Mathematics - 2012-03-22 (page 1 of 4)

t.b.

12:02 AM

There we go! :) It happens here from time to time that people say they apply the l'H rule when they're simply taking the derivative. It's a bit like applying Hahn-Banach to a Hilbert space...

Skullpatrol

Is it true l'Hospital didn't even discover the rule?

@PaulSlevin Hi

He sponsored it.

hello

Hi

hello, first time here!

Paul Slevin

12:04 AM

does anyone here know what a set-theory tree is

Skullpatrol

@XabierDomínguez Welcome.

Brian M. Scott

@PaulSlevin I know the usual meaning.

Paul Slevin

im a little confused about the definition of height

t.b.

@Skullpatrol There are different versions of the story. This and this.

Brian M. Scott

@PaulSlevin There are slightly different versions of tree; what’s the definition that you’re using?

Skullpatrol

12:07 AM

@tb Thanks!

Paul Slevin

it says here that its the least $\alpha$ such that the $\alpha$ level of tree is empty... but apparently its the same as saying the least ordinal bigger than the height of any element...

my definition is a poset T where every set of predecessors is well ordered by the order

Tree (set theory)

In set theory, a tree is a partially ordered set (poset) (T, 1 has an antichain of cardinality ω1 or a branch of length ω1. Tree (automata theory) Following definition of the tree is slightly different formalism of the tree compare to above introduction. For example, each node of the tree is a word over set of natural numbers(\mathbb{N}), which helps this definition to be used in automata theory. A tree is a set T \subseteq \mathbb{N}^{*} such that if t.c \in T, t \in \mathbb{N}^{*} and c \in \mathbb{N} then, t \in T and for all 0 \leq c' . The element of T are known as nodes...

Skullpatrol

@tb so it seems likely that Bernoulli discovered the rule

Brian M. Scott

Okay; that’s probably the most general definition. Hang on a minute; I’m going to have to reload the page to get MathJax to work.

Paul Slevin

I just dont understand how this can be. surely if we have the 0th level of the tree being empty, then the height of the tree shoul dbe 0. but that characterisation says that the height of the treee is then at least 1...

Brian M. Scott

Level $0$ isn’t empty (unless the whole tree is): it contains the root of the tree.

Paul Slevin

12:11 AM

i have $o(x) = $ order type of $\{y \mid y < x \}$ and the $\alpha$th level of the tree is $\{ x \mid o(x) = \alpha\}$

Brian M. Scott

That’s correct. And if $x$ is the root of the tree, then $\{y:y<x\}=\varnothing$ and therefore has order type $0$.

Paul Slevin

but if the tree has a minimal element, then the 0th level is empty

so how can the whole tree be empty

Brian M. Scott

No, the $0$-th level contains the minimal element and nothing else.

Paul Slevin

how can it? the 0th level means that the order type of everything less than the minimal element is 0, which is true, cos there's nothing less than the minimal element

Brian M. Scott

In the Wikipedia illustration that you inserted above, Level $0$ of the tree on the left is the set $\{\epsilon\}$. Level $1$ is $\{0,1\}$. Level $2$ is $\{00,01,10,11\}$.

Paul Slevin

12:14 AM

surely the 1th level contains the minimal element

Brian M. Scott

@PaulSlevin Exactly: Level $0$ contains the elements with no predecessors.

Paul Slevin

ahhhhhhhhhhhhhhh

Brian M. Scott

@PaulSlevin Absolutely not: the set of predecessors of the minimal element does not have order type $1$, because it’s empty.

Paul Slevin

ok that makes sense. being stupid.

everytime I learn a new definition lots of alarms go off in my head, i have to go and turn them all off

thanks :)

Brian M. Scott

No problem!

Paul Slevin

12:16 AM

however

I am still unsure of the characterisation

Brian M. Scott

Of the height of the tree as a whole?

Paul Slevin

why the heigh of the tree = the smallest $\alpha$ biger than the heigh of every element

Brian M. Scott

Because that’s exactly the same ordinal as the supremum of the heights of the elements. Oops: I meant of the heights $+1$.

Split it into two cases.

Case 1: There are elements of maximal height, say $\alpha$.

Case 2: There is no element of maximum height.

