User talk:Musictheory2math

A new version of Goldbach's conjecture

Nanas lemma: If $\mathbb{P}$ is the set of prime numbers and $S$ is a set that it has been made as below: Put a point on the beginning of each member of $\Bbb{P}$ 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.$($Nanas Dadile is my parrot and I like it so much.$)$

This lemma is a base for finding formula of prime numbers.

There is a musical note on the natural numbers that it can be discovered by the formula of prime numbers.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 a base for 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; Yours Sincerely, Alireza Badali Sarebangholi'

Theorem $1$: For each natural number like $a=a_1a_2a_3...a_k$ that $a_j$ is $j$_th digit for $j=1,2,3,...,k$, 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 I want say philosophy of This lemma is a base for finding formula of prime numbers: However we loose the induction axiom for finite sets but I thought that if change space from natural numbers with cardinal $\aleph_0$ to a bounded set with cardinal $\aleph_1$ in the real numbers then we can use other features like axioms and important theorems in the real numbers for working on prime numbers and I think it is a better and easier way.Alireza Badali 20:11, 22 July 2017 (CEST)

I see. Well, we are free to use the whole strength of mathematics (including analysis) in the number theory; and in fact, analysis is widely used, as you may see in the article "Distribution of prime numbers".

But you still do not put four tildas at the end of each your message; please do. Boris Tsirelson (talk) 11:16, 19 March 2017 (CET)

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 I was unable for proving the Nanas lemma but you proved it seemly and guided me honestly, that in principle you gave me a new hope to continue.

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.

Assume $S_1$={ $a/10^n$ | $a\in{S}$ for $n$=$0,1,2,3,...$ } & $L=\{(a,b)\,|\,a,b \in S_1$ & $0.01 \lt a+b \le 0.1$ & $\exists m \in \Bbb N,\, a \times 10^m,\,b \times 10^m$ are prime numbers $\}$

Theorem: $S_1$ is dense in the interval $(0,1)$.

If $p,q$ are prime numbers and $n$ is the number of digits of $p+q$ and $m=$max(number of digits of $p$, number of digits of $q$), let $\varphi : L \to \Bbb N,$ $\varphi ((p,q)) = \begin{cases} m+1 & n=m \\m+2 & n=m+1 \end{cases}$

Theorem: For each $p,q$ belong to prime numbers and $\alpha \in \Bbb R$ that $0 \le \alpha,$ now if $\alpha = q/p$ then $L \cap \{(x,y)\,|\,y=\alpha x \}=\{10^{-\varphi ((p,q))}(p,q)\}$ and if $\alpha \neq q/p$ then $L \cap \{(x,y)\,|\,y=\alpha x \}=\emptyset $ and if $\alpha = 1$ then $L \cap \{(x,x)\}$ is dense in the $(0,1) \times (0,1) \cap \{(x,x)\,|\,0.005 \lt x \le 0.05 \}$.

Definition: Assume $L_1=\{(a,b)\,|\,(a,b) \in L$ & $b \lt a \}$ and $F$ be a curve contain all of points of $L_1$ such that for each $\alpha \in \Bbb R$ that $0 \lt \alpha \lt 1 ,$ $\{(x,y)\,|\, (x,y) \in F \} \cap \{(x,y)\,|\, y= \alpha x \}$ has exactly one point and let $E=(0.007,0.005)$ be a base for homotopy groups!

$V:=\{(a+b)\times 10^m \,|\, a,b \in (S_1 \cap A) \setminus L, \, \exists m \in \Bbb N, \, (a+b)\times 10^m$ is an odd number.$\}$.

A conjecture: For each natural number like $t=t_1t_2t_3...t_k$, now $1)$ if $t_1=5 ,6, 7, 8, 9$ and $\forall m \in \Bbb N $ $t \neq 5 \times 10^{m-1}$ then there are two points in the $L \cap \{(x,y)\,|\,x+y=2 \times 0.00t_1t_2t_3...t_k\}$ like $(a,b),\,(b,a)$ such that $0.00t_1t_2t_3...t_k=(a+b)/2$ & $10^{k+2} \times a,\,10^{k+2} \times b$ are prime numbers and if $t$ is a prime number then $a=b=0.00t_1t_2t_3...t_k$ and $2)$ if $t_1=1, 2, 3, 4$ or $\exists k\in \Bbb N$ $t=5\times 10^{k-1}$ then there are two points in the $L \cap \{(x,y)\,|\,x+y=2 \times 0.0t_1t_2t_3...t_k\}$ like $(a,b),\,(b,a)$ such that $0.0t_1t_2t_3...t_k=(a+b)/2$ & $10^{k+1} \times a,\,10^{k+1} \times b$ are prime numbers and if $t$ is a prime number then $a=b=0.0t_1t_2t_3...t_k$.

