User talk:PBadali

Welcome!

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Speedy deletion nomination of User:PBadali

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Useful Pages given by D.Lazard

welcome
Quick introduction to Wikipedia
How to write a great article
Ten Simple Rules for Editing Wikipedia, an essay from PLoS
Identifying reliable sources (general advice)
Wikipedia's Manual of Style
Wikipedian
sign
discussion pages
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Wikipedia:Questions
ask a question on your own talk page.

Square root of x formula. Symbol of mathematics.

If you are interested in mathematics-related themes, you may want to check out the Mathematics Portal.
If you are interested in improving mathematics-related articles, you may want to join WikiProject Mathematics (sign up here or say hello here).

Thank you Professor Daniel Lazard. Alireza Badali (talk) 20:36, 4 December 2017 (UTC)[reply]

December 2017

Welcome to Wikipedia and thank you for your contributions. I am glad to see that you are discussing a topic. However, as a general rule, talk pages such as Talk:Group theory are for discussion related to improving the article in specific ways based on reliable sources and the project policies and guidelines, not for general discussion about the topic or unrelated topics, or statements based on your thoughts or feelings. If you have specific questions about certain topics, consider visiting our reference desk and asking them there instead of on article talk pages. Thank you. –Deacon Vorbis (carbon • videos) 13:27, 5 December 2017 (UTC)[reply]

Of course there was a careless on

(Q_{1},\star _{Q_{1}})

that correct form is:

let

Q_{1}=\{{m \over n}\,|\,m,n\in \mathbb {N} \}

, it is clear

(Q_{1},\star _{Q_{1}})

is a group with:

{\begin{cases}\forall m,n,u,v\in \mathbb {N} ,\,{m \over n}\star _{Q_{1}}1={m \over n}\\({m \over n})^{-1}={m^{-1} \over n^{-1}}\qquad m\star m^{-1}=1=n\star n^{-1}\\{m \over n}\star _{Q_{1}}{u \over v}={m_{1} \over n_{1}}\star _{Q_{1}}{u_{1} \over v_{1}}={{m_{1}\star u_{1}} \over {n_{1}\star v_{1}}}\quad {\text{if}}\,\,{\begin{cases}{m \over n}={m_{1} \over n_{1}},\,\,{u \over v}={u_{1} \over v_{1}},\,\,{mu \over nv}={m_{1}u_{1} \over n_{1}v_{1}}\\{\text{gcd}}(m_{1},n_{1})={\text{gcd}}(m_{1},v_{1})=1\\{\text{gcd}}(u_{1},n_{1})={\text{gcd}}(u_{1},v_{1})=1\end{cases}}\end{cases}}

I mean is first simplification of fractions and then calculation like

{44 \over 12}

that gets

{11 \over 3}

.

Alireza Badali (talk) 16:08, 7 December 2017 (UTC)[reply]

Pairing function

You may wish to look at Cantor's pairing function. JRSpriggs (talk) 01:18, 7 December 2017 (UTC)[reply]

Thank you. Alireza Badali (talk) 13:07, 7 December 2017 (UTC)[reply]

Dense sets

Hey Alireza! Know that for those two problems, showing that r(ℙ) is dense in (0.1,1) and {p/q | p, q ∈ ℙ} is dense in ℝ⁺ are the hard parts, and if I understand you correctly, those are the parts you have already solved. The step from there into ℂ is trivially easy, and it does not require reinvoking the prime number theorem but somehow in ℂ; it is much easier than that. Getting the solution on your own would be of greatest value to you, but if you need some more help, then feel free to ping me. Cheers! -- ToE 03:02, 8 December 2017 (UTC)[reply]

Thank you, yes I need help, and let me say a narrative never I understood the theory of statistics and probability maybe because of I didn't have a good master or about complex functions that I have problem yet but under Professor Megerdich Toomanian's training I have understood Topology pretty, anyway, I want say master is very important for discernment of correct way. Alireza Badali (talk) 15:53, 8 December 2017 (UTC)[reply]

OK. First, is it accurate that you already have a proof for r(ℙ) being dense in (0.1,1)? I'm not practiced in number theory, and while I understand the prime number theorem enough to "know" this is true in a "handwavy fashion", I am not in a position to critique such a proof. (I don't need to see your proof; I just want to know where we are starting from.) -- ToE 21:30, 8 December 2017 (UTC)[reply]

Theorem $1$ : For each subinterval of $[0.01,0.1)$ like $(a,b),\,\exists m\in \mathbb {N}$ that $\forall k\in \mathbb {N}$ with $k\geq m$ then $\exists t\in (a,b)$ that $t\cdot 10^{k+1}\in \mathbb {P}$ .