Paul Slevin

so im trying to show height (T) = $\sup \{ o(x) +1 \mid x \in T \}$

Ok ill give that some thought

Brian M. Scott

In Case 1, $\sup\{\operatorname{ht}(x)+1:x\in T\}=\alpha+1$.

In Case 2, $\sup\{\operatorname{ht}(x)+1:x\in T\}=\sup\{\operatorname{ht}(x):x\in T\}$.

Paul Slevin

12:21 AM

ok i see that

is it possible that you never get an empty level ?

Brian M. Scott

Not if the tree is a set.

You’d have to have Level $\alpha$ non-empty for every ordinal $\alpha$, so you’d have a proper class.

Paul Slevin

of course

Thanks so much. I will ponder this more tomorrow

have a good night

Same to you!

Good night, Paul

Bye @PaulSlevin

12:24 AM

@PaulSlevin Night Paul.

@tb I don't see how to show that a sufficient condition is that $g$ is injective. In fact, I'm not even sure about it.

And I'm getting sleepier by the minute.

t.b.

A sufficient condition for what?

Skullpatrol

I think Newton's cradle has me hypnotized ...

Matt N.

For $T$ to be surjective. Necessarily, $g$ has to be injective. So I thought maybe that's sufficient (seeing as it doesn't have to be surjective)

It's probably "obvious".

Xabier Domínguez

i'm leaving, it is awfully late in Spain, even for our standards... good night to everybody, will be back soon

Matt N.

@XabierDomínguez Night Xabier.

@AsafKaragila Sounds like a plan.

t.b.

12:30 AM

No I didn't. If $g$ is injective then it is a homeomorphism onto its image. Now use Tietze.

Skullpatrol

@tb May I post one more GIF please?

Matt N.

Thanks for the hint.

t.b.

@BrianMScott I hope you don't mind me asking you this: Can you recognize this result? It is rather intriguing and I wanted to see what this statement in enormously greater generality is but unfortunately Fremlin doesn't give any hint.

Brian M. Scott

Let me stare at that for a few minutes.

t.b.

The proof is not that hard, but it is a bit tricky.

(If it helps I could also post the proof)

Here's the proof given by Fremlin.

Jeff

12:47 AM

This chat is distracting to me. I gotta go. Bye all.

t.b.

Bye Jeff

Matt N.

I think I should go too. I can't see how to use Tietze to get what I want.

Brian M. Scott

It looks like a kind of $\Delta$-system result, but it doesn’t look familiar. I’ll take a look at the proof now.

Matt N.

And I'm too tired to think straight.

t.b.

@MattN I don't remember whether $g: X \to Y$ or the other way around. In any event given a function $f$ on $X$ you want to find a function $F$ on $Y$ such that $F \circ g = f$. Since $g$ is a homeomorphism onto its image (which is closed by compactness) you can put $\tilde{F} = f \circ g^{-1}$ on $g(X)$. This is a continuous function on a closed subset, now get $F$ by Tietze.

Matt N.

12:53 AM

Too tired. Sorry : /

t.b.

@BrianMScott Thank you!

Skullpatrol

@MattN Don't be sorry just get some sleep.

t.b.

@MattN No problem :) Then go to bed...

Matt N.

I hope no one answers this question in the meantime.

t.b.

I certainly won't

Brian M. Scott

12:57 AM

No, I don’t recognize the result, though it feels as if I ought to.

Skullpatrol

@MattN Is it worth the torture of trying to stay awake?

Matt N.

@Skullpatrol Yes : )

t.b.

@BrianMScott Thanks for trying! It definitely is not a nice treat of him to say that it can be found in many textbooks but hard to recognize...

Skullpatrol

@MattN Wow, you must value your work in math a lot.

Brian M. Scott

@tb I noticed that! I’m wondering if I might have seen it and simply failed to recognize it in this guise.

t.b.

1:02 AM

It strikes me as somewhat bizarre and he uses it to construct awful measure spaces. I was hoping to get some more insight by locating the disguised other versions...

Matt N.

Not a good troll.

Skullpatrol

@tb " it can be found in many textbooks but hard to recognize..." sounds like "I have a secret and you have to try and find out what it is."

t.b.

