Right so we know $G(\sqrt a)=G(\sqrt[3]b)=H$, that's the hint

with $G=\Bbb Q(\sqrt a+\sqrt[3]b)$ and $H=\Bbb Q(\sqrt a,\sqrt[3]b)$

@AkivaWeinberger that's not a bunch! It's two ;P

I wasn't going that way, but it's fine. So either $\sqrt a \in G$, in which case we're done, or else $\sqrt a\notin G$, in which case $[G(\sqrt a):G]=2$.

Similarly, either $\root3\of b\in G$, in which case we're done, or else $[G(\root3\of b):G]=3$. (why?) So where does this leave us? ... BTW, it's important that 2 and 3 are prime.

If $\sqrt{a} \in G$, then deg = $6$?

well, if $\sqrt a\in G$, then $\root3\of b\in G$ as well, and we have $H$.

and we know already that deg(H) is 6

But we have $H\subset G$, so we're done.

Ooh!

I wish I could use emojis in here. I'd have a wow expression or maybe a heart eyes emoji for that one

LOL ... I get your sincere happiness.

I think I understand. I'm gonna let it percolate and write out some stuff.

Does the other way I did it in my post actually work though? Or was I just spouting nonsense?

I didn't read it, sorry :P

Brute force is to be avoided as much as possible. Dimension of vector spaces is very powerful. Use it. (And multiplicativity of degree of field extension!)

Hi Demonark.

Hey there!

Heyo

@ALannister: I don't see how you've done anything with all that algebra. How did you get $\sqrt a$ as a rational function of $c$? That formula is full of $a$, $b$, and $c$.

Sorry I disappeared there. I have a bad battery and my laptop accidentally unplugged.

All right then. I shall. Thankses, Ted.

Well, that will work, I think, if you again use a geometric series formula.

You know that fruit flavored sparkling water is bad for your teeth.

Not the plain kind, though. It's the fruit flavoring.

@Blue: The derivative is $x^4/(1+x^2)$, isn't it?

BTW, why was Huy asking for money?

Did someone steal his account and try to run a Nigerian scam on you all?

no, I just need money

can you send me some?

Hi tern/anon/whomever ...

No, I can't really spare any.

:(

sends Huy a dime ... postage due

That happened to my brother-in-law once though - they got into his email account and emailed everybody saying he went to the Phillippines for the weekend (he's Canadian) and he was stuck there and needed to be wired money.

Oh, rats, the factorials. Yeah, I messed up, Blue.

All his son's friends got the message, and they were like "Why is your dad in the Phillippines?" and he was like "He's not!"

@ALannister: are you someone I know and you changed your name or are you new?

I've had many names. Been on here since 2013.

so I'll probably know some of your earlier names

do you remember them and want to tell me?

I like how the person pretending to be @Huy is pretending to know people in this chat.

LOL @PVal

who are you? can you send me some money, please?

Not everybody liked my old name

@Blue $\sum_{n = 0}^\infty (-1)^n x^n/n!$ is $e^{-x}$. That's probably relevant.

I mean not everybody liked me that much when I went by that name.

hands Huy so much monopoly money that the paper can be sold for real money

@ALannister: = ?

I have a coin for 2 Argentinian pesos somewhere

hi chat

It's the equivalent of like 12 cents or something

When I was maybe 3 years old, I tried to buy something in a store with Monopoly money

If you differentiate your sum and add it to the original sum you should get an exponential series (because you recover the odd terms)

@Huy Are you joking, Huy, about the money?

@JasperLoy: no, I never joke about money.

@Huy I can send you all my Bitcoins

LOL

I think, actually, a generalization of the first derivative criterion follows from Taylor's Theorem.

if you have one, that would be nice already. @Akiva

The website's not working so I'll have to use snail mail

@TedShifrin you mean the Taylor error formula?

(Joking. Also I don't have any Bitcoins)

@Huy If I were rich, I would definitely help you, because I know you well enough. But I am in need of money myself.

If $f^{(k)}(a)=0$ for $k=0,\dots,n$, and $f^{(n+1)}(x)\ge 0$ for all $x\ge a$, then $f(x)\ge 0$ for all $x\ge a$.

