12:01 AM
I thought the hindu God of CS was Dr. Chandrasegarampillai from 2010 Odyssey Two?

shrug

I assumed Srivatsan was a buddhist... Not that it matters much. I know nothing about religion.

I assume that I don't care what he really is...

I hope my assumption doesn't offend you, Srivatsan.

@Srivatsan yeah, but more interesting is the deleted answer..

12:02 AM
Didn't you go to sleep Asaf?

From this assumption I can prove that I don't care if he is Hindu, Buddhist, Self immolating monk, etc. etc.

Hi tb!

Hi Matt

@Matt Oh noes! I'm sleep chatting again!!! The horror!
Hi tb.

Hi Asaf

12:03 AM
You might enjoy the peculiarity in my last answer on

A self immolating monk?

@Matt They're a religion unto themselves these days, didn't you know?

@Matt Quite.

Though I wouldn't want to be responsible for their evangelism program.

@HenningMakholm Oh, somehow I thought it went to the Poincaré comment (which I would label similarly)
@AsafKaragila Well, nice try, but still not amsplain :)

12:07 AM
@tb Well, I don't have a bibtex compiler installed in my head yet. I only have plain LaTeX compiler, and some set theory compiler. Also I have installed that emacs parenthesis highlighting for Lisp writing.
It's the contents of that reference I'd think you'd find intriguing, though. :-)

@AsafKaragila I barely consumed the votes today. I ran out of them only an hour ago.

@Srivatsan Very surprising! Hung over from last night's partying? :-)

@AsafKaragila Yeah, but in good Asaf style I felt like complaining first

@tb Touche. ;-)

12:09 AM
@Srivatsan nah, it was my mistake.

@Matt Not at all. Assumptions are fun.
tb: Whatever I asked you and Henning, I posted in the main site. Guess who shot it down quickly... =/

@AsafKaragila How bizarre. However, at some point in the near future (a few more answers of yours) I will be surprised if I'm not surprised by mindboggling conclusions in the absence of choice.

@tb Yeah, it's all too bizarre but you get used to it.
Although a decomposable continuum is very bizarre.

Okay, this has to be put to a stop. I am going to bed, and I will not chat from my iPhone at least until the morning :-)
Goodnight!

12:15 AM
G'night, Asaf

@tb I'm was expecting a negative answer; but what I didn't imagine was that the answer could never be positive...
Esp. with Henning's partial solution for rationals.

I bet that the answer to this one is no. I couldn't come up with an example, yet. I always have to discard a few points in my examples...

"asked Dec 29 '11" -- so even posts from a few days ago now get a qualifying year!
I'm used to considering things with a year in the date to be old.
@Srivatsan Just goes to show how strong the Intermediate Value Theorem is, despite being so everyday.

@HenningMakholm In this case I don't see how applying a canned theorem let's me make progress skill-wise. I needed to work out the details of that proof.

@HenningMakholm Agreed.
@Matt Well, don't forget that knowing and applying canned theorems is a separate skill worth mastering.

12:26 AM
@Matt Fair enough, but the skill the exercise is meant to train might be that of recognizing that the canned theorem applies.

@HenningMakholm well, if you go for canned theorems, you could as well go for the canned theorem "continuous with compact support implies uniformly continuous".

@tb Thank you.

That wouldn't have been canned yet at the point in a course where proving that very theorem was posed as an exercise.

I'm not in the mood for this argument.

Yes you are.

12:31 AM
Will the Lebesgue number lemma give another proof of this? // My guess is that Matt's proof implicitly invokes this argument anyway.

@tb Never mind editing your comment while I'm typing a reply to it : )

@Matt You had me utterly confused. Anyway, the proof looks fine up to that point but it's too late for me now, too.
(and sorry for editing/deleting)

Don't be I wasn't serious.
Well then, good night folks.

Good night, Matt!

@tb Don't stay up too late then. : )

12:42 AM
@Matt I'll try :)

1:12 AM
Assume we have a function
H(x) = f(x) + g(x)
How do we show that H'(x) = f'(x) + g'(x) ?
I guess, I am asking how do we show that differentiation is a commutative operation

Write down the difference quotient defining the derivative: $$H'(x) = \lim_{h \to 0} \frac{H(x+h) - H(x)}{h}$$ plug in $H = f+g$ distribute and use that $\lim (a_n + b_n) = \lim a_n + \lim b_n$.

I wouldn't say commutative. I would say $\mathbf R$-linear.

I thought about that the word commutative is wrong, but could not find a better word. Sorry for my english =)

Sure.

1:17 AM
t.b are you sure?

