Mathematics - 2014-07-07 (page 1 of 3)

blue

00:23

I can't remember why if $n\ge\aleph_0$, the direct product of $n$-many copies of a field has dimension $>n$.

Ted Shifrin

I guess I didn't know that, @blue.

Mike Miller

it's harder to write down a proof than I'd imagined

Ted Shifrin

Oh, Hamel basis ...

Mike Miller

heh

Balarka Sen

00:47

@Mike I proved it, btw.

Why does it always happen with me. Whenever I come people freezes up. This is so not fair.

Alizter

Can anybody rearrange $x^{x^x}=y$?

Balarka Sen

@Alizter Rearrange?

Alizter

for x

Balarka Sen

Oh you mean expressing $x$ in terms of $y$?

Alizter

yes

Balarka Sen

00:57

You'll end up with a lot of Lambert logs I guess.

Alizter

I get as far to $\displaystyle\large\frac{\log y}{x^{x-1}}e^\frac{\log y}{x^{x-1}}=x\log y$ ready for Lambert but that x on the RHS kills me.

Balarka Sen

Perhaps it's not expressible in terms of Lambert logs.

Alizter

I wonder if it is?

c c

@PedroTamaroff with $$f(t,s)=\int_0^{\infty}\frac{x^s}{(1+e^{tx})^2}dx =\int_0^{\infty} -\frac{d}{dt}(\frac{x^{s-1}}{1+e^{tx}}) dx =\int_0^{\infty} -\frac{d}{dt}(\frac{u^{s-1}}{t^s(1+e^{u})}) du $$ $$= -\frac{d}{dt} (\frac 1 {t^s}\Gamma(s)\eta(s))=\frac s {t^{s+1}}\Gamma(s)\eta(s)$$, then we should differentiate in s to find $$\frac{\partial f}{\partial s}(1,0)=\int_0^{\infty}\frac{\log x}{(1+e^x)^2}dx$$

Balarka Sen

@Alizter I don't think it is. Look at the growth rate of those functions.

It's about the same as saying that $\exp(x)$ can be inverted using fractions.

If $y$ is a given constant, then things change though.

Alizter

01:02

Sure but $x^{x^{\textstyle.^{\textstyle.^{\textstyle.}}}}=y$ can be written in terms of $x$ and that has a crazy growth rate

Balarka Sen

@Alizter I am talking about the slowness of the growth.

In any case, i don't know and i am not sure.

you have to precisely define what you mean by 'writing' in terms of something.

i have to go.

c c

Balarka, thanks for the previous explanations

I mix 'sooner' and 'earlier' for some reason

r9m

01:45

@Pedro HBD =)

undergroundmonorail

02:04

I'm playing a game where you have to construct things with a straight-edge and compass. There are some levels where, once you beat it, you get a tool that lets you do that thing for convenience (e.g. an "angle bisector" tool). The level I'm stuck on asks for the incircle of a triangle. I have a solution that I believe should work, but it's not being accepted. Is there a bug in the game or is my understanding wrong? i.imgur.com/12OAIuW.gif

Aha! I solved it. I assumed that the circle would touch the outer line at the same point the angle bisector crossed it, which is not the case.

Pedro Tamaroff

02:23

@DanielFischer

r9m

02:51

Alrighty then !!! .. added a bounty on head ... wanted DOA

user1876508

03:38

Do the reals contain a minimal integral domain?

AndrewG

@Ted naw, not hiding, out eating at mexican restaurants :3

@Ted two lines in $\mathbb{P}^{2*}$ correspond to pencils in $\mathbb{P}^2$, and their point of intersection corresponds to the line common to both pencils, yeah? ...But what's it mean when the two lines in $\mathbb{P}^{2*}$ are parallel? Then their point of intersection is at infinity and...bwuh?

eXtremiity

03:57

What does $\mathbb{P}^{2*}$ represent?

AndrewG

the dual to the projective plane

eXtremiity

Oh.

Alizter

04:40

Damn. Another night without sleep. Oh well, Galois Theory was too interesting anyway.

