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12:00 AM
@AlexanderGruber I upvoted you.
People need to work harder!
 
thanks. and seriously, they do... it must be copying for homework.
 
12:24 AM
@AlexanderGruber Do you know about set theory?
 
12:50 AM
@AlexanderGruber You there?
 
Alexander has left the building ;-)
 
Darn.
 
which to study first, category theory or general algebraic systems ?
 
@nerdy What?
Why are you wondering about studying those?
 
which subject to study first ?
 
1:04 AM
I have been told that Category Theory is something you study after you know a nice deal of mathematics, else it is useless.
 
oh got it
 
What is "general algebraic systems" about?
Rings, fields, groups...?
 
include those structures with a very simple axiom structure, as well as those structures not easily included with groups, rings, fields, or the other algebraic systems. In this field one may consider the general nature of algebraic axioms and how the different classes of them are related.
yes
 
@nerdy That you can study whenever you feel like you're ready.
 
does it require knowledge of lattices and ordered algebraic structures ?
 
1:10 AM
@nerdy Depends on what the course gives.
But you can always look up the definitions! =)
 
1:27 AM
Does anybody have any hints about sending mathematics over email?
 
The frequencies of this scale can perform miracles.
THERE'S A FREQUENCY THAT CAN REPAIR DNA!!!! XD
 
@GeorgeV.Williams I send PDFs
 
1:43 AM
@GustavoBandeira I call bullshit on that one!!!
 
@robjohn, thanks, I hadn't thought of that.
 
@robjohn Rob.
 
@PeterTamaroff yes?
 
@PeterTamaroff Yes! XD
 
@robjohn Why does the Dedekind construction of $\Bbb R$ work out "so nicely" for $\sup$s?
That is:
 
1:45 AM
@PeterTamaroff yo i'm in and out, it's my lady friend's birthday and we are getting ready to go dancing
i do not know much set theory though so it is not much of a loss.
 
If we're given a nonempty collection of real numbers that are bounded above, $ \mathscr C$, then $\sup \mathscr C=\bigcup\mathscr C$
 
@PeterTamaroff Do you have any comment for the image above posted by Gustavo? :-)
 
Note that for any pair of sets, $\sup (A,B)=\bigcup \{A,B\}$ @rob so the "solution" is amazingly natural, and works out perfectly.
@Karl'sstudents Didn't I just comment on that?
 
@PeterTamaroff Comment about hyperlink I meant.
 
@GustavoBandeira i hear there's also a frequency that can make you poop, they sohuld include that at the bottom
 
1:48 AM
@Karl'sstudents Seems like the hyperlink is the image...
 
@PeterTamaroff I meant you usually ask people to use hyperlinks for images.
 
@PeterTamaroff so you've answered your question?
 
@robjohn No. I wonder why it works so well.
@anon said something about "limit ordinals" IIRC and $\sup =\bigcup$, but I have no idea what he was talking about.
@Karl'sstudents Yeah, I know. =)
 
It's a property of initial segments, eh.
 
@Karl'sstudents Hi honey.
 
1:51 AM
Since a set of ordinals is a set of initial segments of some ordinal, you have $\sup=\bigcup$ for ordinals. The case you have here is analogous.
 
@skullpatrol Hi my beloved mathematician. :D
 
@user1 Could you explain what ordinals are?
...
How are they defined?
 
@Karl'sstudents I posted a video link in the Party Zone on "Why is it called Algebra?"
 
@PeterTamaroff You start at zero, and I would refer you to some source on set theory for a formal definition.
 
@skullpatrol I will go to the room now.
 
1:53 AM
@user1 OK, but how do cuts come into play?
 
The property that we need here is that $\alpha\subseteq\beta$ when ever $\alpha<\beta$.
 
@user1 Right.
 
See the analogy with cuts?
 
