Mathematics - 2012-10-25 (page 1 of 2)

hhh

1:07 AM

How do you remember Farkas's Lemma and its proof?

Some intuition thing here, trying to find some intuitive way to it.

robjohn

1:51 AM

My next answer will be my $900^{\text{th}}$.

Gustavo Bandeira

@robjohn Pro Player!

I just wait for the day I'll answer my first decent question!

anon

2:52 AM

what's a monoidal category where coproducts distribute over $\otimes$ called?

I can't follow practically anything ncatlab says

skullpatrol

4:18 AM

hi @LinearMan

infinitesimal

4:44 AM

@skullpatrol Why don't you stop annoying everbody with your greetings?

skullpatrol

hi @JasperLoy

user19161

@skullpatrol Hello.

skullpatrol

@JasperLoy Do you find my greeting annoying?

user19161

@skullpatrol It's alright, but perhaps cut down a little.

skullpatrol

@JasperLoy When somebody new comes in the room I say "hi." How should I cut that down?

user19161

4:51 AM

@skullpatrol Hmm, I only say it to whoever I wish to talk to. But just go by your feel.

skullpatrol

@JasperLoy Good point.

user19161

@skullpatrol Also, don't worry when someone says they find you annoying. We all find others or ourselves annoying at times.

3

skullpatrol

I want to see how many "stars" this gets.

9 mins ago, by infinitesimal

@skullpatrol Why don't you stop annoying everbody with your greetings?

user19161

@skullpatrol I don't do stars, but I don't find you annoying, just a little eccentric. Also, perhaps cut down a little on the huge text and pictures and removal.

skullpatrol

@JasperLoy But that is who I am...

user19161

4:56 AM

@skullpatrol Hmm, OK. I was only expressing what I feel honestly.

skullpatrol

@JasperLoy I appreciate your honest opinion :-)

user19161

@skullpatrol It is hard to be honest in a world filled with lies.

skullpatrol

Indeed.

user19161

@skullpatrol It would be interesting to see who has ignored who in this room! I have not put anyone on ignore.

skullpatrol

@JasperLoy Neither have I.

user19161

5:01 AM

@skullpatrol They don't know what stuff they are missing out on when they ignore skullie. Skullie might just post a proof of the Riemann Hypoethesis in chat one day.

skullpatrol

@JasperLoy You, on the other hand, are a master of complements and quick wits, Sir.

user19161

@skullpatrol I am only a lunatic trying to heal.

5

skullpatrol

@JasperLoy I believe you are just misunderstood by people.

user19161

@skullpatrol I am off, don't sleep too late skullie.

skullpatrol

@JasperLoy “Before diagnosing yourself with depression or low self-esteem, make sure you are not, in fact, surrounded by assholes.”

skullpatrol

5:24 AM

19 mins ago, by Jasper Loy

@skullpatrol I am only a lunatic trying to heal.

“Before diagnosing yourself with depression or low self-esteem, make sure you are not, in fact, surrounded by assholes.”

Nick Kidman

7:48 AM

are possible topolgies in any sense classified?

say w.r.t. certain sets or w.r.t. tyes of topologies like T1, T2, ...

Nimza

9:20 AM

Hi everybody!

Help me please with english :) If I readed some papers but not thoroughly, how should I say about my action? in russian literally "overlooked", but it is bad In english I think

infinitesimal

9:39 AM

skimmed

user19161

@Nimza The past tense of read is read, not readed.

Nimza

thanks, yup :)

user19161

@Nimza You can always ask on ELU where you will get better answers.

Nimza

ELU?

user19161

English SE.

Nimza

9:41 AM

ah, thank you

infinitesimal

@JasperLoy Are you saying "skimmed" is not a good answer?

Nimza

@infinitesimal it's like skimmed milk? Google translate gives only adjective

infinitesimal

@Nimza Skim milk is what is left behind after you skim off the cream.

