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00:00 - 13:0013:00 - 00:00

1:00 PM
@robjohn that's true.
 
@Lord_Farin ahh, never learned about topos theory
 
@TobiasKildetoft It's nice. :)
 
@Lord_Farin not that fine ...
 
@DominicMichaelis Why? (If you care to share with some strangers in a mathematics chatroom, that is.)
 
@robjohn for instance, by considering the real and imaginary parts of $-\log(1-i)$ we are done.
 
1:05 PM
@Lord_Farin one of my best friends stop talking to me and so ...
 
...how childish.
(not to mention their loss, not yours)
 
@DominicMichaelis ... Sad. I presume it was sparked by some fireworks over nothing? (I furthermore second @skull's statement.)
 
Well she suddenly said that I always annoyed her and she doesn't want to talk to me anymore
 
@robjohn that minus sign in front I think it was a kind of automatism.
@DominicMichaelis life is tough. :-)
 
You just gotta be tougher.
 
1:11 PM
@DominicMichaelis Such surprises me. You have never succeeded in annoying me; though perhaps our means and regularity of contact, as well as the usual topics discussed may have contributed to that as well.
 
she studies math
 
Girls generally like the strong silent type.
 
well I will drive home see you later
 
@skullpatrol I haven't seen them flocking around me, though. And most would consider me silent, thoughtful and what else has been thrown at me for the past years...
@Dominic Bye. Have a safe journey.
 
@Lord_Farin maybe it is the "strong" part that is missing :)
 
1:16 PM
thoughtful $\neq$ strong
 
@TobiasKildetoft We should have a survey to make them explain what girls' definition of "strong" is. :P
 
Strong in the sense of being confident.
 
This confuses me. I seem to have helped OP, but he's added some cryptic comments...
 
@Lord_Farin it seems the OP does not really have the required mathematical maturity to tackle the stuff he is learning
his second comment starts with nonsense (talking about elements of $[K:F]$ which is a number)
 
@TobiasKildetoft The "run before you can crawl" syndrome. Yes, I suspected as much. Though, I'm glad I'm not alone in my assessment.
 
1:28 PM
@Lord_Farin there was an extreme example of it recently where someone asked about what it meant when one had a "$i+j = k$" as summation index, and where he had seen it was in connection with graded cohomology
 
@TobiasKildetoft I've fallen victim to this pitfall several times myself, but nothing of this magnitude... The accumulated stupidity and stubbornness of the human race never seizes to amaze, I guess. :)
(And that's directed at myself, too. I'm not trying to burn the person in question.)
 
@Lord_Farin yeah, I can feel that I lack some background in some areas once in a while
but in this case, it does not look so much as lack of knowledge as lack of experience
 
@TobiasKildetoft Agreed.
 
@TobiasKildetoft that $"i+j=k"$ reminds me of the relation between some double sum & stars and bars (combinatorics).
 
Btw, are you (aspiring to become) a group theorist?
 
1:32 PM
(not that I have not done my share of courses before I had the necessary maturity to really grasp the subject involved. For example, a course on Galois theory as the first course I took on my second year undergraduate, and a course on advanced Galois theory as the last course on my second year)
@Lord_Farin to some extend
my current research has been more focused on modular representations of algebraic groups
but I am also waiting for some input on something that will hopefully turn into a project in finite group theory
(and the one paper I have published was about complex characters of finite groups)
 
@TobiasKildetoft Nice. Though group theory and algebraic geometry aren't really among my fortes.
 
@Lord_Farin for most of my purposes, I can practically forget about the algebraic geometry that lies behind the stuff
 
(Which may come off a bit paradoxical as AG is the main field of application for Grothendieck topoi, but I try to shelter safely on the logic side :).)
I'm currently investigating if I can start a PhD in nonstandard logics in December.
 
@TobiasKildetoft Yeah, I'm pretty excited.
 
1:38 PM
Is that the same as what they call "fuzzy" logic?
 
@skullpatrol Fuzzy logic is one type of nonstandard logic. I haven't been filled in on all the details, but it appears the research will be directed towards logic that can be employed in multi-party communication.
 
@Lord_Farin if that communication is taking place on the internet, I doubt any sort of logic will be applicable.
 
@TobiasKildetoft Heh. :)
 
Present company excluded :-)
Hi @amWhy how are you?
 