Actually $\alpha\leq \beta$ is defined to mean $\alpha\subseteq\beta$; yes?
(For cuts, I mean)
 
Yes, same with ordinals btw.
Anyway, if a supremum exists, it must equal the union as a result of this property.
So Dedekind cuts are nice because they have this property and they define a complete linear ordered set.
 
1:55 AM
OK, so $0:=\varnothing$, $1=\{\varnothing\}=\varnothing\cup\{\varnothing\}$, $2=1\cup \{1\}$ and so on ,yes?
That's why $0\subseteq 1\subseteq 2\dots$
 
Yes.
You need the axiom of infinity to guarantee the existence of $\omega=\{0,1,\dots\}$.
 
@user1 OK.
And then?
These are the "small" ordinals?
 
Heh. Well, once we have constructed $\omega$,
we can go on surprisingly far.
 
@user1 Yes, I have been told! =)
 
For instance exponentiation is definable as an ordinal operation.
So we have $\omega^{(\omega^{\dots})}$.
 
2:00 AM
@user1 Wikipedia is telling me ordinals are the "order type" of well ordered set.
 
@PeterTamaroff Yes, one can consider the ordinals as we have defined them here to be representatives of equivalence classes of well-ordered sets.
 
@user1 Hmmm... when did we do that?
 
@PeterTamaroff One can. We did not.
 
@user1 Oh. When you say here you mean Wikipedia, yes?
 
@PeterTamaroff I mean in this chat.
 
2:03 AM
@user1 But when did we do that?
 
@PeterTamaroff We constructed the ordinals.
 
@user1 We have $0,1,2,\dots,\omega$; yes?
And then how do we go further?
 
The same way as you do with finite ordinals.
$\omega+1=\omega\cup\{\omega\}$.
 
@user1 Oh, OK. DERP.
And what would be an example of a well ordered set that has order type $\omega +1$?
 
I am devoid of examples, besides you know just $\omega+1$.
It's attaching a point at infinity to $\mathbb N$.
Also, it's worth noting how we can make it to $\omega+\omega$ without creating a new axiom for it.
We have a family $\{\omega+n:n\in \omega\}$ defined by taking successors.
It has a union. That is $\omega+\omega$.
 
2:12 AM
@user1 Oh, OK.
@user1 Is this correctly stated?
"Let $R$ be a ring. Then $R$ is a division ring if and only if the only left ideals and right ideals of $R$ are $R$ and $\{0 \}$."
 
Yes.
 
@user1 OK. I'll prove it.
 
In fact you just need "... iff the only left ideals of $R$ are $R$ and $\{0\}$".
The corresponding statement for right ideals follows.
 
@AlexanderGruber :P
 
@user1 $I\neq 0$, $a\neq 0\in I\implies a^{-1}a$, $1\in I\implies x\in R,x1=x\in I\implies I=R$.
That is for left ideals.
(assuming $R$ is a division ring)
 
2:19 AM
Looks correct (assuming you have the quantifiers right).
 
@user1 Oh, right. Quantifiers =)
$I\neq 0\implies \exists a\neq 0\in I\implies a^{-1}a=1\in I$
$\implies \forall x\in R, x1=x\in I\implies I=R$.
 
@PeterTamaroff Yes. :)
 
@user1 OK, the proof for right ideals is analogous.
 
@PeterTamaroff Do you have any ideas for the other direction?
 
@user1 Well, if the only nonzero left ideal of $R$ is $R$; then pick $a\neq 0\in R$.
It follows that $(a)=R$.
Thus $\exists x:xa=1$
 
2:25 AM
What's $(a)$ (left or right)?
 
Left.
That's why I wrote $xa$ and not $ax$.
 
Ok, now do the right ideal generated by $a$.
 
@user1 Well, in said case we have the same, $ay=1$ for some $y\in R$.
Now $x=x(ay)=(xa)y=y$.
Done.
 
Yes.
 