Nimza

@infinitesimal nice! thank you

infinitesimal

Try looking up "skim read."

Nimza

9:49 AM

aha, I understanded

infinitesimal

also skim reading...

or browsing....

Nimza

10:18 AM

Do you know some simple generalizations of fractional derivatives?

user19161

No, post on main.

Nimza

I posted on mathoverflow :\

user19161

@infinitesimal Just making a general remark.

Nimza

s

user19161

@Nimza Wrong room!

Nimza

10:32 AM

:D AHA!

user19161

@gustavo Are you comfortable with the definition of limit? Was it done in terms of $\epsilon$ and $\delta$?

Gustavo Bandeira

@JasperLoy I guess so. I got a feeling on what they are, but I feel I'm not completely sure.

user19161

@GustavoBandeira Actually, studying limits is a good way to understand the use of the universal and existential quantifiers!

Gustavo Bandeira

@JasperLoy How?!

I was also reading about them on Halmos Naive set Theory.

user19161

@GustavoBandeira Well, look at the examplea and you will finally understand what "for all" and "there exists" mean!

skullpatrol

10:38 AM

*examplea?

@JasperLoy Is that an exam plea of guilty?

Gustavo Bandeira

There's an eecercise where he asks me to determine the values of a for which $\lim_{x\rightarrow a}f(x)$ exists. They exist for all values, except for -2,0,2. Can I use this notation: $a\neq \{-2,0,2 \}$?

user19161

@GustavoBandeira No, that means that $a$ is not equal to the set.

Gustavo Bandeira

$\forall a\neq \{-2,0,2\}$?

Maybe this isn't so clear.

user19161

@GustavoBandeira No. Why do you need this notation? Just say for all $a\neq -2,0,2$.

Gustavo Bandeira

@JasperLoy Yep. Maybe I could extend and learn/think a little about notation.

This is the last thing I could think: $\{a\in A:a\neq\{-2,0,2 \} \}$

user19161

10:45 AM

@GustavoBandeira Yes, I think you need to understand basic notation first.

user19161

@GustavoBandeira No, all wrong!

Gustavo Bandeira

=/

user19161

You can use $a\not\in\{-2,0,2\}$ to express that it is not equal to any of the three numbers.

skullpatrol

@GustavoBandeira set builder notation

Gustavo Bandeira

Oh, thanks.

user19161

10:48 AM

@skullpatrol He's reading Halmos's book which is more than sufficient.

skullpatrol

@JasperLoy Just trying to help, sorry.

Gustavo Bandeira

Relax. =)

user19161

@skullpatrol No need to say sorry man.

Gustavo Bandeira

Hey, In december I'll start going to math classes.

=)

user19161

@GustavoBandeira Where?

Gustavo Bandeira

10:50 AM

@JasperLoy The university near me. =)

user19161

@GustavoBandeira OK, so you will be studying mathematics there?

Gustavo Bandeira

Yep. =)

I wish december come soon. There's something nice on it: The classes are going to start on december 3, it's the date of my birthday. =)

user19161

@GustavoBandeira Oh, OK.

skullpatrol

You will be born again :-D

Gustavo Bandeira

Yep. =)

And I was on an economics course. Dude, WTF I was doing with my life?!

user19161

10:53 AM

@skullpatrol I hope to be born again soon too.

skullpatrol

Remember we were born to be alive!

user19161

@GustavoBandeira Just consider carefully and choose your own path!

user19161

@skullpatrol Si, si!

Gustavo Bandeira

►

skullpatrol

Cool tune...

user19161

10:59 AM

Hello @zhen. You must have started your school now.

user19161

@skull Maybe now I should say the same thing that you always say ^. =)

Nimza

Hm, if we extend linearly operator $(a+bt)^{\beta} \mapsto \frac{\Gamma(\beta+1)}{\Gamma(\beta+1-\alpha)} (a+bt)^{\beta-\alpha} b^{\alpha}$ what will we receive? Something well-known like fractional derivative?

