x=a^{-1}

wait can i just cheat and do this?

It's said to hold for all choices of a,b,c,d,x, right?

These crazy people at the chess club always ask me when I will play a game.

I'm just there because the beer is cheap.

Yup.

oh ok

that easy then

@JonasTeuwen e4

@HenningMakholm e5

The last time I played a tournament was like 10 years ago. And the one 'crazy chess' with all weird variants.

I like weird chess. And chess that does not last five bloody hours.

Rapid chess is cool.

I suck at chess

I used to be a pretty decent player.

Now I suck at all.

@orange: extreme chess

Fischer Random is fun.

@robjohn Hey I was talking to Jonas. (Hm, do I actually have a chess board in the house? Don't think so).

When you connect the diagonals.

The chess board is in your head.

No need boards.

@HenningMakholm I just play on chess.com. Who needs boards.

It is a memory game for one.

@Eric Chess + Arm Wrestling.

@HenningMakholm I know... I figured he wouldn't answer, so I interrupted :-)

@JonasTeuwen Stefan Zweig convinced me not to try that.

Why?

@robjohn I figured he wouldn't answer too, so I didn't worry about what would happen if I had to go beyond one move.

@OrangeHarvester when we were punchy, we had a game called finger chess, where you shot the pieces at each other by flicking them with your fingers.

I've outsourced my pizza production facilities to Dr. Oetker.

@HenningMakholm >8(

That seems like a good use of chess pieces.

Kick their ass - with the pieces.

@JonasTeuwen He wrote a story in which the protagonist goes crazy from mind chess.

Najdorf once did like 58 games simultaneously blind.

I beg to differ that it makes you crazy.

@robjohn Ahh. Painful, we do that with carrom men.

It is easier to flick them, because they are made to be flicked anyway.

@JonasTeuwen The most I've done is one. A friend and I used to play blind chess while waiting in line for movies.

And if we place them on the edges instead of on the board, they can really fly.

Ah - I've done two - my advisor and his son 8-).

While drinking wine - to add to the 'boss' factor.

And chatting about quantum optics with some other guy.

@robjohn 2. conceded.

@HenningMakholm Ha! you fell into my trap!

You just take the pieces and throw them at your opponent - how hard can it be?

I've always fantasized of secretly punching the opponent on the nose - and then quickly look like you are in deep thought while staring at the board.

'What's wrong?'

Most probably he will say: 'You just punched me on the nose, motherfucker' instead of my preferred one: 'I... don't know...'.

I will say, nothing, and then kick you in the groin later.

Even better.

'What's wrong?'

'Are you okay?' and then when I say 'yes, no problem!' you kick me in the face?

'What about now?'

You have such violent thoughts, mister Orange... I would not be surprised if the court psychiatrist demand the chat logs in say 15 years.

you should be in a position to say 'yes, no problem'. the least will be you on the floor rolling around, worst will be you passed out.

@JonasTeuwen :P

Heya

@OrangeHarvester My brass is made of steel, so no biggie.

@JonasTeuwen That sounds metallurgically unlikely.

My cherries do not care about material science.

Talking about brass. You need quite some brass to give your Nobel lecture in Comic Sans MS.

Who did that?

nobelprize.org/mediaplayer/index.php?id=1871

@OrangeHarvester Serge Haroche

I see.

@JonasTeuwen If you're giving a Nobel lecture, you probably don't need to worry about subtle typographical clues to convince your audience that you're competent.

Some people are just unable to become more awesome.

Yes, I'd insert zombies and pirates. Always makes it better.

Perhaps also slides the 'burn away'.

In high school, I used Comic Sans as the only font for all my reports. :-/

Haha, übertroll.

My French lecturer used it, and when I remarked that it is butt ugly, just like (...) he used it even more.

And looked at me like 'whatcha gonna do about it huh?'.

In those days it was still quite 'shocking' when you would punch the teacher. It might sound strange nowadays, but in those days... hell broke loose if you did that.

Hahaha. My lecturers were simple powerpoint folks, they would whatever default font came with the software, so Arial in the beginning and I guess Cailbri later.

@JonasTeuwen Have you ever punched your teacher?

No, of course not.

I was a good boy.

I only got punched once. I had to write like 120 bloody pages.

Each word in a different colour.

Okay.

I got cured immediately.

Is there any easy way to prove that a multivariable function is continous?

Hahaha. I was excellent in studies, and sports and general discipline like cleanliness etc, but I still found inventive ways to get into trouble always.

@N3buchadnezzar !?!

The only time I got kicked out of class... was in university.

Use norm for both domain and range?

@N3buchadnezzar Composition.

A sum of continous functions is continous, yeah

hey, does anybody here know what a fusion system is?

Yes.

You can fuse things with it.

Usually to wrap things.