3:07 AM
@Charlie not sure if tsundere or defensive on jasper's behalf...
what I had to put up with at work
 
3:48 AM
@anon You must have been under a lot of stress.
My sympathies.
 
@PeterTamaroff +1 Nice job...I could use a few myself :-(
@anon Where is that taken from?
 
the final the students I T/A have to take in a few weeks
 
@anon Are you allowed to bitchslap them if they get that one wrong?
I'm off.
Gotta sleep.
 
4:04 AM
g'night Peter.
 
4:15 AM
Definitely does not have a future in math education.
 
 
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$\frac{1}{\infty}$ of an hour later...
 
hahaha
 
in PSTricks and PGF/TikZ, 25 mins ago, by Karl's students
North Korea failed to launch its missiles due to some bugs in PSTricks.
 
lolol
 
9:41 AM
Hello everybody!
 
Is it true that volume form on hypersurface $f(x)=c$ induced by Euclidean metric is $\omega = \frac{\partial f}{\partial x_k} \frac{1}{|\nabla f|} \bigstar x^k$?
 
10:14 AM
@JayeshBadwaik hi Jayesh, do you remember to what equal determinant $\det (E+a a^T)$ for vector $a$?
Ah, yes, it is $1+a^T a$
 
 
1 hour later…
pen
11:29 AM
Jasper Loy... left?
 
pen
...
He could have gifted me all that reputation
 
You're gifted enough to earn it pal :-)
 
pen
@skullpatrol Err, no!
 
Even I get reputation it should be no problem for anyone else
 
11:32 AM
Believe in yourself.
 
pen
@caveman hey, why do I not see your MSE account?
 
 
2 hours later…
1:13 PM
heloo?
 
hi
@PaulSlevin hi
 
hello!
 
:D
 
does anyone here know anything about integration on manifolds?
 
at least one guy does but he's not here now
 
1:21 PM
ahhhh. it is troubling me
 
you could try asking and if anyone who knows about it sees it they might ping you
 
I was wondering if the integral of a volume form (non vanishing top form) was always non-zero, and if so why
?
In particular I am trying to understand why $\int_M \omega ^{\wedge n} \not= 0$ if $(M, \omega)$ is a symplectic manifold
 
1:42 PM
assuming $M$ compact
 
2:20 PM
does someone know a book where I can find a (elementary, by any chance) proof of Dirichlets simultaneous approximation theorem?
 
@CBenni Cassels - Diophantine approximation
 
Does it contain the standard or the simultaneous version?
also, thanks, @caveman
 
it's a basic consequence of geometry of numbers
Minkowski's theory
 
@anon haha you are so otaku!
 
why do you want a book
 
2:35 PM
Hallo @dominic
 
@Charlie Huhu
 
@caveman is there an comprehensive proof on the internet, @caveman? I tried to find one
 
OK my last question can be ignored it - it has become: If $M$ is an oriented manifold and dim$M = n$ and $\mu$ is an $n-1$ form, and $d\mu$ is nowhere vanishing then why can't we have $\int_M d\mu = 0$ ?
 
also, books sound better as references ;)
 
@DominicMichaelis wie gehts?
 
2:36 PM
Do you know geometry of numbers
 
No, I donk know much about number theory
and diophantine equations
I stumbled over it from analysis
 
the idea is that any convex centrally symmetric body must contain a lattice point if the volume is large enough compared to the determinat of the lattice
 
Basically, it should suffice to say that theorem exists and there exist proofs; However for further knowledge, id like to see a proof.
and I know the minkowski theorem, yeah. I also saw that it can be proved using the pidgeon hole principle, but I just cant make the connection
 
it's not proved by pigeonhole
 
according to wikipedia, it is
 
2:40 PM
ok...
 
