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10:08 PM
@FreeMind You can ask your problem here
 
@FreeMind Is the link private?
 
hi @Studentmath, @robjohn, @Alizter ... Studentmath, the average on my test was about 70.
 
70%?
 
Yup, @Alizter.
 
That quite good no?
 
10:15 PM
@FreeMind I am sorry if I go away for a while. It can be that I am thinking about a problem, it could be that I have been called away from my computer. It could be that moderator duties have called me away. I apologize for that.
 
thought @robjohn still had a job to call him away :D
@Alizter: Don't you want René for a mod? That would be best.
 
Hey @TedShifrin Could you have a look at this exercise? math.stackexchange.com/questions/1052007/…
 
@TedShifrin I thought so, too :-)
@evinda it is hard to decipher what is being asked there.
 
@TedShifrin I haven't seriously thought about anybody being a mod
 
@evinda is it that $f(x,y)=0$ and $g(x,y)=0$?
 
10:16 PM
@robjohn No problem, but thank you for explaining what's going on, because it gives bad feeling when you expect someone to say something but they don't!
@Alizter I have already asked the problem check the link above!
 
@Ted final test?
 
@robjohn Yeah, the link is private, do you create a chatroom?!
 
@FreeMind Hey, this is chat. it is only a little more responsive than texting. People come and go without warning.
 
@Ted Shoot, I forgot to send you an email about the problem I'm having trouble with!!! Do you know which problem I'm referring to when I say #2 on the 2410 Final? Can I share my answers here?
 
@robjohn Ok :)
@robjohn I want to give you the link but not here
 
10:18 PM
no, Exam 3, @Studentmath. The final is in a week.
@teadawg: I could try to look it up.
 
What do you expect the average to be there?
 
@Ted I'll nominate you for mod to force you to stick around.
6
 
Wikipedia says that a solenoid is a topological group and then the article has nothing on groups
 
@FreeMind it is a link to a book? Is the book available online?
 
GRR @Mike.
 
10:19 PM
@robjohn You forced me to public it
 
@evinda: Tell me your actual definition of intersection multiplicity. I don't see that anywhere in all those lines and lines.
I don't know, @Studentmath. I truly expect 20% Ds and Fs.
 
@robjohn It isn't given any further information... :/

I tried to apply the theorem of the site 37 from this link: http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf
 
@robjohn Did you download?
 
In our major courses, that's very high. You tell me that's not unusual for you.
 
@Ted Well, first I'd have to get you elected... and I don't know how many fans you have :)
 
10:20 PM
@FreeMind I did? I was just asking some questions. We are only supposed to create private chat rooms for moderation issues.
 
I had a risk management major crying (literally) that she can't graduate without a C in my class. I tried to tell her that a 33% test average and 40% homework average wasn't going to get her a C, and that she should have been worried about that from day 1.
 
That's a shame - it's not unusual here, nope.
 
I can name a few anti-fans, @Mike. I don't know if you and @Pedro count as "fans."
 
Frienemies?
 
That's such a shame, I don't get what she was expecting you to do. If she scores well on the final, does she have a chance?
 
10:22 PM
@robjohn The solution is long, I want to give you the page to take a look at it!
 
Though if she goes like that, I don't think she will do too well on the final either..
 
@TedShifrin At the sites 36-37 from the link: http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf
there are the identities and a theorem that I tried to apply...
 
@robjohn It also gives you an insight of what I am looking for.
 
@FreeMind I received the link... I have the PDF
 
OK, that's a reasonable definition, @evinda. So you need to look at the quotient of the local ring at $(0,0)$ by the ideal generated by the two polynomials. Have you done that?
 
10:23 PM
@TedShifrin, I'm getting a "pre-requisite courses not met" error trying to register for point-set, even though Dr. Boe cleared me previously -_-
 
@Kaj: Only Julie and Dr. Graham and I have access to override prerequisites. It's because you never took 3200. Remind me tomorrow and I'll give you an override.
 
@FreeMind being a mod, I can not only create private chat rooms, I can read deleted chat comments. :-)
 
@Ted You sent me, among several other things, the spring 2007 final for 2410
 
Exactly. That's why you're my hero every semester around this time :)
 
@FreeMind where is the problem from?
 