@AlexanderGruber Expected to be commercially feasible within a decade or two (and has been so since the 1950s).

I think it would a great relief for many people once incompetence is not so taboo anymore. That you can be openly incompetent without people dissing you.

Mayors, doctors, all can go for their coming out. How wonderful would that be?

The incompetent & proud of it movements will pop up as shrooms in my fridge.

@JonasTeuwen They are already there I think.

Oh really? How blimey. That fills my intestine with lovely rainbows.

So crazy it might make me shit bricks.

And I do not like hemorrhoids!

@JonasTeuwen Hasn't math been at the forefront of the incompetence-is-not-a-taboo movement for decades? It seems to be fashionable to be mathematically incompetent and proud of it.

@HenningMakholm Hah, my promotor addressed that in his inaugural lecture 'often, at parties, when one introduces himself as a mathematician you might get the reply 'I was never good at math' with a sense of pride.'.

(Just my rephrasing).

And now its computers.

He continues with: 'Say you are a writer. And I tell you I was never good at reading? Wouldn't that disqualify myself with you? As if the inability to think logically is something to be proud of.'.

Of course, that made my giggle my cherries off.

As the guy next to me made the remark that he was not good at it.

@JonasTeuwen Why do you care?

Care about what?

How others present them selves

Oh?

At what point did you dismiss the possibility in your reasoning that I actually do not care much about that?

And why?

Is it because I is black?

The chat box is mostly filled up with your comments

I fail to grasp the connection. Did you feel personally attacked by the statement?

No lol I was just currious

What?

I am sure they tell you that more often.

@JonasTeuwen That is offenisive I am sure.

How rare.

Did anyone else do the COMAP Mathematical Modelling Contest? We finished up our report this evening and sent it in (15 min before the deadline).

wow... I thought for sure that there would be someone on here who had done it...

Hi, can someone help me with a quick question? I forget what happens to the index when you take the derivative of a summation

@AlanH Typically nothing...

It depends on the sum, though...

Doesn't the index increase or something?

the lower index

Ummm... let me think for a sec--I think you're right.

@anorton Like the sum from i=0 to n of f(x)

becomes i=1 to n+1? or is it just i = 1 to n? I always forget

If the lower bound changes at all, it will only increment the lower index.

Why should anything happen to the index (unless the index variable is somehow related to variable the function is differentiated with)?

errr i made a mistake: sum from i = 0 to n x^i

@Orange yeah, i made a mistake

@AlanH Okay. Just differentiate the term for an arbitrary $i$. No need to remember complex rules like what happens to the index.

@OrangeHarvester $\sum_{i=0}^{n} i(x^{i-1})$

oh no latex support

so it doesn't change?

@AlanH There is $\LaTeX$ support-- look here: meta.math.stackexchange.com/questions/1088/…

@AlanH see "LaTeX support for chat" pinned on the panel to the right

@AlanH Yes, your expression is true only for $i\geq1$. See anorton's post before mine.

@AlanH $\frac1{(1-x)^2}$

Ah. Thanks Orange and anorton

@AlanH If you have trouble at that site, try this one

@AlanH That is the site that is referenced inside the other

@robjohn On a new browser, I directly go to your page and get links from there. For some reason, your home page seems easier to understand than the post.

@robjohn did you TA 115 a few years ago?

@OrangeHarvester There is more history and stuff in the meta page.

@AlanH Nope... where are you?

I was at LA

@AlanH I used to teach at UCLA, but that was a while ago.

@robjohn Ah, I thought you might have been my TA a few years back.

@AlanH what course was 115?

@robjohn linear algebra

@AlanH Not unless it was at Princeton about 25 years ago... I was a TA there.

@robjohn Oh, nevermind then. I had a TA named rob, then you sent a UCLA link, but nevermind.

That's got to be pretty cool. grad school at princeton

@AlanH It was. I frequently ate lunch with Andrew Wiles.

@robjohn WHAT!?!? That is really cool.

@robjohn !!!! AWESOME

@robjohn can you uh...write me a letter of rec? hahaha

(It took a while to write--that's why it was a bit delayed...)

@robjohn Are you teaching now?

anybody know perl here?

@anorton No. After teaching at UCLA, I went to Apple for 8 years, then back to UCLA to write some software for teaching Logic.

Gotta go walk the dog. BBL

See you later!

thanks @anorton see ya

@HenningMakholm are fusion systems the same thing as the concept of 'controlling G-fusion' (from transfer theory)?

@AlexanderGruber Fillionggonggong.

[i.e., I haven't a clue, why would I have one?]

@HenningMakholm Oh. I guess you must have been joking.

never mind then! thanks

@AlexanderGruber Here's what I know about fusion systems.

anybody here use oeis?