In number theory, Dirichlet's theorem on Diophantine approximation, also called Dirichlet's approximation theorem, states that for any real number α and any positive integer N, there exists integers p and q such that 1 ≤ q ≤ N and : \left | q \alpha -p \right | \le \frac{1}{N+1} This is a foundational result in diophantine approximation, showing that any real number has a sequence of good rational approximations: in fact an immediate consequence is that for a given irrational α, the inequality : \left | \alpha -\frac{p}{q} \right | is satisfied by infinitely many integers p and q. Th...
 
hi , how can tel me properties of Polish and Suslin spaces please
 
There are (as always) multiple approaches
@Vrouvrou have you tried wikipedia for a first overview?
 
yes but in wikipedia there is no proof
 
of what? You were only asking for properties
 
2:44 PM
I am a beginner , so i want to understand I want to understand things well
 
ganz ok
 
tel me for example : R and X are two polish spaces why R*X is Suslin ?]
 
@DominicMichaelis :) good
 
@Charlie Hi.
Wassup?
 
@jayesh hi
@jayesh nothing new
 
pen
2:51 PM
@jay Hello!
 
@pen Hello!
Wassup?
 
pen
@JayeshBadwaik Appearing for the VMC admission test tomorrow, heh.
 
@pen Hmm, Vidyamandir?
 
pen
@JayeshBadwaik Yes.
 
@pen Heh.
Best luck for a peaceful test. :-)
 
2:56 PM
@pen hi
 
lol why is integral x^n always x^{n+1} except when n=-1
 
lol why?
 
because it is log then
 
I gues it's because the derivative of x^{n} is x^{n-1} except when n=0
 
you have $$\int x^n \; \mathrm{d}x= \frac{1}{n+1} x^{n+1}$$ and dividing by zero is not a good idea
 
3:01 PM
nothing d/dx x^n is never 1/x
 
considering only ZFC, can we say that Must there exist distinct real numbers x,y,z,w, all of the same colour (i.e. f(x) = f(y) = f(z) = f(w)) such that x + y = z + w ?
 
@nerdy you need to explain more
 
problem is i dont know :( i quoted from someone i was talking about mathematics, he said that there are somethings that cant be answeared by ZFC
so we cant actually reduce everything to set theory
 
In geometric graph theory, the Hadwiger–Nelson problem, named after Hugo Hadwiger and Edward Nelson, asks for the minimum number of colors required to color the plane such that no two points at distance 1 from each other have the same color. The answer is unknown, but has been narrowed down to one of the numbers 4, 5, 6 or 7. The correct value may actually depend on the choice of axioms for set theory. The question can be phrased in graph theoretic terms as follows. Let G be the unit distance graph of the plane: an infinite graph with all points of the plane as vertices and with an edge...
 
like we think we can
 
3:07 PM
@nerdy you think we can what?
oh
 
reduce EVERYTHING to set theory
 
@nerdy, you are aware of independent statements like continuum hypothesis?
 
unfortunately not :( i didnt even start studying set theory i was just talking about it, i always thought everything could be reduced or answeared by set theory ( even tho it would not be that useful and necessary ) but then he said that not everything can be
then im a bit confused, is his statement true /
 
@nerdy, there are some things you can't prove or disprove in ZFC
@nerdy, but you can always add more axioms
 
i see
someone said that it is pointless to add more axioms becauses we can always find more independent statements no matter how many axions we add
would that even make sense ?
 
3:17 PM
Hm, some smackdown needing to be laid: math.stackexchange.com/questions/33904/…
 
@nerdy, no I don't agree with that
@nerdy, the goal isn't to have a complete theory
 
3:36 PM
@caveman is it?
 