Proof given by @Adayah from stackexchange.com: Without loss of generality (by passing to a smaller subinterval) we can assume that

(a,b)=\left({\frac {s}{10^{r}}},{\frac {t}{10^{r}}}\right)

, where

s,t,r

are positive integers and

s<t

. Let

\alpha ={\frac {t}{s}}

.

The statement is now equivalent to saying that there is

m\in \mathbb {N}

such that for every

k\geqslant m

there is a prime

p

with

10^{k-r}\cdot s<p<10^{k-r}\cdot t

.

We will prove a stronger statement: there is

m\in \mathbb {N}

such that for every

n\geqslant m

there is a prime

p

such that

n<p<\alpha \cdot n

. By taking a little smaller

\alpha

we can relax the restriction to

n<p\leqslant \alpha \cdot n

.

Now comes the prime number theorem:

\lim _{n\to \infty }{\frac {\pi (n)}{\frac {n}{\log n}}}=1

where

\pi (n)=\#\{p\leqslant n:p

is prime

\}.

By the above we have

{\frac {\pi (\alpha n)}{\pi (n)}}\sim {\frac {\frac {\alpha n}{\log(\alpha n)}}{\frac {n}{\log(n)}}}=\alpha \cdot {\frac {\log n}{\log(\alpha n)}}{\xrightarrow {n\to \infty }}\alpha

hence

\displaystyle \lim _{n\to \infty }{\frac {\pi (\alpha n)}{\pi (n)}}=\alpha

. So there is

m\in \mathbb {N}

such that

\pi (\alpha n)>\pi (n)

whenever

n\geqslant m

, which means there is a prime

p

such that

n<p\leqslant \alpha \cdot n

, and that is what we wanted.

Theorem $2$ : Let $\mathbb {P}$ is the set prime numbers and $S$ is a set that has been made as below: put a point at the beginning of each member of $\mathbb {P}$ 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.

Theorem

2

follows from Theorem

1

.

Transfer concepts to

\mathbb {C}

~~and if you aren't capable~~ please let me do it and then I will say you, thanks! Alireza Badali (talk) 23:46, 8 December 2017 (UTC)[reply]

OK, I'll take that as a yes. You do not need to then reassert the prime argument in ℂ, but instead just use Dense set#Properties, specifically from paragraph 3: "The image of a dense subset under a surjective continuous function is again dense." (This is a trivial consequence of Continuous function#Definition in terms of limits of sequences.)

So what is the first mention of ℂ in the problem you presented? It is {tⁱ | t ∈ r(ℙ)}. Do you know enough about Complex arithmetic#Exponentiation to understand what tⁱ, or equivalently, exp(i lnt) does? What is the image {tⁱ | t ∈ (0.1,1)}? Do you yet see the error in the problem as stated? What are two ways the problem could be fixed? -- ToE 01:42, 9 December 2017 (UTC)[reply]

Please consider I'm not free and I can give answer after probably several months but if you can solve it is very good but leave an email to me so I will send you all notes, thanks. Alireza Badali (talk) 11:24, 9 December 2017 (UTC)[reply]

No worries. Once you have the time to study complex arithmetic and understand the behavior of exp(i x), you will see how simple this problem was. Cheers! -- ToE 17:15, 9 December 2017 (UTC)[reply]

Thank you for your guidance and help, but I need time for extension of my theories to

\mathbb {C}

and surely we have more features in

\mathbb {C}

than

\mathbb {R}

especially I have a problem about definition of a production of points that with

\mathbb {C}

this problem will be resolved. Alireza Badali (talk) 14:39, 10 December 2017 (UTC)[reply]

Answered questions by Professor Daniel Lazard

Question: I need to know what is function of this sequence in $\mathbb {N} \times \mathbb {N}$ : $(1,1),(1,2),(2,1),(1,3),(2,2),(3,1),(1,4),(2,3),(3,2),(4,1),...,(1,2k-2),$ $(2,2k-3),...,(k-1,k),(k,k-1),...,(2k-3,2),(2k-2,1),(1,2k-1),(2,2k-2),$ $...,(k-1,k+1),(k,k),(k+1,k-1),...,(2k-2,2),(2k-1,1),...$ Alireza Badali (talk) 15:05, 8 December 2017 (UTC)[reply]

