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10:00 PM
Are there negative Erdos numbers?
@OldJohn No, I don't think it would be undeserved. If you are worried about misrepresentation, I wouldn't be. People who know what an Erdös number means, know what it means :-)
@aDangerousIdea Since Erdös is the root of the tree, It would be difficult to know what that would mean.
If you wrote a paper with his grand father would that be negative?
@aDangerousIdea There might be imaginary Erdös numbers for people who wish they wrote papers with him...
8
@robjohn true!
@OldJohn You could have an Erdös number of 3i :-)
10:03 PM
@robjohn I would be happy with that!
How about -3i
@aDangerousIdea doesn't sound so nice - even though $i$ and $-i$ are pretty much indistinguishable :)
@aDangerousIdea let's not start the left-right, clockwise-counterclockwise discussion
Now then. That's it for today here. Be seeing you!
10:05 PM
bye!
@MattN. l8r
2l8
@Jordan are you alright?
@pourjour Asalaamu aleykum
@OldJohn Do you speak Arabic?
10:08 PM
@Argon badly
@OldJohn 3alaykom osslamo wa ra7mato lahi wa barakatoh
:)
@OldJohn h=7
3=a
@pourjour yep - I understood that one :)
@OldJohn hhh It was long time
we've not talk
10:10 PM
I would have written it "wa alaykum salaam, wa rahmatu llahi wa barakaatuh" (because I have still not mastered the arabic text spelling process)
My Arabic = aiwa
@Argon that is very useful!
Hahaa!
@pourjour yes - did I tell you I was in Abu Dhabi last week?
@OldJohn woow you're traveling a lot!
10:12 PM
@pourjour won by 7 seconds!
@OldJohn "Bukra Fil Mishmish"
It's been a long time that i don't learn a new language.... i began to learn Marathi....but it's kind hard...
@Charlie Vut about Deutch?
Prove it's irrational: $$\sum_{p\text{ prime}} 2^{-p}$$
@Argon mishmish is apricot, I think - but I don't understand the rest :(
10:14 PM
@OldJohn It is apricot - very good!
Bukra = morning
@robjohn hh yep but I've already a hint in the book $h(x)=\frac { f(b)-f(a) }{ g(b)-g(a) } (g(x)-g(a))-(f(x)-f(a))$
@Argon so "in the morning there are apricots" ????
@Argon i really don't have time to improve my german, french, spanish...
@pourjour so what is causing the problem?
@robjohn I use rolle's theorem but I don't get the result
10:16 PM
@OldJohn Yep
@Charlie How about Portuguese?
@cassandra0 If it were rational, there would have to be some repeating pattern to the primes
@Argon I like my portuguese. it's really good. i speak it since i was 9 month, when i started to learn some language
@pourjour So you have some point c in [a,b] where $h'(c)=0$ right?
@Charlie Sounds good :) I need to brush up on it still!
10:17 PM
@Argon I don't know
@Argon Você gosta da língua portuguesa?
@pourjour What?
@pourjour Then $f'(c)(g(b)-g(a))-g'(c)(f(b)-f(a))=h'(c)=0$
@Charlie Traduz Google faz.
@pourjour Divide out
10:18 PM
Eu sei zilch
@Argon sorry I want to send the message for robjohn
@Argon "translates google does" it's written
@pourjour Oh, haha, no prob
@Argon zilch?
@Charlie "Google translate does"
10:19 PM
@robjohn let me try ok?
@Argon in portuguese stays: Google tradutor sabe
@Charlie Traduz Google é horrível!
@Argon o google tradutor é execrável
@Charlie *tradutor
@Charlie HAHA "google translator is execrable"
10:22 PM
@Argon i like letter x in portuguese. has many sounds: ss, z, c
@Charlie cs
That's about it
Or Z
@Argon i like ç best cedilha
@Charlie Cedille! Like French!
Palestine have got their vote at the UN ...
vote for what
10:23 PM
Geez
@Argon yes! but you cannot start a word with this letter
non-member observer state status
@Charlie Watch me: çar
@OldJohn What were they before?
interesting
@Argon nothing
10:25 PM
@Argon you can't, você não pode começar uma palavra com essa letra! é a regra mais absoluta da gramática!
hmmm
@Argon and also don't start a word with double letters
@Charlie Aaron :'(
@Argon do you that there's a poor soul here called:"facebookson"?
@Charlie REALLY?
10:27 PM
@Argon YES
@Charlie In New Zealand, someone is named "Talula Does the Hula From Hawaii..."
@Argon GEEZ
user19161
I like it that although my stars are decreasing my rep is increasing.
@WillHunting You are back to the blue square?
user19161
10:29 PM
@Argon Yes, I think I prefer the blue square, just like amwhy.
:7067886 $h(b)-h(a)=(f(b)-f(a))(g(b)-g(a))-(g(b)-g(a))(f(b)-f(a))=0$
user19161
@robjohn Wow, that spoiled my eyes!
I wanna hang myself
@Charlie why?
user19161
...