This conjecture is a new version of Goldbach's conjecture. This conjecture has two solutions $1)$ Homotopy groups $\pi _{n \gt 1} (X)$ and $2)$ Algebraic methods, but in either case we must attend to two spheres $S^2$.

Assuming correctness the new version of Goldbach's conjecture, it guides us to finding formula of prime numbers at $(0,1) \times (0,1)$, in natural numbers based on each natural number is equal to half of one even number so in natural number main role is with even numbers but when we change space to natural numbers with a point in the beginning of each number like $0.2$ or $0.4$ then main role will be with odd numbers (with a point in the its beginning) because for example $0.400=0.40=0.4$ or $0.5=0.50=0.500$ but however a smaller proper subset of such odd numbers namely $S$ is helpful, but for finding formula of prime numbers we need to all power of the Nanas lemma not what such that is stated in above conjecture namely we must attend to the set $V$ too!

Theorem: $\forall p,q,r,s$ belong to prime numbers & $q \lt p$, $(p,q)$ is located at the direct line $(0,0),10^{-\varphi ((p,q))}(p,q)$ and if $(r,s)$ is belong to this line then $p=r$ & $q=s$.

Let $A=\{(x,y)\,|\,0.01 \lt x+y \le 0.1$ & $0 \lt x$ & $0 \lt y $ & $y \lt x \}$ and $S^2_2$ be a sphere with center $(0,0,r_2)$ and radius $r_2$ and $S^2_1$ be a sphere with center $(0.007,0.005,c)$ and radius $r_1$ such that $S^2_1$ is into the $S^2_2$ now suppose $f_1,f_2$ are two mapping from $A$ to $S^2_2$ such that $1)$ if $x \in A,$ $f_1 (x)$ is a curve on $S^2_2$ that is obtained as below: from $x$ make a direct line that be tangent on $S^2_1$ and stretch it till cut $S^2_2$ in curve $f_1 (x)$ and $2)$ if $x \in A,$ $f_2 (x)$ is a curve on $S^2_2$ that is obtained as below: from $x$ make a direct line that be tangent on $S^2_1$ and then in this junction point make a direct line perpendicular at $S^2_2$ till cut $S^2_2$ in curve $f_2 (x)$.

Let $f_3: L_1 \to S^1,$ $f_3 ((a,b)) = (a^2+b^2)^{-0.5}(a,b)$ and $f_4:L_1 \to S^2,$ $f_4 ((a,b)) = (a^4+a^2b^2+b^2)^{-0.5}(a^2,ab,b) $

It is clear $L_1 \cup \{(x,x)\,|\, 0.005 \le x \le 0.05 \} \subseteq \bar {L_1}$ and maybe $\bar {L_1} \setminus \{(x,y)\,|\, y=x$ or $y=0\} =F$.

Guess $1$: $f_3 (L_1) $ is dense in the $S^1 \cap \{ (x,y)\,|\, 0 \le y$ & $2^{-0.5} \le x \}$.

Let $U: S^2_2 \setminus \{(0,0,2r_2)\} \to \{(x,y,0)\,|\, x,y \in \Bbb R \}$ such that make a direct line by $(0,0,2r_2)$ & $(x,y,z)$ till cut the plane $\{(x,y,0)\,|\, x,y \in \Bbb R \}$ in the point $(x_1,y_1,0)$. Now must a group be defined on the all the points of $S^2_2 \setminus \{(0,0,2r_2)\}$.

Let $G=S^2_2 \setminus \{(0,0,2r_2)\}$ be a group by operation $g_1 + g_2 = U^{-1} (U(g_1)+U(g_2))$ that second addition is vector addition in vector space $(\Bbb R^2,\Bbb Q,+,.)$ and now we must attend to subgroups of $G$ particularly $y=\pm x,\,y=0,\,x=0$

Theorem: Let $\Bbb P$ is the set prime numbers and $K_3 =\{p+q+r\,|\, p,q,r \in \Bbb P \}$ and put a point in the beginning each member of $K_3$ in order to be obtained $N$ so $N$ is dense in the interval $(0.1,1)$ of real numbers. Proof by Goldbach's weak conjecture.

Guess $2$: If $\Bbb P$ is the set prime numbers and $K_2 =\{p+q\,|\,p,q \in \Bbb P \}$ and put a point in the beginning each member of $K_2$ in order to be obtained $M$ so $M$ isn't dense in the interval $(0.1,1)$ of real numbers.

Let $F= \Bbb Q$, so what are Galois group of polynomials $x^4+b^2x^2+b^2$ and $(1+a^2)x^2 +a^4$.

Because of my weakness in algebraic topology, from now on I follow the second longsome way namely algebraic methods!

Now I want find a relationship between $L_1$ & $(S_1 \cap A) \setminus L$.

