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8:01 PM
 
I'll make a meta post about how I don't like balogna.
 
@MikeMiller I call your bluff.
 
I'm waiting for the meta post @MikeMiller so that I can upvote and favorite it.
 
Maybe I will settle for saying in chat that I don't like balogna.
 
Make a meta post saying you don't like meta ? :D
 
8:12 PM
That would be a lie.
 
meta what ?
html ?
?
 
I'm so meta, even this acronymn
 
slow clap
 
8:32 PM
He's so meta, even his acronym
 
@columbus8myhw what's "meta" ??
 
Self-referential, basically.
 
@columbus8myhw then why is it called "meta" ?
 
Like, the meta stack exchange is where you post stuff about stack exchange.
ORIGIN 1980s: from meta- in the sense ‘beyond.’
 
@Hippalectryon basically I should spend my time on finishing the modules for my book, but if I receive some problems for research as yours and then spend time on them (because it's hard not to do it) I'll finish the book in a couple of years. :-))))
 
8:34 PM
Source: my dictionary
 
@columbus8myhw oh ok I see...
@columbus8myhw you got a dic ??
 
I don't know if that was accidental or if you're trying to make an immature pun.
 
40 secs ago, by columbus8myhw
ORIGIN 1980s: from meta- in the sense ‘beyond.’
Aristoteles begs to differ.
 
@Chris'ssis Sure, I want that book to come out too !! (I saw your profile btw :D )
 
@robjohn did you see my L+4?
@Hippalectryon :D
 
8:37 PM
omfg coffee is making me jittery af
 
@robjohn and L+1 becomes interesting when considering $k^a$ instead of $k$, with $a>0$.
 
@DonLarynx Drink flour
 
@hippa bread flower ?
 
As you want.
 
it's not so bad @Hippa but I was just talking to a girl and I had to restrain myself a lot from shaking then I thought I lost my sweater and that ended the conversation lol.
 
8:41 PM
@DonLarynx q_q
 
hello folks. This is the first time I'm using this chat so I hope I'm right here.
 
@Chris'ssis Sorry, I have been dealing with some moderator issues and a post on meta dealing with someone suggesting that high rep users only answer harder, more advanced questions.
 
@robjohn OK. I didn't visit meta for a while. Let me see what happens there.
 
In just after 3 days of relentlessly studying number theory, I can perform mental math effortlessly.
brb 29*42 = 1218
OH FUK YA
 
one question: I am preparing myself for a functional analysis exam and I was wondering why we always assume for weak formulations that the test or bump function has compact support? Is this necessary to make the integral vanish at the boundaries? Are there other reasons? I have no mathematics background and every time I look for it I stumble across things that are way to advance for me...
 
8:55 PM
no math background and yet u r studying something like a 5000-level math course...gl.
 
I'm looking for a, let's say, intuitive reason :-)
@don
 
@DonLarynx That's part of why I drink tea :-)
 
@DonLarynx: Computational science and engineering, so not that far. and it's "introduction to fa for engineers" :-)
 
@Hippalectryon That's a fine suggestion.
 
@robjohn Right, equating reputation with ability is a mistake I think (from META). Besides that, usually a question can be answered in more ways.
 
8:57 PM
@andreee Yes, I believe that if you don't assume compact support then you have to be more careful about what happens at infinity
 
@DonLarynx we don't go into measure theory etc. but yet we still have more insight than the usual engineering student I would say
 
tea is very good for you, @robjohn. It contains things related to GABA, i.e. depressors.
 
@DonLarynx Blood pressure and not tongue depressors, I hope ;-)
 
Very soon, I should have an algorithm for linear Diophantine solns ready in C++.
 
@Kevin thank you!
 
9:01 PM
@robjohn If someone really wants to bring the contribution to the site then if there are users that are faster, say, that user can check which answer he would have answered and check if the answer already posted is correctly stated, can be improved or no, and add comments. Moreover, one can always find possibility of giving answers on site because of the large volume of the questions, and as I previously said, any question can be answered in more ways.
I don't care too much about the reputation, I'm mainly interested in math, not for being remarked by X,Y,Z because I don't care what X,Y,Z thinks of me.
 
r9m
@Chris'ssis Any awesome ideas regarding how to prove $e^x+e^{-x} \le 2e^{x^2/2}$ ? :D @robjohn @DanielFischer
 
@Chris'ssis I bet you care very much that your reputation doesn't drop below 20.
 