Better get ready for school.

Mike Miller

@AndrewG The dual?

Kaj Hansen

05:21

Hmmm...take home test? math.stackexchange.com/questions/858638/…

AndrewG

05:44

@MikeMiller its dual space. The space of lines in $\mathbb{P}^2$. Every point corresponds to a line, a line corresponds to a pencil of lines, etc.

Mike Miller

Cool.

Sawarnik

09:23

@r9m lol...no one will answer you :P :P

@DanielFischer Hey.

Daniel Fischer

Hey @Sawarnik.

Sawarnik

@Daniel Is this answer correct? math.stackexchange.com/questions/858717/…

Daniel Fischer

@Sawarnik No, as user1337 pointed out, $\xi$ depends on $x$. However, since $x^2 \geqslant 0$, there is a $c\in (-1,1)$ such that $$\int_{-1}^1 x^2g''(\xi(x))\,dx = g''(c)\int_{-1}^1 x^2\,dx,$$ since derivatives have the intermediate value property.

blue

09:57

let A be an infinite set, F a field, W the vector space of functions A->F, U the subspace spanned by characteristic functions of subsets of A, and V the subspace of functions with finite support. does dim W = dim U? is dim U > dim V?

blue

10:42

drats, the kernel of $\bigoplus_{2^A}k\to\prod_Ak$ has dimension $2^A$

Balarka Sen

Heya @blue

blue

heya

Balarka Sen

Fun things happen if one considers Fix_Gal(L/K)(A). For example, you are really looking at Gal(L/K) acting on A by automorphisms so as the very first thing you get is a representation Gal(L/K) --> Aut(A) and if you let A to be some direct product of copies of Z_p then you get a representation Gal(L/K) --> GL_m(F_n)

@Karl

Plouffe's copy of Artin

Karl Kronenfeld

11:10

@blue I think if $|F|$ is at a certain sweet spot (near $|A|$) then you get the cardinality of $W$ to exceed the cardinality of $U$, let alone dimensions.

Ted Shifrin

13:05

@AndrewG Andrew, first of all, parallel makes no sense in the projective plane. What you are thinking is parallel in $\Bbb R^2\subset\Bbb P^2$, which, of course, means that the lines intersect in a point on the line at infinity. In the case of such lines in the dual plane, say $\xi_1=0$ and $\xi_1=\xi_0$, these would correspond to the pencils of lines through $[0,1,0]$ and $[1,-1,0]$, respectively. And the intersection point $[0,0,1]\in\Bbb P^{2*}$ corresponds to the line $x_2=0$ joining those.

Balarka Sen

Hello professor @TedShifrin

I am trying to find a way to prove that $\log z \not \in \overline{\Bbb C(z)}$ in a geometric way. Probably one way is to note that the Riemann surface $w = \log(z)$ consists of a bunch of (countably infinitely many) sheets over $\Bbb C$ and is being cut through the branch from $z = 0$ towards infinity and pasted upwards so that a point near $z = 0$ which tries to traverse a clockwise loop winds continuously upwards in clockwise sense.

This implies the monodromy must be an infinite cyclic, i.e., $\Bbb Z$. Perhaps one can draw the conclusion by noting that $\Bbb Z$ has no composition series? It's still solvable in the classical sense though.

Mats Granvik

13:54

What does the 2*Pi*I/Log(-p-1) periodicity mean in this wolfram alpha calculation?
http://www.wolframalpha.com/input/?i=-p*%28-%28p+%2B+1%29%29%5E%28n+-+2%29

broken link

wolframalpha.com/input/?i=-p*%28-%28p+%2B+1%29%29%5E%28n+-+2%29

broken link again

Ted Shifrin

@Balarka: Doesn't it suffice to say that any rational function has a finite degree?

Mats Granvik

this is the expression I entered in wolfram alpha: -p*(-(p + 1))^(n - 2)

Ted Shifrin

@Mats: I can't figure out what you're talking about.

What are $p$ and $n$?