10:24 PM
@robjohn Ok :) , that's good :D , page 58 ( in pdf reader )
 
@TedShifrin Could you explain me further what do you mean by the quotient of the local ring ? :/
 
@FreeMind let me look
 
@robjohn page 48 of the book , 58 of pdf reader
 
What is $S_\infty$
as a symmetric group
 
Oh rats, @teadawg. I see that this was not the ultimate version of the exam. There are problems missing. Ugh. I'll have to find the actual file at school tomorrow.
 
10:25 PM
@Alizter You're going to need to clarify what you mean by $S_\infty$
 
That's what is on p. 37 of Fulton, @evinda. You need to know some commutative algebra to read his book. But think about taking $k[x,y]/\langle f,g\rangle$ and try to what its dimension (as a vector space over $k$) is.
 
@MikeMiller Symmetric group $S_n$ as n gets really big
for example Z/nZ becomes the circle group
 
Anyhow, @teadawg, what was your question?
@Kaj: I don't think I've ever been heroic.
 
The direct limit is the term for what you want. And that's not isomorphic to the symmetric group of any set - the symmetric group of an infinite set is uncountable, while yours is countable. m
 
@Ted If it's not the final version, is it alright if I post my question in chat?
 
10:28 PM
BTW, @Kaj, how did the analysis exam go?
@teadawg: I don't mind your posting ... That was a course 7 years ago, anyhow. And I won't ever teach it again :P
 
The circle group is not the direct limit of $\Bbb Z/n\Bbb Z$, the rational points on the circle group are.
 
But I need to try to get you a complete version of the test ... What you're asking about is a standard calculus question, right?
 
Isomorphic to $\Bbb Q/\Bbb Z$.
 
@MikeMiller Yes I think it is $\Bbb R / \Bbb Z$ (the circle group)
 
Also, @Ted, if it's no trouble, after they're all done with it and handed and graded and whatever is needed and if it's okay, I would like to practice on your final :)
 
10:29 PM
Well it certainly went better than the first @TedShifrin. I'm 100% focused on doing well on the final right now.
 
@robjohn Just notice that, the physical stuff does not matter, the main question is the one I asked in MO
 
@FreeMind doesn't equation $(39)$ answer the whole thing? That is what I was trying to get you to when suggesting the Fourier Transform. I thought there must be some simpler looking solution, but that is just using Fourier Inversion.
 
@Mike: I shouldn't waste so much time on you if you're a frienemy.
Good, @Kaj. Make me proud :)
 
It's not, unless your notion of "really big" is not actually given by a definition you or I know, in which case you're not asking a math question that I can help with...
 
There was a horribly bungled answer on main last night, @Kaj. If $A\subset]\Bbb R$ has the property that $A\cap A' = \emptyset$ ($A'$ the set of limit points), prove that $A$ is countable.
 
10:30 PM
I think I mean direct limit
 
@robjohn But on equation 39 we're moving on complex plane from -inf to +inf
 
@Ted It's a surface of revolution, I'd assume so.
 
but everytime I say direct limit I see inverse limit then I get confused
 
Right, @teadawg ... So what's your issue?
 
@TedShifrin So, can we not do it as I did in my post?
 
10:31 PM
The direct limit of Z/nZ is the rational points on the circle group, isomorphic to Q/Z.
 
probably Balarka now that I think of it
 
I do not have the patience to look at all that, @evinda, sorry.
 
@robjohn I'm looking for a real approach, the person who proposed this question told me that, there are bunch of ways on the internet, I want something else, but simpler from you! :((((
 
@MikeMiller OK so do we have a direct limit of $S_n$
 
@robjohn Now, you see what I'm really talking about!
 
10:31 PM
Yes. It doesn't have a name that I know.
 
@Ted I'm just looking for verification on the answer I've gotten. "Let R denote the plane region bounded above by $y=e^{-x}$, below by $y=0$, and lying between $x=0$ and $x=1$.
Oops
 
It's isomorphic to the group of bijection $\Bbb N \to \Bbb N$ that fix all but finitely many terms.
 