10:31 PM
@robjohn the exercises are not working..simply not right...
user19161
@Charlie Do you mean you can't solve them?
@Charlie !!!
@WillHunting yes :'(:'(
user19161
Or is there a mistake in them?
@robjohn yep I found it now how can I calculate the $lim \frac{sin(x)-x}{x^{3}}$ when x tend to 0
user19161
10:32 PM
It's OK, I never do exercises in books.
user19161
But I think I will do some in future.
i feel useless when i can't solve my exercises...
a trash
@Charlie, I feel like that too when I couldn't solve
@Charlie me too I have the same feeling
user19161
It's OK, math is not everything. Have a beer.
10:34 PM
i don't drink.. but i would like to have junk food
@WillHunting I thought you didn't drink?
user19161
@OldJohn You are right! It's just an expression!
@WillHunting :)
:'( infinitely
Alcohol is not a cure, alcohol sux
user19161
10:35 PM
But I think it's nice to have a beer once in a while. I think I should get some later, since it's Friday.
@JohanLarsson :D
@user46225 hi
user19161
@user46225 Welcome to this chat!
Brian Scott gave me a nice answer to this question: math.stackexchange.com/questions/247528/… but there's another thing I don't understand which I adressed in the comments, could you take a look?
10:36 PM
math.stackexchange.com/questions/247610/… <-- have I left out any necessary details? I think not, but I'd like to be sure.
@Argon are you happy that i'm goint to kill myself?
2
I think it would be a bit pointless to open a new question for such a small thing
@Charlie No!!!! That's why I put !!!!!!!!!!!
user19161
@user46225 Hmm, you can also wait for him to respond.
@Argon oh
@Argon i'm counting the days to 8/12. my favorite jewish holiday is yom kippur. everyone should commemorate it. it's beautiful day. should be universal holiday.
10:37 PM
@WillHunting yes... but I'm anxious to finish understanding the proof!
user19161
@user46225 Ah OK, maybe someone here will help you then. I am only a banana...
@Will: quite the mathematical banana, judging by your profile!
user19161
@user46225 I only aim for low hanging fruits = trivial questions!
@Charlie It is a very special and odd day. I always feel different on Yom Kippur.
user19161
These days, I aim for ground fruit! =)
10:40 PM
@Argon interesting
user19161
By the way, which fruits grow on the ground?
@Argon what about pessach?
watermelon!
@Charlie ...And not from hunger :)
@Charlie One of my favourite holidays!
@Argon hahahahahahahhahahahahahahhah
10:41 PM
although it would be awesome to have watermelon trees
@Charlie I love Matzah a lot :), so I don't mind not having bread. It's a really fun holiday!!
@Argon my favorite holiday is Christmas
@Charlie What do you do?
user19161
@user46225 Hahahaha! Maybe I should call myself a watermelon then!
@Argon everyone is together and it's such a good energy
10:43 PM
:)
Excuse me gentlemen, out for dinner!
@Will: well, you found yourself a profile description then ;)
@Charlie Have a yummy one!
@Argon thanks!
Right - off to celebrate with my Palestinian brothers - g'night all
10:45 PM
@user46225, what is the definition of discrete?
who starred my creepy messages???
geez
@OldJohn Good night
@Charlie I just have a bad attitude, and no one to really vent to and I know it shouldn't be here, I know what I need to change and how, just something I need to work on
user19161
@user46225 I see you are interested in algebraic number theory and galois theory.
@cassandra0: a topological space is discrete if every point has a neighborhood which intersects just the point
10:45 PM
@Charlie Wadafack?
@user46225, oh I understand thank you
@cassandra0: in this case $V\simeq \mathbb{R}^n$ with the usual topology
I can intuitively see that a discrete subgroup must be a lattice
@cassandra0 It means the space is not going around with slutty clothes and stuff.
(I mean, isomorphic as vector spaces, and it inherits the usual topology of $\mathbb{R}^n$)
10:46 PM
lol
@Will: at least that's what I've been studying as of late :)
user19161
@PeterTamaroff HAHAHAHAHA, never expected that to come from Pedro.
so how would I prove this, hmmmm clearly 0 is isolated
by some epsilon ball
@user46225 How is $\Bbb R^n$ discrete?
user19161
@user46225 Gave you a few votes to get you up and moving! (We can't do too many or it will be reversed automatically the next day!)
10:48 PM
@Peter: didn't say it was, no?
@user46225 You said, "In this case..."
@Will: thanks! :)
if $v \in V$ then $v \mathbb Z \subseteq V$
I misunderstood.
@Peter: Ah, I see. "in this case" = "in the aforementioned post", sorry :)
10:50 PM
perhaps we should consider the smallest nonzero vector $v \in V$, by the structure theorem $V \simeq v\mathbb Z \times V'$
the problem is we need some decreasing measure to prove this would terminate
@cassandra0: I've seen proofs of this theorem by induction (e.g. in Milne's ANT), and what you describe works perfectly for the base of the induction, if I understood you correctly
I found those proofs really awful, then I went to Neukirch and found this one which is slicker, but there are details I don't understand :\
@pourjour can you use L'Hopital?
