User talk:Musictheory2math

Painting Theory

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 I was following the formula of prime numbers and 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 (talk) 09:16, 6 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 fact 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 (talk) 09:16, 6 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 (talk) 09:16, 6 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, 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 inverse of $a_1.a_2a_3a_4...a_n$ in $T$.

Theorem: $T$ is dense in the $(0.1,1)$.

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

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 (talk) 09:16, 6 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 (talk) 09:16, 6 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 $S_1$={ $a/10^n$ : $a$ is in $S$ for $n$=$0,1,2,3,...$ } and $T_1$={ $a/10^n$ : $a$ is 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 Choice axiom 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 $W$={ $z+a$ : $z$ is a integer number and $a$ is in $S_1 \cup T_1$ }.

Theorem: $W$ is dense in the set of rational numbers and also real numbers. Proof by the Choice axiom and the Nanas lemma

Assume $W_1$={ $(a_1+a_2)/2$ : $a_1$,$a_2$ are in $W$ and $|a_1-a_2|<1$ }, $W_2$={ $(a_1+a_2+a_3+a_4)/4$ : $a_1$,$a_2$,$a_3$,$a_4$ are in $W$ and $|a_i-a_j|<1$ for $i,j=1,2,3,4$ }, $W_3$={ $(a_1+a_2+...+a_5)/5$ : $a_1$,$a_2$,...,$a_5$ are in $W$ and $|a_i-a_j|<1$ for $i,j=1,2,3,4,5$ }, $W_4$={ $(a_1+a_2+...+a_8)/8$ : $a_1$,$a_2$,...,$a_8$ are in $W$ and $|a_i-a_j|<1$ for $i,j=1,2,3,...,8$ } and $W_5$={ $(a_1+a_2+...+a_10)/10$ : $a_1$,$a_2$,...,$a_10$ are in $W$ and $|a_i-a_j|<1$ for $i,j=1,2,3,...,10$ }

Conjecture: The set of rational numbers is equal to $W_i_1$

And if one of these conjectures is true, a BIG Revolution shall happen in the Mathematics, Because, a NEW and USEFUL definition of rational numbers is obtained, anyway, 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 and $a$ is 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 (talk) 09:16, 6 May 2017 (CEST)

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. 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 $S$.Alireza Badali (talk) 09:37, 6 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. 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 $k=0,1,2,3,...,n$ , I define the natural number $t=a_1a_2a_3...a_ka_{k+1}...a_n00...00$ such that I have inserted $k$ up to $0$ 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 therefore $0.a_1a_2a_3...a_ka_{k+1}...a_{n}00...00b_1b_2...b_r$ 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.

Alireza Badali (talk) 09:37, 6 May 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.Alireza Badali (talk) 10:55, 6 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". 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) 10:55, 6 May 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 theorem.Alireza Badali (talk) 10:55, 6 May 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 theorem order-able means every set is EXACTLY a line. Sincerely yours Badali

About ?!: The set of real numbers is dense in the WHOLE of Mathematics( WHOLE and not set ).

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) 10:55, 6 May 2017 (CEST)

From Leonard Huang

I don’t mean to pour cold water on what must be an exciting mathematical discussion, but the excessive editing of this user page swamps the Recent Changes section with irrelevant updates, which makes it hard for anyone to track changes made to articles. This is, after all, an encyclopedic website, not a mathematical blog. Leonard Huang (talk) 23:43, 22 April 2017 (CEST)

But my English is so weak and I try become better and secondly I am busying to making a theory and each time that be necessary edit my pages this is the reason of high number of editing. and if you have any question I am ready to answer. Alireza Badali (talk) 16:14, 23 April 2017 (CEST)

Hi Alireza. I understand that English is not your forte, but that is not my point. My point, rather, is that this is an encyclopedic website, so I do not consider it appropriate for you to be discussing your own theories here, especially when these discussions do not directly benefit the health of the encyclopedia. It is perfectly alright for you to have a user page that describes your area of expertise and some of your mathematical accomplishments, but when you start treating your user page like a research blog, all that people see under the Recent Changes section are updates to your user page and not updates to the articles herein. If you would like to discuss your theories with the mathematical community at large, please visit either Mathematics Stack Exchange or MathOverflow instead, where you may find that some of us are users. Leonard Huang (talk) 20:08, 23 April 2017 (CEST)

I thank you for your guidance, but I want discuss with Professional Mathematicians.Alireza Badali (talk) 21:03, 23 April 2017 (CEST)

MathOverflow is a question-and-answer site for professional mathematicians only. You will find famous number-theorists like Andrew Granville and Terence Tao who can answer your questions. Leonard Huang (talk) 21:13, 23 April 2017 (CEST)

Okay, I am going there!Alireza Badali (talk) 22:06, 23 April 2017 (CEST)

Here

And from now on I shall be absent here mostly and sometimes probably once a week I shall come back here.

Alireza Badali (talk) 19:02, 6 May 2017 (CEST)