Charlie

12:14 PM

Buenos dias, Good afternoon, guten abend!

@JayeshBadwaik Hello

Jayesh Badwaik

@Charlie hi..

Charlie

@JayeshBadwaik How are you?

Jayesh Badwaik

@Charlie good. good. (laughing right now ;) )

Charlie

@JayeshBadwaik Why?I wanna laugh too!

user19161

Hey @charlie @jay!

Charlie

12:19 PM

@JasperLoy Hey!

Jayesh Badwaik

@JasperLoy Hi!!

@Charlie hmm, show of hands ;-)

Charlie

@JayeshBadwaik :D

is that so bad?

user19161

Wow, I got 2+2 stars, yum yum!

Jayesh Badwaik

@Charlie no, its good. really good.

Charlie

@JayeshBadwaik hmm :P

Jayesh Badwaik

12:21 PM

@JasperLoy Now its 3 + 3.

Charlie

@JayeshBadwaik did you listen?

Jayesh Badwaik

@Charlie barely audible, will have to get my headphones for it.

user19161

Do you guys have any idea why I seem to be the only one who has problems with using Chrome in chat?

Jayesh Badwaik

@JasperLoy no, because I don't use chrome. get firefox already!!

user19161

I know many others have problems with Chrome on the main site, but for chat no.

Charlie

12:22 PM

@JayeshBadwaik yup... i told you...it's terrible...

@JasperLoy Maybe is the device between the keyboard and the chair :P

user19161

@JayeshBadwaik I keep vacillating between the two. Currently back to Chrome. Because I want the latest flash. And also because Firefox keeps crashing on one my favourite sites which I shan't name!

user19161

@Charlie Yes, my aura is too strong...

Charlie

@JasperLoy hahahahaha

@JayeshBadwaik i will make it better.Promise.

user19161

Now probably the site that keeps crasing has some bad code, but then Chrome doesn't crash on it.

Jayesh Badwaik

@Charlie hmmm, no its not terrible, just inaudbile yet, probably if i can listen to it at more volume, it will be better. :-)

@Charlie :-)

user19161

12:24 PM

It's a public holiday here tmr.

user19161

Hari Raya Haji.

Charlie

@JayeshBadwaik :D

Jayesh Badwaik

@JasperLoy we had a one yesterday.

Charlie

My next holiday is the "the Dead's day"

user19161

My friend will be back next month, but my books I asked him to get will only be here in March as he is shipping all his things via cargo ship.

Jayesh Badwaik

12:25 PM

@JasperLoy we call it bakr id, bakr means goat. goat is offered for sacrifice and then eaten.

skullpatrol

Hi gang!!!

Jayesh Badwaik

@Charlie halloween. hmm, when is thanksgiving?

@skullpatrol hi nam.

user19161

@JayeshBadwaik OK. I might go vegetarian in future, I don't know. For now, I try not to eat things I think are killed too inhumanely.

Jayesh Badwaik

@JasperLoy hmm, visit jc bose institute in kolkata and you might stop eating plants too.

Charlie

@JayeshBadwaik it's not halloween... is a day that people remember the beloved people that died,it's religious... about thanksgiving... we don't have a thanksgving day, like in USA, e.g...

user19161

12:27 PM

@JayeshBadwaik What do they do to plants there?

Jayesh Badwaik

@JasperLoy there are public display experiments that show how plants also have "Emotions" when they react to various stimuli like music, caffeine etc. It is a "rumor" that people who have given up meat after going to slaughterhouses tried to give up eating after going to that institute.

@Charlie okay. isn't that the same thing?

user19161

@JayeshBadwaik Let me tell you what I think.

Charlie

@JayeshBadwaik what is the same?

