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.