(like, contribute to it)

An exercise I've been given is to show that $\rm f(x),g(x)\in{\bf Q}[x]$, $\rm f(x)g(x)\in{\bf Z}[x]$ implies $\rm a_ib_j\in\bf Z$ for the coefficients $\rm a_i$ of $\rm f$ and $\rm b_j$ of $\rm g$, for every $\rm i,j$. Equivalently, $$\rm f(x)g(x)\in{\bf Z}[x]\implies f(x)g(y)\in{\bf Z}[x,y]\iff f(x)\frac{g(x)-g(y)}{x-y}\in{\bf Z}[x,y].$$ I was wondering if formal derivatives could be of help here. Any insight? (I am, of course, looking for a beautiful solution, rather than a dirty one.)

In unrelated thoughts, $\LaTeX$ should have built-in comic sans support for quoting dumb statements.

maybe that's too confrontational...

@anon Hi.

Hmm, perhaps using three formal variables could help.

yo

@anon Would this be helpful: Let $L/K$ be a field extension

can anyone help me with a division problem? Without enumerating all possibilities, is there a way to determine how many 2-digit numbers are divisible by X such that the first digit is not divisible by X, and neither is the second digit?

then $\alpha \in K$ is integral iff its minimal polynomial has coefficients in $\mathcal{O}_K$

@AlexanderGruber hi.

hi ben

@AlexanderGruber are you a graduate student alex?

I will be next year.

ah ok.

when is next year?

august-ish

ah ok. So you' re a senior now? Is that what americans call it?

@anon Do you know the usual proof to show that $g,f$ have coefficients in $\Bbb{Z}$?

nope

but I'm assuming it's dirty

No it's very nice.

Yeah, I'm a senior in undergrad.

@anon Let $m,n$ be respectively the smallest positive integers so that $mf(x)$ and $ng(x)$ have coefficients in $\Bbb{Z}$.

but I did an unrelated degree before this, I went back to school. So I'm not sure what that makes me.

@AlexanderGruber Right. I'm just about to get into third year in like a few weeks. The system in australia begins in february.

that sounds nice.

Third year grad?

@AlexanderGruber Why do you think I'm a graduate student ? :D

didja see this one over @ MO?

@anon Then the coefficients of $mf(x)$ have no common factor, similarly for $ng(x)$

@anon No not yet. If you want to know more about that sort of stuff there is a chapter in Procesi's book.\

@anon Ok from here if we can show that $mn = 1$

then $m = n = 1$ yes?

@anon Ok suppose that $mn > 1$. Then the fundamental theorem of arithmetic tells us that there is a prime $p$ that divides $mn$.

@BenjaLim because I usually have no idea what your questions are about. :P

the product of primitives is primitive, so mnf(x)g(x) would have to be primitive, so mn=1

@anon $mf(x)ng(x) = mnh(x)$

reducing mod $p$ the right hand side is zero.

But then $mf(x) ng(x) = 0$ in $\Bbb{Z}/p\Bbb{Z}[x]$ implies that one of them is zero

i.e. $p|mf(x)$ say, contradiction.

@anon Hence $mn = 1$.

@anon implying that $ m = n = 1$.

@AlexanderGruber I'm usually interested in comm. algebra/ algebraic number theory

@anon I can't probably give the MO guy an example.

@BenjaLim That would explain why we don't see each other much on main.

@robjohn hahahahahhahahahahahahahahahahahahahahahaha

@robjohn I don't usually have complex analysis/ real analysis as my favourite tags :D

The people I mainly see on main are Georges Elencwajg, QiL, YACP, Jacob Schlather, etc

:(

@BenjaLim I only see some of them when handling flags...

Right.

@JasonBourne Hence the high-rep.

@AlexanderGruber

@anon

@BenjaLim @BenjaLim @BenjaLim @BenjaLim @BenjaLim @BenjaLim

@anon How did you do that?

Ctrl+c, Ctrl+v

@anon @anon @anon @anon @anon @anon @anon @anon @anon @AlexanderGruber @AlexanderGruber

what have I done

can anyone help with my division question at all?

@JasonBourne mid-August, your rep really took off

@robjohn That was the time I started using the chat.

@JohnSmith where?

Without enumerating all possibilities, is there a way to determine how many 2-digit numbers are divisible by X such that the first digit is not divisible by X, and neither is the second digit?

@anon So is it ok now?

(or perhaps a recursive approach etc)

@BenjaLim is what okay?

@BenjaLim algebraic number theory looks fairly cool. i don't know what i think about commutative algebra.

@anon the proof of gauss' lemma.

i like it when things are nonabelian.

@AlexanderGruber right. All the things or most of the things I deal with are commutative :D

sure

@AlexanderGruber On the other hand my group theory is not very strong. I haven't really studied things like Sylow theory.