@skullpatrol Hello! I'm okay :-|
=D
@skullpatrol How are you doing?
 
1:51 PM
@amWhy Fine thanks.
:-D
 
Does anybody know what tag should be used for a question concerning cardinality?
 
In what context?
 
I'm about to ask why the natural numbers and integers have the same cadrinality.
cardinality*
 
@SujaanKunalan I would go for [elementary-set-theory].
 
@SujaanKunalan you might want to check if that has not been asked before (I have a feeling it must have)
 
2:03 PM
Ok, thanks. Alright, I'll double check if its been posted.
 
@SujaanKunalan How about 74820?
 
Thanks! That looks good.
 
@SujaanKunalan Glad to help :).
 
and note the tags :-)
1 hour ago, by skullpatrol
 
2:23 PM
Shalom
@skull hi
 
@Charlie hi
 
@lord_f how are you doing today?
@skullpatrol how are you?
 
@Charlie Fine thanks, how are you?
 
@skullpatrol fine ....
 
2:37 PM
:-/
 
:-|
(:-|)-<-<
(:-D)->-<
 
@skullpatrol :-/-<--<
 
@Charlie (:-D)-/-<
 
Hello
 
2:48 PM
@cyclochaotic hello
 
(:-D)-/-<

(:-D)-\-<

(:-D)-/-<

(:-D)-\-<
 
Does anyone know how to get Mathematica to expand $2^{n+3}$ into $8*2^n$?
 
@skullpatrol (:-/)-<--<
 
@Charlie Less worse. Losing some steam using FPS.
 
@cyclochaotic Have you tried asking in their chat room?
 
2:55 PM
@Lord_Farin oh!
 
@cyclochaotic AFAIK it can't be done.
 
@skullpatrol no, do I have to join mathematica.stackexchange.com?
 

 Wolfram Mathematica

Welcome! This is the main Mathematica chat room for mathematic...
 
@skullpatrol ty
 
3:01 PM
@skullpatrol I figured a work around. `Assuming[k \[Element] Integers && k > 1,
FullSimplify[Mod[2^(1 + k) + 2^(3 + 2 k) - 3^(2 + k), 2]]]`
 
good
 
o brother, Mathematica gets stuck on this Mod[2^(5 + 4 k) + 4^(1 + k), 3]
 
3:18 PM
@skullpatrol ORLY?
Hi @dominic
 
@Charlie hi
 
@DominicMichaelis so...wassup?
 
@cyclochaotic That is clear because of it doesn't know if k is an integer
 
@cyclochaotic why use mathematica for that anyway?
 
@Charlie nothing :/
 
3:27 PM
@DominicMichaelis did you try that thing?
 
treid it once didn't work, trying it again
 
@DominicMichaelis I am using Assuming for k
 
@cyclochaotic it is 0
 
@DominicMichaelis ok :)
 
@TobiasKildetoft Im using Mathematica so I don't have to do every Mod by hand
 
3:33 PM
@cyclochaotic I doubt if mathematica will be able to give you general results for arbitary k
 
@TobiasKildetoft agreed
 
@Chris'swisesister hi
 
@Charlie Hello!
 
@Chris'swisesister how have you been doing?
 
I was sleeping. How about you? :)
 
3:45 PM
@Chris'swisesister I'm fine,.a bit lost, but I'm fine :)
 
@Charlie What happened?
 
@Chris'swisesister I took sometime to rest...
But I don't know what to do
 
@Charlie In a way I understand you because sometimes I also feel myself lost.
@Charlie I always want to do something, but I'd like to find the power to rest and do absolutely nothing for some time.
 
@Chris'swisesister I don't know if I should stay sometime without doing anything ....even if I want to
 
3:59 PM
Oh......
 
4:45 PM
Hi @skull
 
Hi @Charlie
 
@skullpatrol :?
 
5:07 PM
2 hours ago, by skullpatrol
(:-D)-/-<

(:-D)-\-<

(:-D)-/-<

(:-D)-\-<
 
5:38 PM
Hi @Charlie
 
@skullpatrol hi, Skull
 
@skullpatrol hehehehe
 
@Charlie rating please
 
5:53 PM
@skullpatrol 10.0
 
 
1 hour later…
7:19 PM
hi folks
I need help in algebraic structure
 
@pourjour what sort of algebraic structure?
 