I said it's not
 
It is $nx^{n-1}$ and when $n=0$ it is not of much use for integrating $1/x$
 
yeah
 
Note that $\frac{\mathrm{d}}{\mathrm{d}x}n\left(x^{1/n}-1\right)=x^{1/n-1}$
 
I do't understand what that is about
 
3:44 PM
So consider $\lim\limits_{n\to\infty}n\left(x^{1/n}-1\right)$
its derivative should be what?
 
if you commute d/dx with lim you get 1/x
that's cool
 
I have used that as a substitution for log in some integrals with good result :-)
 
wow
I can't imagine doing that
I wish I was good at integrals, I never paid attention in school because I though it was just differentation backwards
but these things Chris' sister comes up with make me regret it
 
@caveman It is just differentiation backwards, but that doesn't mean it's easy
 
what I meant was there's these integrals you do which has no antiderivative
 
3:50 PM
That's because there is no algorithm to do them in general.
or are you talking about computing definite integrals where there is no antiderivative?
 
computing definite integrals
 
like $\int_{-\infty}^\infty e^{-x^2}\,\mathrm{d}x=\sqrt\pi$?
 
yeah!
 
hi
please how to prove that projection of Suslin sets along Polish space are measurable
?
 
@Vrouvrou That might be a question better suited to the main question site. Unless someone happens to be here who knows.
 
4:01 PM
Hi @peter , what have you been doing?
 
@Charlie Just came back from uni.
 
@PeterTamaroff good good
 
@Charlie What you doin'?
 
@PeterTamaroff now, I'm cleaning a few stuff
 
@Vrouvrou I cleaned up one of your questions (LaTeXified the image)
check to make sure there are no mistakes
@Charlie GN attack: "some stuff" or "a few things"
@WillHunting: do you have any active accounts other than chat?
if you're still listening.
 
4:16 PM
He hasn't been seen for 3 days now.
 
@robjohn thanks, I believe it's some.
 
May be he is watching without being logged in.
 
@JayeshBadwaik his chat account is logged in (or hasn't yet noticed that he is not there)
 
@robjohn Ahh.
 
He's fine
Like he always said to be.
 
4:23 PM
@Charlie but his math account is gone :-(
He said he was going so many times, and I suggested other avenues. I had hoped this was the same, but I guess not.
 
Hi.
Who is "he"?
 
@MattN. Jasper Loy
There is a new Jasper Loy, but it is not he afaik
 
@robjohn Right. I recently saw another Japser Loy account. Did you see that too? And if so, is it himself with a new account or just someone random pretending?
Right. In this case ignore my previous comment : )
Sorry. I read after I type : D
 
@MarianoSuárez-Alvarez: hey there.
 
pen
@jay thanks! what do you think about VMC?
@Charlie Hey, I was gone.
 
4:30 PM
@robjohn yes,.it's gone, he's focusing on his real problems now, taking care if himself. he said he has more time now. I hope that he can get better
 
@pen I don't know man, 7 years is a long time. So many things can change.
 
pen
@JayeshBadwaik Oh, you've been there? Nice :-)
How was it? I hope not much has changed.
 
@pen No, I have not been there. I heard it was good 7 years ago, now I do not have a slightest clue.
 
pen
@JayeshBadwaik How is FIITJEE?
Oh, and I am not a fan of competition, since I always lose.
 
@pen I think ones in Delhi and Mumbai are good. Others not so much.
 
pen
4:35 PM
@JayeshBadwaik OK, so how about VMC vs. FIITJEE? I'm a little inclined towards VMC.
I don't really want to appear for IIT JEE (it's a waste of time IMO). Just doing this for fun.
Instead of spending the first 20 years of life “preparing” for this exam, I might just master a thing and start researching on it.
 
@pen Ehh, then this stuff does not matter I think. Talk to the people at the centers, see which one is more "accommodating" and less "militaristic".
By the way, its sad but true, but even if you want to go into basic sciences or master a field, you might have to perform well in IIT JEE (or ISEET as it is now called I think), else you might not get even a decent institute in India.
 
pen
@JayeshBadwaik That one is definitely VMC.
 
@pen Good.
 
pen
@JayeshBadwaik Who says I want to study in India? :-)
 
@pen hehe ;)
 
4:41 PM
@pen Then you are much better off. :-) Then follow my advice about accomodating.
 
pen
@JayeshBadwaik Yup!
 