Answer given by Professor Daniel Lazard:

This sequence is

\mathbb {N} \times \mathbb {N}

, ordered by the Graded lexicographic order, that the order obtained by comparing first the sums of the two coefficients, and, in case of equality, by comparing the first coefficients. This is the standard order for proving that

\mathbb {N} \times \mathbb {N}

and

\mathbb {N}

have the same cardinality. This is may be defined by the recursion:

(a_{1},b_{1})=(1,1),

and

(a_{n+1},b_{n+1})={}

if

b_{n}=1

then

(1,a_{n}+1)

else

(a_{n}+1,b_{n}-1),

for

n > 0

.

Question: What is function of this sequence: $2,3,3,4,4,4,5,5,5,5,6,6,6,6,6,...,k,k,k,...,k,...$ that $k$ repeats $k-1$ times. Alireza Badali (talk) 14:09, 11 December 2017 (UTC)[reply]

Answer: This may be rewritten, using the ceil function,

f(n)=\left\lceil {\frac {1+{\sqrt {1+8n}}}{2}}\right\rceil .

Who's a mathematician?

ToE $\color {red}{!}$ are you a mathematician? and if yes why you said me you don't know prime number theorem? Alireza Badali (talk) 19:28, 10 December 2017 (UTC)[reply]

I am not a professional mathematician, though I enjoy dabbling. Regarding my earlier remark, I've never formally studied much number theory, and while I am familiar with the prime number theorem and its implications, I am not comfortable critiquing proofs invoking it. -- ToE 19:55, 10 December 2017 (UTC) I've moved your question here, as RD/MA isn't really an appropriate venue for such a question. I hope you don't mind. Cheers![reply]

ToE, everybody you are, you are helping others hence you are great

\color {red}{!}

Alireza Badali (talk) 13:52, 11 December 2017 (UTC)[reply]

My thought line on the formula of prime numbers

According to important theorems in number theory and distribution of prime numbers I think there doesn't exist any polynomial $p:\mathbb {R} \to \mathbb {R}$ include such points $(n,p_{n})$ that $p_{n}$ is $n$ _th prime and by taking in mind that prime number theorem follows from normal definition of primes in terms of factorization to primes because logarithm function is inverse of the function $f(a)=a^{n}$ and primes formulas are as logarithmic functions but ideally there exists an unique formula as an infinite series that generates primes simply. Alireza Badali (talk) 18:34, 16 December 2017 (UTC)[reply]

Are these groups isomorph ...

Hey again, Alireza! Is [[WP:RD/MA#Are_these_groups_isomorph_to_%7F'"`UNIQ--postMath-00000056-QINU`"'%7F|this]] another problem from that same professor? How far along have you gotten and where are you stuck? In particular, given the initial group of assumptions, is it clear to you that you have a group isomorphic to (ℤ,+)? Do you see what is going on well enough to describe the isomorphism? -- ToE 00:48, 20 December 2017 (UTC)[reply]

Hello, no that Professor said bye and went angrily and I don't see him (her) again, indeed I want make a group on

\mathbb {N}

isomorph to

(\mathbb {Q} ,+)

by sequences in

\mathbb {N} \times \mathbb {N}

for Goldbach's conjecture, I'm not far only I want make an algebraic and topological structure for Goldbach, and if those groups are isomorph to

(\mathbb {Z} ,+)

so what are generators of those groups? Alireza Badali (talk) 07:15, 20 December 2017 (UTC)[reply]

Cool. (I'm 0 for 2, thinking that your density questions were self generated and this one was posed to you. Ha!) I suggest that you add a note to the problem stating that this is your own construction and explaining what you are seeking to do. That will aid those who can help. -- ToE 11:12, 20 December 2017 (UTC)[reply]

I was thrown out from stackexchange sites because of my theories when finally I said number theory should be moved towards homotopy theory so nobody likes my notes only I am only and only, alone standing against the wind! Alireza Badali (talk) 13:56, 20 December 2017 (UTC)[reply]

Wording

Hi Alireza! I hope you don't mind, but I undid your edits on WP:RD/MA regarding your wording on the now archived question. That reference desk really should be reserved for mathematical questions, and I think I can help you here better.