Grr...
 
@teadawg: I like people who Grr. :P
 
(It's also an example of a countable group that has nothing to do with $\Bbb Z$, as you requested earlier...)
 
10:34 PM
@MikeMiller Where can I find literature on this?
 
@evinda: What is your final answer to the question?
 
I dunno. What do you mean by "this"?
 
Who when where the group of bijections you just talked about
 
Dunno.
 
@TedShifrin The last steps are the following:
$h_1(x,y)=h(x,y)-f(x,y)=3y^2-xy^2+2x^4$

$h_2(x,y)=2f(x,y)-xh_1(x,y)=2x^4+2y^2-3xy^2+x^2y^2$


$I(P,h_2\cap h_3)=I(P,h_2\cap y)+I(P,h_2\cap (1+2x-x^2))$

$I(P,h_2\cap y)=4$

Is $I(h_2\cap (1+2x-x^2))=0$?
 
10:36 PM
That set of bijections is a subset of $\Bbb N^{\Bbb N}$ which has cardinality $\aleph_0$
 
Part A: "Find the volume of the solid obtained by rotating R about the line $y=1$." The answer I got: $$\pi\left(\frac32-\frac{4e-1}{2e^2}\right)$$
 
and because it is infinite it must be countable
cool
 
$\Bbb N^{\Bbb N}$ has the same cardinality as $\Bbb R$
 
I don't want an answer, @teadawg, as I don't have the work here. Write down the integral you did.
@evinda: I'm looking for a positive integer. What is your answer?
 
The actual reason is that you can write it as a countable union of finite sets
 
10:37 PM
@MikeMiller Wait I am confused
 
@Ted Alright, give me a second to format it into MathJax
 
Oh wait
 
The set of bijections $\Bbb N \to \Bbb N$ is uncountable.
 
What does $\mathbb{N}^{\mathbb{N}}$ mean?
 
But when you put in the restriction "... that fix almost everything" it's countable.
 
10:38 PM
$|\Bbb N^{\Bbb N}| = | P(\Bbb N^2)$
 
@Kaj, if you gave me an answer, I missed it in all this hullaballoo. :P
 
K gotchya
@Khallil Set of mappings from N to N
 
So @Alizter is joining @beginner and Balarka as a prodigy ...
 
All mappings. So they can be injective, surjective or both, @Alizter?
 
@Khallil All mappings as far as I know
 
10:39 PM
I hope they don't finish their theses before I do, @Ted....
 
@MikeMiller So my logic is that a mapping can be though of a a set of ordered pairs therefore the powerset of pairs of N is uncountable thus maps are uncountable
 
Or me, @Mike.
 
cool
 
$$V=\pi\int_0^1\left(1-\left(1-\frac{1}{e^x}\right)^2\right)\,\mathbb{d}x$$
 
@teadawg1337 What is wrong?
Expand that
and integrate
 
10:40 PM
Is there any reasonable way to see for which z in (-2, 2) the sum $\sum_{j=0}^\infty |((z^2 + \sqrt{z^2 - 4} - 2)/2)^j + ((z^2 - \sqrt{z^2 - 4} - 2)/2)^j - 2|$ converges?
 
First of all, @teadog, $e^{-x}$ is easier than $1/e^x$. Second, it's right ;)
@Alizter: Leave him alone. He already did. I wanted to see the set-up.
 
@TedShifrin I was just helping. There was not "Do not touch sign" ;)
 
LOL, sorry, @Alizter. He's confirming answers with me, but I didn't want numerical answers.
 
I've been trying to find the spectrum of an operator forever, and I think I can get something meaningful if I can figure out whether or not that converges.
 
@TedShifrin With the calculations I did, I find $4$.. Is it right? :/
 
10:43 PM
@Ted It's a habit of mine to turn any expression with a negative exponent into a fraction
 
@Alizter No, this is not a correct argument... There is an injection $\Bbb N^{\Bbb N}\to \mathcal P(\Bbb N \times \Bbb N)$. Not all subsets of the latter define a mapping. The better reason is that we know the latter is in bijection with $\Bbb R$, and as the cardinality of $\Bbb R$ is the cardinality of $2^{\Bbb N}$, we have an injection $\Bbb R \to \Bbb N^{\Bbb N}$, there is a bijection between the two by Cantor-Schoeder-Berenstein.
 