Alireza Badali 22:21, 8 May 2017 (CEST)

From Nanas Lemma

Assume $H$ is a mapping from $(0.1,1)$ on $(0.1,1)$ given by $H(x)=1/(10x)$. And 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 digit 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 $(0.1,1)$.

Theorem: $S×S$ is dense in the $(0.1,1)×(0.1,1)$. Similar theorems are right for $S×T$ & $T×T$.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).

"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: Let $(X,T_1),\,(Y,T_2)$ be topological spaces and $H$ be a homeomorphism from $X \to Y$. If $C$ is a dense subset of $X$, $H(C)$ is dense in the $Y$ necessarily.

Proof: Since density is a property which only depends on the topology, this is true, namely, suppose $U$ is a nonempty open subset of $Y$, then since $H$ is bijective, we can rewrite $U \cap H(C) = H(H^{-1}(U) \cap C)$, however, $H^{-1}(U)$ is open by continuity of $H$, and nonempty since $H$ is surjective, therefore, $H^{-1}(U) \cap C$ is nonempty since $C$ is dense; and therefore, $U \cap H(C) = H(H^{-1}(U) \cap C)$ is also nonempty. (So, this proves only using that $H$ is continuous and bijective, it is actually possible to refine the proof to work only assuming that $H$ is continuous and surjective - in that case, $U \cap H(C) \supseteq H(H^{-1}(U) \cap C)$.)

Another proof: Let $y\in Y$; for proof of every neighborhood $N$ of $y$, $N\cap H(C)\neq\emptyset$, take $x\in X$ such that $f(x)=y$, then $f^{-1}(N)$ is a neighborhood of $x$ and therefore $f^{-1}(N)\cap C\neq\emptyset$. So, $N\cap H(C)\neq\emptyset$.

Let $D=\mathbb{Q} \cap (0.1,1)$

Theorem: $D$ and $S$ are homeomorph by the Euclidean topology.

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 by the Euclidean topology.

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

Theorem: $T_1$ is dense in the interval $(0,1)$.

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 $z=0,1,2,3,...$ } & $G=\mathbb{Q} \setminus W$

Theorem: $W$ and also $G$ are dense in the $\mathbb{Q}$ and also $\mathbb{R}$.

A conjecture: For each member of $G$ like $g$, there are two members of $W$ like $a,b$ in the interval $(g-0.5,g+0.5)$ such that $g=(a+b)/2$.

Of course it is enough that the conjecture just is proved for the interval $(0,1)$ namely assume $g\in{(0,1) \cap {G}}$.

An important question: Is there any proper infinite subset of $S$ such that it is 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)$$?$ This is a suitable definition for information security and coding theory!

Alireza Badali 22:21, 8 May 2017 (CEST)

Other problems

1) About set theory: So many years ago I heard of my friend that it 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". Boris Tsirelson (talk) 16:46, 8 April 2017 (CEST)

So is there any set with cardinal between $\aleph_0$ and $\aleph_1$ and not equal to $\aleph_0$ and $\aleph_1$, and also for $\aleph_1$ and $\aleph_2$ and so on, I think so the well-ordering theorem answers this question.Alireza Badali 22:21, 8 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 should go to the Wikipedia more and I think the well-ordering theorem results each set is exactly a line. Yours Sincerely Badali

2) About ?!: 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 with a property too, 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 offer 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 phenomenon. Fifth dimension is thought with more speed than light.

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. _ I believe graph theory (with hyper-graph & multi-graph & ...) is the best and latest mathematical theory forevermore that it includes all mathematical theories and mathematical logic and even mathematical philosophy, now I need to know whether any graph has been defined contain all natural numbers and also all concepts of number theory or that has been any graph defined contain all natural numbers as vertices with satisfying below rules for definition edges: 1) Theorem division algorithm 2) One of beauties of mathematics, Fermat's last theorem (and I guess somehow this theorem is an equivalent for induction axiom and maybe by another condition): The equation $x^n+y^n=z^n$ has no solution in nonzero integers if $n \ge 3$.

9) Mathematical logic is the language of Mathematics(There is not any difference between language of Mathematics and English literature or Persian literature or each other language, because both Mathematical and literary just state statements about some subjects and each one by own logical principles, of course language of Mathematics is very elementary till now compared with literature of a language and must be improved.) and Mathematical philosophy is the way of thinking about Mathematics, always there was a special relationship between Mathematical philosophy and Mathematical logic, but nobody knows which one is first and basic, and there is no boundary between them, however, weakness of Mathematics is from 1) disagreement between the Mathematical philosophy and the Mathematical logic(Of course whatever this disagreement is decreased, equally Mathematics grows.) and 2) weakness of own Mathematical logic, but music can be help to removing this problem, because of everything has harmony is a music too, In principle I want say that whatever this disagreement between expression and thinking is decreased equally Mathematics grows.