Jayesh Badwaik

@Charlie halloween at least in some place is celebrated as the festival of loved ones who are dead right?

user19161

Life is a spectrum and plants fall somewhere on the lower end. Nonetheless, it is still a life form. However we need to eat to survive. So we try to draw the line somewhere. Different people draw the line differently. I think going vegetarian is good enough already. I might do that some time in the future.

user19161

12:31 PM

@JayeshBadwaik Really? Not that I know of.

Jayesh Badwaik

@JasperLoy from wikipedia: "Most scholars believe that All Hallows' Eve was originally influenced by western European harvest festivals and festivals of the dead with pagan roots, particularly the Celtic Samhain.[6][7][8] Others maintain that it originated independently of Samhain.[9]"

skullpatrol

October 1 is All Souls' Day.

user19161

@JayeshBadwaik Hmm, OK.

Jayesh Badwaik

@skullpatrol hmm, nice :-)

Charlie

All Souls' Day

All Souls' Day commemorates the faithful departed. In Western Christianity, this day is observed principally in the Catholic Church, although some churches of Anglican Communion and the Old Catholic Churches also celebrate it. The Eastern Orthodox Church observes several All Souls' Days during the year. The Roman Catholic celebration is associated with the doctrine that the souls of the faithful who at death have not been cleansed from the temporal punishment due to venial sins and from attachment to mortal sins cannot immediately attain the beatific vision in heaven, and that they may b...

user19161

12:33 PM

@jayesh I see on TV how they slaughter pigs. It is still quite cruel. They are electrocuted but often they are still conscious after that and then slaughtered.

Charlie

@skullpatrol For catholic church is november 2nd

@JasperLoy O.o

skullpatrol

Oops I meant November 1

2 mins ago, by skullpatrol

October 1 is All Souls' Day.

The day after Halloween...

Charlie

@skullpatrol We don't have halloween here

skullpatrol

What!!! no trick or treat???

Charlie

@skullpatrol No.

user19161

12:36 PM

Closevoting math.stackexchange.com/questions/220780/…

Charlie

@JasperLoy crazy question

user19161

@Charlie Yes.

Jayesh Badwaik

@JasperLoy yup, and hence, what matters is the intention and the manner. You are going to kill someone to be a part of the food-chain. The only thing is you have to be dignified about it. In original Indian tradition, I don't know if it is followed now, the animal to be slaughtered would be calmed down first and then slaughtered in a quick manner, so that he/she feels as least pain as possible.

Charlie

@skullpatrol only when i was 7 my school did a stupid party.... but it's not part of our calendar

skullpatrol

@Charlie All the ghostly skulls warms skullpatol's heartless soul :-D

Charlie

12:38 PM

@skullpatrol hehehaha

In june we have more "important" parties, junine parties

Food is good!

user19161

@JayeshBadwaik I hope the animals are slaughtered humanely for the Haj.

Jayesh Badwaik

@Charlie food? where? Me needs good food.

Charlie

@JayeshBadwaik In the junine parties

@JayeshBadwaik take a look

Jayesh Badwaik

@Charlie yumm... yummy

Charlie

@JayeshBadwaik I love paçoca

it's a , some sort of a candy made with peanut

Jayesh Badwaik

12:43 PM

@Charlie hmm. nice.

Charlie

it's very soft and delicate, if your hold with strongly it becomes a manioc

t0.gstatic.com/…

really tasty

and you definetively can't speak with this in your mouth

Another delicious candy is brigadeiro

Jayesh Badwaik

looks like a rumball

Charlie

it's made with chocolate

Jayesh Badwaik

hmm. looks nice. I

skullpatrol

I check my calender and it says:
All Saints' Day (Christian) = Nov 1
&
All Souls' Day (Catholic) = Nov 2.

Charlie

12:47 PM

@JayeshBadwaik easy to make!

condensed milk, chocolate in powder , a bit of butter

Jayesh Badwaik

@Charlie :-)

@skullpatrol I thoughts $\text{Catholics} \subset \text{Christians} $

Charlie

@JayeshBadwaik try it someday

@Charlie I will.