Mats Granvik

@TedShifrin -p*(-(p + 1))^(n - 2) has periodicity $\frac{2 \pi i}{\log(-p-1)}$

Ted Shifrin

This comes from the complex exponential. What are $p$ and $n$?

Mats Granvik

13:59

p and n are just variables

Ted Shifrin

Not integers or positive ...

Mats Granvik

this comes from a transform I have invented

n is integer, I forgot

Ted Shifrin

So you need to make Mathematica know that or else it's going to use the complex exponential. If you have an integer for an exponent, then there's no multivaluedness here.

Mats Granvik

So the periodicity does not apply in my case?

But continuity is a chimera, as someone said.

Clear[t, n, k, i, p]
p = 2
coeff = {p^0, -p^1, p^1, p^2, p^3, p^4, p^5, p^6, p^7, p^8, p^9, p^10,
p^11, p^12, p^13, p^14, p^15, p^16};
nn = Length[coeff]
cc = Range[nn]*0 + 1;
Monitor[Do[Clear[t];
t[n_, 1] := t[n, 1] = cc[[n]];
t[n_, k_] :=
t[n, k] =
If[n >= k,
Sum[t[n - i, k - 1], {i, 1, k - 1}] -
Sum[t[n - i, k], {i, 1, k - 1}], 0];
A4 = Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}];
A5 = A4[[1 ;; nn - 1]];
A5 = Prepend[A5, ConstantArray[0, nn]];
cc = Total[
Table[coeff[[n]]*MatrixPower[A5, n - 1][[All, 1]], {n, 1,

"p" is the input
"cc" is the output

Bob

@TedShifrin can you suggest a book for probability

Ted Shifrin

14:23

@Bob: There are a bunch available on-line that seem pretty good. The standard undergraduate texts seem to be Pitman and Ross. I'm learning/teaching the stuff for the first time this fall, using Ross.

@Mats: You use the word periodicity, but that's really wrong. $\log z$ is a multivalued function. It's not a matter of period. But when you have an integer exponent $n$, $z^n$ is unambiguously defined.

Bob

@TedShifrin can you provide me some link to download the pdf copy

Ted Shifrin

No, but you can google Probability + any of Ash, Grinstead/Snell, Bertsekas and Tsitsiklis.

Mats Granvik

15:01

@TedShifrin If I can transform $5 x+\frac{1+7 x}{1+6 x}$ into $\frac{1+x}{1-5 x}$, is it progress?

More generally the transform, transforms:
$$-\frac{1+p (-2+x) x+p^2 x^2}{-1+p x}$$ into $$\frac{1+x}{1+x+p x}$$

Balarka Sen

15:39

@TedShifrin I am talking about $\overline{\Bbb C(z)}$ not $\Bbb C(z)$.

Balarka Sen

15:49

Well, perhaps having infinite number of galois conjugates (equivalently, sheets) is just enough. I haven't seriously thought about it.

If it is true (which it obviously is, I just don't know how to prove it), then we get transcendence of infinitely many (and Zariski dense!) elts $\log(\zeta)$, $\zeta \in \Bbb C$ for free, by Hilbert's theorem.

nullgeppetto

17:26

Hi MathSE people! May you help my soul, please?

Ted Shifrin

18:07

@DanielF: Question for you that's arising in a colleague's work, and I'm feeling stoopid. What's an example of a harmonic function on the unit disk, continuous up to the boundary, whose harmonic conjugate is not continuous up to the boundary? What conditions will rule this out?

Daniel Fischer

Oy, @Ted, I'm not sure that is possible. Gotta look closer at the Poisson integral whether I can see a way.

Ted Shifrin

I somehow thought it would be a well-known regularity result if it were true in general. I found a comment on SE earlier that suggested there are examples. I'll see if I can find that.

@DanielF: Here's the comment.

nullgeppetto

18:31

Ok, I know I'm one step before I should get banned as a spammer, but I really need an answer quickly! mathoverflow.net/questions/173524/…

Balarka Sen

@nullgeppetto MO is for research questions, AFAIK.

nullgeppetto

@BalarkaSen, yes you're right, but I was desparate...

