Mathematics - 2015-04-30 (page 1 of 5)

Disciple of Barney

00:00

so if $f$ is the homeomorphism use the metric $d(x,y) = |f(x)-f(y)|$

columbus8myhw

00:37

Get a homeomorphism from $\overline{\mathbb R}$ to $[0,1]$ (such as $\dfrac1{1-e^x}$) and use the metric on $[0,1]$.

($y=\frac1{1+e^x}$ is rotationally symmetric around $(0,\frac12)$, I believe. Not that it matters much.)

*I meant $\dfrac1{1+e^x}$ in the first thing.

Incurrence

01:32

@Disc Can you help me understand the fundemental theorem of finitely generated abelian groups?

Disciple of Barney

what is there to understand? @Incurrence is there something lacking in the things you read. It just classifies them

Incurrence

I understand for finite groups

But for infinite groups I have trouble

Disciple of Barney

What in particular?

Incurrence

I suppose the cyclic nature of them

Disciple of Barney

That means nothing to me

Incurrence

01:35

So if a subgroup is generated by some elements, we say that we take every word in the alphabet of those elements

Which reminds me of the free group

And the definition of a finitely generated abelian group is that there exists finitely many elements in the group, such that every element in the group can be written as some linear combination of them

If this is the case, we say that these elements are the generating set

So if we want to generate $(\Bbb Z, +)$ we can just take linear combinations of $n_1 \times 1$ where $n\in \Bbb Z$

So $1$ is the generating set of $\Bbb Z$

If I wanted to setup a generating set for $\Bbb Q$, I am not sure how I would go about this. I know it will fail, but pretty much I want to setup $x_i\in \Bbb Q$ to fail it. I am just choosing the denominator as coprime values right?

And hence I will be able to inductively add elements to this generating set forever

Hence there is no generating set

Disciple of Barney

Every group has a generating set

Not necessarily finite though

Incurrence

Okay, so what is the generating set of $(\Bbb Q,+)$?

Disciple of Barney

Generating sets are not unique

Incurrence

What is one form of the generating set of $(\Bbb Q,+)$?

Disciple of Barney

$\{1/n \mid n \in \Bbb Z_{>0} \}$, $\Bbb Q$, the set of all rationals less than 1, etc

Incurrence

01:45

Okay

Thanks, I'll go work on that. I think I'll stop asking questions here until I have spent a few days on this

I think I was underprepared for this part of the course

Disciple of Barney

Were you able to figure out the direct product thing

user143442

@DiscipleofBarney I don't understand what you mean with 'pull the metric'

Incurrence

Nope

I couldn't get it to split across the second monomorphism or the first epimorphism

(for the direct product)

Disciple of Barney

@user You since the space is homeomorphic just treat it as if it was $[0,1]$, and use the metric there, I gave you an explicit example

Incurrence

But I'll go do more textbook

Actually I'll take a break from Algebra and do some complex analysis and functional analysis, talk later guys

Disciple of Barney

01:50

Okay, maybe mess around with more explicit examples., see yah

user143442

02:08

@DiscipleofBarney so if $f:X \to Y$ is an homeomorphism and $X$ is metric then $Y$ is metric too?

user143442

I've been looking for that result and I think it's not true

Mike Miller

morning

Alex Wertheim

Evening, @MikeM :)

Mike Miller

how's things

Alex Wertheim

Better than yesterday - can manage to eat dinner tonight. You?

ᴇʏᴇs

02:20

Are you guys gonna eat together

Mike Miller

seems unlikel

he had the stomach flu, hence the comment

that's good to hear, @Alex

Alex Wertheim

Not unless Mike has a teleportation device, lol.

Mike Miller

finishing up this talk

Alex Wertheim

Cool, what on? I'm excited to actually get to come to some of these next year.

ᴇʏᴇs

I thought you were chatting on here while simultaneously giving a math talk right now lol

Mike Miller

02:24

lol @ᴇʏᴇs

@AlexWertheim The title is "The dimension barrier in topology"

the diff between low dimensions and high dimensions

I'm thinking strongly about making this into a blog post

Alex Wertheim

Ah, I believe I was here when you were describing it before. Good stuff.

MSE blog, you mean? Or do you have your own?

ᴇʏᴇs

@MikeMiller I would subscribe, like, follow and read your blog every day

Mike Miller

I don't have my own, I would make it

OK, I'm done

The punchline: why is high dimensional topology easier than low? because 2 x 2 < 5

Alex Wertheim

That'd be cool. I bet a lot of people here (myself included) would enjoy that.