I don't see anything obvious, @Xindaris. Have you posted that question on the main site?
@evinda: No, I think it's rather larger.
You should get $4$ for $f=x^2+y^2$ and $g=x^3+y^2$, for example.
@teadawg: If it's also your habit to expand things that are factored, please break the habit as soon as possible :P
 
@TedShifrin So, this $I(h_2\cap (1+2x-x^2))=0$ should be wrong.. Could you tell me why?
 
@MikeMiller I don't get how you get $\Bbb R \to \Bbb N ^{\Bbb N}$ surely its the other way round?
 
No, @evinda. I don't have the energy to do so.
 
10:45 PM
Why there's an injection $2^{\Bbb N}\to \Bbb N^{\Bbb N}$?
 
Hello @MikeMiller
 
@TedShifrin, I'm in the middle of a math club officer meeting, but I intend to look at it very shortly.
 
@evinda: That is correct, but I have no idea why that's relevant. $1-2x+x^2$ doesn't have $(0,0)$ as a zero.
 
Hello @Ted, @Kaj
 
If you know that $N^N \to P(N^2)$ is an injection and $P(N^2)\to R$ is a bijection surely by function composition we have $N^N\to R$ is an injection.
 
10:47 PM
LOL, say hi to everyone, @Kaj.
 
I will run for mod.
 
hi @Pedro.
 
I will ban all integral questions for life.
6
JK.
 
I am a bit frustrated about trying to talk math on my phone, @Alizter. I will let someone else take over.
 
@TedShifrin I haven't yet. The spectrum I'm trying to find is that of $S_l + S_r$ on $l_2[0, \infty)$, which I know I can turn into the spectrum of what I actually want.
 
10:47 PM
'm
 
@Alizter What's your problem?
 
Oh, yes, we do have that that is an injection, @Alizter. But to get a bijection by CSB we need an injection borh ways.
 
@evinda has a cool algebra problem for you, too, @Pedro. She wants the height of an ideal in a local ring.
 
@TedShifrin Really? OK.
 
Take the local ring of $k[x,y]$ at the origin. What is the height of the ideal $\langle y^2+x^4+x^5,y^2-x^5+x^6\rangle$?
 
10:49 PM
Hello!!
 
@MikeMiller So how do we get the second injection $R \to N^N$
 
Maybe I've forgotten the words. Depth? Height? Dimension of the quotient vector space.
 
@PedroTamaroff Showing $\Bbb N^{\Bbb N}$ is uncountable
 
@Alizter: Have you done $\{0,1\}^\Bbb N$?
 
Hmm, I just realized something. When I'm confident with an answer, it tends to be slightly incorrect. However, if I question an answer I've gotten, it tends to be correct
 
10:50 PM
@TedShifrin, what do you mean by $\subset ]$ ?
 
Hi all
 
Heya @DanielR
 
@Alizter That's easy. $2\leqslant \aleph_0$ so $\mathfrak c=2^{\aleph_0}\leqslant \aleph_0^{\aleph_0}$.
 
I mean it's a typo @Kaj.
 
@Alizter, you can use Cantor diagonalization
 
10:51 PM
@Pedro: Cardinal arithmetic not allowed.
 
Does anyone happen to have a reference for the result that the subgroups of $\mathbb{Q}^2$ cannot be classified?
 
Very similar to showing the set of infinite binary strings is uncountable.
 
Thank you @PedroTamaroff @KajHansen and @MikeMiller
 
Oh fine, I get no thanks. :D
 
@TedShifrin Well, you can look at it that way. I just says "sequences of $0,1$s inject into sequences of natural numbers."
In fact, @Alizter, equality holds.
 
10:52 PM
I think thats what Mike was trying to show
 
I wish I remembered how to permalink correctly. I asked about $\{0,1\}^\Bbb N$ above :P
 
But I got stuck understanding why there is a bijection $R \to N^N$
 
Don't, @Alizter. See why there's a bijection $[0,1]\to \{0,1\}^\Bbb N$.
well, not quite a bijection ... AGH.
I prefer an injection.
 
