constant polynomials don't :D

nice one

@anon 21 years old.

yes I am 21

Is the main page not refreshing for anyone else?

Never mind, I had to restart my browser.

@robjohn Weird, I swear I had almost 1,6k rep.

Is that right?

I think so but then, why it do not appears explicit in the texts?

@PeterTamaroff hola!

@leo Leo. ¿Que tal?

@PeterTamaroff todo bien! vos?

un poco ocupado con el final de semestre

@leo Bien. Casi casi de vacaciones.

si yo tambien

@leo Que tenes este semestre?

@PeterTamaroff poquitos. Solo Lógica y Geometría Diferencial.

@leo Ah. Que tal es logica?

Vieron ZFC?

@PeterTamaroff No. Es un curso básico. Vimos cálculo proposicional, de predicados, y de ahí hasta un poco de teoría de modelos

seguiamos este

@leo Ah, OK.

@leo Nice! Bajando

je je

@leo estas?

@leo leo le

@PeterTamaroff now yes

@JasperLoy cool

@JasperLoy yes indeed =)

@leo oH

Look at this

@leo

@PeterTamaroff reglada es que se puede aproximar por una escalonada?

@leo Sips. En realidad que es el limite uniforme de una sucesion de escalonads

Creo tiene que ver con Lebesgue y la medida y eso

@PeterTamaroff ok

@PeterTamaroff sip. No veo claro porque sigue tu afirmación sobre la $s''$

@leo EN cual de las pags?

@PeterTamaroff en la segunda

con sup lo usual es correrse hacia abajo...

Side comment: It's really cool to me that even though I know very little Spanish (that is Spanish, correct?), I can still read much of the photocopied papers. Math is truly a universal language! :)

@leo No queda claro como estoy construyendo $s''$?

@anorton It is not photocopied...

Well... ok

Photographed.

@anorton =)

:)

@PeterTamaroff si no queda claro

@leo Hmm. A ver...

@leo $s''$ es el maximo de $f$ sobre $(\alpha,\alpha+\delta)$

@JasperLoy I can assure you it isn't derived from anon. :P

It is cool isn't it, 3 guesses what que es absurdo means :)

oh wow...

I was under the impression that anon was short for anonymous user... oops. (sorry anon)

@AlexJBest Huh?

I was expanding on anortons comment that maths is universal, by saying it's cool that even though I have no idea how to speak/read spanish I can still tell what that means.

@AlexJBest Oh, sure!

@PeterTamaroff A ver si te entiendo. $\epsilon\gt 0$ es fijo. Existe un $\delta\gt 0$ tal que $|\max f([\alpha,\alpha+\delta])-f(x)|\lt\epsilon.$ Entonces tu defines $s'':[\alpha,\alpha+\delta]\to \Bbb R$ por $$s''(x)=\max f([\alpha,\alpha+\delta]).$$

sí eso está bien

@leo Sipi!

@JasperLoy Hmm, let's a see.

@PeterTamaroff seems good to me

voy a comer =)

@leo Buen provecho.

@PeterTamaroff gracias

:\ I need quite a bit more to cap today...

I got 22 rep yesterday (which was pretty good for me)

:(

@GustavoBandeira Sometimes, I think in chat, your total rep (here and Mma, etc) is shown. That would be 2775.

@GustavoBandeira try saying a few lines together and we can see.

@JasperLoy what is cap?

everybody talks about cap

@leo when you get 200 rep in a day not counting bounties and acceptances

@leo that's as much as you can get that way.

@robjohn oh

thanks

@robjohn: What was this thing about all these questions and answers that got deleted?

there is no cap on bounties and acceptances

@Thomas we are still discussing it. I believe they will probably be undeleted sometime.

@amWhy (this is a crazy idea, but bare with me) Is it possible that this question is implying that $X=\{e\}$?

We have that 1) $x:gxg=g$, 2) $x$ is unique, and 3) $\cdot$ is an associative operation. $gxg=g$ implies $g(xg)=g$. The only way this is true is if there is an $e$ such that $xg=e$. However, $xg=e$ implies $x=g^{-1}$. Yet, we required that $x$ is unique! That means that we can't have $x$ constantly changing for each $g$, yeah? Hence, $x=g^{-1}=e$ and $X=\{e\}$. Trivially, $X$ is a group.

@Thomas a user asked that all of their questions be deleted and they were deleted. The mods are talking it over and hopefully, things will be restored soon.

@robjohn: Ok thanks.

@Thomas 48 questions and 78 answers were deleted.

@Limitless $x$ depends on $g$; so it is not unique "universally".

@peoplepower That's kinda what I am saying.

@Limitless That's how it is defined. ;)

The only way that $x$ can be unique is if we restrict $g=e$.

For all $g$.

It is defined to depend on $g$.

$\forall\exists$ is different from $\exists\forall$.

@Limitless I think the claim is not true

I think it's false as well. It seems impossible to produce idempotents.

I think you are both right.

I think $X=\{e\}$ is the only case where it happens to be true.

Thomas posted what I was thinking :-(

@leo :)

@Thomas anyway, I've started to type, so there it goes

@leo: Yeah, I have done that lots of time myself...

@Thomas +1 there by the way

Is anyone available for a topology chat? I would ask a question, but I really need to have a discussion regarding homotopy for a couple of minutes.