I mean are you sure about $$\lim (a_n + b_n) = \lim a_n + \lim b_n$$

As sure as I can be :)

$$a_n = \cos^2(x) \qquad b_n = \sin^2(x)$$

You need both limits on the right to exist, of course.

1:20 AM
There's no $n$ in your sequences. Note that you're assuming that $f$ and $g$ are differentiable at $x$ otherwise the right hand side of $H' = f' + g'$ doesn't make sense.
(presumably you meant $x = n$).

Well the clock is 2:20 am here, I guess i can afford being a tad sloppy

Sure, so do you have the argument, now?

My point is just that I know for an example that

$$\lim_{n \to \infty} (cos^2(n) + \sin(n)^2) = 1$$

But I am not sure if

$$\lim_{n \to \infty} \cos^2(n) + ( \lim_{n \to \infty} \sin^2(n) )\; = 1$$\begin{align*}H'(x) & = \lim_{h \to 0} \frac{H(x+h) - H(x)}{h} \\ & = \lim_{h\to0} \frac{(f(x+h) - f(x)) + (g(x+h) - g(x))}{h} \\& = \lim_{h\to0} \frac{f(x+h) - f(x)}{h} + \lim_{h\to 0}\frac{g(x+h) - g(x)}{h} \\ & = f'(x) + g'(x)\end{align*} Or rather, the sum of the last two limits are hard for me to imagine being equal to one. 1:27 AM The point is that the last two limits don't even exist. If one knows or can prove that the sum of two limits are additative, the argument follows quite easily as you show =) But if f and g are differentiable, then the limits involved in their derivatives do exist, and limits that do exist are additive. How can the first, but not the second exist? Again, for\lim(a_n + b_n) = \lim a_n + \lim b_n$to hold we really need the two guys on the right to exist. I always show my students$a_n = n$and$b_n = -n$and stuff like that. I feel rather dumb now... Sorry 1:29 AM It happens. So lets assume a sum of limits do exist, that does not mean that the limit of the individual parts exists? If two out of$\lim a_n$,$\lim b_n$and$\lim (a_n+b_n)$exist then so does the third. So it is not a commutative operation, if$ \lim a_n + \lim b_n $exists, then$\lim (a_n + b_n)$exists. But on the other hand if$\lim (a_n + b_n)$exists, that does not imply that$\lim a_n + \lim b_n$exists? =) Correct -- as your 0,1,0,1,.../1,0,1,0... example shows. Math can sometimes be strange 1:55 AM @N3buchadnezzar Not at all. @N3buchadnezzar Any nice object can typically be broken down into bizarre sub-objects. What one usually expects is that when you assemble an object from nice parts, then the resulting object will also be nice. [Even that is false in many cases, but that is another issue.] For example, imagine taking a mirror and smashing it; let's say you ended up with two weird-looking pieces. You did start with something nice, but smashing destroyed the niceness. And just because the two pieces join together to form the original single nice mirror, we do not expect that the pieces themselves are so nice. And the master smiled and started sipping his tea again. 2:11 AM Last year was a pretty good year. There are a lot of "new" people up in the ranks. :) @JM Hey JM. Hey Sri. :) 2:23 AM #64 woo @DylanMoreland ?? Woo, +508! :p @Srivatsan The list that J.M. posted. Oh, right. Out of curiosity, did anyone here vote to reopen the big-list mispronunciation question? I was the last vote for reopening the Tikhonov-Alaoglu thread, but I didn't vote for the mispronunciation thread. 2:33 AM @tb Well, I agree that the Tikhonov-Alaoglu thread must be open. I don't understand why the other one is being reopened now. I left a comment in Jonas's post here: meta.math.stackexchange.com/questions/3409. Snooow! @Srivatsan I'm not sure if it must be open, I'm not a big fan, honestly. But I thought that I'd like to see how it works out and I'd like to see a really good informed answer to that. Hm, ok. My feelings are summarised by Jonas already. =) I tried searching the various obvious Greek spellings of Alaoglu but never found more than a thousand results. That makes me think it's an unusual or mutated surname... @ZhenLin It's the Turkish name Alaoğlu, originally. Ah. Then the ğ must be basically silent. 2:41 AM @ZhenLin yep. @ZhenLin Given that A.'s parents were of Greek nationality and he was born and raised in Canada, no one can tell for sure how to pronounce it from just reading the name. Usually can't go too far wrong if you trace etymology... Well, have you ever tried to pronounce Italian-sounding last names of Americans? No way to tell... What are some examples? Special props to Mariano for keeping his ranking constant between years :) The reputation league pages have recently (?) changes to showing the logged-in user's ranking at the top of the table. I keep going "woo! I'm number 1!" and then "oops, it's just .." 2:52 AM Barbagelata (“frozen beard”) :) 3:07 AM My family name means "kill". But it's an ancient family name, so it's probably been long semantically bleached... 3 hours later… 5:47 AM I found a .djvu copy of Hartshorne with a clickable table of contents. 