@MattN Neither did the English speaking Wikipedians, it seems, but at least they found a photo.

Matt N.

@tb : )

I think I'm half way on my way to a sleeping disorder.

Going to bed now. Good night!

Brian M. Scott

@MattN That’s only a problem if the disorder wakes up!

t.b.

1:05 AM

Good night!

Skullpatrol

Sleep well

Brian M. Scott

G’night!

Matt N.

@BrianMScott : )

Don't stay up all night.

t.b.

I don't know. Won't make promises.

Matt N.

I wonder why I keep saying that... (exactly : ))

Brian M. Scott

1:06 AM

I’m probably going to go have a lie-down pretty soon, actually, though I doubt that I’ll sleep soon.

t.b.

Then the cold must be really bad (it's quite early)... I hope you get better very soon! Have some good tea!

Brian M. Scott

@tb I may end up doing just that, though at the moment I’m concentrating on water.

t.b.

Can't hurt either!

Tim

@BrianMScott Hi Brian, take care!

Brian M. Scott

@t.b.: The more I look at (b) of that theorem, the more it appears to me to be designed to work with the $G_\delta$ topology on $\{0,1\}^\kappa$ for large-ish $\kappa$.

@Tim Thanks!

t.b.

1:12 AM

That might be to the point. The $G_\delta$-topology is what you used here, right? Let me look at how he applies it in his book.

Eric Gregor

hey @tb, i put up an answer to my question: math.stackexchange.com/questions/122642/…

you'd asked me to show it to you when i thought i had it

Brian M. Scott

@tb That was a very special case of a $G_\delta$-topology; in my comment here I was just thinking of restricting countably many factors in $\{0,1\}^\kappa$ to get basic open sets.

t.b.

@BrianMScott I'm impressed... That sounds as if it is exactly what is happening there.

Tim

@tb Hi t.b., by "his book", are you talking about "Atomic and nonatomic measures "?

Brian M. Scott

@tb That’s my kind of topology! :-)

t.b.

1:21 AM

@BrianMScott Here's one example where the Theorem is used.

Brian M. Scott

I’ll look in a sec; I’m putting the finishing touches on an answer.

t.b.

@Tim No. Fremlin's measure theory.

Hi, by the way.

Tim

@tb Okay, I have that ebook.

t.b.

@BrianMScott No hurry! I'll be here for quite a bit more, I believe.

Brian M. Scott

I had that impression from Matt.

Tim

1:23 AM

@BrianMScott I just remember that drinking water boiled with ginger makes me feel comfortable when I am not feeling well. Maybe you already know.

t.b.

Yes. Not sleeping too much these days. That some electrician guy seems to tear down one wall after the other in the morning (when I usually sleep) for about a week now doesn't really help.

Eric Gregor

@tb, did you have a chance to glance at my answer? it's quite short

t.b.

@EricGregor I've read and voted on it (and azarel's answer). Yes, that's it.

Eric Gregor

great, thank you!

t.b.

Didn't you have a question on manifolds earlier today?

What was it, you never actually stated it, unless I've overlooked it.

Eric Gregor

1:28 AM

oh, yes i did

@tb, this question poses a theorem at the start, and i don't see if it's obvious

3

Q: Tautological vector bundle over $G_1(\mathbb{R^2})$ isomorphic to the Möbius bundle

Let $V$ be a finite dimensional vector space, and let $G_k(V)$ be the Grassmannian of $k$-dimensional subspaces of $V$. Let $T$ be the disjoint union of all these $k$-dimensional subspaces and let $\pi:T\rightarrow G_k(V)$ be the natural map sending each point $x \in S$ to $S$. Then $T$ ...

t.b.

If you don't see it it isn't obvious for you, I guess :)

Eric Gregor

i don't see how to show that T has a unique smooth structure...

t.b.

Let me take a look.

Brian M. Scott

@tb No, it wouldn’t! And I suppose that if one can’t sleep, one might as well do something productive, though I occasionally descend to mere vegetation.

t.b.

@BrianMScott In my experience, the worst state that you can get to under these circumstances is that you are in the state of mere vegetation and hallucinate that you're actually productive :/

@EricGregor The important part is what comes afterwards: "... with $\pi$ is a bundle projection and with the vector space structure..."