Balarka Sen

So that question would probably be locked before getting answers.

nullgeppetto

I give a link to my first MathSE question, here...

What do you suggest to me to do?

Balarka Sen

@nullgeppetto Have patience.

nullgeppetto

18:34

@BalarkaSen :)

Please tell the same to my supervisor!

Daniel Fischer

@TedShifrin Ah, it is possible. As long as you don't want an explicit example, take a biholomorphic mapping of the unit disk to $\{ x+iy : x < 0, \lvert y\rvert < e^x\}$. The imaginary part extends continuously to the boundary, the real part doesn't.

nullgeppetto

@BalarkaSen, it's kinda irrelevant, but I feel like my (probably stupid) questions get lost after some minutes, as newer ones appear...

Balarka Sen

@nullgeppetto Yes, new questions appear in front of all the questions when you open MSE up.

nullgeppetto

@BalarkaSen, :), yeah, I know.. So, that's why I act like that, which is probably wrong, I know...

Ted Shifrin

Interesting @DanielF. I was just down in the library looking at PDE books and didn't find any results ...

Balarka Sen

18:41

@DanielF Terrific complex analysis by RV

Ted Shifrin

I'm not understanding your example yet, @DanielF.

Modulo knowing what the Riemann mapping looks like, @DanielF, I guess it's plausible, but I don't have a nailed-down proof.

Daniel Fischer

@TedShifrin You need a holomorphic map on the unit disk such that the imaginary part extends continuously to the boundary, but the real part doesn't (or vice versa). A biholomorphic mapping between the disk and some other domain extends continuously to the nice enough boundary points. So I took a domain with piecewise smooth boundary such that it is unbounded in one direction, but the imaginary part has a limit as the real part tends to $-\infty$.

Ted Shifrin

Right. I get that :) I see that the real part can't extend continuously. But is it clear that the mapping is sufficiently well-behaved that the imaginary part must?

Daniel Fischer

Map it to a curvilinear triangle with one angle zero in the plane by a Möbius transformation. I'm almost sure the Riemann mapping extends to a homeomorphism of the closures then. Gotta look up Painlevé-Warschawski to be sure, though.

Ted Shifrin

OK, I get it. We're looking at an annulus, of course, with piecewise smooth boundary. I wonder if there is any bounded counterexample like yours. (Actually, what we're discussing is a Dirichlet problem on an annulus with $u=0$ on one boundary and $u=1$ on the other. Then the question is whether the harmonic conjugate of $u$ (modulo period) is continuous up to the boundary.)

Balarka Sen

18:57

@PedroTamaroff

Daniel Fischer

Ah, well, @TedShifrin, not Painlevé-Warschawski, but Carathéodory.

Ted Shifrin

Yes, that name has come up several times today.

Daniel Fischer

@TedShifrin The solution for the Dirichlet problem on an annulus with these boundary values is $c\log \lvert z-a\rvert - d$, where $a$ is the centre of the annulus, the conjugate is basically $\arg$, which has no problems with the boundary circles, but has a problem on some radius (or other curve connecting the boundary circles).

Ted Shifrin

19:17

Right, that's why I said mod periods. Of course, our annulus isn't a round one :) Thanks for thinking !

Daniel Fischer

@TedShifrin I'd think that if you cut it open along a "radius" and map it conformally to $[0,1]\times [0,2\pi]$, then the solution of your Dirichlet problem is the real part of that biholomorphism, and the map extends continuously to the boundary.

> Although it is known that the set of irrational numbers is greater than the set of rational numbers, is there any usefulness/applications of this fact outside of mathematics?

People have the strangest questions.

robjohn

@DanielFischer Don't invite irrational people to your party. They will eat you out of house and home. Invite nice rational people; they are much more, uh, countable.

Ted Shifrin

Well, @DanielF, I suppose constructivists are bothered by such sorts of philosophical questions :) hi @robjohn

@robjohn: They're still damned expensive.

robjohn

@TedShifrin hey there...

Daniel Fischer

Constructivists, deconstructivists, I'm not a fan of French postmodernism.