2012 is already great. @DylanMoreland wonderful... Are cross-references clickable as well? No :( Oh, I didn't mean to spoil your 2012! Anyway, djvu things are usually OCR'ed, so quickly searchable and that can be tremendously useful, especially with a book like Hartshorne. Indeed. But to get to certain sections I had gotten in the habit of memorizing page numbers Is this really mathematics? I had a quick glance at that paper and it didn't appear too mathematical to me, but admittedly, I didn't understand much of it. 5:57 AM I wish the OP had given any sort of hint as to what this principle is. 6:34 AM @DylanMoreland Which djvu viewer supports things like bookmarks, hyperlinks etc. I am using the viewer from lizardtech, several times I've seen djvu file, which should be bookmarked (according to descrition) but I did not see bookmarks there. 7:07 AM thanks for the edit @tb @MartinSleziak I use DjView. It is the best of a bunch of bad options. Thanks a lot Dylan. The LizardTech viewer might have more features—features I'd really like—but it always does funny things to my setup. In fact, the LizardTech (which made that viewer) changed its name to Celartem in the meantime - probably the company was sold. Now they have already a new name, which I do not remember. 7:29 AM I wish I could turn off that "someone has posted an answer already!" thing Can anyone tag a question on meta as faq, or should this be done by mods only. Is there some policy on this? @MartinSleziak Only mods can do it. This is the point of the red tags. @tb So I should tag a question as faq-proposed or write a comment? I thought that this one shoud be a faq: meta.math.stackexchange.com/questions/3286/… 7:47 AM @MartinSleziak I think this is something worth consideridering. On the other hand, it might be better to have as few (faq)'s as possible. It could be worth considering extending your "why should we accept answers" to "why should we accept answers and how to accept answers" or something to that effect. Hopf of the morrow to you, and whatnot. @tb Well I do not like idea of combining questions. Morning, Asaf. New year's greeting or are you running out of permutations? @MartinSleziak I see. I don't feel strongly either way. Maybe flag the question for mod attention? @tb Latter, of course. tb: I have already tagged that questions as faq-proposed and wrote a comment there. But if you think that is not good thing to do, feel free to edit/comment. ok, I'll flag the question too Thanks for your advice! 7:56 AM So what's up? If procrastination is a bad thing, why does it feel so good? 2 @MartinSleziak I'll tell you why it's not a bad thing, but later. Asaf I see your answer on 1+1=2 has already score of 29. Gold badge is at 100? Yeah. He won't reply to my comment either! :\ Some SE linkage indicate that he's about 13-14 years old. He probably didn't understand too much of my answer, and I want to improve it. I'm not sure how though, I tried to write more but I ended up writing additional 700+ words (the current answer is about 500), so I decided to save it on my hard drive and not post the edit. I should think more closely about it, what and how I want to write. I don't care writing 700 words, but I don't want to write 2000. No one will read a 2000 words monologue. I think the first sentence In a very "raw" sense the symbol 2 is just a shorthand for 1+1. There is really not much to prove there. is understandable for 13-years old guy. 8:07 AM I discussed it here with Henning and Zhen yesterday, I should probably add some more about what is a mathematical proof (and what is a formal proof), and why do the symbols are mostly irrelevant while the fact that 1+1=2 says something about the axioms instead. You seem to put a lot of effort into your answers... I think I've seen something like keep improving it, until you think it's perfect in some meta thread. @MartinSleziak Yeah, I do. I think they help me a lot later when I explain things to my students in class. I also enjoy talking about mathematics, so it's a win-win :-) You teach things that would fall under here? Or something more advanced too? I'm a TA in an introductory course in set theory and logic. The next semester I'll TA calculus to engineering students. I never taught calculus. 