10) The Goldbach`s conjecture is a way for finding formula of prime numbers!

11) Whether the formula of number of topologies on a finite set is equivalent with the formula of prime numbers.

12) Whether always there is a topological space in each Mathematical proposition that embraces all of content of proposition?

13) Only algebraic theories or algebraic structures join cardinal of a set with its members and however make us so stronger but restrict the field of our act.

14) Universe is a wave toward truth! so lying is swimming in the opposite direction of the flow of water instead that universal wave helps honest man!

15) Most our knowledge about algebraic structures on numbers is result of normal addition and multiplication (of course I don't understand why $\sqrt 2 \times {1 \over \sqrt 2}$ is $1$ in order to how being multiplied this two number that result to be $1$ and this can be resulted from $1)$ another operation like $÷$ in definition algebraic structure is needed namely an algebraic structure with three operation, normal addition & multiplication & division as $(\Bbb R ,+,×,÷)$, or from $2)$ axiom of choice ) but I want define a group on the natural numbers for this, I must obtain rule of sequence $a(n)$, $a:\Bbb N \to \Bbb Q,$ by $a(n)={n \over k(n)}$ such that $k:\Bbb N \to \Bbb N$, $k(n)$ is the number of digits of $n$ so $a(1)=1,\,a(2)=2,\,...,\,a(9)=9,\,a(10)=5,\,a(11)=5.5,\,...,\,a(100)=33.\bar 3 ,\,a(101)=33.\bar 6 ,\, ... $, in principle I believe this sequence is a fundamental concept in mathematics and mathematical logic, so find rule of $a(n)$. _ It is a truth: It is possible that prime numbers don't have any rhythm between natural numbers like number $π=3.14159265358979323846264338...$ namely there won't any formula of prime numbers forevermore but in this case above sequence describe its reason in mathematical logic and of course discussion of prime numbers will continue forevermore but never a effective proposition can't be presented however maybe be asked what about prime number theorem but this theorem don't discuss about own prime numbers directly and discussion is about all natural numbers not only about prime numbers! Of course my reason for possible of absence of rhythm is this that cardinal of set contain all numbers as $0.a_1a_2a_3...$ that $∀i∈N,a_i=0,1$ is $\aleph_1$! _ But how must think about infinity? by axioms? by accepted properties? by dreamy imagination? ... who knows? ... but finally must think because without infinity a consequence or result isn't artistic! and every property or proposition is a result of infinity and infinity has all properties concurrent and this is definition of infinity, something has everything! But formula of prime numbers is a definition of infinity and the best! and key of gateway of eternity is prime numbers and its formula or the same unique music! Of course the field $(\Bbb R,+,\times )$ is an axiom or in fact is a definition that shouldn't ask about it but this $(\Bbb R,+,\times )$ is coming from where? it defined before than everything and every concept is a result or extension from it however, whether Axiom of Choice is enable for expression reason of $\sqrt 2 \times {1 \over \sqrt 2}=1$. _ Summation of digits of a natural number with consideration number of zeros in middle of number and not in its beginning (with partition by number of zeros) is related to divisors of that number particularly for prime numbers. _ relation between normal addition and multiplication has created all propositions in number theory. _ How can it be possible that each uncountable group has some countable subgroups certainly however, cardinal isn't an algebraic concept!, about finite groups this fact follows from algebraic structures on finite sets with a natural number as a cardinal and from properties on natural numbers in number theory because the own numbers have algebraic structures but about infinite ordinal numbers problem is basically different in principle I want to know whether infinite ordinal numbers have algebraic structures as only a set of course the number of finite groups with cardinal $n$ is a specific number like $m$ but about infinite ordinal numbers the number of groups is infinite. in principle I need to know whether there is an algebraic structure on sets with infinite cardinal as an axiom so being answered some questions about infinite ordinal numbers like continuum hypothesis! so can result cardinal concept from algebraic structures of course particularly for infinite sets and this is a way for understanding infinite ordinal numbers and answering to this question answers continuum hypothesis. only in algebraic structures attention is on members of set and this is attention to members of infinite sets and being resulted cardinal from it! in order to algebraic theories are key of understanding infinite ordinal numbers, from understanding members of a set being realized cardinal of set and this is absolutely logical so being said cardinal is a algebraic concept and result of algebraic theories although it hasn't been mentioned in algebraic structures and attention to members even about infinite sets is attention to cardinal so answering to infinite ordinal numbers is in algebraic structures in order to from knowing algebraic properties being reached to realizing of sets. _ A basic system like decimal system must be existed as an axiom and another systems must be defined based on it, otherwise concept of numbers is obscure.

16) Induction axiom is unable for discovering formula of prime numbers. _ Induction axiom is a direct result of formula of prime numbers.

Alireza Badali 22:21, 8 May 2017 (CEST)