Eric Gregor

1:35 AM

@tb yes, i know. i wrote "..." to signal that without writing it. sorry

Brian M. Scott

@tb I’ve done that a time or two. Usually, though, I just vaguely remember having had what seemed a good idea at the time.

t.b.

Let's hope this happens to me this time... :)

@EricGregor So, requiring that $\pi$ is a bundle projection means exactly what? For each point downstairs you have a small neighborhood $U$ such that $\pi^{-1}(U) \cong U \times V$. Now $U$ must have the smooth structure inherited from $G_k$ and the second requirement in that sentence determines the smooth structure on the fibres, no?

Eric Gregor

isn't $\pi^{-1}(U)$ supposed to be congruent to $U\times \mathbb{R}^k$?

t.b.

Yes, sorry I forgot that the vector space on which $G_k$ is built was already called $V$.

Skullpatrol

@tb Do you mind me asking how many hours you've slept in the last 48?

t.b.

1:46 AM

@Skullpatrol I don't, and I don't know.

Skullpatrol

Less than 3?

t.b.

No, it was certainly more :)

Eric Gregor

@tb i think i understand. my dumb followup question is, why is it obvious that the vector space structure on $V$ determines a unique smooth structure. i am also short on sleep, so forgive me if this is trivial

t.b.

But $V$ is given, no?

Eric Gregor

@tb your answer seems so straightforward i'm afraid i'm losing some piece of rigor

@tb, yes, $V$ is given

t.b.

1:48 AM

So you already have a smooth structure on $V$, no?

Now every $k$-dimensional subspace inherits one from $V$ because it is a smooth submanifold.

But yes, there's nothing really deep here.

Andypandy

5+5

t.b.

1

Eric Gregor

@tb maybe that is my confusion. i didn't assume that $V$ has a smooth structure to begin with. i know it has the "standard" smooth structure, but i didn't know it was the only one or that we could assume it is fixed

or is the smooth structure on $V$ fixed by the smooth structure on $G_k(V)$

t.b.

First of all: on $V = \mathbb{R}^n$ for $n \neq 4$ there is a unique smooth structure on $V$. So the only trouble could arise for $G_k(\mathbb{R}^4)$. I don't know enough about 4-manifolds to say anything sensible here, but I would assume that in an exotic $\mathbb{R}^4$ not all $k$-dimensional subspaces are smooth submanifolds.

In any event the "standard" structure is intended.

Eric Gregor

@tb, that's interesting. is it fair to say that $V$ inherits its smooth structure from $G_k(V)$, assuming this is endowed with a smooth structure?

4-manifolds are weird

if all of the $k$-dim linear subspaces of $V$ are endowed with compatible smooth structure, then presumably so is $V$. is this ok reasoning?

@Skullpatrol, that is hypnotic!

t.b.

2:02 AM

@EricGregor I don't think "inherit" is the appropriate word here.

Eric Gregor

@tb, perhaps $V$ "sires" the smooth structure of $G_k(V)$?

t.b.

Perhaps...

But I think the key here is to recall that $V$ is a vector space, thus a Lie group and only in standard $\mathbb{R}^4$ addition is smooth.

Eric Gregor

can i really just brush aside the 4-dim case?

if i understand you, @tb, you are saying that this is the only place where something could possibly go wrong

that unsettles me

4-manifolds scare me. and they're important

please forgive me if i'm testing your patience

Skullpatrol

@BrianMScott Have you ever heard of the book "Algebra: Structure and Method" by Brown, Dolciani, et al and if so what is your opinion of it?

t.b.

@EricGregor I'm not entirely certain how meaningful the question actually is but it might be an interesting one to ask. But you are talking to the wrong person here. I only know about 4-manifolds from hearsay.

Sorry, I didn't mean to sound impatient, I just can't help here :)

Eric Gregor

2:15 AM

ok, thanks @tb

@tb, there is a problem with the problem! see the errata, pg 122: math.washington.edu/~lee/Books/Smooth/errata.pdf

@tb: ": The stated conditions are not suﬃcient to guarantee the uniqueness of
the smooth structure on T . Instead, show that $T$ can be identiﬁed with a subset of $G_k(V ) \times V$ ; show
that $T$ is a smooth subbundle of this trivial bundle, and use this to deﬁne the smooth structure on it."

t.b.