Don't worry about achieving perfect English. I'm sure that everyone understood what you meant and didn't have any issue with your wording.
If you would like to correct it after the fact, then you are welcome to do so. Your question is archived at Wikipedia:Reference desk/Archives/Mathematics/2017 December 23#Can I take apply to the University of Cambridge. Policy for editing reference desk comments is similar to that for talk pages which is discussed at WP:Talk page guidelines. Generally, if you want to edit a comment of your own so that the meaning changes in any way, then you should make it clear by ~~striking through~~ the old text and underlining the new text. That way it is obvious that there was a change, and responses made to the original form of your comment won't appear so confusing. For a simple grammar or spelling error, you don't need to fix it because talk pages are not intended to be held to the same standards as our articles. But if you want to fix a minor error like that, you are free to do so without bothering with the striking and underlining, since that wouldn't change the meaning.

So go ahead and correct your wording if you would like, and let me know if I can help you any more with this. -- ToE 05:44, 31 December 2017 (UTC)[reply]

Thank you so much I corrected it. Alireza Badali (talk) 07:07, 31 December 2017 (UTC)[reply]

Account blocked

Hi, PBadali. I'm sorry, but I've blocked your account indefinitely. On your userpage, you say that your account has been hacked. Our standing policy for accounts is one user per account, and so standard operating procedure for handling hacked accounts, to which multiple people have access by definition, is to block the account. That's not the only reason I've blocked your account--the events in the threads above certainly played a part in the decision--but it's the most salient one at this point. I'm going to drop a standard block template just after this, which will provide directions on appealing the block, but I wanted to provide a more detailed rationale first. Writ Keeper ⚇♔ 19:33, 5 January 2018 (UTC)[reply]

No it may be a mistake only, there is no need to block. Alireza Badali (talk) 19:57, 5 January 2018 (UTC)[reply]

What's a mistake? Writ Keeper ⚇♔ 20:04, 5 January 2018 (UTC)[reply]

I thought my account has been hacked but now that recall exactly no it hasn't been hacked. Alireza Badali (talk) 20:22, 5 January 2018 (UTC)[reply]

But don't you want free my account?! Alireza Badali (talk) 20:33, 5 January 2018 (UTC)[reply]

January 2018

You have been blocked indefinitely from editing for abuse of editing privileges.

If you think there are good reasons for being unblocked, please read the guide to appealing blocks, then add the following text below the block notice on your talk page: {{unblock|reason=Your reason here ~~~~}}. Writ Keeper ⚇♔ 19:34, 5 January 2018 (UTC)[reply]

This user's unblock request has been reviewed by an administrator, who declined the request. Other administrators may also review this block, but should not override the decision without good reason (see the blocking policy).

PBadali (block log • active blocks • global blocks • contribs • deleted contribs • filter log • creation log • change block settings • unblock • checkuser (log))

Request reason:

I thought my account has been hacked but now that recall exactly no it hasn't been hacked.

Decline reason:

I am declining your unblock request because it does not address the reason for your block, or because it is inadequate for other reasons. To be unblocked, you must convince the reviewing administrator(s) that

the block is not necessary to prevent damage or disruption to Wikipedia, or
the block is no longer necessary because you
1. understand what you have been blocked for,
2. will not continue to cause damage or disruption, and
3. will make useful contributions instead.

Please read the guide to appealing blocks for more information. PhilKnight (talk) 21:22, 5 January 2018 (UTC)[reply]

If you want to make any further unblock requests, please read the guide to appealing blocks first, then use the {{unblock}} template again. If you make too many unconvincing or disruptive unblock requests, you may be prevented from editing this page until your block has expired. Do not remove this unblock review while you are blocked.

This user's unblock request has been reviewed by an administrator, who declined the request. Other administrators may also review this block, but should not override the decision without good reason (see the blocking policy).

PBadali (block log • active blocks • global blocks • contribs • deleted contribs • filter log • creation log • change block settings • unblock • checkuser (log))

Request reason:

I am here just for math activities but incidentally I wrote some notes but indeed I don't consume my time and capability for politics!