8:14 AM Most of what we teach is elementary and naive set theory, and basic propositional/predicate logic. I enjoy teaching set theory, although it's a very elementary course. (It's for education majors - I hope this is the correct English term.) If I'll go to do my Ph.D. in Jerusalem I may get to TA something better. By something better you probably mean more advanced set theory...? Ok, I'm off to school. See you guys later! I made a weird mistake in writing an answer Well, in my university any advanced undergrad courses have no exercises class. In Jerusalem they all have. So I can perhaps TA courses like topology, logic, etc. 8:26 AM Thanks for the chat Asaf, I guess not many people will be at work already, so this chatroom is probably my best chance for social interactions... ;-) @QED what kind of mistake? I was trying to solve xP(x,y)=0,yQ(x,y)=0 and I know there is a solution with x=0, so I factored it out but I did the same for y=0, which isn't actually a solution not good :) But these things happen: In a recent answer I had to modify a silly counterexample at least 5 times until it was correct, so... aw I wrote a better answer than the other guy buy his was accepted already 8:30 AM Post it anyway. yeah I did I don't really understand why people post "see wikipedia link" as answers.. I don't understand that either. I find that rather annoying. What I find even more astounding is that people upvote that stuff. Recently an answer got 4 upvotes and the link didn't even remotely answer the question posed. yeah lol It's better as a comment, but really - we should just close questions that are well answered by wikipedia (with a link of course) Well, I'm not sure if everybody would agree with that. what do you think? I can't see any reasons against it 8:40 AM I think the problem with that is where to draw the line. I don't see a clear-cut criterion to decide on that. That is: I don't like questions that can be answered by the most obvious Google search. On the other hand, many questions reveal some specific confusion that can be much better addressed by someone answering. that's true but presumably the moderators can decide I think generally the point on the circle$(\frac{\sqrt{A}}{B},\pm\frac{\sqrt{B^2-A}}{B})$corresponds to an irrational angle. like the radians wouldn't be any rational multiple of pi Does this exaust the algebraic deg. 2 points on the circle of rational angle? basically, what values of sin(pi q) for q in Q are quadratic irrationals This doesn't answer your questsion, but according to this the only$n$for which$\cos{\frac{2\pi}{n}}$is algebraic of degree 2 are$n = 5,8,12$while it is rational for$n=1,2,3,4,6$. 5 hmm I think that does answer my question! @AsafKaragila "In what class were you given this question?" - In none. I am just reading a textbook on set theory and computation and sometimes I stumble across the problems I do not know how to solve or material I do not fully understand. I see :-) I was hoping it was an assignment given by a professor, and you could surprise him with the reference. 8:56 AM @QED well,$\sin{\frac{\pi}{5}} = \sqrt{5/8-\sqrt{5}/8)}$$$(x^5 - 10y^2x^3 + 5y^4x, 5yx^4 - 10y^3x^2 + y^5) = (1,0)$$ @Daniil If it's a book, then the assumption of the axiom of choice is quite probable. that's not degree 2 but it's a quadratic in x^2 so it's a square root of a square root as you showed (if you factor out x) @AsafKaragila It is, actually. However in this part of the book they do not deal with cardinals arithmetics yet. I thought I had to use Cantor-Berstein because the question was introduced in the same chapter under the C-B theorem. @tb, I think that's a mistake in the paper Would quite like to look at this paper.. 9:00 AM @Daniil I see. Well, either way. The result does require some choice. that (p-1)/2 is curious oh not really oh wait There is also the same question about the square (the square on the plane is divided into 2 sets, prove that at least one of them is equinumerous to the whole square). 4x^2+2x-1 @tb, this is interesting! @Daniil This can be reduced back to the line segment. Since there is a bijection between the line and the square, this gives a bijection from the partition of the square to a partition of the line, which we already solved. it's the pentagon! of course duh lol 9:03 AM @QED Gentlemen, you can't fight in here! This is the War Room :D @AsafKaragila yeah, true. But I still can not understand how to solve this problem without cardinal arithmetics. (In fact, the only thing about cardinal arithmetics they covered so far in this book is that if A is infinite and B is countable then$|A \cup B| = A$) @Daniil This too requires some choice. @tb, now I have to add this to my answer... but it looks very difficult, thanks :P @Daniil Or to have the definition of infinite has having a countably infinite subset. Which book is that? 9:06 AM @QED still want to have a look at that paper? By definition: set A is infinite if it contains a strict subset B, such as A equinumerous to B. no @Daniil Yes, this is equivalent to having a countably infinite subset. This is known as Dedekind infinite. Morning. Hello 9:08 AM It is possible to have very strange sets which are not with bijection with any finite set (thus infinite) but still have no proper subset which is equinumerous to them. Yeah. The book is: mccme.ru/free-books/shen/shen-logic-part1.pdf (A. Shen, Lectures on mathematical logic and algorithms) @QED okay :) Nevertheless this is nice. Didn't know that. @Daniil Oh, I can't read Russian :\ @AsafKaragila I am perfectly willing to accept the