@user46225, how do we show that this induction terminates?
@cassandra0: well... I didn't really finish reading that proof, since I found it ugly, and then I found this one which is fundamentally different, I think :P
I guess we could show that if $\bar v \cdot \mathbb Z^n$ is a dim $n+1$ lattice in $\mathbb R^n$ then we can find points arbitrarily close to the origin
therefore the lattice must have dimension $\le n$
hm
no this is wrong
what's ugly about this proof is the structure theorem is just ISOMORPHISM
2
so we lose the geometry when we just look at it
we have to view the group through the isomorphism to see the geometry
10:58 PM
well, I'm more algebraically-oriented myself, maybe that explains why I like it best ;P
this is geometry
user19161
@cassandra0 Wow, SHOUTING.
what's this about finite index, isn't $\Gamma = \Gamma_0$?
that's what you want to prove
Brian has answered to me in a comment, I have to digest it though
they never prove $q = 1$ do they?
11:01 PM
well, yes
you end up having $r\leq m$
and also $m\leq r$, so...
well
that's not enough
okay, that's not really what you prove
but they are the same up to scaling
ah gamma is not equal to gamma_0, because you might choose a basis for it which is say double the basis for gamma
then it misses out half the points gamma has
yes, exactly
the point is you can just start with that, then fix it
so they use te fundamental parallelogram (of gamma_0) to fix it: by seeing which points of gamma lie inside that
so they do $\Gamma_0/q \subseteq \Gamma \subseteq \Gamma_0$
@cassandra0: gee, I don't have that much intuition about what is going on
11:16 PM
@user46225, which part?
(or really, I barely have any intuition... :P)
I'll draw you a diagram
so you can see what I see
well, your last three lines were quite mysterious to me
thanks a lot!
@WillHunting hahaha, I hadn't seen your message. That earned cassandra's message a star!
First of all here's Gamma in green (which we don't yet know is a lattice) with Gamma_0 (generated by the basis {a,b} chosen arbitrarily) i.imgur.com/oEDGw.png
er I labelled a wrong it should be the one nearest the origin
Now here's Gamma_0/q in red i.imgur.com/fb6xA.png
(I couldn't be bothered drawing much more of it)
q=3
I'm following you
11:26 PM
So you see why $\Gamma_0/q \subseteq \Gamma \subseteq \Gamma_0$? And this has been proved without using that $\Gamma$ is a lattice
but it (actually only need the $\Gamma_0/q \subseteq \Gamma$ bit) immediately implies that $\Gamma$ is a lattice
hmmm, I don't know, I'm not convinced
that's ok but do you see why Gamma contains the lattice Gamma_0 like in my picture?
Gamma_0/q, no?
Gamma_0/q in red contains Gamma
but Gamma_0/q comes later
just think about Gamma in green and Gamma_0 in pink in the first picture
yes, that I see
11:32 PM
the construction of Gamma_0 ensures that it's a lattice contained inside Gamma
user19161
Hey @limitless! Have you done your homework?
so the next step is to take the fundamental paralellogram with sides a and b
user19161
@cassandra0 It's parallelogram.
@WillHunting, lulz. No. I have yet to get around to it.
I've been laying here pondering why I have a headache.
@Will: it's "fundamental p." in my notebook :P
11:34 PM
do you know what that is?
yes, I do
"fundamental mesh" is what Neukirch calls it
user19161
@user46225 The SE day starts at 0000 GMT, I might be able to give you some more votes then. =)
the number of points in the fundamental mesh in my example is 3 (the origin and two points along the way to b)
yes, I see
you don't count the points on the outside edges of the fundamental domain because of 0 <= x_i < 1
11:39 PM
@Will: oh, please just upvote what you think it's worthy! thanks in any case :)
so do you see why Gamma_0/3 is the red lattice?
by S/3 I mean {(x/3,y/3)|(x,y)in S}
Hi everyone!
QED??
a subgroup of a lattice is a lattice
A and B are events satisfying P(A) = P(A|B) = 0.6. Are A and B disjoint or not? What is P(A_complement |B)?
11:45 PM
why does it prove that Gamma is a lattice?
I would say they are disjoint, and P(A_complement|B) = 0.4.?
Let H be a lattice (so it's freely generated and has finite dimension) and G a subgroup of H, then G is a lattice since it's also freely generated
(H = Gamma_0/q)
@Jordan You said you know what needs changing and you know how to change it. What do you think needs changing?
now I understand. but isn't it the same as what Neukirch does?
@aDangerousIdea I just need to study more
11:52 PM
well yeah that's the proof I thought you didn't fully understand
@Jordan That is what I said at the beginning, all you can do is study hard.
Yeah! thanks for the geometric insight. Now I understand the proof better (I "only" understood it algebraically)
are there any aspects you still don't get?
not at the moment, no
thanks a bunch!
11:56 PM
well, I gotta go now
see you all!
Bye!
hi @argon
@cassandra0 Hi!
11:58 PM
:)

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