@DanielFischer I don't know what you mean.
@r9m hmmm
 
@Chris'ssis you couldn't chat
 
@Chris'ssis That's the threshold for participating in chat ;)
 
9:06 PM
@DanielFischer lol, then you're right! :-)
@robjohn Good point.
@r9m I suppose you have an awesome way ... right ...?
 
r9m
@Chris'ssis I'm not sure if you can call it awesome or sth .. but I like the one I have :)
 
@r9m it's an even function whose derivative is $0$ only at $0$
 
@r9m Can you share it now or later?
 
Thinking to add a bounty on a question, as I don't think the accepted answer is quite right.
3
A: Limit of Ratio of Chebyshev Polynomials

ryagamiFirst thing we should note is: $$\lim_{n\to\infty}\frac{U_n(x)^2}{U_{n-1}(x)^2+U_n(x)^2}=\lim_{n\to\infty}\frac{1}{\left(\dfrac{U_{n-1}}{U_n}\right)^2+1}$$ So, we look at $\displaystyle\lim_{n\to\infty}\frac{U_{n-1}}{U_n}$. Wikipedia gives the formula: $$U_n=\frac{\left(x+\sqrt{x^2-1}\right)^{n...

(see my comment at the end.) can anyone give a second opinion? would rather not offer a bounty if i'm being thick
 
r9m
@robjohn yes that works too !! :)
@Chris'ssis here :)
rofl
 
9:11 PM
...yep, just realized i'm probably being thick. nvm
 
@r9m How did you think of that? I think you have a mistake there.
 
r9m
@Chris'ssis I'd give 80% credit to flawr's comment :-) I saw it an hour ago and thought meh .. after reading his comment it hit my 'ead like a bell :P
 
Linear Diophantines and their solutions are pretty cool. I'm glad I learned them.
 
r9m
@Chris'ssis where ? :o
 
@r9m No, I misread it. It's perfect. I initially saw that $\pi^2$ inside the brackets.
 
r9m
9:16 PM
@Chris'ssis okay! :D
 
@r9m Well, I think you'd love a question very much ... let me find the link
 
@DanielFischer It is a given a binary tree that simulates a (not necassary binary) ordered tree. I wat to write a recursive function that takes as argument a pointer to the root of a binary tree and that prints the degree of the ordered tree (The degree of a node is the number of child nodes of the nodes. The degree of a tree is the maximum from the degrees of the nodes).

I have written the following but I am confused now and I don't know how I could call the function recursively.

`Algorithm(pointer A){
 
@r9m See the problem $385.$ (Solution by the author.) here ssmr.ro/gazeta/gma/2014/gma1-2-2014-continut.pdf. It's simply too much art in there ...
 
@evinda Find degree of current node. For each child, recur and compute the degree of the subtree. Take the maximum.
 
r9m
@Chris'ssis Awesome !! :D
 
9:24 PM
@r9m I love that terribly much. :-) Your way here reminded me of that one.
 
r9m
@Chris'ssis :D
 
Hello!! We have that $$y^2 \mid 2^6 \cdot 3^3$$ which are the possible values for $y$ ??
Do we take all the powers of $2$ and $3$ that are divisible by $2$??
@r9m
 
r9m
@MaryStar yes :)
 
@DanielFischer So is it like that?


`Algorithm(pointer A){
q=A, deg=0;
if (q==NULL) return 0;
if (q->lc!=NULL) deg++;
if (q->rs!=NULL) deg++;
if (A->lc!=NULL and A->rs!=NULL){
return max(deg, Algorithm(A->lc), Algorithm(A->rs) );
else if (A->lc!=NULL) return max(deg, Algorithm(A->rs) );
else if (A->rs==NULL) return max(deg, Algorithm(A->lc) );
else return 0;
}
 
9:42 PM
http://gyazo.com/3ad0dd8eb543055657c094d0b953033a

$\angle DAC = 56^\circ$, and $\angle CBD = DAC$, so $\angle CBD = 56^\circ$.
$\angle CMD = \angle DAC*2$, so $\angle CMD = 56*2 = 112^\circ$.
$\angle CMD = \angle BMA$, so $\angle BMA = 112^\circ$.
So, $\angle BAD = 180-(41+112) = 27^\circ$.
So, $\angle BAC = 27 + 56 = 83^\circ$.

Can anyone explain me why this is wrong ?
 
