User talk:Musictheory2math

A way for finding formula of prime numbers

Main theorem: If P be the set of all prime numbers and S be a set has been made as below: Put a point on the beginning of each member of 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 theorem is an introduction for finding 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 "a way for finding formula of prime numbers". Boris Tsirelson (talk) 22:10, 16 March 2017 (CET)

Dear Professor Boris Tsirelson , in fact finding formula of prime numbers is very lengthy. and I am not sure be able for that but please give me a 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)

'I have special thanks to Professor Boris Tsirelson for this beauty proof. Sincerely yours, Alireza Badali Sarebangholi'

Now I want say one of results of the main theorem:

Theorem 1: For each natural number like a=a(1)a(2)a(3)...a(k) that a(j) is j_th digit in the decimal system there is a natural number like b=b(1)b(2)b(3)...b(r) such that the number c=a(1)a(2)a(3)...a(k)b(1)b(2)b(3)...b(r) be 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)

And I want say philosophy of "A way for finding formula of prime numbers " : However we loose the well-ordering axiom and as a DIRECT result we loose the induction axiom for finite sets but I thought that if change SPACE from natural numbers with cardinal countable to a bounded set with cardinal uncountable in the real numbers then we can use other TOOLS like axioms and important theorems in the real numbers for working on prime numbers and I think this is better and easier.

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)

This several lines are less important: →→→→→

I believe rectangle is the best for a figure (and even concept like multiplication at natural numbers) Now I want go to the (0.1 , 1)x(0.1 , 1) in the Euclidean page.( Euclidean is the best every where) Now we have more tools to do.(my mind is sequences in the Euclidean page)

now I define a mapping H from (0.1 , 1) to (0.1 , 1) by H(x)=(10x)^(-1) thus H is continuous and descending.

Theorem: T=H(S) is dense in the (0.1 , 1).

T={ 2^(-1) , 3^(-1) , 5^(-1) , ... }={ (10^(n-1))xp^(-1) : p is in P and n is number of digits of p} T is a interested set for its members because of, a member of S like 0.a(1)a(2)a(3)...a(n) that a(j) is j-th its digit in the decimal system for j=1,2,3, ... ,n is basically different with inverse of a(1).a(2)a(3)a(4)...a(n) in T.

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.

Theorem: for each point in the (0.1 , 1)×(0.1 , 1) like t=(x,y), if t(n)=(x(n) , y(n)) be a sequence such that limit of t(n) be t and x(n) and y(n) are sequences in the S or T then limit of x(n) is x and limit of y(n) is y.

now I divide the (0.1 , 1)×(0.1 , 1) to three areas one the line y=(10x)^(-1) two under the line namely V and three top of the line namely W. Obviously each point in V like t=(x , y) has a dual point like u=((10x)^(-1) , (10y)^(-1)) in W , PARTICULARLY if x be in T.

Now, I define a continuous mapping from V to W like G by G(x , y)=((10x)^(-1) , (10y)^(-1)) thus G keeps the topological properties. Therefore each topological property in V like important theorems for example middle amount theorem and main axioms can be transferred by G from T to S for the first coordinates. In fact I want work on rational numbers and then transfer to the set of S.

←←←←←

Musictheory2math (talk) 15:00, 30 March 2017 (CEST)

"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)

Dear Professor Boris Tsirelson, your help is very valuable to me and I think we can make a good paper together of course if you would like.

Musictheory2math (talk) 16:47, 27 March 2017 (CEST)

Thank you for the compliment and the invitation, but no, I do not. Till now we did not write here anything really new in mathematics. Rather, simple exercises. Boris Tsirelson (talk) 18:50, 27 March 2017 (CEST)

But do not you think this way about prime numbers be new and for the first time.

Musictheory2math (talk) 14:23, 30 March 2017 (CEST)

It is not enough to say that this way is new. The question is, does this way give new interesting results? Boris Tsirelson (talk) 21:03, 30 March 2017 (CEST)

Let D={ q : q is a rational number and q is in (0.1 , 1) }

Theorem 2: D and S are homeomorph.

For each member of D like w=0.a(1)a(2)a(3)...a(k)a(k+1)a(k+2)...a(n-1)a(n)a(k+1)a(k+2)...a(n-1)a(n)... ( a(k+1)a(k+2)...a(n-1)a(n) repeats ) that k=0, 1, 2, 3, ..., n , I define the natural number t=a(1)a(2)a(3)...a(k)a(k+1)...a(n)00...00 such that I have inserted k up to 0 in the beginning of t , now by the induction axiom and theorem 1 , there is the least number in the natural numbers like b(1)b(2)...b(r) such that the number a(1)a(2)a(3)...a(k)a(k+1)...a(n)00...00b(1)b(2)...b(r) be a prime number and therefore 0.a(1)a(2)a(3)...a(k)a(k+1)...a(n)00...00b(1)b(2)...b(r) is in S.

And now by Professor Boris Tsirelson`s valuable help , my theory is completed, but now I need to some Mathematicians at Topology and Number theory for continue, and of course we must walk at the Edge of Darkness.

Topology is the strongest and broadest Mathematical theory because of limit of Topology is continuous and even some Mathematicians say Topology is the Mathematical philosophy, any way, we can transfer the Real numbers properties and Rational numbers properties that mostly are Topological, to the set S, that S is made by prime numbers DIRECTLY and is so similar to the set of prime numbers, therefore this means more FEATURES.

Theorem 3: D and T are homeomorph.

But a big problem is there: where is the rule of this homeomorphism? thus we find another better IT SEEMS.

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