@EricGregor Thanks! The sloppiness is probably hidden in the isomorphism $\pi^{-1}(U) \cong U \times V$ I wrote

Eric Gregor

Say the total space $E$ of the bundle consist of pairs $(X, x)$, where $X$ is a $k$-plane in $\mathbb{R}^n$ and $x$ is a vector in $X$. Then there is an obvious projection map $\pi (X, x) = X$. Then I think we can define a vector space structure on the fibers over $X$ by $a(X,x_1) + b(X,x_2) = (X, ax_1 + bx_2)$, with $a$ and $b$ real.

Then can we just construct local trivializations and use unique product smooth structure on $G_k(n) \times V$, to get a unique smooth structure on $E$?

t.b.

Yes, this sounds good.

(but I thought the exercise sounded good, too, so you should take that with a grain of salt).

Eric Gregor

3:11 AM

If $G$ is an abelian Lie group, it is a fact that the Lie algebra of $G$ is abelian. I know you can show this (rather easily) with basic representation theory. But according to Lee this should be rather easy to show if one uses the fact that inversion, $I: G\to G$ is a group homomorphism of , and that the pushforward $i_*:T_e G \to T_e G$ is given by $i_* X=-X$.

can someone help me proceed?

is the point to consider something like $x g x^{-1}$?

Kannappan Sampath

3:43 AM

Hmph! I think I can break 5k. Let me see.

Skullpatrol

@KannappanSampath Why does your Area 51 profile say you have 151 reps while your user profile says you have 5.5k reps?

Kannappan Sampath

@Skullpatrol The user profile here is sum over all rep points from all sites.

Skullpatrol

@WillHunting Hi

user19161

@Skullpatrol Woof!

Skullpatrol

I like the new gravatar Big Dog

user19161

3:52 AM

@KannappanSampath And the flair is sum over all sites with 200 rep.

user19161

So the flair total and the chat total are not the same. QED.

user19161

In fact one can simply join many sites just to get 101 points, and then his chat total will be huge.

Skullpatrol

@WillHunting Me and Maria go back like a baby and a pacifier...

@WillHunting youtube.com/watch?v=PkFUP3tL9o8

Skullpatrol

4:43 AM

@robjohn Hi Rob how are you?

Skullpatrol

5:08 AM

@BrianMScott @robjohn sorry for bothering you :-(

Matt N.

7:47 AM

Quick good morning everyone.

Asaf Karagila

You too!

Matt N.

On your way to uni?

Asaf Karagila

Just woke up, so I want to shower first but then I'm gonna head out.

robjohn

@Skullpatrol bothering? I was simply not at my keyboard.

Asaf Karagila

Hi, @robjohn you got to the 20k!

Skullpatrol

7:53 AM

@robjohn I forgot the question. Oh ya, how did your blood tests go?

robjohn

@Skullpatrol I didn't have any. The doctor decided to wait until June.

Skullpatrol

@robjohn All that waiting for nothing?

robjohn

@Skullpatrol pretty much. Well, I did get to pay them for the visit ;-)

Skullpatrol

;-))

@robjohn Do you have time for a quick simple high school question?

robjohn

@Skullpatrol sure

Skullpatrol

7:59 AM

I want to factor 14x^2 -17x + 5

The factors of 14x^2 are (x, 14x), (2x, 7x) and the negative factors of 5 are ( -1, -5)

I want to know why I have to consider (-5, -1) also for negative factors of 5?

user19161

@Skullpatrol Because order matters.

user19161

By fixing the order of the terms in x, you have to change the order of the constant terms so that you get all possibilities.

Matt N.

@Skullpatrol I don't understand what you mean. The factors of $14x^2 -17x + 5$ are $(2x -1)(7x -5)$.

user19161

@MattN See above.

Matt N.

@Skullpatrol So there is no factor $x$ as you wrote there.

user19161

8:09 AM

@MattN He is not doing the factorisation yet. It is just a step to consider all possibilities.

user19161

Of course later one does rule out some of the possibilities thus obtaining the correct factorisation.