LOL. A deep theorem.

Mike Miller

the only downside is I have to draw to write this post

there's no way to get around it

Alex Wertheim

02:28

You're not an artist like Ted, @Mike?

Mike Miller

If I'm a topologist, I have to be... but I'm not so much

ᴇʏᴇs

You can't use TikZ?

Mike Miller

Nope

I don't know how

ᴇʏᴇs

:(

Mike Miller

The other issue is that doing this properly takes explaining handlebodies, which take a couple months to actually grasp

So I will probably lose some folks when doing that in 5 minutes

ᴇʏᴇs

02:30

Nice, a couple months worth of blog posts

Alex Wertheim

All the more reason to start a blog!

Mike Miller

Hahaha

Alex Wertheim

Lol @ᴇʏᴇs, same thought :)

Mike Miller

At least, it took me a couple months to grasp

And by grappling lots of examples

ᴇʏᴇs

You should post an example every week or something

Mike Miller

02:31

It's not that exciting on its own sadly

Alex Wertheim

Explaining hard things well is always exciting, especially if they build to something cool.

Mike Miller

Fair enough!

Okie doke, done

pjs36

02:52

You're done writing the blog up already? That was quick! (I kid, I kid)

Mike Miller

:p

user1667423

03:37

Anyone have a closed form for the generalized harmonic numbers in terms of special functions (like digamma)?

22

Q: Do harmonic numbers have a “closed-form” expression?

One of the joys of high-school mathematics is summing a complicated series to get a “closed-form” expression. And of course many of us have tried summing the harmonic series $H_n =\sum \limits_{k \leq n} \frac{1}{k}$, and failed. But should we necessarily fail? More precisely, is it known th...

Harmonic number

In mathematics, the n-th harmonic number is the sum of the reciprocals of the first n natural numbers: This also equals n times the inverse of the harmonic mean of these natural numbers. Number n such that the numerator of is prime are 2, 3, 5, 8, 9, 21, 26, 41, 56, 62, 69, 79, 89, 91, 122, 127, 143, 167, 201, 230, 247, 252, 290, 349, 376, 459, 489, 492, 516, 662, 687, 714, 771, 932, 944, 1061, 1281, 1352, 1489, 1730, 1969, ... (sequence A056903 in OEIS) Harmonic numbers were studied in antiquity and are important in various branches of number theory. They are sometimes loosely termed harmonic...

Remember me

05:21

@Disciple are you online?

Disciple of Barney

playing team fortress 2 at the moment, whats going on?

Remember me

i have a doubt....@DiscipleofBarney

Disciple of Barney

You know you could just say what your doubt is... @Rememberme

Remember me

i have to prove that AX=Y will have solutions if and only if the row rank of A and the row rank of the augmented matrix are equal@DiscipleofBarney

@Disc i was thinking that the augmented matrix will be m*n+1

so the row space will be $\beta=(\alpha_1,\alpha_2,.......\alpha_m)$ of the matrix A

Now since the augmented matrix also contains m rows i was thinking will the row space be the same?@DiscipleofBarney

Remember me

06:04

can someone please check that i am right or not

Regret

I would, but I don't know much about matrices :v

Remember me

you know about ranks etc@Regret

Regret

nope! can you explain it to me?

Disciple of Barney

What do you mean the same? they have a different amount of coordinates in the space @Rememberme

Remember me

but it also contains m rows right @DiscipleofBarney

Disciple of Barney

06:08

@Rememberme When you augment a matrix, you add a coordinate so they can not be exactly the same

Remember me

so how should i prove it....what i thought was the augmented matrix will also contain m rows

Disciple of Barney

@Rememberme could you give the verbatim question, I don't think yours makes sense, or is wrong

Remember me

AX=Y Prove that this system has solutions iff the row rank of A is equal to the row rank of of the augmented matrix of the system@DiscipleofBarney

Eric Stucky

@MikeMiller: Please do it! I've always been astounded at the claim that 5+ dimensions is easier than 4- but I'm a course or two short of getting into technical details. I'd love to see your exposition :)

Remember me

Hi@EricStucky long time no see

@disc according to this is what i am saying that the rows of the augmented matrix will be the same as that of the matrix A