@Alannister you gotta sneak in the monopoly money with real money to create the illusion of having more money than you actually do

Yes @ALannister.

of course 3 year olds probably dont have money

heya, Eric.

@JasperLoy: that's very nice of you. if I was rich, I would help you too.

How's the Ric argument, Eric?

I'm actually 24, Asians just look a bit younger.

@EricSilva nope. And if I remember correctly, it was one of those math workbooks for kids.

@Huy By the way, I really take it that you are not joking, and I am also not joking with you.

hi @Ted, I actually got stuck on it

I could perform one of those scams that makes you think I'm giving you free money when in reality I'm stealing from you

my brain's been going at snails pace the past couple days

@ALannister: now what was your name

Eric: Maybe I should make Escargots à la Bourgignonne with your brains :P

appetizing yet terrifying

I put a comment on the question which should give it away, Eric :P

You mean you are a Nigerian prince, @AkivaWeinberger and you want to give me millions of dollars? But in order to get it, I have to send you $1000 so you can pay the international transaction fees? No problem. Done.

Well, anything with enough butter and garlic (and fresh parsley) has to be good :P

i wont look

Well, it doesn't seem like the OP or the first answerer got it from my hint, but ... :P

ill redo that classic exercise and see what happens

@Huy Hello I am the IRS we have discovered bad things in your account please send your credit card number so we can sort it out

@Blue: Did you see my suggestion?

@Akiva: we don't call it IRS over here.

DogAteMy: No wonder your parents ran away when you came home!

I have a solution to the Riemann hypothesis, but I need $500 in order to publish my solution. If you send me $500, I will promise to share half of the millennium prize with you.

@ALannister Did that happen with twitch at some point? People could "donate" money, but get the money back if they asked quick enough... Then the transaction fees would mess up the streamer?

@Huy Well then who are we

@Dair dunno. It was a thing going around in people's inboxes since the late 90s

When did @Dair show up?

But to this day, some people still fall for it.

pls identify yourself @ALannister

@Huy no.

By the way, someone posted a solution to P=NP to the arxiv not too long ago. It's apparently by a well-respected professor, but obviously it hasn't gone through the required peer review yet

Hi @Ted Interesting day we're having, right?

If you're referring to news, @Dair, every day is a disaster.

Oh, @Ted, I was just thinking of the weather...

@akiva It's by a guy with like 2 other preprints, and only one is in a related area.

@ALannister: pls

If the sun disappeared it would be a dis-aster

LOL, oh.

Are you still in SD, @Dair?

Yup.

The people who need to know know.

I leave tomorrow early morning.

ok

@PVAL "Hey guys can someone graph this function for me? Let $f(x) = 0$ if the Riemann hypothesis is true, $1$ if it isn't. Thank you so much guys"

It did slightly drizzle on me when I walked to meet someone for lunch, @Dair.

I've got some math to do and some dinner to fix.

Wait... what about coffee? Lol. I guess that is out of the question.

G'day.

I thought we were planning that in Berkeley, @Dair? When do you get there?

i cant keep up with the chat

I get there the 22nd.

@Dair All his coffee got turned into theorems by that machine

Ohhhh .... And I leave to come back here on the 22nd. You confused me in our last discussion.

good night kids

time to sleep @BalarkaSen

Bye, Huy.

who have ever thought to invent Catalan numbers

welp. rip. my bad.

Yes, @Balarka, as usual, it's way past your bedtime.

@AkivaWeinberger A comathematici... oh wait

Actually it's earlier than my bedtime.

one of these days...

If you decide you wanna have coffee, @Dair, next time you're back here, for sure.

@Kirill: Eugène Catalan

I probably won't go back to Berkeley for another year.

My sleep cycle finally cycled behind my bedtime.

A penny for the sleep cycle.

@Ted Well, I'm not sure where I'm going to end up after this semester since I graduate this semester haha.

Oh, right. Well, keep me posted!

Ok.

@Blue: Did you find it?

Heh

Today's Pythagorean triple day: 8/15/17

@Blue Told you something like e^-x works.

Tomorrow Wednesday is Debian's birthday.

Who?

@Huy they are named after him

And next Wednesay is my birthday, one week after Debian's.