Skullpatrol

@WillHunting That's what I first thought, but order does not count for the terms in x.

user19161

@Skullpatrol See above carefully.

user19161

You don't have to change the order of everything. Just changing some orders will take care of all possibilities.

user19161

That is because multiplication is commutative. (2x-1)(7x-5)=(7x-5)(2x-1).

Skullpatrol

8:12 AM

@WillHunting When you said "By fixing the order of the terms in x," that is where order does not matter?

user19161

@Skullpatrol Yes.

Skullpatrol

@WillHunting But after that step order does count?

user19161

For example, (2x,7x) with (-1, -5) is the same as (7x,2x) with (-5, -1).

user19161

@Skullpatrol The idea is to generate all possibilities.

user19161

You need to do exactly what is needed for that, no more no less.

Skullpatrol

8:16 AM

The factors of 14x^2 are (x, 14x), (2x, 7x) and the negative ordered factors of 5 are ( -1, -5) and (-5, -1)?

user19161

@Skullpatrol Well, I initially supposed the way you wrote the factors as an ordered pair means you want them to be used in the factorisation.

user19161

So (a,b) with (c,d) would correspond to (a-c)(b-d) I assume in your method of factorisation.

user19161

While (a,b) with (d,c) would correspond to (a-d)(b-c).

user19161

Or maybe you want it the other way round, the idea is the same.

user19161

So I have tried to guess your thought processes and extrapolated from there.

Matt N.

8:22 AM

@Skullpatrol Yay btw, this is the first time that I see you ask a maths questions. I like.

Skullpatrol

@WillHunting You have guessed correctly :-)

user19161

@Skullpatrol With great power comes great responsibility.

Asaf Karagila

With great trolling comes a great headache.

Skullpatrol

shrug

@robjohn Would you like to add anything?

@WillHunting Could you tell me if this is right? Factoring 14x^2 -17x +5 ---> listing all possible factors of 14x^2 and all possible negative factors of 5 i.e. for 14x^2 run down the list {1,2,3,...14} and for negative factors run down the list {-1, -2, ..-5} wil this generate all possible combinations? In the form (px + r)(qx + s).

@MattN What do you think about this?

Daniil

8:41 AM

Sup

Skullpatrol

Yo

@Daniil Factoring 14x^2 -17x +5 right now.

Matt N.

@Skullpatrol Hm... I've never done it like this. But I think when listing the factors of $5$ you need to bear in mind whether the middle term $-17x$ has a positive or negative sign. If it's positive you'll have to list positive factors of $5$. Or something like that.

Skullpatrol

@MattN Yes that is correct. I am listing the negative factors of 5.

Matt N.

@tb Thanks. I don't think I'd've remembered to use Tietze here. Now I've come across Tietze many more times than I thought when I first saw the theorem : )

Skullpatrol

Because, as you say the linear term is negative.

Daniil

8:48 AM

Uh, one root is 1/2 and the other is 20/24

I think

Skullpatrol

@Daniil It's not an equation equal to 0. It is an expression that needs to be factored.

Daniil

Well you can use root to factor it

14(x-1/2)(x-20/24)

3

Skullpatrol

@Daniil Fox News strikes out again

@anon hey whatz up?

anon

9:04 AM

what the... how did you know to ping me the exact same minute I woke up?

Skullpatrol

@anon Lucky guess ;-)

Ilya

@Dan: hehe. it's even worse than 147% of voters

Skullpatrol

@anon I was trying to figure out why when factoring 14x^2 -17x +5 order does not matter for the quadratic term's factors i.e. factors of 14x^2 are x, 14x and 2x, 7x. While the negative factors of 5 are -1, -5 and -5, -1. My question was why do we include the order in the constant term and not in the quadratic term?

anon

observe that (2x-1)(7x-5) = (7x-5)(2x-1); only the order of the "constant terms" relative to the "quadratic terms" (as you call them) is important. you could, if you wanted, fix the order of -1, -5 and then go over the different orders of x,14x and 14x,x and 2x,7x and 7x,2x.

Skullpatrol

9:21 AM

@anon That explains it perfectly, thank you.

Asaf Karagila

@Daniil That's nuts.

anon

faux news is like that all the time

Kannappan Sampath

Hi all of you.