@JayeshBadwaik yay!

@Charlie bookmarked!

12:52 PM

@JayeshBadwaik :DDD

@JayeshBadwaik What about the indian desserts?

sperners lemma

hello

skullpatrol

hi

Jayesh Badwaik

@Charlie there are many, I don't know how to make most of them yet.

sperners lemma

@skullpatrol, any number theory

Charlie

@JayeshBadwaik haha isn't there any you think it's tasty and interesting?

skullpatrol

12:56 PM

@spernerslemma no

sperners lemma

come on

Charlie

come off

skullpatrol

@spernerslemma I'm still struggling with what a "number" is...

Jayesh Badwaik

@Charlie There are lots, so many, I would have to think! Gulabjamun, Rasagulla. But the main thing is every store has its own variety of desserts except for some few famous ones. So, even if I tell you a name, you might not find it! For example, there is a bengali sweet that I know by the name of "mother india". I don't know what the real name or the original dish is .

sperners lemma

natural numbers

{1,2,3,4,5,6,...}

skullpatrol

12:58 PM

That^ explains nothing

sperners lemma

@skullpatrol, natural numbers

Charlie

@JayeshBadwaik Rasagulla! Funny name!!

Hey

Ra's Al Ghul

alphabet
A, B, C...

12:59 PM

@BenjaLim hey!

sperners lemma

ignored

BenjaLim

@JonasTeuwen Too much homological algebra atm....

Jayesh Badwaik

@Charlie ras=juice, gol= round. juicy round ball .

BenjaLim

@Charlie homological algebra!

Charlie

@BenjaLim YUP!

@JayeshBadwaik hahaha

BenjaLim

1:02 PM

@Charlie homological algera

Charlie

@JayeshBadwaik Brazilian cuisine is really tasty and quite easy to cook.

this is good!

Jayesh Badwaik

@Charlie ohh. Here, you have many different dishes, some of them quiet intricate, some of them dead simple. Every household has his own method of making things as well.

Charlie

@JayeshBadwaik hmm very nice!!! I really want to eat it!

Jayesh Badwaik

@Charlie :-) I am going out now. Be back later.

Charlie

@JayeshBadwaik Me too.See you later, Jayesh!

skullpatrol

1:10 PM

Are you a number theorist?
I apologize if I have offended you.

user19161

@skullpatrol No need to keep saying sorry man. The people who should say it never say it and the people who should not say too many.

skullpatrol

12 mins ago, by sperners lemma

ignored

Jayesh Badwaik

@skullpatrol i am ignored too, for one flimsy reason, guy seems to have low threshold for pain. don't worry one bit.

anyway, going out, bye

skullpatrol

@JayeshBadwaik later pal :-D

8 hours ago, by skullpatrol

“Before diagnosing yourself with depression or low self-esteem, make sure you are not, in fact, surrounded by assholes.”

@JasperLoy What did you think^

user19161

@skullpatrol I have many things to say about some of your sentences but it would exceed the page, so I rather not say it. =)

skullpatrol

1:23 PM

@JasperLoy Could you at least give me your first impression Sir?

user19161

@skullpatrol Well, I understand what you want to mean by it, that's all.

skullpatrol

The sentence is referring to those who starred your comment:
" I am only a lunatic trying to heal."

user19161

@skullpatrol Ah, my sentences have many hidden meanings too...

skullpatrol

@JasperLoy That's why I said most people misunderstand you.

8 hours ago, by skullpatrol

@JasperLoy I believe you are just misunderstood by people.

Matt N.

2:28 PM

@JayeshBadwaik By who?