@JasonBourne What is CNY?

i know jacob schlather. he goes to my university.

@JohnSmith If $n$ is divisible by $x$ and one digit is divisible by $x$, so is the other.

sylow theory is awesome

@AlexanderGruber that guy is quite pro.

hmmm

@AlexanderGruber where did you learn it from?

He's a good drinker too.

@AlexanderGruber Come here and we will get absolutely maggot

is this a generalizable (sorry if my english makes no sense here) concept/

@JasonBourne Please erase the comment asking if i'm excited about some ethnic celebration.

I got started on sylow theory in algebra classes, then improved on it during a research program, then really learned it fully when i started doing independent research.

@AlexanderGruber You're pro doing independent research only in third year.

if i have one number, and want only one subnumber number to be divisible by X (but no other part)

@JasonBourne doesn't matter. I am not excited about it. The only part that is exciting is you get money.

@BenjaLim thanks. :) I wish it was in more than just group theory, though.

@JasonBourne youtube.com/watch?v=DoQkShfckF8

i was thinking about trying to pick up some topology or homology theory

@AlexanderGruber ahhhhh you haven't studied algebraic topology?

not officially.

@JohnSmith So you can count out all numbers whose first digit is divisible by $x$ then count the remainder which are divisible by $x$

ah that stuff is really cool.

topological groups look badass

@AlexanderGruber But I struggled a bit in last sem's algebraic topology.

@AlexanderGruber Let me tell you what is badass: comm. algebra!

lot's of diagrams all over the place

algebraic topology also has lots of diagrams everywhere

even more

would that only work for two-digit numbers correct?

I do love diagrams.

@JohnSmith yes

@AlexanderGruber Example of proof by diagram: math.stackexchange.com/questions/259239/…

What got you into commutative algebra? like, was there a particular problem or question you had, or a particular chapter in a book?

@AlexanderGruber I guess the people in my university were all algebraic geometers and stuff

@AlexanderGruber But I became really addicted to it

@JasonBourne No.

@AlexanderGruber If I have trouble with group theory I'll come and ask you :D

@AlexanderGruber You must have studied things like brauer groups and frobenius reciprocity ?

@BenjaLim Please do :)

Yeah, I know representation theory.

maybe my approach is wrong, i will think on this

@AlexanderGruber where did you learn abotu brauer groups from?

@BenjaLim Character Theory of Finite Groups, by Isaacs.

ah ok. woaah I heard that book is hard.

It's not too bad. he's a good writer.

@JohnSmith $\left\lceil\frac{9}{n}\right\rceil$ would be the number of first digits divisible by $n$

@robjohn would an approach like that extrapolate to arbitrary len number? if i have a number and i want only one subnumber divisible by X (and no other subnumber)?

@AlexanderGruber At the moment I think I am tending towards number theory/ AG

so $10-\left\lceil\frac{9}{n}\right\rceil$ would be the number of first digits not divisible by $n$

@JohnSmith no, it would not. This one even has a bit of counting yet to go

@JasonBourne Btw in buddhism one should not be attached to a particular culture.

@JasonBourne Because it causes en.wikipedia.org/wiki/Avidy%C4%81_%28Buddhism%29

for example 3651 only has one subnumber 36 divisible by X=4, nothing else is (not 3651, not 3, not 6, 5, 1, 36, 365, 65, 651, 51)

i am trying to make recursive rule

@AlexanderGruber When do you usually hang out in chat?

@JasonBourne Why do you hate me?

@BenjaLim Whenever I remember it exists. Usually evening.

@BenjaLim Is the practice of not being attached to a particular culture a culture in itself?

@BenjaLim have you ever read any of Mirela Ciperiani's papers?

@AlexanderGruber What time zone are you in?

@OrangeHarvester no.

EST, it's 10:30 pm here

right.

@AlexanderGruber who is mirela?

an algebraic number theorist / arithmetic algebraic geometer

No.

But I have heard of poeple like Matt Emerton, Qing Liu, Pete Clark, etc

which university are you at?

Ok bye guys! I'm off!

later ben

Later.

@JasonBourne My point actually was one who is truly not attached does not have any problems. But one who is attached, will see the tradition of non-attachment as an attachment itself. In which case, telling him to be not attached is futile.

@JasonBourne Its okay. I am not here to argue. :-)

I am not serious about such things. I just want to have fun. If it is not fun for you, it is not fun for me.

@JasonBourne Probably.

@JasonBourne Not yet. I am on Rings now. Will start Factorization soon.

@JasonBourne Ahh. Most of the book publishers have offices in your place. I thought, International versions will be available there too.

I am going for a bath now.

@JasonBourne okay. Kind of sad. I say, you should just download and print stuff yourself.

Get a laser printer with refillable toners. :-)

@JasonBourne cool.

Yes. Bye.