@Thomas Had a stupid moment. =D

@Limitless: :)

We are having so much confusion on that question.

I actually found idempotents.

You are right, I think, @Thomas.

Every point such that $x^3=x$ is idempotent, and such a point exists.

All right. I made a mistake after all :(

@what you say me?

@Thomas what mistake?, if you mistake I mistake

@JasperLoy Yeah, I think I am wrong.

yes, @Limitles. xg=xg does not imply x=g^-1; it does not even imply inverses or a neutral element exists.

:-) @Thomas what you posted is not what I was thinking after all

@leo: Yeah, I was wrong after all. The $x$ isn't unique for $g = 0$.

@leo: Yay.

I saw the identity matrix instead 0

@anon My logic was completely wrong. I was trying to see what satisfied the theorem, not proving the theorem.

I like how everyone is stumbling for proofs when the OP is not interested in proofs.

@anon I was going to do a proof as comment if I found one, but it seems no one has discovered any references (probably because it's wrong).

@anon well, if the claim is false we must say why

@JasperLoy I've studied some of this stuff pretty hardcore, but you're right.

I need to try to work deeper on it.

@thomas I'm wrong as well

@leo: Oh well :) good try though.

@JasperLoy I'm kinda confused about the equivalence relation question, though.

@leo yes, that is true.

@anon but perhaps in this case the claim is true

@JasperLoy The issue is that I have no clue what is confusing you.

@JasperLoy I know that.

@JasperLoy My argument is messy in format because it's hard to get at what I'm trying to say.

Like.

I think we can cook a counter example by writing multiplication tables

just as play Sudoku

My point was to take a look at the bigger picture.

@JasperLoy To observe that $a+d$ and $b+c$ are just integers in and of themselves.

Hence we're just working with the statement $x=y$ with $x,y \in \mathbb{Z}$.

@JasperLoy See, that's where I am confused

Why can't we just simplify the problem?

@JasperLoy: what is this... you haven't capped yet?

we need someone that write a program to construct multiplication tables and find a counter example

@JasperLoy This is where I am confused. How does noticing that $a+d$ and $b+c$ are just integers not reduce the problem to showing that $x=y$ and, as a result, showing that $=$ is an equivalence relation in that context?

How is it that showing that $=$ is an equivalence relation in the context of $x=y$ not imply that $=$ is an equivalence relation in the context of $a+d=b+c$?

I am completely and totally missing your critique, even though I am struggling to understand it.

My math tutor with an M.A. reviewed that for me and said it checks out, but you may very well be right.

Yes, she did.

Can we assume that $x\sim y\iff f(x)\sim f(y)$ is an equivalence relation for any function $f$ without asking ourselves why?

Ah.

Wait.

I think anon just made it click.

Ah.

Yes! Okay.

@anon, I see exactly what Jasper meant!

You're saying the issue is in the details!

Like

It boggles my mind how so many people can mix up the order of quantifiers in Lubins post. He makes it perfectly clear... It's a veritable trainwreck!

@JasperLoy I'm mulling this over in my mind. Anon's point is very interesting.

@anon yes. I thought that in the Clayton's example if you assume $\exist ! x$ for all $g$ that implies for each $g$ exist a unique $x$

@JasperLoy Yeah, I am taking my time. I am just worried I might annoy you all with my issues.

I was wrong evidently

@JasperLoy I hope that doesn't happen.

leave it to bill to ruin my fun trying to prove it

Bill to the rescue?

but it is enough to find an example where $\exists ! \forall $ implies $\forall \exists !$

\exists

@leo \exists

@anon =)

owe me a soda

Morpheous is calling!

@anon But why?

@JasperLoy How so?

That is a nice argument indeed, proving the uniqueness of idempotents.

@PeterTamaroff http://en.wikipedia.org/wiki/Jinx_(children's_game)

Upvote me, mortals.

@anon Fair enough. To what name shall I mail it to?

@JasperLoy Oh, no! My dog has suddenly combusted and my pen is out of ink. How horrible! My pen is out of ink!

@JasperLoy I'm not.

I love it when some question produces deletes, like this one

Were you around for the local-Taylor-series-are-all-polynomials one?

@anon Ah?

I will find it.

@anon Oh, yes!

I like that one.

Okay.

@anon Nope

?

@anon Bye byes. I have to rest.

@PeterTamaroff this one

yeah, I told myself I was going to go to sleep 5 hours ago.

I am the easiest person for me to fool.

@PeterTamaroff Ya es tarde en la Argentina

@leo 1 am

@anon 5 hs ago? What time is it there?

@anon I will now proceed to close ACIII... I hope I'm successful.

here are almost 22

5 hrs ago was 5pm here. however, I skipped sleeping last night to finish a maple project I started 10 hours before the due time

also to rewatch death note as procrastination

@anon did you succeed?

Mostly. The final question I ended up calculating would only be about 0.3% of my final grade but would take me over half an hour of tedious hand calculations and inevitable headbanging debuggings and pleadings with maple.

So I skipped that one.

@JasperLoy Outside of the maths, how's life?

@JasperLoy I'm sorry to hear that. On the bright side of my life, I am going to have a lot of spare time . . .

@JasperLoy I feel you. I have been playing with that problem and analyzing the structure of my argument to death. I'm not going to annoy you with it, though.

@anon I see. Often programming projects demands time