User talk:Musictheory2math

Painting theory$(Edge$ $of$ $Darkness)$

Nanas lemma: If $P$ is the set of prime numbers and $S$ is 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 lemma is the Base of finding formula of prime numbers.

$($So many years ago, a Night I saw a dream, in this dream I saw completed proof of formula of prime numbers clearly but after than waking up, I forgot everything except starting point this lengthy paper, namely the Nanas lemma that of course Nanas is my parrot and I like it so much, and anyway, bellow notes are myself theories that I have made them newly, after than Professor Boris Tsirelson`s beauty proof for the Nanas lemma.$)$Alireza Badali 22:21, 8 May 2017 (CEST)

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 the Base of 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.Alireza Badali 22:21, 8 May 2017 (CEST)

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'

Theorem $1$: For each natural number like $a=a_1a_2a_3...a_k$ that $a_j$ is j_th digit in the decimal system 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.Alireza Badali 22:21, 8 May 2017 (CEST)

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)

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. Of course graph theory and hyper graph theory are so better than topology, but I personally prefer topology for this painting theory, and in principle, each mathematical theory is a graph or a hyper graph.

Now, assume $H$ is a mapping from $(0.1,1)$ on $(0.1,1)$ given by $H(x)=1/(10x)$.

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 its digit in the decimal system 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 $(0.1,1)$.

Let $D$={ $q$ | $q$ is a rational number & $q\in(0.1,1)$ }

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.Alireza Badali 22:21, 8 May 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 is new and for the first time.Alireza Badali 22:21, 8 May 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)

Dear Professor Boris Tsirelson, I thank you so much for your valuable helps to me and I owe you because more than $10$ years I could not prove the Nanas lemma, and a day accidental came here and saw you and however in the beginning I did not have a good behavior with you but you with kindness guided me and I must say that in principle you gave to me a new hope to continue.

Assume $S_1$={ $a/10^n$ | $a\in{S}$ for $n$=$0,1,2,3,...$ } and $T_1$={ $a/10^n$ | $a\in{T}$ for $n=0,1,2,3,...$ }.

Theorem: $S_1$ and also $T_1$ are dense in the interval $(0,1)$. Proof by the Axiom of Choice and the Nanas lemma

