,...,k$ that $i\neq j,$ $p_i\neq p_j$ we have $r_1(p_i)\times r_1(t_i)=r_2(n+i)$ or $r_3(n+i)$.

This conjecture is an equivalent to Grimm's conjecture.

Alireza Badali 23:30, 20 September 2017 (CEST)

From Badali 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 $T$.

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 $S×T$ & $T×T$. Musictheory2math (talk) 17:40, 25 March 2017 (CET)

"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. Boris Tsirelson (talk) 18:53, 25 March 2017 (CET)

Theorem: Let $(X,T_1),\,(Y,T_2)$ be topological spaces and $H$ be a homeomorphism from $X \to Y$. If $C$ is a dense subset of $X$, $H(C)$ is dense in the $Y$ necessarily.

Proof: Since density is a property which only depends on the topology, this is true, namely, suppose $U$ is a nonempty open subset of $Y$, then since $H$ is bijective, we can rewrite $U \cap H(C) = H(H^{-1}(U) \cap C)$, however, $H^{-1}(U)$ is open by continuity of $H$, and nonempty since $H$ is surjective, therefore, $H^{-1}(U) \cap C$ is nonempty since $C$ is dense; and therefore, $U \cap H(C) = H(H^{-1}(U) \cap C)$ is also nonempty. (So, this proves only using that $H$ is continuous and bijective, it is actually possible to refine the proof to work only assuming that $H$ is continuous and surjective - in that case, $U \cap H(C) \supseteq H(H^{-1}(U) \cap C)$.)

Another proof: Let $y\in Y$; for proof of every neighborhood $N$ of $y$, $N\cap H(C)\neq\emptyset$, take $x\in X$ such that $f(x)=y$, then $f^{-1}(N)$ is a neighborhood of $x$ and therefore $f^{-1}(N)\cap C\neq\emptyset$. So, $N\cap H(C)\neq\emptyset$.

Let $D=\mathbb{Q} \cap (0.1,1)$

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.

Theorem: $D$ and $T$ are homeomorph by the Euclidean topology.

Let $T_1$={ $a/10^n$ | $a\in{T}$ for $n=0,1,2,3,...$ }.

Theorem: $T_1$ is dense in the interval $(0,1)$.

Assume $T_1$ is the dual of $S_1$ so the combine of both $T_1$ and $S_1$ make us so stronger.

Let $W$={ $±(z+a)$ | $a\in{S_1 \cup T_1}$ for $z=0,1,2,3,...$ } & $G=\mathbb{Q} \setminus W$

Theorem: $W$ and also $G$ are dense in the $\mathbb{Q}$ and also $\mathbb{R}$.

A guess: $\forall g\in G,\,\exists n\in\Bbb N,\,\exists w_1,w_2,w_3,...,w_n\in W\cap (g-0.5,g+0.5)$ such that $g=(w_1+w_2+w_3+...+w_n)/n$.

Of course it is enough that guess just be proved for the interval $(0,1)$ namely assume $g\in{(0,1) \cap {G}}$. And if this guess is true then a new definition of rational numbers will be obtained.

Alireza Badali 22:21, 8 May 2017 (CEST)

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. Musictheory2math (talk) 20:57, 2 April 2017 (CEST)

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". Boris Tsirelson (talk) 16:46, 8 April 2017 (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 (talk) 23:34, 8 April 2017 (CEST)

See Continuum hypothesis. Boris Tsirelson (talk) 07:27, 9 April 2017 (CEST)

Also, did you try Wikipedia? Visit this: WP:Continuum_hypothesis. And this: WP:Project Mathematics. And WP:Reference desk/Mathematics. 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 further and I think the well-ordering theorem results each set is exactly a line. Yours Sincerely Badali

$2)$ A wonderful definition in the mathematical philosophy (and perhaps mathematical logic too): In between the all mathematical concepts, there are some special concepts that are not logical or are in contradiction with other concepts but in the whole mathematics a changing occurred that made mathematics logical, I define this changing the curvature of mathematical concepts.