Decline reason:

I'm sorry, you stated your account has been hacked and so, as per WP:COMPROMISED, it is not eligible for unblock consideration. You are free to create a new account and start editing with that. Yamla (talk) 13:19, 7 January 2018 (UTC)[reply]

If you want to make any further unblock requests, please read the guide to appealing blocks first, then use the {{unblock}} template again. If you make too many unconvincing or disruptive unblock requests, you may be prevented from editing this page until your block has expired. Do not remove this unblock review while you are blocked.

This user's unblock request has been reviewed by an administrator, who declined the request. Other administrators may also review this block, but should not override the decision without good reason (see the blocking policy).

PBadali (block log • active blocks • global blocks • contribs • deleted contribs • filter log • creation log • change block settings • unblock • checkuser (log))

Request reason:

I changed the password so may you free my account. Alireza Badali (talk) 10:07, 10 January 2018 (UTC)[reply]

Decline reason:

We have no way of knowing whether this message was posted by the original owner of the account or not. Consequently, we have no way of knowing whether it is true or not. Unless you already had a committed identity prior to this block - which as far as I can see, you did not - there is no way that we can unblock your account. Yunshui ^雲 _水 11:02, 10 January 2018 (UTC)[reply]

If you want to make any further unblock requests, please read the guide to appealing blocks first, then use the {{unblock}} template again. If you make too many unconvincing or disruptive unblock requests, you may be prevented from editing this page until your block has expired. Do not remove this unblock review while you are blocked.

This user's unblock request has been reviewed by an administrator, who declined the request. Other administrators may also review this block, but should not override the decision without good reason (see the blocking policy).

PBadali (block log • active blocks • global blocks • contribs • deleted contribs • filter log • creation log • change block settings • unblock • checkuser (log))

Request reason:

I have a way for knowing that I am the true guy, email of this account was (Redacted) that at the first days I changed it to another email and I can send a code message to confirmation for you. Alireza Badali (talk) 14:29, 10 January 2018 (UTC)[reply]

Decline reason:

This account is not eligible for unblock consideration. There's no way to get this account unblocked. You must start a new account. I have revoked talk page access from this account. Yamla (talk) 15:20, 10 January 2018 (UTC)[reply]

If you want to make any further unblock requests, please read the guide to appealing blocks first, then use the {{unblock}} template again. If you make too many unconvincing or disruptive unblock requests, you may be prevented from editing this page until your block has expired. Do not remove this unblock review while you are blocked.

United Countries or briefly UC

UC is the countries including Britain, Canada, Australia, all other European countries, Russia and the countries India, Pakistan, Afghanistan, Tajikistan, Iran, Azerbaijan, Armenia, Turkey, Iraq, Syria and Yemen with a new universal Internet line located at Europe and a great army. Alireza Badali (talk) 10:25, 10 January 2018 (UTC)[reply]

I am Alireza Badali Sarebangholi, I want confess that I get used to make theory in mathematics and theorizing is a tendency at me but it isn't good if I judge about a people because my information probably be wrong or defective hence I reject my previous opinions about Jewish people and I apologize at Jewish people please forgive me! of course Jewish people are diligent and intelligent and wealthy and it isn't bad. 89.45.63.85 (talk) 22:32, 21 January 2018 (UTC)[reply]

@WritKeeper:, @Yamla:, @Yunshui: or @PhilKnight: Can we have a short block on this IP for block evasion? Beyond My Ken (talk) 22:50, 21 January 2018 (UTC)[reply]

Blocked. Also, that was a strangely racist statement made by 89.45.63.85. --Yamla (talk) 00:37, 22 January 2018 (UTC)[reply]

Note in general, it's okay for this user to come back and set up a new account; this account is blocked because it's compromised. But... I'd rather they didn't post weirdly racist statements here. --Yamla (talk) 00:38, 22 January 2018 (UTC)[reply]

Not the first time it's happened: [1]. Beyond My Ken (talk) 02:47, 22 January 2018 (UTC)[reply]

I hate racist at each form because I believe human dignity is higher than such silly thoughts and only God knows who is better (am I better? certainly no because I know I am who and I have done many shameful works!), but only I theorized mentally and indeed I mean was just a message (North America isn't my choice!) that I would transmit it with a reasonable excuse but I didn't think contention be expanded to this extent! and I am sorry for I have accused Jewish people and I hope they will forgive me! 130.255.240.214 (talk) 18:29, 22 January 2018 (UTC)[reply]