@r9m So, are the possible values for $y$ the following??

$$y^2=1, 3^2, 2^2, 2^2 \cdot 3^2, 2^4, 2^4 \cdot 3^2, 2^6, 2^6 \cdot 3^2$$

$$ \Rightarrow y=\pm 1 , \pm 3, \pm 2, \pm 2 \cdot 3, \pm 2^2 , \pm 2^2 \cdot 3 , \pm 2^3, \pm 2^3 \cdot 3 \\ \Rightarrow y=\pm 1 , \pm 3, \pm 2, \pm 6, \pm 4 , \pm 12 , \pm 8, \pm 24 $$
 
@r9m while working on the modules (problems and solutions) for my book, lots of new ideas come to mind, to explore other and other things. This slows me down to a certain extent, but then I have better problems and solutions (that is the benefit).
 
r9m
@MaryStar yes! :-)
 
@r9m Ok!! Thank you!!! :-)
 
r9m
@Chris'ssis okay :)
@MaryStar welcome
 
9:46 PM
@r9m even what you showed me gave me some new ideas ... :-))) However, I don't intend to add inequalities in my book.
 
r9m
@Chris'ssis okay !! cool :D
@Chris'ssis if you start adding inequalities too there will be no end to it .. it's a disaster
 
@r9m hmmm, I don't know if I showed you this one ...
$$(1+x)^{1-x}\le 1+x-x^2, \space x \in [0,1]$$
 
r9m
@Chris'ssis yes I think you did :-)
 
@r9m You managed to prove it?
 
r9m
@Chris'ssis I don't remember .. I'll try again anyway
 
9:52 PM
@r9m as you wish. I only wanted to reminded it to you (because it's simply too nice).
 
r9m
@Chris'ssis bernoulli's inequality
 
@r9m Why nvm? What did you write? I was away.
@r9m that should work, right? Still, this is not the inequality I wanted to give you though ...
 
r9m
@Chris'ssis ya .. okay which is it then ?
 
how can I compute the $\limsup \cos(n)$?
 
@Gato: Hint: Can you prove that for $\theta/2\pi$ irrational, the set of points $e^{in\theta}$, $n\in\Bbb Z$, is dense in the unit circle?
 
10:01 PM
@r9m I don't find it now. Maybe later.
 
r9m
@Chris'ssis okay :) I was watching Alexander (movie) :-)
 
@r9m hmmm
 
@TedShifrin I saw a proof about it but do it alone I don't think I can.
 
r9m
@Chris'ssis yes :)
 
10:04 PM
@Chris'ssis (This film is not historical correct but anyway ^^)
 
@r9m It's a very nice movie, I saw it when I was abroad (in 2004) at an amazing quality (video and sound).
 
r9m
@TedShifrin I never understood the need to drag the unit circle in this .. its just another closed set like $[0,2\pi]$
@Chris'ssis :D
 
historically*
 
@Gato I don't remember the details. I saw it a long time ago.
 
I think of it more naturally on the circle, but of course we can do $\Bbb R/\Bbb Z$.
 
r9m
10:06 PM
@Gato who cares about historical accuracy? .. I was watching Angeline :P
 
@TedShifrin To prove the density I need the result for additive subgroup of R, right?
 
In addition, @r9m, I would rather use powers of $e^{i\theta}$ than think about addition formulas for trig functions.
 
@r9m Lol,, I prefer Rosario Dawson!
 
I don't understand, @Gato.
 
r9m
@TedShifrin I think just continuity of $\cos x$ suffices .. $\{m+\pi n| m,n \in \mathbb{Z}\}$ is dense in $\mathbb{R}$ anyway .. no need for any properties of trig function other than it's periodicity
 
10:10 PM
@TedShifrin To prove that the set points $e^{i\theta}$ is dense I need that additive subgroup of $\Bbb{R}$ are dense or in the form $\alpha \Bbb{Z}$
 
r9m
@Gato okay :-)
 
oh, that's not the way I would approach the problem at all, but ok.
 
Quick question regarding the freshman's dream lemma. The question is this one: math.stackexchange.com/questions/127353/… It is pretty straightforward for when the field in question is of order $p$, prime.
We wind up with $f(x)=a_0+a_p x^p+a_{2p}x^{2p}+\dots+a_{kp}x^{kp}$, which by fermat's little is $f(x)=a_0^p+a_p^px^p+a_{2p}^px^{2p}+\dots+a_{kp}^px^{kp}$, and by the freshman's dream is $=(a_0+a_ox+a_{2p}x^p+\dots+a_{kp}x^{(k-1)p}) ^p$
 
@TedShifrin In any case we need that the set $\{ma+mb\}$ is dense if $a/b$ is irrational no?
 