$3)$ Most important question: The formula of prime numbers what impact will create on mindset of human.

$4)$ About physics: The time dimension is a periodic phenomenon. Fifth dimension is thought with more speed than light.

$5)$ Artificial intelligence is the ability to lying without any goofs of course only for a limited time.

$6)$ Each mathematical theory is a graph or a hyper graph. _ I believe graph theory (with hyper-graph & multi-graph & ...) is the best and latest mathematical theory forevermore that it includes all mathematical theories and mathematical logic and even mathematical philosophy, now I need to know whether any graph has been defined contain all natural numbers and also all concepts of number theory or that has been any graph defined contain all natural numbers as vertices with satisfying below rules for definition edges: 1) Theorem division algorithm 2) One of beauties of mathematics, Fermat's last theorem (and I guess somehow this theorem is an equivalent for induction axiom and maybe by another condition): The equation $x^n+y^n=z^n$ has no solution in nonzero integers if $n \ge 3$.

$7)$ 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 relation 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.

$8)$ The Goldbach`s conjecture is a way for finding formula of prime numbers. Each odd number is a sum of three primes (Goldbach's weak conjecture has been proved by Peruvian Mathematician, Harald Andrés Helfgott) and if each even number be a sum of two primes so from prime numbers we can make all natural numbers by summation of two or three prime numbers but from another way than previous factorization to primes (but it is higher than natural number concept as a factorization to primes.) and it means a new definition of natural numbers and with this overview to numbers Goldbach conjecture is an equivalent to induction axiom and normal definition of natural numbers and formula of prime numbers.

$9)$ Find rule of this sequence $a(n)$, $a:\Bbb N \to \Bbb Q,$ by $a(n)={n \over k(n)}$ such that $k:\Bbb N \to \Bbb N$, $k(n)$ is the number of digits in $n$ 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 ,\, ... $, in principle I believe this sequence is a fundamental concept in mathematics and mathematical logic.

$10)$ But how must think about infinity? by axioms? by accepted properties? by dreamy imagination? ... who knows? ... but finally must think because without infinity a consequence or result isn't artistic! and every property or proposition is a result of infinity and infinity has all properties concurrent and this is definition of infinity, something has everything! But formula of prime numbers is a definition of infinity and the best! and key of gateway of eternity is prime numbers and its formula or the same unique music!

$11)$ How can it be possible that each uncountable group has some countable subgroups certainly however, cardinal isn't an algebraic concept!, about finite groups this fact follows from algebraic structures on finite sets with a natural number as a cardinal and from properties on natural numbers in number theory because the own numbers have algebraic structures but about infinite ordinal numbers problem is basically different in principle I want to know whether infinite ordinal numbers have algebraic structures as only a set of course the number of finite groups with cardinal $n$ is a specific number like $m$ but about infinite ordinal numbers the number of groups is infinite. in principle I need to know whether there is an algebraic structure on sets with infinite cardinal as an axiom so being answered some questions about infinite ordinal numbers like continuum hypothesis! so can result cardinal concept from algebraic structures of course particularly for infinite sets and this is a way for understanding infinite ordinal numbers and answering to this question answers continuum hypothesis. only in algebraic structures attention is on members of a set and this is attention to members of infinite sets and being resulted cardinal from it! in order to algebraic theories are key of understanding infinite ordinal numbers, from understanding members of a set being realized cardinal of set and this is absolutely logical so being said cardinal is a algebraic concept and a result from algebraic theories although it hasn't been mentioned in algebraic structures and attention to members even about infinite sets is attention to cardinal so answering to infinite ordinal numbers is in algebraic structures in order to from knowing algebraic properties being reached to realizing of sets.

$12)$ A basic system like decimal system must be existed as an axiom and another systems must be defined based on it, otherwise concept of numbers is obscure. _ Summation of digits of a natural number with consideration number of zeros in middle of number and not in its beginning (with partition by number of zeros) is related to divisors of that number particularly for prime numbers.

Alireza Badali 22:21, 8 May 2017 (CEST)