Ted Shifrin

19:29

Let's face it, @DanielF. You're just not a fan of French.

Speaking of which, what's become of our French lads?

robjohn

@TedShifrin but far less so than the others.

Ted Shifrin

Maybe you have cheaper friends than I, @robjohn :D

Daniel Fischer

@TedShifrin Oh, I am. I like good food. And occasionally a good wine.

Ted Shifrin

These days you just settle for a good whine, just like @Mike :D

user18481

19:47

Hi, why Frenchman are famous for analysis in Math?

Mike Miller

Huh @Ted?

@user18481 Presumably because they have done very good work in analysis.

@DanielF I think the best postmodernists were American or English.

Balarka Sen

Gah

@DanielFischer You beat me to it.

Daniel Fischer

@MikeMiller "best postmodernists" has a definite smell of oxymoron.

Mike Miller

@DanielF Perhaps you and I have different taste.

Daniel Fischer

Sure.

Chris's sis

19:54

Greetings

Balarka Sen

@Chris'ssis Do you mean you are greeting me or that it is greetings to everyone or that you greet regardless of whether I take it or not or that you are greeting simply to be greeted back?

=P

Chris's sis

@BalarkaSen I didn't say "Greetings" specifically to a certain person. So ... :-)

Shisui

Yoroshiku minna!

Balarka Sen

@Chris'ssis I was joking. A similar joke is in the beginning of The Hobbit. You might not have read it though.

Chris's sis

@BalarkaSen I know I know :-)

Shisui

19:58

I found this interesting problem on Facebook. Don't ask why I was there. I don't even know. $$ \displaystyle \lim_{n \to \infty} \dfrac{\log(n!)}{\log(n^n)} $$

Chris's sis

@BalarkaSen by the way, when will the last part of the Hobbit appear?

Balarka Sen

@Chris'ssis I have only read the books.

Daniel Fischer

@Shisui Do you know Stirling's formula?

Balarka Sen

@Shisui Stirling.

Gah, @DanielF beat me to it. Again.

Shisui

@DanielFischer It sounds familiar ...

Chris's sis

19:59

@BalarkaSen Ahhh ... (I was thinking you watched the movies)

@Shisui Cesaro-Stolz

Balarka Sen

@Chris'ssis The hard way.

Chris's sis

@BalarkaSen lol, is that hard?

Balarka Sen

$\log(n!) = n\log(n) + O(\log(n))$ and be done.

Chris's sis

@BalarkaSen Maybe it is hard for you. Using this, any kid in high school can finish it.

Balarka Sen

@Chris'ssis Yes, it is hard for me.

I was not speaking in general.

@Chris'ssis Well, respective to Stirling...

Ah, well. Both are elementary -- one is a bit too tedious to me. I won't go in mathematical philosophy again.

Chris's sis

20:05

@BalarkaSen I'd probably use that too, but for a beginner I'd recommend Cesaro-Stolz.

Balarka Sen

@Chris'ssis I know. I just don't do soft analysis. Asymptotic approximations are my thing.

Chris's sis

(in this case, Cesaro-Stolz theorem combined with the mean value theorem)

@BalarkaSen Sure, that's OK.

Balarka Sen

Heya @TedShifrin

Ted Shifrin

hi @Balarka

@Balarka: Your Stirling is in error

Balarka Sen

@TedShifrin Heh. error

$\log(n!) = n\log(n) - n + O(\log(n))$

Ted Shifrin

20:09

Correctamundo.

Balarka Sen

What I always like to think is that prime number theorem is really striling with primorials.

Chris's sis

Let me propose a new beautiful question ...

Pedro Tamaroff

@BalarkaSen Hello.

You called.

Chris's sis

$$\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{k+n} \frac{\log(k+n)}{(k+n)^2} $$

Balarka Sen

@PedroTamaroff Yes. I solved your problem.

Pedro Tamaroff

20:21

The one about solvability being preserved under extension?

Balarka Sen

Yes.

Suppose G is extension of solvable group N by solvable group K.