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

Let $W_1$={ $z+a$ | $z$ is a integer number & $a\in{S_1 \cup T_1}$ } and $W_2$={$±(z+a)$ | $a\in{S_1 \cup T_1}$ for $z=0,1,2,3,...$}. '''Theorem''' $2$: $W_1$ and $W_2$ are dense in the set of rational numbers and also real numbers. Proof by the Axiom of Choice '''Conjecture''' $1$: For each rational number like $t$ that $t\notin{W_1}$, there are two members of $W_1$ like $a,b$ in the interval $(t-0.5,t+0.5)$ such that $t=(a+b)/2$. '''Conjecture''' $2$: For each rational number like $t$ that $t\notin{W_2}$, there are two members of $W_2$ like $a,b$ in the interval $(t-0.5,t+0.5)$ such that $t=(a+b)/2$. Conjectures $1$ and $2$ might be proved by the Axiom of Choice and theorems $1$ & $2$ and $...,$ of course this is enough that conjectures just is proved for the interval $(0,1)$ namely assume $t\in(0,1)$. And if these conjectures are true, a big revolution shall begin in the mathematics, because a new and useful definition of rational numbers is obtained, to wit, there is an unique algebraic equation that begets all of rational numbers from the prime numbers. It seems I can not find a homeomorphism from $D$ on $S$ '''('''or from rational numbers located at the $(0,1)$ on $S_1$ and even from the set of rational numbers on { $z+a$ | $z$ is a integer number & $a\in{S_1}$ } ''')''' but if a continuous mapping is found, we must make some topological spaces by common topology, into the $(0.1,1)×(0.1,1)$ of euclidean page such that some topological properties '''in''' important theorems or main axioms( particularly around the sequences ) is transferred to the set $S$, and most important section of our works is in here '''('''of course we can work on the $(0,1)×(0,1)$ or even on total of euclidean page, each one that is easier and better.''')'''. And now I must say that the formula of prime numbers is equal to an unique painting in the $(0.1,1)×(0.1,1)$ or $(0,1)×(0,1)$ and or euclidean page. Alireza Badali 22:21, 8 May 2017 (CEST) =='"`UNIQ--h-1--QINU`"' About ''Painting theory'' == I believe rectangle is the best for a figure (and even concept like multiplication at natural numbers), now I want go to the $(0.1,1)×(0.1,1)$ 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 are 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 $t_n=(x_n,y_n)$ is a sequence such that limit of $t_n$ is $t$ 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 1) the line $y=1/(10x)$ 2) under the line namely $V$ and 3) top of the line namely $W$, obviously each point in $V$ like $t=(x , y)$ has a dual point like $u=(1/(10x),1/(10y))$ in $W$, particularly if $x$ or $y$ are in $T$. Let $G$ is a mapping from $V$ on $W$ given by $G(x,y)=(1/(10x),1/(10y))$ thus $G$ keeps the topological properties, so 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 $S$.Alireza Badali 22:21, 8 May 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. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 18:53, 25 March 2017 (CET) Theorem: $D$ and $S$ are homeomorph. 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? so we must find another better it seems. Theorem: $D$ and $T$ are homeomorph. Alireza Badali 22:21, 8 May 2017 (CEST) =='"`UNIQ--h-2--QINU`"'Other problems== '''1)'''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 order. Is this right? and please say by who and when.Alireza Badali 22:21, 8 May 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". [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 16:46, 8 April 2017 (CEST) ::Dear Professor Boris Tsirelson, I thank you so much. But today I can not see that, I need few time for perception. ::But what about the "Axiom of Continuum" and the "Axiom of the Lack of Continuity", so should be proved that one of them is <big>'''wrong'''</big>. what is your idea?Alireza Badali 22:21, 8 May 2017 (CEST) :I do not know such axioms. What do you mean? [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk: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 $\aleph_1$ and $\aleph_2$ and so on I think that one of these axioms does not match with the order-able theorem.Alireza Badali 22:21, 8 May 2017 (CEST) :See [[Continuum hypothesis]]. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 07:27, 9 April 2017 (CEST) :Also, did you try Wikipedia? Visit this: [https://en.wikipedia.org/wiki/Continuum_hypothesis WP:Continuum_hypothesis]. And this: [https://en.wikipedia.org/wiki/Wikipedia:WikiProject_Mathematics WP:Project Mathematics]. And [https://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics WP:Reference desk/Mathematics]. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk: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 theorem order-able means every set is exactly a line. Sincerely yours Badali '''2)'''About '''<big>?!</big>''': The set of real numbers is dense in the whole of mathematics( whole and not set ). '''3)'''A wonderful definition in the mathematical philosophy (and perhaps mathematical logic too): In between the all of mathematical concepts, there are some special concepts that are not logical or are in contradiction with other concepts but in the whole of mathematics a changing occurred that made mathematics logical, I define this changing the curvature of mathematical concepts. :So many years ago in bachelor course my dear friend Mr. '''''<big>Aziz Darbandi</big>''''' and me were in one of our usual gaps in the faculty of chemistry of the University of Tabriz, and when I said this definition he stood up and said go to wc and fire these papers right now and I did, of course that time he was studying functional analysis and I was studying group theory because of algebraic theories just relate cardinal a set with its members. and '''''<big>Aziz</big>''''' if you see this please contact with me by '''''paintingtheory@yahoo.com''''' '''4)'''About mathematical logic: Complex set theory: It seems in different sections of mathematics we connect sets to each other by function concept that of course the own function is a set too with a property, but whether function concept has itself importance as a independent and basic axiom like Axiom of Choice or not, because in mathematics every work is occurred by function, in principle the Axiom of choice or natural numbers(or real numbers and or complex numbers) or function concept are dense in the whole of mathematics, and I want refer that a set theory is made just like complex numbers that set concept is real line and function concept is imagine line. Of course I think that the own perception is a function of own human not more and not less. '''5)'''Most important question: The formula of prime numbers what impact will create on mindset of human$?$ Alireza Badali 22:21, 8 May 2017 (CEST) =='"`UNIQ--h-3--QINU`"' Goldbach's conjecture is wrong! == '''Please just insert your comments here!''' '''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. '''Nanas lemma''': If $P$ is the set of prime numbers and $S$ is 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. Proof by Professor [[User:Boris_Tsirelson|Boris Tsirelson]] is [[User_talk:Musictheory2math#Painting_theory.7F.27.22.60UNIQ-MathJax1-QINU.60.22.27.7F_.7F.27.22.60UNIQ-MathJax2-QINU.60.22.27.7F_.7F.27.22.60UNIQ-MathJax3-QINU.60.22.27.7F|here]]. Assume $S_1$={ $a/10^n$ | $a\in{S}$ for $n$=$0,1,2,3,...$ } Theorem: $S_1$ is dense in the interval $(0,1)$. Proof by the Axiom of Choice and the Nanas lemma '''A conjecture''': For each rational number like $t$ with a finite sequence as $t=0.a_1a_2a_3...a_m$ in the interval $(0.1,1)$ such that $t\notin{S}$, thus there are two members of $S_1$ like $a,b$ that $t=(a+b)/2$. This conjecture is an equivalent for Goldbach's conjecture, but what will happen when $t\to{1}$, in principle I want make a set of numbers as $v=999...999a_ra_{r-1}a_{r-2}...a_3a_2a_1$ that $a_j$ is j_th digit in the decimal system & $9$ repeats $k$ times for natural numbers $r,k$ such that for some natural numbers like $n$ we have: $n!|v$, so the Goldbach's conjecture is wrong!

Alireza Badali 23:51, 11 May 2017 (CEST)