User talk:Musictheory2math

Edge of Darkness

Main theorem: If P be the set of all prime numbers and S be a set that 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 Base of 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 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 "This theorem is Base of 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)

Topology is the strongest and broadest Mathematical theory because of limit of Topology is continuous as far as some Mathematicians say Topology is the Mathematical philosophy, anyway, 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.

now I define a mapping H from (0.1 , 1) on (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 interesting 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.

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

These 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)

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 or y are in T.

Now, I define a continuous mapping from V on 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 on S. In principle, I want work on rational numbers and then transfer to the set S.

Theorem: 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)... that a(k+1)a(k+2)...a(n-1)a(n) repeats and 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.♥ But there is a big problem: where is the rule of this homeomorphism? thus we MUST find another better IT SEEMS.

Theorem: D and T are homeomorph.

←←)←←)←←

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)

Assume D(1)=T U { [a(1)b(1)+a(2)b(2)+a(3)b(3)+...+a(n)b(n)]×n^(-1) : a(j) , b(j) are in T for j=1, 2, 3, ... , n & for each n in the natural numbers }

  A guess: D=D(1)

If this Conjecture be true, a BIG Revolution will happen in the Mathematics. Because of one obtain a NEW and USEFULL definition of rational numbers and two THIS theorem: The set of real numbers is dense in the WHOLE of Mathematics( WHOLE and not set ).

In principle, I will be making D(2) , D(3) and so on such that D(1)=D(2) , D(2)=D(3) and so on and Finally result D=D(j), and then I will define a homeomorphism from D=D(j) on S.

Alireza Badali (talk) 19:30, 9 April 2017 (CEST)

Other problems

About the Set theory:

So many years ago I heard of my friend that has been proved that each SET is order-able with total ordering. Is this right? If YES, please say very slowly, BECAUSE OF my brain is exploding. 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)

Dear Professor Boris Tsirelson, I thank you so much. I am going to DIE from wonder. But today I can not see that, I need few time for perception.

But how about the "Axiom of Continuum" and the "Axiom of the Lack of Continuity", SHOULD therefore be proved that one of them is wrong. what is your idea?

Alireza Badali (talk) 21:17, 8 April 2017 (CEST)

I do not know such axioms. What do you mean? Boris Tsirelson (talk) 22:57, 8 April 2017 (CEST)

My mind was this: Is there something between of the cardinal of natural numbers and the cardinal of real numbers? and also between 2^(N(0)) and 2^(2^(N(0))) and so on I think that one of these axioms does not match with the order-able TEOREM.

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 must go to the Wikipedia more. And I am unable in the SET theory, but, I think this TEOREM order-able means every set is EXACTLY a line. Sincerely yours Badali

Alireza Badali (talk) 09:04, 9 April 2017 (CEST)

BUT not more than this

About the Mathematical philosophy (and perhaps Mathematical Logic too):

A wonderful definition: In between the all of Mathematical concepts, there are Special concepts that are not logical OR are in contradiction with other concepts but in the whole of Mathematics a changing occurred that make Logical the Mathematics, I define this changing the curvature of Mathematical concepts.

Alireza Badali (talk) 18:14, 13 April 2017 (CEST)

Required Books

 Please send to: musictheory2math@outlook.com

Introduction to Analytic Number Theory, Authors: Apostol, Tom Mike

Some Books About the "Modern number theory"

Alireza Badali (talk) 15:28, 15 April 2017 (CEST)

THE THEY

THEY do not allow me to continue my job, do not allow contact with Mathematicians or Mathematical websites, THEY remove my emails and do not let my Emails be sent or when I find a good website THEY do not let it be opened. under these conditions I do not know what I must do.