Disciple of Barney

06:16

Where is it in Hoffman and Kunze

Remember me

Hoffman Kunze

Disciple of Barney

Where in Hoffman and Kunze @Rememberme

Remember me

Vector spaces:- computations concerning subspaces

last question@DiscipleofBarney

@DiscipleofBarney I think what i am doing is right the number of independent rows in the augmented matrix will be the same as the number of rows in A and due to this the row space will be the same and hence the row rank

Disciple of Barney

That is what you are suppose to prove

Except they are not the same exact space

Remember me

i am supposed to prove that the row rank of A is the same as the row rank of the augmented matrix and i do that by saying that the number of rows in both of them are same .....AM i right @Disc

Disciple of Barney

06:31

No, because they have the same number of rows when there is no solution

Remember me

SO how do i prove it@Disc

Disciple of Barney

I am in the middle of playing a game, I can't think of exactly how I would go about it. I would probably reduce to row echelon and when something is not a solution you will have extra stuff on the zero rows, that shouldn't be there making the rank of augmented larger. I would have to put more time than I care to to make sure that actually works. @Rememberme

Incurrence

@Rememberme If the augmented matrix has a higher rank, then it is linearly independent, and hence there is no solution

Hi @Will

Will Hunting

@Incurrence Hi Alex.

Remember me

@Incurrence i am saying that the rows are equal in both of the matrix is it right?

Will Hunting

06:44

@Rememberme All you know is that the number of rows are the same, yes.

Incurrence

@Rememberme I don't know what this means, and I have to leave in a few min

Remember me

yes @WillHunting

Incurrence

The number of rows are the same yes

You have an additional column

Will Hunting

@Rememberme Your sentences do not make sense at all. I think you are still not clear about writing proofs in general and should return to Hammock.

Incurrence

For a standard [xyz|a] augmentation

If this new column is dependent on the others, it is solvable and the rank remains the same after augmentation

if the new column is independent the rank will be increased, but it will not be solvable

Okay I have to go now, cya later :)

Remember me

06:46

cya

Will Hunting

@Rememberme Reading these three lines, I do not think you are ready to study linear algebra yet.

Remember me

@WillHunting you said you wont make any comments about what i do but still you are .... and infact i did hammock and please dont interfere

Will Hunting

@Rememberme Sorry, I won't interfere then, goodbye.

Disciple of Barney

@Rememberme Figure it out?

Will Hunting

@DiscipleofBarney I think that if you want to help him, you should comment on what he originally wrote. The first line already does not make sense.

Remember me

06:59

@disc i really cant

Disciple of Barney

@Rememberme Okay just wondering. You should figure it out.

@WillHunting Are you talking about the rows on the matrices being equal?

Will Hunting

@Rememberme I am sorry if my words have hurt you. I won't talk to you again. Thanks for your encouragement.

Remember me

If the row rank is more what happens does it mean there is one row extra

Will Hunting

@DiscipleofBarney "i was thinking that the augmented matrix will be m*n+1" this already does not make sense.

Disciple of Barney

@WillHunting Yah

Will Hunting

07:01

@DiscipleofBarney "so the row space will be $\beta=(\alpha_1,\alpha_2,.......\alpha_m)$ of the matrix A" this also does not make sense.

Disciple of Barney

Yup

Remember me

i mean the augmented matrix will m by n+1 that means m rows and n+1 coloumns

Will Hunting

@DiscipleofBarney So there is no point in going further.

"since the augmented matrix also contains m rows i was thinking will the row space be the same" again does not make sense.

Remember me

@DiscipleofBarney it does make sense that what hoffman kunze says

Disciple of Barney

@Rememberme You should acually write out what row rank is, do some examples, and then do some examples with augmented matrixes, it seems like you have some misconception that points to you either not spending enough time/ wrongly convincing yourself you do understand.

@Rememberme the problem in Hoffman and Kunze does make sense, I was just reading it wrong.

Remember me

07:04

See i told you

i am not making sense to you

Disciple of Barney

@WillHunting What brings you back? Once you go math chatroom you cant go back? :D

Will Hunting

@DiscipleofBarney Well, no idea. =) But I think if you really want to help him, you should encourage him to study math properly, and I don't think he is and if he continues like that he will just be wasting his time.

Remember me

I am studying math properly others can understand what i say why cant you @Will hunting

Will Hunting

@Rememberme I already gave you an example of what you are writing wrongly.

@Rememberme Others do not set high standards for their students, maybe. I do.