@robjohn By the way, any idea?
 
@Ted That was the only question I had, since I'm a bit rusty with surfaces of revolution. I find double integration and triple integration much more useful
 
10:54 PM
@TedShifrin Do you renember the exercise about proving that $cl(A)_{\Vert \cdot \Vert_1 }=C([0,1],\Bbb{R})$ where $A=\{f\in E: f(0)=0\}$ you said draw a picture but I don't know how can I do this (to find a suitable sequence (g_n)). Can you help? Thanks
 
@TedShifrin And then another injection the other way?
 
@robjohn If you found something, let me know, in whatever way.
 
Well, that's in the next course, @teadawg :P
 
@robjohn Do you want my gmail?
 
Yes, vaguely, @MarcGato. It's way too exhausting being here.
 
10:55 PM
of course here $\Vert \cdot \Vert _1=\int_0^1 $
 
I'll take that as a no for my reference request :P
 
Not me, @DanielR.
 
@TedShifrin I would first reduce the ideal.
 
googling for "classifying subgroups of Q^2" doesn't turn much up
 
@DanielRust What you said is pretty tragic.
Poor $\Bbb Q^2$.
 
10:56 PM
Well, go for it, @Pedro ... and help @evinda. I get $6$ for the answer.
So what did you try, @MarcGato?
 
@TedShifrin lol sure too much people need your help ;p
 
So, @MarcGato, take your continuous function and choose $x_0$ very close to $0$. Join $(0,0)$ to $(x_0,f(x_0))$ with a line segment, and use $f$ by itself to the right of $x_0$. Show it works.
 
I took $f\in E$. I define a function $(f_n)\n ge 2$ as $f_n(t)=f(t)$ for $t\in [1/n,1]$ and I am not sure for the part $[0,1/n]$ something like $f_n(t)=f(1/n)t$ we have $f_n /in A$ I think...?
 
You don't need a sequence, @MarcGato, although one is OK. You only need to show that you can get somebody in $A$ as close as you want to a given function $f$.
 
@TedShifrin Okay, I'm currently reading your 'hint'.
 
11:02 PM
@MikeMiller What is the direct limit of $A_n$?
 
@MarcGato, yes, the formula you wrote is exactly right.
 
@FreeMind I can get the email address from your profile, if you have one there (only mods can see it).
 
Given $\epsilon>0$, you just need to choose $n$ big enough so that $\max_{x\in [0,1/n]} |f(x)|/n<\epsilon$ or something like that.
OK, time for me to eat dinner and go to a concert. Y'all misbehave without me.
 
cool and thanks. Bon appétit @TedShifrin
 
@robjohn :D , ok!
 
11:07 PM
(et bon "amusement" au concert!).
 
@TedShifrin Coolio! I've had enough of single-variable calculus, I've pretty much "mastered" it. I'll start watching your lecture vids tomorrow (I was too sick over the weekend to do much of anything), I promise
 
Now that I'm out of my meeting, I think I've solved your problem @TedShifrin
I'd certainly be interested in running my solution by you once you're back. :D
 
@KajHansen What was the problem?
 
@KajHansen Yes, what was the problem?=
 
@Alizter, Consider $A \subset \mathbb{R}$ with the property that $A \cap A' = \emptyset$, where $A'$ is the set of limit points of $A$. Prove that $A$ is countable.
 
11:22 PM
So prove that an $A$ that includes non of its limit points is countable
?
 
Yes
 
For example the rationals are such an A
 
Indeed!
Actually, that's problematic @Alizter
 
oops why?
 
Every rational is a limit point in the set of all rational numbers.
 
11:26 PM
ok
Now I am having trouble imagining A :P
 
For any $x \in \mathbb{Q}$ and $\varepsilon > 0$, we can find a $y \in \mathbb{Q}$ such that $|x - y| < \varepsilon$.
So every $x$ is a limit point.
 
hmm yes
Such a set would be open
Maybe the cantor set?
no
I am contradicting myself
 
@Kaj hrm, every $a\in A$ is isolated point, right?
 