@Blue Well, uh, you mean cos(x) I guess.

@AkivaWeinberger Debian is a Linux distribution.

@Blue: My way works too, I think, to get rid of the factorials, if you differentiate more times.

The founder of Debian, Ian Murdock, committed suicide a couple years ago. Rest in peace.

Yeah, those are terms in the $\cos$ expansion, Balarka.

Anyhow, I'm gone for now. I have packing to do.

@Akiva Oh, I get it, it's pythagorean triple day becasue $8^2 + 15^2 = 2017^2$.

See you on 22nd, @Ted.

More likely 23rd.

I can bug people more constructively after my exams.

All the better.

Wait @Balarka, you are in SD?

Not that kind of "see you," @Dair.

I live on the antipode of San Diego.

Well, at least somewhere near the antipode.

Not many people know the word antipode.

Aren't we roughly in the same hemisphere, Balarka? So antipode seems unlikely.

antipodesmap.com

Did you use that link to prove Borsuk-Ulam, @PVAL?

LOL

Antipode of San-Diego is just east off the coast of Madagascar in the Indian Ocean.

Who knew they had internet there?

Balarka makes the best puns in this chat, while Ted makes the best dinner.

Fun fact, Taiwan's old name is Formosa. There is also a Formosa Province in Argentina. These two places are exactly antipodes of each other.

How do you know they have internet?

Oops

I'm in the antipode$^2$ of myself

@Jasper: I defy you to prove that Balarka's puns are anything but excruciating.

@Dair Balarka is posting from there.

@AkivaWeinberger Fun fact: the word apparent has two meanings which are opposite of each other's.

Really? How?

wai am i a meme all of a sudden

I think @Hippa's been too busy and has gone out of the meme business.

@BalarkaSen Isn't that what you've always wanted

@TedShifrin If something is apparent, it can be true, or it can be false. QED.

Apparent means either "obviously true" or "seemingly true"

Yeah, I don't agree opposites.

@Blue "are clearly positive." lol.

@Blue: I still like my idea to differentiate until you get a geometric series.

Anyhow, I'm gone.

What's the question?

@Ted How can you differentiate until a geometric series? Trailing factorials will be in the end.

@Blue

Alright, gonna hit the bed

see ya

Fixed points of differentiation... Interesting. Can you can find a fixed point of the differential operator that ins't $0$ or the original function itself?

cya @BalarkaSen @Ted

@Dair Are you asking about solutions y'=y?

@PVAL-inactive I guess I was, now that I think about it... you can't lol...

Well the solutions to that are all Ce^x=y

it seemed a lot more interesting with the word "fixed point" in there...

where C is an arbitrary constant.

if you differentiate ye^(-x) where y is such a solution, its easy to see that all solutions have to be of that form.

y'=y is probably the easiest non-trivial example of a linear ode with constant coefficients.

so you should see how to deal with similar things more generally in any 1st course on ODEs.

@Ted Im pretty sure this Ricci problem is just counting things

@PVAL-inactive I have lol. I just brainfarted lmao.

@TedShifrin: Is this not a consequence of the Hairy Ball theorem? Once the differential has rank $1$, you can just smoothly assign that vector to the points on the sphere. math.stackexchange.com/questions/2394540/…

I forgot so much the ODE stuff though. haha. I remember making a lot of ansatz.

Can anybody check that I am not missing something here?

I've got a really vague question. What statements are provable?

Actually, I'm more interested in a slightly different question. What arithmetic statements are provable?

It seems like there's some sort of consensus among mathematicians as to what constitutes a valid proof.

I'm not completely sure whether you're trolling or not.

I mean, I think it's generally agreed that a statement is provable if and only if there exists a formal proof of it in a "good" theory.

But, there are "arithmetic" statements which are known to be unprovable but still true. (Godel's second incompleteness theorem)

Yeah, I admit that what I'm saying does sound a lot like trolling...

Right.

I guess I am confused what you're asking. It seems like you're not asking about what constitutes a proof, but instead what constitutes a good theory. Am I right?

@Faraad: Well, a line field, yeah. That's what PVAL explained in his comments.

@EricSilva: That's what my comment said, with a bit more explicitude.

Yes, that's right.