Jayesh Badwaik

@MattN. sperners lemma

sperners lemma

only for a short time

because you kept making it beep at me

Jayesh Badwaik

@spernerslemma I apologized for the mistake, it was not intentional. Its glad to see its not so though. :-)

sperners lemma

3:17 PM

I wonder why no one will help me with my algebra question

what are modules actually about?

they seem to be like a weird version of ideals

I don't understand them at all

anon

3:42 PM

modules are where rings act on abelian groups with the obvious associativity and distributivity conditions, much like groups acting on sets as symmetries (automorphisms) or fields acting on vector spaces as scalars (which is actually makes vector spaces modules over fields, a special case)

@spernerslemma in a sense, since End(G) is a ring under pointwise addition and composition as multiplication when G is an abelian group, we can encode a ring R's action on G via a map R->End(G) (the kernel of the map would be the annihilator). in a module this is more or less implicit but suppressed.

if I is an ideal of R, then in particular it is an abelian group (under addition) that absorbs ambient multiplication by R, so ideals are special cases of modules.

sperners lemma

thanks a lot!

that is so simple

I didn't realize vector spaces were modules too

$$\frac{R}{x^n R} \to \frac{x^m R}{x^{m+n} R}$$

do you know about this R-module isomorphism?

why x^m is cancelled

anon

Are you considering R an R-module over itself or something?

sperners lemma

R as an R-module

user19161

I don't understand what it is you don't get about modules.

user19161

You have not stated the area of doubt.

sperners lemma

3:54 PM

I can't understand or prove this R-module isomorphism

that's my working on it so far math.stackexchange.com/questions/217555/…

also I guess it's not (L/N)/(M/N) = L/M

it's something about quotienting by kernels

anon

The quotient map $x^mR\to x^mR/x^{m+n}R$ is given by $x^mr\mapsto x^mr+x^{m+n}R$. You can check that $a(rx^m+x^{m+n}R)=(ar)x^m+x^{m+n}R$ and $(r+s)x^m+x^{m+n}R=(rx^m+x^{m+n}R)+(sx^m+x^{m+n}R)$, so this map is indeed an $R$-module homomorphism.

It'd be easier to check why a generic quotient map $R\to R/I$ is an $R$-module homomorphism and notice that $x^mR=(x^m)$ is an ideal.

user19161

With the great anon answering sperner, the rest of us can keep our mouths shut and listen.

sperners lemma

a x^{m+n} R = x^{m+n} R?

anon

Similarly, the map $R\to J= (a):r\mapsto ra$ is an $R$-module homomorphism into the ideal $J=(a)$. Set $a=x^m$ and $I=x^{n+m}R\le x^mR=J$ (so $I$ is an $R$-submodule of $J$), and we can compose to get $R\mapsto J\mapsto J/I$ as an $R$-module homomorphism. The map is given by $r\mapsto x^mr+x^{m+n}R$, which you can verify has kernel ideal $K=(x^n)$ so that by the isomorphism theorem $R/K\cong (J/I)$, or for your purposes $\frac{R}{x^nR}\cong \frac{x^mR}{x^{m+n}R}$

sperners lemma

I don't see how :(

anon

4:05 PM

@spernerslemma Sort of, because $ax^mR\subseteq x^mR$ for all $a\in R$. Technically it can be a proper ideal so you have to "expand" back out to get an actual coset/translate of $x^mR$.

sperners lemma

but why isn't x^m R = R then?

anon

@spernerslemma It is, in $R/R$.

sperners lemma

oh.. so is this a x^{m+n} R = x^{m+n} R (mod <something..>)

anon

But we're talking about $x^mR$ so let's go up to there.

if $K$ is an ideal of $R$ and $a+K\in R/K$ is an additive translate / coset of the ideal $K$ in the quotient space $R/K$, then for any $x\in R$, the set (and therefore the element in the quotient) $x(a+K)$ is equal to $xa+K$ modulo $K$.