Skullpatrol

@KannappanSampath Hi

Asaf Karagila

@anon Reminds me of that website I once saw by some Texan republican claiming that Linux is funded by al-Qaeda, and if we use Linux instead of Microsoft products then the terrorists win.

Ilya

9:27 AM

I had a weird dream today

anon

that's hard to believe

Kannappan Sampath

Two conversations mix well.

Ilya

@AsafKaragila if you use Linux => Microsoft loses => [...] => terrorists win

Kannappan Sampath

:-)

Ilya

@Matt
thanks

Kannappan Sampath

9:29 AM

What, How did I get pinged as @Matt!!!

Ilya

you know, picking me about grammar makes you Matt :D

@Asaf: white hair in your beard?

Asaf Karagila

Yeah, I have one :\

Kannappan Sampath

See, I was not at all serious. I thought you would not mind it... I am sorry!

anon

@KannappanSampath replies to your comments automatically ping you, regardless of whether your name is in them

Ilya

@KannappanSampath too late, in 7 days you'll become Matt

you will hear your phone ringing now

anon

9:31 AM

it has begun.

Kannappan Sampath

@anon But, that comes with an orangy shade over my name, irrespective of the user typing my name out! : )

I don't follow this conversation any more.

Ilya

@Matt you have
only 7 days...

Kannappan Sampath

@Ilya @Matt How is Additive combinatorics going?

^Fail

Ilya

@Looser hehe,
doesn't work

Kannappan Sampath

Hmm.. I am no geek! : (

anon

9:34 AM

the mind boggles

Ilya

@Kan come on, you're mathematician - look at my messages to you. What is common?

Kannappan Sampath

Any way, I got to go. Bye!

@Ilya They are two line blocks...

Ilya

yes

and surprisingly, that's it. Use Shift+Enter

Skullpatrol

Shift
enter

Kannappan Sampath

Let me try.

Good, @Skull

Yay! I got it!

Ilya

9:36 AM

nice

Kannappan Sampath

Thank you @Ilya

Anyway, Bye once again

Ilya

@KannappanSampath you're welcome. See you in 7 days (muahahaha)

Skullpatrol

Good
teacher
that
Ilya
@Ilya

anon

wonder what happens
:3914945 if I make two replies [test]

Ilya

@anon only me was pinged (I huess)

anon

9:39 AM

yeah, I know

Ilya

@Skull: how did you turn letters from black to grey!!!!!1111eleleelvenevn its awesome

(removed)

Skullpatrol

;-)

anon

$\color{Silver}{\text{(removed)}}\quad\color{Green}{\text{(removed)}}\quad\text{(‌removed)}$

Rajesh D

Hi folks

Daniil

D8

Skullpatrol

9:41 AM

@RajeshD Hi

@anon hehe, bold :D

does now

asks

the

brown
c
o
w

anon

@Ilya test

Ilya

9:43 AM

@anon whom?

anon

dangnabbit. somebody get the comment # associated to this comment.

Ilya

never mind

I've got the Chinese visa!

Skullpatrol

@anon 3914988

anon

@anon I'm going to try this one first.

@anon and this one second

both of them pinged me

Skullpatrol

@anon Stop pinging yourself in public

Ilya

9:45 AM

@anon ping yourself! just do it!

Rajesh D

@rajesh

nope

Skullpatrol

@@@@@@@@@@@@@@

Rajesh D

it isn't

anon

@RajeshD type ":3915004 test"

Rajesh D

@RajeshD test

@RajeshD

Ilya

9:48 AM

this room has become too egocentric

Skullpatrol

not me

Rajesh D

@Ily just to satisfy yours

Skullpatrol

n
o
t

m
e

Asaf Karagila

@AsafKaragila Look at me, I'm a strange loop.

Ilya

@AsafKaragila without for?

Rajesh D

9:50 AM

Do While

Skullpatrol

end

Rajesh D

@Skull : are you a software engineer ?

}

Skullpatrol

@RajeshD No I'm just a softy at heart.

Rajesh D

and Hardy at ????

Asaf Karagila

@RajeshD He's got a thick skull.

When you think of a real insult, let me know.

Transcript for

$Mathematics$ Mathematics