Remember me

i am just saying that an augmented matrix is an m by n+1 matrix is that what you mean i am saying wrong@WillHunting

Will Hunting

07:09

@Rememberme Before we can make sure you understand, we need to be sure you are clear, and being clear means writing down every sentence clearly.

It is very important to be precise in math. That day when you talked about n points in a plane, we don't even know what you were asking.

@Rememberme Yes, just one example. The following two sentences also do not make sense like I said.

Remember me

and what i think row rank is the dimension of the row space of a matrix A now this is wrong @WillHunting

Disciple of Barney

What you were saying wasn't making sense. Maybe I am just feeling extra grumpy, but I also think you put an inconsiderate small amount of thought in some of the questions you ask, solutions you give, and how you ask them. I basically have to pry every detail out of you. @Rememberme

Will Hunting

@Rememberme What I am saying is that first you need to express your ideas clearly in writing.

Remember me

But what i say is right it is an m by n+1 matrix

Will Hunting

@Rememberme Yes, that is right, but different from what you originally said.

Disciple of Barney

07:12

@Rememberme The row rank is the dimension of the row space, I am not sure though you understand what the row space is, at least from some of the things you were saying.

Will Hunting

@Rememberme So first you must write your sentences clearly before asking for help.

@Rememberme Yes, that is also right.

Disciple of Barney

Just because two matrices have the same number of rows does not mean their row space are equal, or even their row rank @Rememberme

Remember me

yes i understand the row space it is the space spanned by the vectors

$(\alpha_1,\alpha_2,.......\alpha_m)$

where

$\alpha_1=(A_{i1},........,A_{in})$

now i think i am right @DiscipleofBarney

Disciple of Barney

How about you right up a full proof @Rememberme

Will Hunting

@Rememberme Yes, you should write up a full proof and then ask. Then it is clear whether you know or do not know.

Remember me

07:18

fine let me write.....but tell me something first if the augmented matrix has a row rank more than A then what does it mean does it mean that there is a row with all 0's and one 1 or something else @Disc

Will Hunting

@Rememberme Again, it is not clear what you are asking here.

Remember me

okay let me write a proof straight away and then see what happens

Disciple of Barney

@Rememberme It means it has row rank more than $A$... At least how you phrased the question, it does not mean what you said. As Will says, I am not entirely clear as to what you are saying. You should be able to figure whatever it is out though

Will Hunting

@DiscipleofBarney I don't play any computer games.

Disciple of Barney

@WillHunting I don't play to many games, I have only recently started getting into playing computer games again, and I don't even play that much, couple hours a week... maybe

Remember me

07:30

@DiscipleofBarney i have got a brilliant idea for the question(at least in my perspective)

2

Samuel Yusim

Hey guys, I'm trying to prove that $\{B_r(x) : r \in \mathbb{Q}, x \in \mathbb{Q}^n\}$ is a basis for the euclidean topology on $\mathbb{R}^n$ but I feel like this is one of those problems where I just don't know the trick I'm supposed to know to easily solve the problem.

Disciple of Barney

Well, fully prove it using that brilliant idea if you can @Rememberme

Samuel Yusim

Any advice?

$B_r(x)$ is of course the open ball of radius $r$ around the point $x$.

Remember me

the idea is to do this thing in two different cases that is first if i say that AX=Y and the this matrix Y is a matrix has all zeroes in its column then the augmented matrix will be just the matrix A.
Second case:-
If the matrix Y contains some non zero values lets say then i can show that these two matrix are row equivalent to each other and if they are row equivalent to each other then voila!!!!!!they have the same row space right @DiscipleofBarney

Samuel Yusim

I can do similar problems, but I just can't come up with a nice way to deal with the fact that the important values need to be rational

Remember me

07:37

Am i right @DiscipleofBarney

Samuel Yusim

like if I have an open set $U$ then I want to pick an arbitrary point $x$ in $U$ and show that I can fit an element of my supposed basis around it which is entirely in $U$, so I'm trying in particular to squeeze this basis element inside of the particular $B_r(x) \subset U$ which I already know must exist, but doesn't necessarily have a rational $r$ or $x$.

Disciple of Barney

@SamuelYusim The idea is you can put rational points arbitrarily close to a real point, so any ball around a real point you should be able to put a rational point so close that some ball will be contained in the ball around the real point and contain the real point. (although it sort of sounds like you get this)

Samuel Yusim

honestly I might have figured it out from just typing my question out

Disciple of Barney

Thats good, I have done that quite a few times @SamuelYusim

Remember me

@disc am i right????????????