That's one approach, yes, @Gato.
 
10:14 PM
@TedShifrin Can we avoid this?
 
My difficulty comes when the field in question is of order $p^r$. The freshman's dream still holds for $q=p^r$, but not all of my terms will have $p^r$ as a factor of the exponent*.
 
But @JMoravitz every element is still a $p$th power, so isn't that good enough?
I prefer to suppose there's an open interval $U$ in the circle that contains no power of $e^{i\theta}$, @Gato, knowing that the circle has finite measure.
 
I'd think so, but does the freshman's dream hold for $(x+y)^p = x^p + y^p$ when the order of the field is not in fact $p$ but instead $p^r$?
 
No.
Hmm ...
 
This idea need to renamed from the freshman's dream to the freshman's fact. Becuase it is true in the minds of most freshmen.
 
10:20 PM
Unfortunately so @Kevin I see it alot myself too, even in the calculus classes.
 
Cute problem: show that $S^1$ is not the additive group of any unital ring.
 
@TedShifrin This is very interesting, i need to learn about this one.
 
@JMoravitz: You should still be able to write $\sum a_kx^{pk} = \sum b_k^p x^{pk} = \left(\sum b_kx^k\right)^p$.
 
@TedShifrin Do you know where can I find this proof?
 
No, @Gato. I figured it out many years ago :)
 
10:25 PM
Does someone of you know how we check if a point of an elliptic curve is of finite or infinite order??
 
I'm going to pull off a @Chris'ssis here and give you guys a very interesting problem.

Find solutions $x, y \in \Bbb{Z}$ such that $(2n-1)x + (2n+1)y = 1$ and $n \in \Bbb{Z}$.
 
@TedShifrin theres the problem again though that if the freshman's dream doesn't hold for $p$ when the order of the field is $p^r$, then that step still seems invalid, yet it seems like the obvious choice. (and in fact is for when $r=1$), it seems we would need to have it be $\sum b_k^{p^r} x^{kp^r}$
which is definitely not the case
 
@TedShifrin ah! Vous êtes peut être ,donc, e seul qui connait cette preuve ;P
 
It does hold mod $p$, @JMoravitz.
Je m'en doute sérieusement, @Gato.
@JMoravitz: You don't care what order the field has. You just care that $p=0$ in it.
 
oh? >_< I think I momentarily forgot that. Of course, $\mathbb{F}_4\equiv \mathbb{Z}_2\times \mathbb{Z}_2$...
 
10:34 PM
NONONONONO, that last statement is wrong.
The RHS is no field.
 
wasn't it isomorphic to the Klein 4 group?
oh right, the additive group of $F^4$ was
 
Additively, yes, not multiplicatively.
 
hi @N3
 
anyone know some basic python?
 
10:36 PM
@N3 shoot
 
from visual import NOPE
 
omfg my problem is so cool. @Chris'ssis I definitely know how you feel now when everyone except a few ignore your problems.
 
@DonLarynx scratches head
 
@DonLarynx lol, I'm terribly tired. Thank you for you problem and my eyes hurt like hell. ;)
 
10:37 PM
@N3buchadnezzar where the hell is line 1
 
@DonLarynx ? :p
 
Where is line 1 @N3buchadnezzar it says the error occurs in line 1.
 
@TedShifrin Un jour vous me montrerez la preuve? :D
 
@TedShifrin shame on me lol
 
10:41 PM
Et bien: tu n'es qu'un chat :P
 
@DanielFischer Could you take a look at it? math.stackexchange.com/questions/1121009/…
 
@DonLarynx Yeah, that is just my IDE being a ****.
 
Lord Farin + Lord Oeuf + Lord Eau = Lord Gato
 
@TedShifrin Oui, d'ailleurs j'ai trois chats
 
Et moi j'adore les chats :)
 
10:42 PM
@TedShifrin Im glad you like us
 
Miaou
 
Not you, @Kevin.
 
Sigurd, Ragnar et Batman. @TedShifrin
 
My mistake everyone!@!!! $n \in \Bbb{N}$
 
@Hippalectryon LOL
 
10:42 PM
@N3buchadnezzar Point out line 1 for me please.
 