AC, can you please explain to me how to solve this problem without cardinal arithmetics? That's really cool 9:12 AM @Daniil Eventually this amounts of cardinal arithmetics without identifying them as such. Let me think about it a little bit. @AsafKaragila Ok. Sorry for taking your time. @Daniil It's alright, my time is worthless :-) I have prepared a folder called "MSc Thesis" and any day now it should be filling itself with my thesis :-P Either that, or I'll actually have to write it during the next semester. I'll go run a shower, I'll think about this some more and edit my answer when I have a solution. Should be later today, I expect. Jonas Meyer should not have gone through the trouble to write a canned answer. link? 9:23 AM I don't really understand doesn't matter :2864911 Why? why what @Daniil Is there no additional information of the partition? Just a general partition? @QED Would you think I'm a Ph.D. student? (I'm somewhat flattered, I guess) I dunno all the whiskey I guess Haha!! :-) 9:30 AM @AsafKaragila No, just general partition. @AsafKaragila I thought you were a PhD student too, because you cite and talk about graduate-level texts. @Daniil Interesting. Seems like a very hard question, if I knew all the contents covered before this question I'd probably be of more help. I'll keep thinking about it. @Daniil I am a grad student... I also research in a field which has been somewhat low on activity over the past 30 years, so it's easy to get around there. @AsafKaragila Well, it does not have to be very formal, I think. @Daniil I don't see any other argument than cardinal arithmetics. @AsafKaragila Ok, thanks. Don't worry about it, I'll ask my teacher after the winter break. It's just that the cardinal games is such a simple argument, it distracts you from seeing other possible options. 9:41 AM @tb So spelling out why it's uniformly continuous on the whole space if it is on its support is the right thing to do? Well. I'm going to head out now, I gotta swing by my office before heading to Jerusalem. Sweet dreams : ) See you folks when I've got wifi on the bus (I hope!) Bye Asaf. Have a good ride, Asaf. Hm, this book has a really interesting Cantor-Bernstein proof. Ok, I'll be going; have a great day, everyone. 10:09 AM and you I've a Q in HCF and LCM , May I ask? ? ya ask How do I prove the equality : X*Y = HCF(X,Y) * LCM(X,Y) ? take a look at this @NikhilBellarykar where? 10:24 AM I have posted a link, can you see that? @NikhilBellarykar Thanks , lemme check :) @NikhilBellarykar That proof is really above my head :( syntax'es are really eluding me like d∖a∧d∖b well hold on a sec d\a means d divides a and d∖a ∧ d∖b means d divides both a and b @MrAnubis ,that's a really nice one gcd * lcm First prove gcd(p^i,p^j) lcm(p^i,p^j) = p^i p^j 10:36 AM ok.. for a prime number p then you just combine it for all the different primes and use multiplicative nature of gcd right @QED ? yes @NikhilBellarykar k cool I was wondering whether covering systems of integers could be put to more general uses other than proving that there exist infinitely many numbers NOT of some type does there exist some literature on this with EXAMPLES? @ all ? all r zzzleeping in the complex plane it seems.. 10:52 AM I don't know @MrAnubis, did you get it yet? @QED no :( , but reading a googled article. Did you prove$\gcd(p^i,p^j) \text{lcm}(p^i,p^j) = p^i p^j$? @QED Can you please use TEX, those syntax'es gives me headache He does. ? I think no such numbers exist math.stackexchange.com/questions/95774/… 11:03 AM @QED sorry (was reading that article) , Now understood it :) @QED Thanks for your good concern , I appreciate it :) 11:28 AM Must every automorphism of a vector space have some polynomial equation it is a root of? Or some other subspace I can associate to the automorphism. Finite-dimensional vector spaces? Sure. But otherwise choose your favourite automorphism with infinitely many eigenvalues... Assume it has none. And of course this is not f.d. :-) What field are we thinking of? General at the moment. Hmmm. Well, if I remember correctly, it's a general fact that if$\alpha$is a linear operator and$p$is a polynomial such that$p(\alpha) = 0$, then any root of$p$is an eigenvalue, and vice-versa. But this is most meaningful in the algebraically closed case. 11:39 AM Does such polynomial exist always? As I alluded to above, if$\alpha$has infinitely many distinct eigenvalues, then there can't be such a polynomial. Hm, no, roots of$p\$ need not be eigenvalues. But the eigenvalues are always roots.

Then the useful implication still holds. :-/
I am trying to generate a decomposition of the space from an automorphism.
Argh. No! I am doing it all wrong again :-)
I already concluded two weeks ago that this implication is impossible.
I am going to stare at the road vacantly while thinking about the solution. See you later, and thanks!

Bye!