Disciple of Barney

07:45

@Rememberme Didn't you read what I said: fully prove it! You will see what is right or wrong when you do that, maybe you are half right and half wrong, or onto a pretty good idea but doesn't work, maybe it is spot on... It is good to understand your full proofs, if you made and actually good proof you wouldn't need my confirmation. I don't want to read all your half-assed attempts

Plus I have already explained why that doesn't exactly work

to you

Will Hunting

@DiscipleofBarney Why aren't you sleeping yet? It's late there.

Disciple of Barney

Where am I? @WillHunting

Will Hunting

@DiscipleofBarney I thought you were in the US.

Don't worry, I don't check people's IP addresses, LOL.

Disciple of Barney

Haha, that is what @Incurrence did

Will Hunting

Why did he do that?

That's a little scary.

Disciple of Barney

07:54

I had been screwing with him, making him think I was in one of his classes and he tricked me. He was like "Can you check my blog post" and then his stats showed him where I was. I did say it was a VPN though...

Will Hunting

Oh OK. I don't use a VPN these days, even though many sites are blocked here.

Anyway I really should not come here, bye.

Disciple of Barney

Bye

Remember me

@DiscipleofBarney there goes my proof

Disciple of Barney

It didn't work ?

And why?

Remember me

it did work and it s right i am sending you a pic

there we go

@DiscipleofBarney

@DiscipleofBarney how is it............. fine? i feel it is

i have showed that it would be inconsistent is it okay#@DiscipleofBarney

@Balarka please check my proof

Balarka Sen

08:15

@MikeMiller I think it really does prove it for in general $M$.

@Rememberme I don't even know what you are trying to prove.

Disciple of Barney

@Rememberme It looks pretty good, I don't think claim one is necessary. More detail in case three would be nice, and I don't think case one is necessary

Remember me

thanks so i am right.....

thats what i was trying to tell you from the first @DiscipleofBarney

Disciple of Barney

By more detail, I mean expand on how certain things imply certain things (even if they are quite simple). Yes @Rememberme It doesn't sound much like what you were trying to tell me, in fact it basically looks like what I told you.

You have it though, if you think you understand it all and how each step is done then it is good @Rememberme

Incurrence

@Rememberme Can you please define right now clearly, what does rank refer to?

Disciple of Barney

@Incurrence Hey. Did your analysis? And when is that project for algebra?

due

Incurrence

08:25

@DiscipleofBarney Analysis study was alright. I was just recovering heaps of the basics to make sure I don't have any hole in my understanding(did first 50 pages of my complex text)

Project for algebra is 5 questions(with many subparts) + the presentation

And that is due on Wed 6th

Disciple of Barney

Okay, so about a week

Incurrence

Yep

Balarka Sen

what is going to be the topic of the presentation?

Incurrence

@BalarkaSen Central extensions

Balarka Sen

ah

Disciple of Barney

08:27

What is the other one you are doing (free groups, or braid groups I think)

Incurrence

With an example of $H_3(\Bbb Z)$

Balarka Sen

you know what group (co)homology is? cool!

Incurrence

@DiscipleofBarney We only have to present one, and then prepare another for unknown reasons, and I am learning some of all of the topics

@BalarkaSen I do not

Disciple of Barney

Its the Heisenburg group @BalarkaSen

Balarka Sen

:p

oh, blah.

Incurrence

08:28

I will in 6months :)

Balarka Sen

I'd have done free groups and braid groups if I were there, but that'd involve alot of non-algebraic stuff.

Disciple of Barney

Its a five minute presentation

Balarka Sen

@Incurrence Good for you, 'cause I've never studied group co/homology.

Incurrence

"Topics from groups, rings, fields, algebraic number theory, category theory & homological algebra, with applications to quantum algebras. Offered in odd numbered years only."

Disciple of Barney

Although I probably would have done free group or, the tensor products of groups

Balarka Sen

08:31

i don't even know what tensor product of groups are.

Incurrence

"Algebraic structures & their representations of importance to current mathematical physics research: Lie algebras & superalgebras; quantum groups & algebras; Hopf & quasi-Hopf algebras; affine & Kac-Moody algebras. Illustrative applications to knot theory."

@BalarkaSen You don't want to

Disciple of Barney

Haha

Balarka Sen

what the heck is it about? we already have a symmetric monoidal structure on $\mathbf{Grp}$.