Ce sont des noms bizarres, même pour des chats.
 
@TedShifrin Nope you said chats. That's us.
 
Sigurd and Ragnar étaient des guerriers de la mythologie nordique.
 
@Gato: First you can deduce that there's a neighborhood $V$ of $1$ so that $1$ is the only element of $\{e^{in\theta}\}$ in $V$.
 
@N3buchadnezzar Tuples are "immutable". You can't assign to components of tuples.
 
10:45 PM
@DanielF: are they ineta-able?
 
@TedShifrin ineffable
 
LOL
censors @DanielF
 
Perhaps too American a slang
 
@DanielFischer What is the work around?
 
10:47 PM
@TedShifrin it's almost immediate on a drawing ^^
@Hippalectryon ça va?
 
@Gato oui, et toi ?
 
@DonLarynx Here
 
@N3buchadnezzar That depends on what you want to achieve. It will probably involve using a type other than tuples.
 
oui! @Hippalectryon Tu es bien silencieux.
 
@Hippa, comme toi, il se fâche que je ne lui donne pas la preuve.
 
10:48 PM
@Gato J'ai du travail :-)
@TedShifrin comme moi ?
 
@Hippa ne fait guère de boulot.
 
@DanielFischer I want to change the list in the tuple with another list, is there a better type to use than tuple?
 
says the one retiring next year
 
nope, this year
 
even worse
 
10:50 PM
Je ne vous manquerai pas. :)
 
@TedShifrin I guess it's because in most of french books there are solutions with exercises.
 
I don't believe in such a thing, @Gato :P
 
@TedShifrin Why not? or perhaps the book I have lol
 
Because my view is that people who have solutions look at them.
 
What about a book that give you solutions... and you have to find the problems :3
Difficulty 90/10
 
10:52 PM
@N3buchadnezzar Can't do that with assignment to the tuple component. You could use a list or a dictionary instead of the tuple maybe. Or say test[1] = (range(1,11),test[1][1],test[1][2]), so copy the components you want to keep.
 
@TedShifrin I guess thats the point of solutions..
 
That's why I don't publish solutions to most of my problems, @N3.
 
@TedShifrin You are right, this is a very bad habit !
@Hippalectryon Let's start this with Ted's solution of my exercise haha
 
It makes people think you learn mathematics by memorizing solutions.
 
@Gato What's your exercise ?
 
10:54 PM
Yep and think that memorizing is the same as thinking..
@Hippalectryon "I prefer to suppose there's an open interval $U$ in the circle that contains no power of $e^{i\theta}$, @Gato, knowing that the circle has finite measure."
pour prouver la densité de $e^{i n\theta}$
 
Chebichev
don't tell me that's not how we spell it. There are so many ways e_e
 
C'est un nom russe, mes enfants.
 
plaît-il ?
Tchebychev*
 
@TedShifrin mes enfants you sound like the ogre in the Petit poucet or whatever :3
 
Parfait :P
 
10:57 PM
>:c run for your life
He drags kids by teaching mathematics, then eats them
 
@Hippalectryon pq parles tu de Tchebychev?
 
@Gato C'est pas similaire à la densité de $\sin(nx)$ ? Je sais qu'avec Tchebychev cette dernière peut se montrer.
 
@Hippalectryon Oui on peut projeter et ça revient à cela. Par Tchebychev tu entends polynôme ?
 
Comment utilises tu cela?
La preuve de Ted semble vrmt belle aussi mais il faut que j'essaye de la travailler..
 
11:02 PM
Je ne sais pas ce que c'est qu'une mesure finie :/
 
@DanielFischer Switching to a dictionary worked, thanks
 
@Hippa: $S^1$ has finite arclength.
 
@TedShifrin What is $S^1$ ?
 
$\{z\in\Bbb C: |z|=1\}$. :)
 
@Gato Je peux me tromper, mais ça doit être du genre : $\{p\theta + 2q\pi \mid p\in\mathbb{N}, q \in\mathbb{Z}\}$ dense dans $\mathbb{R}^+$ (ce qui est plus classique)
 
11:06 PM
C'est équivalent, bien sûr, @Hippa.
 
Où utilises tu les polynômes ici ?
 
Moi, pas.
 
@N3buchadnezzar I was away sorry. Where is "here"?
 
@Gato Bonne question :D je sais qu'une version plus générale utilise la formule générale des polynômes de T., mais ici pas besoin.
 