Yes @Studentmath
 
@Pedro So are you running?
 
11:32 PM
@MikeMiller Yes.
I'm thinking about what to write.
"Hello. Yes, you know me. I'm pure awesome. Vote me. Thanks."
 
Hmm no @KajHansen I don't know enough topology to solve this
 
Vote for Pedro.
 
I've been slacking off on self-study for the past few days, I need to get back on track... I'm falling a bit behind with topology...
 
I had an idea of maybe splitting A up into smaller parts and then showing the union is countable.
But I have no idea if that will work
or how to do it
 
That's along the lines of what I'm doing @Alizter. I'm currently working out some kinks.
 
11:34 PM
oh really?
Because I remember that this is a way to prove countability of thingies
 
hahaha :)
 
@Kaj well, every neighborhood of $a\in A$ contains some rational number, that's also known
 
But then again I know a limit points worth of topology
 
@Alizter More like a null set :P
 
Voting opens next week right?
 
11:36 PM
If we look at the neighborhood in $\Bbb R$ of course
 
Yes @Studentmath. That was my first solution. I'm working on an alternative contrapositive proof right now.
 
Ah, cool
My topology isn't that bad :P
 
I remember when Pedro almost killed me for mixing up Contrapositive and Contradiction.
 
My Algebra apperently is. I am stuck on this question, this time I think it's actually true..
 
@Studentmath Same question?
 
11:37 PM
I like algebra. I could help if you aren't too far in
 
@Alizter nah, that one was false
 
Wow its been 8 hours since school finished and I am still wearing my bag.
 
@Pedro Look at past elections for inspiration.
 
@MikeMiller I didn't know they kept them.
 
...How is that possible @Alizter?
Wow.
 
11:39 PM
This one is rather nice. Let $P$ be a sylow-p subgroup of $G$. We look at some $A$ subgroup of $N(P)$, and we know that $o(A)=o(N(P)/P)$. I want to show that we can conclude that $PK=N(P)$
 
@KajHansen I just kind of raided the fridge and sat on my bed when I came home
 
hahaha
 
@Pedro copy-paste what the previous winners wrote, just change the name
 
My back feels really light now :P
 
@Pedro They do; you can find links to them in one of the recent meta posts.
 
11:41 PM
So, I managed to use definitions and see that $N(P)/P=\{Pg| g\in G, gP=Pg\}$
 
@MikeMiller When does one study limit points?
 
Never, God willing.
 
What I need to show is, then, that for any $a_1,a_2\in A$ so that $a_1\neq a_2$ $ a_1P\neq a_2P$
 
@TedShifrin i am a prodigy, yay!
 
I have no idea how to do that though. I should reach a contradiction, I am hoping. I have no idea how.
 
11:43 PM
And now the question I was working on earlier has been posted on the site, here, along with most of the details.
 
hi @kaj @mike @teadawg1337
 
@beginner hello!
 
@beginner nice to meet you
 
i am trying to learn the chinese remainder theorem now
 
You are the same age as my sister
approx
 
11:45 PM
hi @alizter you are my fellow prodigy :)
 
@beginner I wouldn't fall in that category. How old are you?
 
I would've been a prodigy if I'd been given proper attention in early schooling...
 
@Alizter im in grade 5
for a few more days :)
 
@beginner I have literally no idea what age that corresponds to
 
good hehe
 
11:47 PM
@Alizter ~10-12
 
10 or 11?
I guessed right
OK so my sister is still younger than you
 
that would be against the rules, i am 13
 
@beginner You can say anything here nobody cares (within reason)
 
apparently it is illegal to chat here if you arent 13, even though i am not in the US
 
@Pedro: I think you might want to spend more time on math itself and on life, but you have my support if you want to run.
 
11:48 PM
@Ted Welcome back(?)
 
@TedShifrin Mod work is not that time consuming.
 
Oh, really?
 
@TedShifrin ted retires from being a professor and becomes a mod maybe?
 
Also, $o(N(P)/P)$ is the number of sylow-p subgroups in $N(P)$..
 