Incurrence

When I take algebraic topology next semester, it is a 6xxx course, where the first number usually refers to the year you are in :-)

@BalarkaSen The definition is as follows:

"the condition or quality of being sad." - Tensor product of finite groups( aka sadness )

Balarka Sen

lol

Incurrence

08:33

Well the tensor product of non-abelian groups is very local

www-irma.u-strasbg.fr/~loday/PAPERS/…

Disciple of Barney

They look interesting, don't really know anything about them. Looks like semidirect with two actions. I guess they came up in some higher dimentional van Kampen theorem.

Balarka Sen

ooh.

Incurrence

And since I don't know pretty much any module theory, it was inaccessible to me[only because I can't appreciate it in that form, which is better documented by far]

Balarka Sen

no, no, i don't think that's ad hoc at all.

$\pi_3(X; A, B) \cong \pi_2(A, C) \otimes \pi_2(B, C)$ looks very interersting.

Incurrence

I can't see where this $\pi$ map is defined

Seems to be a group action?

Balarka Sen

08:36

it's not a map : it's a functor $\mathbf{Top} \to \mathbf{Grp}$

google homotopy groups

Incurrence

Ok

Wait is this going to lead me down a rabbit hole

yep

Balarka Sen

:p you'll learn about them when you do algebraic topology nevertheless

Incurrence

So it refers to the third homotopy group

Will Hunting

Hey @Incurrence.

Incurrence

Hi @lliW

Gotta go

Will Hunting

08:39

@Incurrence Are you busy now? Shall we talk in the other room?

Balarka Sen

i thought the best one could do for higher homotopy groups is homotopy excision.

Incurrence

Girlfriend needs me to weigh some rocks lol

Will Hunting

@Incurrence OK, bye.

Incurrence

Doing sample testing for her thesis

Will Hunting

@DiscipleofBarney I consider case 1 proof wrong though.

Disciple of Barney

08:43

@WillHunting Formally it isn't great, and maybe there is some misunderstanding on rm's part

Will Hunting

But I think I should keep my mouth shut. He doesn't like me, LOL.

Balarka Sen

he probably didn't do the proof-writing bit of Hammock, thus the horrible logical reasoning.

Will Hunting

I think I can't read what he writes anymore, so maybe I should press ignore.

Disciple of Barney

I was laughing when a few days ago you (I am guessing it was you) when I mentioned him supposedly completing Apostol but either expressing himself terribly or did not know about "density" and he was like "who starred this"

Will Hunting

I am not sure what that question about n points was.

Maybe he will say two different things and then say they are the same, and after that you don't know what he wants to say.

There is no point trying to guess what he is thinking.

Disciple of Barney

08:49

I am not sure either, I feel like we ended up figuring out what it was, but I am not sure.

Will Hunting

It's very dangerous to guess his thoughts and say they are right when they are actually wrong.

Balarka Sen

i feel like i don't care about his questions anymore :p

Will Hunting

At this stage, we must be strict and say that case 1 above is wrong.

Disciple of Barney

I have been having a similar feeling

Will Hunting

A matrix with a column of zeros added is no longer the same, and why should the rank be the same immediately?

Balarka Sen

08:51

i was doing this from the very beginning

Will Hunting

If you want to help, be thorough. Otherwise, don't say anything and do more damage.

Disciple of Barney

@BalarkaSen Answering questions?

Will Hunting

I don't think someone who has been through Hammock and Apostol thoroughly will express himself like that.

Balarka Sen

@DiscipleofBarney no, i mean, being strict while checking his proofs.

Will Hunting

Especially after doing all the exercises, which I seriously doubt.

Disciple of Barney

08:54

@WillHunting Probably right. Past few days my caffeine levels have been a lot lower than usual, and my patience is thin

Will Hunting

It's my bad but I can no longer tolerate his mathematics.

Disciple of Barney

And I already told him multiple times it didn't make sense, so I just didn't care anymore

Regret

Who are you guys talking about?

Disciple of Barney

@Regret You

Balarka Sen

lol

Regret

08:55

Oh no. :v

Will Hunting

He seems to take it badly though, so I should just stfu.

I think he should ignore me too. I don't wanna upset him either.

I think maybe he thinks my math is not good enough to help him too, LOL.

Maybe I am the one being stupid and not getting what he says.

Transcript for

$Mathematics$ Mathematics