User talk:Musictheory2math

$Edge$ $of$ $Darkness$ (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.

$($Nanas is my parrot and I like it so much.$)$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)

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 $(0.1,1)$.Alireza Badali 13:49, 17 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). Boris Tsirelson (talk) 18:53, 25 March 2017 (CET)

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 help to me and I owe you because more than $10$ years I could not prove the Nanas lemma, but you proved it seemly and guided me honestly that in principle you gave me a new hope to continue.

Assume $S_1$={ $a/10^n$ | $a\in{S}$ for $n$=$0,1,2,3,...$ } & $T_1$={ $a/10^n$ | $a\in{T}$ 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 the dual of $S_1$ so the combine of both $T_1$ and $S_1$ make us so stronger.

Let $W$={ $±(z+a)$ | $a\in{S_1 \cup T_1}$ for $z=0,1,2,3,...$ } & $W_1$={$(w_1+w_2)/2$ | $w_1,w_2\in{W}$ & $|w_1-w_2|\lt{1}$} & $G$={ $g$ | $g$ is a rational number & $g\notin{W}$ } '''Theorem''' $2$: $W$ and also $G$ are dense in the set of rational numbers and also real numbers. '''Conjecture''': $W_1$ is equal to the set of rational numbers. Of course it is enough that the conjecture just is proved for all rational numbers located at the interval $(0,1)$, namely assume $g$ is a rational number located at the $(0,1)$ then should $g\in{W_1 \cap (0,1)}$ If this conjecture is true, a big revolution shall happen 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 located at the interval $(0,1)$ from the prime numbers. '''Therefore''' if a continuous mapping from $D$ on $S$ is found, we must make some topological spaces by the common topology, into the $(0.1,1)×(0.1,1)$ of euclidean page such that some topological properties particularly around the sequences is transferred to the set $S$. And now I must say that the formula of prime numbers is equal to an unique painting in the $(0.1,1)×(0.1,1)$. Alireza Badali 22:21, 8 May 2017 (CEST) =='"`UNIQ--h-1--QINU`"' About ''Painting theory'' == Theorem: $C=S×S$ is dense in the $(0.1,1)×(0.1,1)$. Similar theorems are right for $C=S×T$, $C=T×S$ & $C=T×T$.Alireza Badali 13:49, 17 May 2017 (CEST) :"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. Theorem: $D$ and $T$ are homeomorph. ''An important question'': Is there any real infinite subset of $S$ such that be dense in the interval $(0.1,1)$ of real numbers? Assuming $P_1$={ $p$ | $p$ is a prime number & number of digits of $p$ is a prime number } that $P_1$ is called the set of ''primer numbers'', now whether $P_1$ is dense in the $(0.1,1)$$?$ 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. '''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. '''6)''' About physics: The time dimension is a periodic function. '''7)''' Artificial intelligence is the ability to lying without any goofs of course only for a limited time. '''8)''' Each mathematical theory is a graph or a hyper graph. Alireza Badali 22:21, 8 May 2017 (CEST) =='"`UNIQ--h-3--QINU`"' A new version of Goldbach`s conjecture == '''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]]. '''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. 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 Assume $Y$={ $y$ | $y=0.y_1y_2y_3...y_m$ for each natural number like $m$ & $y_1,y_2,y_3,...,y_m=0,1,2,3,...,9$ & $y_1≠0$ } & $U$={ $(a+b)/2$ | $a,b\in{S_1}$ } are multi-sets. '''A conjecture''': $Y=U \cap (0.1,1)$ This conjecture is a new version of Goldbach`s conjecture, but the question is for the number $999...999000...000$ that there are $k_1$ up to $9$ and $k_2$ up to $0$ for some $k_1,k_2$ is belong to natural numbers, is this new version of Goldbach`s conjecture true or in principle is the Goldbach`s conjecture is true$?$

Assuming correctness the new version of Goldbach`s conjecture, it is a way for finding formula of prime numbers, because it states behavior of prime numbers.

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