I need to prove that $\forall (a,b) \in I^2) a*b=e^{\ln(a).\ln(b)}$ the law $*$ is commutative and associative
 
@pourjour @What is $I$?
You mean $I\times I$?
@pourjour Do you know what that means?
 
$I=]0,+\infty[$
@PeterTamaroff not sure
 
7:23 PM
You ought to prove $a*b=b*a$ and $(a*b)*c=a*(b*c)$
 
@pourjour is that first part meant to be the definition of *?
 
@TobiasKildetoft I guess so.
@pourjour You mean $\exp(\log a\cdot \log b)$, yes?
 
@PeterTamaroff yes
@TobiasKildetoft * is composition law
 
@pourjour (then putting a $\forall$ seems odd)
 
@pourjour OK, can you show $a*b=b*a$? That is not hard.
 
7:27 PM
@PeterTamaroff yes commutative is easily proved
how about associative
 
@pourjour Good. How?
 
Hi'll
 
@PeterTamaroff because ln(a).ln(b)=ln(b).ln(a)
@cyclochaotic hi
 
@pourjour Well $(a*b)*c=(e^{\log a\log b)}*c= \exp(\exp({\log a\log b})\log c)$ yes?
Whilst $$a*(b*c)=a*(\exp(\log b\log c))=\exp(a\exp (\log b\log c))$$
 
@PeterTamaroff a,b, c are infinite dimensional Tensors? :)
 
7:34 PM
@cyclochaotic No idea.
 
@PeterTamaroff thanks peter :)
 
@pourjour I guess you have to find out if it is associative or not, yes?
Oh, wait!
@pourjour I wrote it all wrong!
 
@PeterTamaroff what is wrong
 
$a*(b*c)=(a*b)*c=\exp(\log a\log b\log c)$
Because $\log\exp (\log b\log c)=\log b\log c$ and $\log\exp(\log a\log b)=\log a\log b$.
 
@PeterTamaroff log a +log b +log c?
 
7:37 PM
I missed the definition of $a*b$.
@cyclochaotic No, not sums, products.
 
@PeterTamaroff no problem I found it
 
7:47 PM
how can I prove that A has no inverse by just calculating $A^2$ and $A^3$
 
@pourjour well, what do you get if you calculate those?
 
hold a momment
 
If we have associative and commutative then why not socialative?
 
@skullpatrol no. i saw arturo's talk (which was nice, his exposition is as good in life as in SE) but I didn't get a chance to meet him (got distracted by women)
he's got a paper coming out though that i intend to read, he's attacking this problem on capable groups from an interesting standpoint
 
@TobiasKildetoft this the result
 
7:53 PM
he looks suave as hell, by the way. learning to rock a vest like him is now high on my to do list.
 
@pourjour so since it is nilpotent, it cannot be invertible
 
@pourjour: Now what would be $A^{-3}A^3$?
 
what does this mean "nilpotent"
 
@pourjour that some power of it is 0
 
7:55 PM
@TobiasKildetoft and so
 
@pourjour see the hint by Julian
 
@JulianKuelshammer I think we can't calculate $A^{-3}$ becasue det is = 0
 
@pourjour but if A was invertible, it would be the third power of the inverse of A
 
@TobiasKildetoft do you mean if A was invertible then there exist $A^{-1}$ then $(A^{-1})^3$ also exist which leads to a contradiction is that true?
 
@pourjour yes.
 
8:01 PM
if $A^{-1}$ existed then $A^2=A^{-1}A^3=A^{-1}0=0$, a contradiction
 
@anon couldn't understand the part where $A^{-1}0=0$ do you mean that all matrices multiplied by 0 are = 0?
 
yes
 
ok
 
8:26 PM
@DominicMichaelis, I can't let your personal post go without comment. It is a very long time since college for me...still, there are three fairly likely reasons for your friend to speak as she did: (A) she has a new boyfriend and does not wish your presence to mess that up; (B) she wants you as a boyfriend and is trying to get you to speak up and stop being iffy; (C) you really annoy her, and your apparent friendship is just do to the artificial closeness of a college campus.
All three are quite possible. I'm not entirely sure how you would find out more, unless you have good friends in com
Just saw you gravatar drop in. Hi there!
 
For my part, I don't know how i would just be friends with a woman, unless I were also friends with her husband, but times change. And, in fact, I have typically not gotten along that well with wives of friends.
Do you have any other friends in common with her?
 