@Karl I think there is an easier way to explain your answer.

Oh, and hi, @Faraad. Hope you had a productive summer.

Right, cool problem

@TannerSwett Hm, I'd be interested.

The key thing, IMHO, is to view curvature as a self-adjoint map on $\Lambda^2 TM$, @EricSilva.

The reason I bring this up is that I don't know of any foundational theories that I'm convinced prove exactly those statements that I'd intuitively consider "provable".

PA is too weak, because Goodstein's theorem definitely feels "provable", but PA doesn't prove it.

@PVAL-inactive It's an old answer, and I don't plan on improving it. Go ahead and write a better one.

I guess only when $n=3$ do we have $n=n(n-1)/2$, huh?

Yup

I've been reading this paper by Allard on varifolds and man this stuff is so technical

@TannerSwett Good. Worthy of a question imo.

ZFC is too strong. All of its axioms do seem pretty unobjectionable, but "pretty unobjectionable" isn't really a good standard, I don't think.

I opined that GMT and varifolds were a bit much given where you were in math, @EricSilva, but different people have different tastes.

Allard is quite a character, though.

I actually think I'm getting a lot out of it now

I'll be writing an expository paper on the Allard regularity theorem

@TannerSwett Indeed. One would expect a "tighter" "fit" for the best theory of arithmetic.

I'm not sure what, exactly, to ask, though. "What theory captures the notion of what we consider 'provable'"?

I think I've come back around to enjoying it after I took a little break to work on Bryant @Ted

ohhey ted and others

In any event things are going good now

Does that sound like a clear and specific question? I don't want to get closed as unclear.

Now that you're at the end of it all, would you still have taken Riemannian stuff over this or no?

Like I know you Neves had his reasons, so I'm just asking for your personal preference

I mean yeah

@Daminark Neves wouldve done geometric flows but I was the only one with Riemannian background

@TannerSwett What I would do. Title: "A quest for theories that meet our expectations" Somewhere in body: "Given that the set of provable statements for a theory varies with the theory itself, while we have a fairly fixed sense for which statements ought to be provably within the theory, there seemingly is a best theory for, e.g., arithmetic which is in the middle between PA and ZFC. Has this sort of problem been studied before?"

Maybe next year more people are gonna know about it so you can do that sorta thing

Is Neves specifically into Riemannian geometry?

(I don't really know any other types to contrast with but... yeah)

@KarlKronenfeld What would you think of replacing that last sentence with "Does such a theory exist, and if so, which one is it?"

He's a geometric analyst and has worked in mcf and yamabe stuff, along with the willmore stuff @Daminark

@TannerSwett The question becomes less answer-able, because there is some degree of subjectivity underlying "best" theories. It probably makes the question clearer though.

Ah, I see, that's cool

It's pretty dope stuff, the things he's told me abt it has made me p sure I wanna pursue geometric analysis of some sort

By the way, I've noticed a certain paradox I find kind of disturbing.

Nice

"Theorem". No theory can satisfy us as adequately capturing what we consider provable.

"Proof". Suppose we have a theory T, and that we think it's clear that a statement is provable if and only if T proves that statement.

Well, clearly we think that T does not prove that 0 = 1. So we think that T is consistent. And if T captures our notion of what's provable, then T must prove "T is consistent".

But that contradicts Gödel's incompleteness theorems.

Heh

Hm, must 'T is consistent' be considered to be provable, or just true?

Good question.

Let me try another version of the opening assumption.

Suppose we believe that P proves all statements we believe, and that all statements proved by P are true.

Um, P was a typo for T there.

Then, you're not addressing the statement of the theorem.

Certainly it wasn't a well-stated theorem in the first place.

The epistemological gap between true belief and statement S such that we have the belief that S is provable still remains.

I certainly feel like in order for a theory T to be a "good theory", then T must prove all arithmetic statements that I believe, and it also must prove only those arithmetic statements which are true.

Have I just ruled out the possibility of a "good theory" existing?

Yes.

Though, you may have a sufficiently non-standard view of what constitutes 'arithmetic' I guess.

What I mean by "arithmetic statement" is a statement in first-order logic where all quantifiers range over the integers and all primitive predicates are computable functions.

Does that sound like the standard definition?