Pedro Tamaroff

Ok.

It suffices you show that if $G$ has a normal subgroup $K$ such that $K$ and $G/K$ are solvable, so is $G$. I take it you did this.

Balarka Sen

Then consider the canonical map G --> K. elts of G^(n) are products of commutators of pairs of elements of G^(n-1)

@PedroTamaroff Yes.

Pedro Tamaroff

Ok.

Balarka Sen

Now note the homomorphisms take commutators to commutators.

Pedro Tamaroff

20:24

Yes.

Balarka Sen

So the image of G^(n) in K is a subgroup of K^(n)

Since K^(k) = 1 for some k, G^(k) = 1

And you have solvability by definition.

This seems like an important result, yet though somewhat easy. I can now prove, for example, that semidirect product of solvable groups is solvable -- which is a fun fact!

Pedro Tamaroff

@BalarkaSen This looks fishy.

Balarka Sen

@PedroTamaroff Hmm. Which step?

Ted Shifrin

mr @Pedro: So, glad to see you've survived your horrid aging process :)

Pedro Tamaroff

@BalarkaSen Well, no, it's wrong.

Balarka Sen

20:29

@PedroTamaroff Explain where I have messed up.

Pedro Tamaroff

That would show that if $G$ is any group, and $H$ a solvable group, and $\eta:G\to H$ and epimorhism, $G$ is solvable. But we always have an epimorphism $G\to 1$; the trivial group.

You know that $\pi(G^{(i)})=1$ for some $i$.

That doesn't mean $G^{(i)}=1$.

@TedShifrin Hehe. I got my combinatorics tests back. I passed, as I thought, but barely. Prof. said I should pass my make up exam "properly" i.e. >50%.

You need to use $K$ and $G/K$ are both solvable.

Ted Shifrin

Maybe you should stick to analysis, @Pedro :D

I'm terrible at discrete math ... that's why teaching probability is skeering me.

Pedro Tamaroff

@TedShifrin I really want to be able to read Stanley's "Combinatorics and Commutative algebra." =D

@TedShifrin >:(

Ted Shifrin

Oh, Stanley did all sorts of beautiful interplay between combinatorics and algebraic geometry (combinatorial hard Lefschetz theorem)

You'll be fine, @Pedro.

Balarka Sen

@PedroTamaroff Hmm.

Pedro Tamaroff

20:32

@TedShifrin I don't know what the Lefschetz theorem is, but I'm really interested in combinatorics.

Even though I'm a bit slow at it. =)

Ted Shifrin

It's about the cohomology (intersection theory) structure of algebraic varieties, @Pedro.

Pedro Tamaroff

@BalarkaSen Do you see it?

@TedShifrin OK. Dunno about it. At any rate, yet again I missed a silly problem.

John Doe

HI everyone

Ted Shifrin

Did I ever tell you about my test grades when I took abstract algebra my second year in college, @Pedro? They went 100, 77, 58.

That was first semester.

Balarka Sen

@PedroTamaroff I am seeing it.

Pedro Tamaroff

20:34

@TedShifrin Oh. Did something happen to you there...?

Ted Shifrin

Yeah, I sucked at algebra, @Pedro. But I survived and learned it eventually.

John Doe

Drugs? @TedShifrin

oh

Pedro Tamaroff

@BalarkaSen For the proof, suppose you have solvable series for $K$ and $G/K$. Then show you can patch them up to get a solvable series of $G$.

It is not hard.

Ted Shifrin

Nope. I didn't do too well with group actions and Sylow Theorems at first. And I messed up a linear algebra proof on that third test, too ... :(

Balarka Sen

@PedroTamaroff I am trying to stay out of solvable series

Pedro Tamaroff

20:34

And indeed solvable series (instead of derived subgroups) work better here.

@BalarkaSen Don't. The more tools you have, the better!

Balarka Sen

@PedroTamaroff Exactly the same thing my professor says.

John Doe

What uni were you at for undergrad? @TedShifrin

Ted Shifrin

MIT, @JohnDoe. Many centuries ago.