And try to do stuff in the field you currently are interested in, even if your schools do not appreciate it as of now. Do try to maintain a decent/good record in school at the same time. You will have to balance it.
 
pen
@JayeshBadwaik I'm pretty good at school. Got 10 GPA :-)
 
is there any sense in saying that people tried but werent sucessful to prove the 5th euclid axiom ( elements ) from the first 4 axioms ?
 
@pen Ahh, no worries then.
 
4:43 PM
doesnt an axiom need not be proved ?
 
pen
@JayeshBadwaik Yeah, but sometimes, it's all a headache. It's a waste of time at school; the real thing is at home.
 
@pen Hmm, yup. Hence, I said, you will have to balance/manage it. :-)
 
pen
@JayeshBadwaik Hard to do... but who knows? :-D
 
@pen You are in CBSE right?
 
pen
@JayeshBadwaik Yup!
The worst thing is the projects. They add absolutely nothing. I go to the Internet, type the thing out, and copy it in my file. Once I just printed it and I was scolded for not doing the handwritten thing. What a mess!
 
4:47 PM
@pen My brother was in CBSE too and I thought his schooling was very healthy.
 
pen
@JayeshBadwaik CBSE is effed up nowadays.
 
@pen Hmm. Ohh, handwritten? Seriously? My brother did everything on computers I think.
 
pen
@JayeshBadwaik He was a lucky dude.
The people who don't have a computer at their home literally cry all the time.
 
@pen But he did not copy/paste. ;-)
 
pen
@JayeshBadwaik LOL, I don't either. My friends do though. I pick lines out of the main article. I'm so picky, that I am left with only 5 sentences. Then I have to add the useless stuff! :-P
 
4:50 PM
@pen Hmm, try doing it honestly, you might find that you might learn a lot more. seriously Even if it is not math, there is much to be learned.
 
pen
@JayeshBadwaik Learned nothing till now. I think I knew all of it already. They just give you stuff you already know. -_-
 
@pen Hmm. I think my brother might have been lucky then, his projects were really good.
There was even a student math/science talk kind of thing in his school.
 
@JayeshBadwaik of course Jayesh's brother ;)
 
@Charlie Bleh. :P
 
pen
@JayeshBadwaik Really? He must have been into a great school!
@Charlie Amen, amen! :')
Has he passed out?
@JayeshBadwaik He must be studying advanced number theory too. 8-)
 
4:54 PM
@pen Naah, he is not interested in math too much I think.
 
pen
@JayeshBadwaik Physics?
 
@pen A little. He is more interested in programming right now. He had c++,python etc since 8th std, and is loving it. So, may be computer science, but then he does not like too much math, so we'll have to see.
 
What was the reaction of 18th and 19th century people about the 5th euclidh axiom ?
 
pen
@JayeshBadwaik Thanks, but please don't make me start hating my school.
They have prescribed an Adobe Flash book for 9th grade! -_-
 
@nerdy They said, "To hell with it" and invented riemann geometry.
@pen Hahaha. Even these guys had something similar I think. I remember only the good parts from the past. ;-)
 
4:58 PM
i heard that they tried to endlessly prove the 5th axiom from the first four
but couldnt do it
 
pen
@JayeshBadwaik Programming has a very good market value.
 
My qiuestion is : Why the fuck did they try to prove the 5th axiom from the first four axioms ?
if they are all axioms
 
pen
@nerdy Because we use axioms to prove other axioms. That's the only choice we have left.
 
really ?
but arent axioms something not need to be proven
 
pen
Axioms are so obvious, that we have to use other obvious statements to prove the given obvious statement.
@nerdy exactly.
 
4:59 PM
oO
but what about the first axiom
doesnt have previous statements
 
pen
We always use some or the other axiom while proving a theorem/lemma/axiom.
 

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