@beginner Tell me your age and if anybody asks say its base 16
 
11:49 PM
Thanks, @teadawg, going to a concert soon.
 
Hey @TedShifrin
I think I've solved your thingie.
 
@Ted I know, I was just checking :P
 
You have my vote too Mr. @Pedro.. according to this sampling you will have 100% of the votes
 
@TedShifrin Your thingie is interesting.
 
Hi @Kaj ... It wasn't hard. There was just a horrid non-answer.
 
11:50 PM
@TedShifrin What is a good example of such an A?
 
@Pedro You might want to ask a mod before you say that.
 
@Ted I managed too! (which made me happy, since I am unsure of my topology. But I got it in the easy-way)
 
@beginner Ok :)
 
@MikeMiller Say what?
Oh.
I've asked.
 
Indeed. My first proof was super fast. I'm now working on a contrapositive proof that's a bit trickier.
 
11:50 PM
@beginner What are you learning?
 
@Ted Looks like I might learn some "tensor triangulated geometry" next quarter, too, which I'm sure you love :)
 
@Alizter balarka has me doing set theory, but i am doing linear algebra and number theory for fun
 
How bout $\Bbb Z$, @Alizter?
 
@TedShifrin a music concert??
 
@TedShifrin in R?
 
11:51 PM
@beginner: you have a 14 yr old charting your education?
5
yes, 5 pianos :)
 
@TedShifrin maybe hehe. he says set theory should come first
 
hi all. Does anyone read Courant's calculus before ?
 
@TedShifrin oh piano that makes more sense, i was thinking rock music
 
Ah @Ted music. What are you seeing?
 
Good book, @XingdongZuo
 
11:53 PM
@beginner Have you got any books on math?
 
@Mike: I have no idea — earthly or otherwise.
 
@beginner I think the point is that if you're looking for advice, Balarka may not be the best person to talk to.
 
Google The Five Browns @Alizter
 
@Alizter my brother does math, so he lets me have some when i see him, but he lends them to other people aswell so i dont get them all the time.
@MikeMiller why not?
 
@TedShifrin I am reading it now. Some friends said it is old fashioned. As far as I read, it seems just have a little different "face" on the proof. But ideas are identical to the lectures I attend.
 
11:55 PM
@Ted It's categoey theory. That's what Balmer calls the theory of tensor triangulated categories (the most common example of which are derived categories).
 
Ugh @Mike.
 
@beginner He might throw the simple stuff and confuse you quickly.
 
@XingdongZuo, sometimes old fashioned is better.
 
@XingdongZuo Mathematics doesn't change so much.
 
That's what I was looking for @Ted
 
11:56 PM
@MikeMiller he caught me the riemann zeta function: $$\zeta(s)=\sum \limits_{n=1}^\infty \frac{1}{n^s}$$
 
He will be like: OK addition. Well you could think in terms of additive groups. Which in Galois theory ...
 
@Alizter he did do that lol. he said something about modules
@Alizter when talking about modulo arithmatic i think
and said it was chapter 13 of dummit and foote lol
 
he wont stop about dummit and foote
 
i have the pdf, but it is hard
 
yeah...
 
11:57 PM
It's like two students ranting about their proffessor
 
it covers injection,surjection and bijection in a small paragraph
 
@Studentmath Except I am older
 
You're younger than me, makes you quite young
 
@beginner I recommend: Book of proof. It goes over set theory, logic and proving things. I think it is a good start. It is not too difficult to read.
 
Hey, @Ted! How've you been?
 
11:58 PM
im younger than teds cloths lol
@Alizter thats actually what balarka recommended weirdly enough
 
Also, can anybody recommend a good Analysis book that I can read and try exercises from to get a solid foundation in the subject?
 
@beginner I recommended it to him ages ago
 
Balarka4mod
 
Hi @Khalil
 
Nice choice, @Alizter! Hammock's style of writing is really accessible!
 
11:59 PM
@Khallil make it a hashtag
 
@Alizter oh, well that worked out well then, the oldest prodigy to the younger and than the youngest hehe
 
oh yeah @Khallil is another one of my recomendees of that book
 
smacks @Kaj
 
@TedShifrin LOL
 

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