Balarka Sen

@TedShifrin lived for almost 300 years, @JohnDoe

Pedro Tamaroff

@TedShifrin Well, hehe. In this test, first I showed that if $P$ is a poset with $n$ elements, then it has either a chain of $\sqrt n$ elements, or an antichain of $\sqrt n$ elements. I proved that nicely. Then second item was showing that if we have $50$ distinct sets one could pick $8$ such that in no set is the union of another two in such collection. I didn't realize $\sqrt 50>7$. =/

Balarka Sen

20:37

61 is only his 'official' age, to cover up the fact that he is one of the Wise that live in Orthnac.

Ted Shifrin

The second sounds like pigeonhole ...

Pedro Tamaroff

Well, yes.

Balarka Sen

It is pigeonhole.

John Doe

How many of those years has he been studying maths?

Pedro Tamaroff

Using pigeonhole, one can show that in any graph on $n$ vertices $\chi(G)\alpha(G)\geqslant n$, @TedShifrin.

Ted Shifrin

20:38

It's the deep math that always trips us up, @Pedro. Like $50>49$.

Pedro Tamaroff

Then take the comparability graph of the poset $P$, and profit.

@TedShifrin Exactly.

John Doe

Do you know if Indiana state university has a good maths department? @TedShifrin

Balarka Sen

@TedShifrin Partial orders are freaking hard.

John Doe

Where did you do your postgrad studies? @TedShifrin

Ted Shifrin

Indiana State? As opposed to Indiana Univ Bloomington?

I know no one at Indiana State. That proves nothing, but it is not well known.

Berkeley, @JohnDoe.

I only think about them if they arise in some interesting context, @Balarka. I'm not a very set-theory oriented person.

Balarka Sen

20:41

@TedShifrin I was talking about $50 > 49$

Ted Shifrin

LOL, oh ... Nah, that's a linear order, not a partial one.

Balarka Sen

I always have trouble with pronunciation. People laugh at me when I say "Cayley's theorem"

Pedro Tamaroff

KAI LEES.

Balarka Sen

Kelly, yeah. I was told that by someone.

@PedroTamaroff Stop pronouncing Japanese.

John Doe

Where is everyone else studying @PedroTamaroff and @BalarkaSen ?

Ted Shifrin

20:43

Kai is ambiguous, @Pedro.

Pedro Tamaroff

@TedShifrin "K"?

Balarka Sen

@JohnDoe I am studying nowhere.

Ted Shifrin

no, K like the letter ...

Pedro Tamaroff

@JohnDoe I'm in Argentina, as my profile says.

John Doe

Room is basement is an acceptable answer...streets also @BalarkaSen

Balarka Sen

20:44

When I said "quintic" people positively laughed at me.

John Doe

@PedroTamaroff What uni?

Ted Shifrin

Better to be laughed at positively than negatively!

Balarka Sen

I pronounced as KOAINTIC.

John Doe

Room in basement that is

Balarka Sen

@TedShifrin Heh.

Ted Shifrin

20:45

kwin = koain?

Pedro Tamaroff

@JohnDoe UBA.

Balarka Sen

@TedShifrin No.

c c

I'd have said "queentik", but maybe "keentik" is good

John Doe

@PedroTamaroff You following the world cup?

Balarka Sen

@cc that's the correct spelling.

queentik.

Pedro Tamaroff

20:46

@JohnDoe Slightly.

Balarka Sen

@PedroTamaroff Messi is too messy.

c c

Angel is too angelic

Balarka Sen

Angle is not an angel, for finding sines can be a tangle.

Ted Shifrin

rolls all six eyes

Balarka Sen

@TedShifrin Great spiders of Mirkwood!

John Doe

20:49

My grandmother asked me who I was talking about when she read my grade 4 book it said 'a cute angel'...true story...I just shouted 'fucking context'

Balarka Sen

Are you serious, @Alexander?

Pedro Tamaroff

@BalarkaSen It's a cool problem

Balarka Sen

@PedroTamaroff Well, I can't disagree.

Transcript for

$Mathematics$ Mathematics