User talk:Musictheory2math

Theorem 1

A way for finding formula of prime numbers

If P be set of prime numbers and S be a set contain numbers that has been made as below: On the beginning of each member of P put a point like 0.2 or 0.19 then S={0.2 , 0.3 , 0.5 , 0.7 , 0.11 , ... } Now I assert that S is dense in the interval (0.1 , 1) of real numbers. This theory is a introduction for finding formula of prime numbers.

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 , first, "A relationship between algebraic numbers and transcendental numbers with rational numbers" was wrong. and I thank you for your guidance. and second, Are you sure that is provable? 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.

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 thank you Professor Boris Tsirelson , for guidance. Now I want say one of results of the 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.

And now I want to 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 another important theorems in the real numbers for working on prime numbers and I think this is better and easier.