@TedShifrin Oui, so the first step is to prove that there's a neighborhood $V$ of 1 so that 1 is the only element of the sequence $e^{in\theta} in $V$ and then?
 
11:09 PM
I have another interesting problem for everyone.

Ted pays $\$1.43$ for some apples and pears. If pears cost $\$.17$ each and apples cost $\$.15$ each, how many of each did he buy?
Next, does Ted even like apples and pears?
 
I will try to prove your step tomorrow :-)
 
smacks @Don
 
@Ted I love number theory. Your suggestion to take only 4 classes has worked out well.
 
Then you should end up with lots and lots of disjoint copies of $V$ with a certain property, @Gato.
 
@Hippalectryon Lol je me disais bien.
 
11:09 PM
Ted steals apples and pears
 
@Don: Do not misquote me. I told you to take no more than 3, and 2 is preferable.
 
@Ted but I need to take 4 to be considered a full-time student.
 
Well, 2 were CS.
 
@DonLarynx Right above my message._.
 
@Hippa: You are a very bad influence on baby Hippa.
 
11:11 PM
@TedShifrin That's his problem.
I'm not telling him to do anything.
 
Et le nôtre aussi.
 
Mute him.
 
I should muzzle you first.
 
OK, dinnertime ... Bubye.
hi/bye @RobertC
 
11:13 PM
@TedShifrin I will think about it and tomorrow we will discuss about them ;p. Thanks and bon appétit!
 
Bonne nuit, @Gato.
 
@Hippalectryon Travail bien, moi je vais dormir lol.
 
@Gato Bonne nuit :-)
 
@Hippalectryon Je ne sais pas comment tu fais pour dormir si tard avec la prépa..
 
Je dors beaucoup le WE
 
11:15 PM
pourquoi pas oui c'est assez bien négocié. Bonne nuit!
 
@N3buchadnezzar I'm still working on it
 
Hello =) do you have any idea for this: http://math.stackexchange.com/questions/1121041/their-product-is-a-cubic-of-a-rational-number-x-minus-x?noredirect=1#comment2287327_1121041 ???

Thanks in advance!
 
@N3buchadnezzar: tuples are immutable, but you are trying to mutate a tuple...
Hence the error.
:)
"So like strings, tuples are immutable. Once Python has created a tuple in memory, it cannot be changed."
Does this answer your question?
 
@DonLarynx Daniel helped me half an hour ago ;)
 
I want to write an algorithm that takes as input an AVL-tree and gives as output two AVL-trees, that contain together all the keys of the initial tree and in addition the keys of the one are greater than the keys of the other AVL-tree.
The algorithm's time complexity should be $O(\log n)$ where $n$ is the number of nodes of the initial AVL-tree.
Could you give me a hint how we could do this?
 
11:33 PM
Hey @TedShifrin
 
@r9m I finally had some time to work on another answer.
 
Ant
hi! quick question for you guys :) If the closure of a set A in defined as the smallest closed set that contains A, can we say that on R (with the euclidean distance) the closure of {0, 1} (this set consist of two points) is [0, 1]? Or is it something else?
 
@Ant $\{0,1\}$ is a closed set
 
@Ant You're thinking about the convex closure, it seems. =)
Le chat is in the chat.
 
11:49 PM
@PedroTamaroff Oh and I need Pedro's knowledge!
 
Really?
 
yep, about infinite group: does an infinite group contains an accumulation point?
 
Well, it depends on the group and on the topology and on the context of all this.
@Gato Where is this question coming from?
 
sure, I am taking $T=\{e^{i\theta}:\theta\in[0,2\pi)\}$ so it will be an infinite subgroup of $\Bbb{R}$.
 
@Gato That's not a subgroup of $\Bbb R$, rather of $\Bbb C$.
That's $S^1$, which is a compact subgroup of $\Bbb C$. Every point of $S^1$ is an accumulation point. You can prove this.
 
11:56 PM
Pedro prove that $S^1$ is not the underlying additive group of a unital ring. Go.
 
@PedroTamaroff Sorry yes $S^1$. In fact I would like to prove that $exp{in\theta}$ is dense. If I prove that Every point of S^1 is an accumulation point it would be the same no?
 
@Gato No.
Just because a set is such that every point is an accumulation point, it doesn't entitle an arbitrary subset to be dense.
For example, $\Bbb R$ has this property, but say $\Bbb Z$ is not dense.
 

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