Jayesh Badwaik

@JasperLoy Is your gravatar a beach from where you live?

user19161

4:09 PM

@JayeshBadwaik No, all my avatars are from avatarsdb.com except the blue one which I produce myself.

sperners lemma

@anon, hmm that seems true but a bit weird. I mean surely x(a+K) = k(a) = ka + K (mod K) so why even add K back?

oh

Jayesh Badwaik

@JasperLoy produce? you use some standardized color palette or something? ;-) :P

anon

IOW, every element of x(a+K) is an element of K away from being a member of xa+K, and vice versa

sperners lemma

it's so that equality without writing "mod K" is still equality mod K

anon

@spernerslemma because formally, the quotient space is constructed as cosets or additive translates of the ideal

user19161

4:10 PM

@JayeshBadwaik Yes, I use GIMP, choose 100 by 100, then colorify with steelblue, then save as a gif to minimize file size.

sperners lemma

I think I understand it! thank you

so now I have a homomorphism (the composition of R --> x^m R --> x^m R / x^{m+n} R) from R --> x^m R / x^{m+n} R with kernel x^n R

Jayesh Badwaik

@JasperLoy gif? why? png! tiff!

user19161

@JayeshBadwaik It has the smallest file size according to my experiment.

sperners lemma

so image(phi) = domain/kernel, x^m R / x^{m+n} R = R / x^n R!

yes!!

Jayesh Badwaik

@JasperLoy hmm. okay.

anon

4:14 PM

yup.

user19161

@JayeshBadwaik But perhaps changing some variables would alter the order of the sizes.

Jayesh Badwaik

@JasperLoy its not as much about the size. I think png is better

user19161

@JayeshBadwaik But in this case, it's just an avatar ain't it?

Jayesh Badwaik

@JasperLoy yup, so, I guess it is a personal choice. But what I prefer is, small images -> png. Lossless images -> tiff. Supported everywhere, portable everywhere, no worries. If I become a photographer, then I might start using some image format for absolutely glorious detail.

sperners lemma

I wrote the proof here math.stackexchange.com/questions/217555/…

hmm now that I look at what i've wrote, it doesn't include the bit I was confused about

anon

4:25 PM

your confusion was about how the quotient space is an R-module to begin with essentially, since you noticed that multiplying a coset by an element of R does not give you another coset, but rather a subset of another coset, so there is an implicit "expansion" required, or alternatively you need to think in terms of sets modulo K generated by say the translates and scalar multiples of cosets.

sperners lemma

yeah I am really grateful that you fixed me on that

now I can try to do the theorem which depends on that

anon

IMO it's actually a good sign that you were having a block on that, as that indicates you have a concrete grasp of the machinery underlying the arithmetic of sets, or at least your intuition tracks it accurately, since this subtle point about coset arithmetic I don't think is highlighted generally.

@spernerslemma do you mean the other way around, the theorem which the isomorphism depends on, namely the isomorphism theorem domain / kernel =~ image?

sperners lemma

it's this x^n R / x^n+m R theorem that gets used later in something about local fields

user19161

4:40 PM

@anon That sounds very deep.

Jayesh Badwaik

5:03 PM

rekonq FTW.

@anon I know its kind of silly, but I think "please see" text from the title is kind of ugly.

sperners lemma

> The p-adic fields or any finite extension of them are characteristic zero fields

I thought they had characteristic p

well I guess not, p isn't 0

skullpatrol

hi @spernerslemma

skullpatrol

5:32 PM

Are you ignoring me?

Matt N.

I think I just missed both Michael and Brian : /

On SE not chat, that is.

skullpatrol

Matt N.

Poor me.

sperners lemma

wow people are upvoting an answer I gave on a combinatorics question

It must be because of my nickname

Jayesh Badwaik

@spernerslemma or the answer must be really good. :-)

Matt N.

5:48 PM

@spernerslemma Of course. Like me, everyone reads Spermer's lemma.

2

sperners lemma

why would people upvote that??

Matt N.

Who knows.

I got a mysterious star : D

Oops, got to go!

sperners lemma

maximal ideals are prime in an integral domain? or is that true in any ring?

oh, "Every maximal ideal is in fact prime" but we need an integral domain to say "prime ideals are maximal"

confusing that this theorem has an unused hypothesis :S

skullpatrol

Jayesh Badwaik

6:31 PM

@Jonas read 3.14 in here

Argon

6:43 PM

Is it actually better to always rationalize denominators?

skullpatrol

@Argon What do you mean?

Argon

@skullpatrol e.g. $$\sin \frac{\pi}{4} = \frac{1}{\sqrt 2}$$

Should I write instead $\frac{\sqrt 2}{2}$?

skullpatrol

Yes.

Argon

Why?

skullpatrol

This is called the "simplest form."

Argon

6:48 PM

Why is $\frac {\sqrt{2}}{2}$ simpler then $\frac {1}{\sqrt{2}}$?

skullpatrol

Because of the fractional exponent in the denominator.

Argon

So? What's the problem?

skullpatrol

It may become hard to work with in further manipulations.

Argon

It's never really been a problem for me

Jim_CS

Can we write the order of an element in group theory as |g|...or is that notation only allowed for the group itself...ie |G|?

sperners lemma

6:52 PM

@Jim_CS, I've seen that notation before

I think it could be confusing if you put a group structure on objects that have cardinality, or points of a space with a metric though

skullpatrol

@Argon If you were trying to form a common denominator and you had to work with a square root or an integer which would you prefer?

sperners lemma

I think ord(g) is better, but ord_m(n) for order in the group Z/mZ is confusing...

Argon

@skullpatrol I see. However, it is often more concise and is easier to work with.

skullpatrol

@Argon It's easier to keep the denominator as simple as possible.

Jim_CS

@spernerslemma: didnt know I could use ord...cheers mate

sperners lemma

7:07 PM

@Jim_CS, you can use absolutely anything as long as you define it

skullpatrol

@Argon This is only a "guideline" not a rule.

29 mins ago, by Argon

Is it actually better to always rationalize denominators?

skullpatrol

7:23 PM

Rationalizing the denominator

sperners lemma

9

Q: Is $n \sin n$ dense on the real line?

Is $\{n \sin n | n \in \mathbb{N}\}$ dense on the real line? If so, is $\{n^p \sin n | n \in \mathbb{N}\}$ dense for all $p>0$? This seems much harder than showing that $\sin n$ is dense on [-1,1], which is easy to show.

WOW.

Good evening

hello

@spernerslemma hi :)

I have one question on basics of topology

sperners lemma

oh cool

I know basic topology

you can ask it

Wannabe Mini Mathematician

7:38 PM

$\tau > \pi$

Nimza

@spernerslemma any orientable closed connected surface has a character $a_1 b_1 a_1^{-1}b_1^{-1} \ldots a_n b_n a_n^{-1} b_n^{-1}$, right? Under character I mean a sequence of labels on edges of polygon to be glued together

Jim_CS

I often see that when someone is showing that U is a subgroup of G, they will demonstrate the inverse and identity properties but then just say 'and associativity follows as U is a subset of G'....why does associativity 'follow' automatically?

sperners lemma

ssorry I don't know anything about characters or stuff like that, that's not what I thought basic meant!

Nimza

:(

sperners lemma

@Jim_CS, if a,b,c are elements of G they satisfy (ab)c = a(bc). If a,b,c are elements of U then they are elements of G.

Jim_CS

7:43 PM

Cheers mate

Jim_CS

7:54 PM

If U and V are subgroups of G, show that if V is normal in G, then UV = {uv | u in U, v in V} is a subgroup of G. For the identity axiom here I said "As U and V are subgroups I (identity of group G) is in both of them. So take uv = II = I. Hence there exists an identity element in UV". Does that look ok?

sperners lemma

what is identity group?

I think you are doing the right thing, but that term sounds weird

basically the identity i of G is in U and V so it's in UV (as ii = i like you said)

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