8:41 PM
One but not a very good common friend
 
@DominicMichaelis, alright. It is harder if you lack sufficient information. And i do not have much more to offer than to suggest those three possibilities. About all MSE can do for you is this: if you do not already have an email address for Charlie, who is your age, you may email me if you like (use the gmail) and I can forward your expanded discussion to Charlie, who then may elect to email you directly. Since she popped in, same with amWhy, who is more graduate student age...
 
@will yeah she already talked to me :)
 
@Dom why do you have a picture on Facebook with green dragonfly wings tied on?
 
@Will it was the end of high school we always have a week where the graduates dress in a special way, like pyjamas and such stuff
 
9:01 PM
@amWhy While that is a sad I sympathies. Highschool sucked, I wish I dropped out and started college or studying mathematics earlier, instead of staying for the full 4 years learning essentially nothing.
 
@Dom, alright. I understand how circumstance might put people together for a while. I knew several people in college, more than I do now. I can see someone beginning to avoid you, if you do actually annoy her. the part i don't get is why she told you that you annoy her. Do you think that you pressed her for such a statement, forced her in some way to comment about you?
 
no I really don't think so
 
@Dom, then my three possibilities are really still open. I have errands to do now and will drop off Chat. You are welcome to email me, but note we are eight or nine hours different. Charlie and John Wordsworth are in closer time zones. I just feel that she chose that course of action to accomplish something that would not otherwise happen, but it spoils it for her if she needs to reveal what it is she wants. If all she wants is your absence, that is fairly easy to accomplish. Bye for now.
 
@Will Bye and thanks :)
 
@WillJagy Hi will
bye will
@DominicMichaelis hi, doms
Hi @anon
 
9:16 PM
hello
 
@anon how's your day?
 
alright
 
gooods
Once a friend stoped talking to me
out of nothing
he went to another school and didn't even told me
 
hmm
 
yes...
I found him in my uni
but now I don't talk to that motherfucker again
 
9:22 PM
I see.
 
@Charlie hi
 
@DominicMichaelis :)
Hi @AlexJBest
 
@Charlie Hello Charlie, how's exams?
 
@AlexJBest ENDED
":)
 
@Charlie WHAT NOOOO, WHY YOU MAKE ME JEALOUS!
 
9:33 PM
@AlexJBest ;)
@DominicMichaelis do you have my email, Dom ?
 
@Charlie Well congratulations I guess, got any plans for what to do with your freedom?
 
I don't know I have you on facebook that should be equal or =
 
Is there a moderator around?
 
@AlexJBest no plans
 
@robjohn Please ping me when you return.
 
9:34 PM
@DominicMichaelis :D
 
have it thanks :)
 
@Potato okay
 
@robjohn Bill Dubuque has returned under another account.
 
key ideas
 
Indeed.
 
9:36 PM
@Potato Fair enough, hope you have a nice rest then :)
 
@AlexJBest thanks :)
 
"Eternal return (also known as "eternal recurrence") is a concept that the universe has been recurring, and will continue to recur, in a self-similar form an infinite number of times across infinite time or space." - Wikipedia
2
 
@AlexJBest not for a long time, but it's being refreshing
 
@anon someone should grab the name Eternal Return. It sounds like an apt name...
 
@Charlie Good to hear, I'm reallllllly looking forward to finishing now.
 
9:39 PM
@robjohn Anyway, I thought you and the moderation team might like to know. I don't really have any skin in the game, but you know, rules and all that.
 
"And this is my ultra sexy stylish homepage" hahahahahah @AlexJBest
@AlexJBest I know that feeling!
 
@Charlie It's pretty stylin' don't you think?
 
@AlexJBest yeah
 
Can someone help me with this post, please? math.stackexchange.com/questions/405240/…
 
@Potato Thanks.
 
9:43 PM
@Charlie I've been making a list of things to do when I'm done but it's still annoyingly empty :/
 
@AlanH what about it? any nontrivial element a will have order not equal to 1 but a divisor of p^n hence of order p^m for some 0<m<=n, hence a^(p^(m-1)) will have order p.
 
@AlexJBest I'm reading about math
 
@Charlie Already on there :), what are you reading?
 
@AlexJBest number theory
 
@Charlie Ohh cool, specifically?
 
9:45 PM
then tomorrow i want to study set theory
 
@robjohn Is there a way to constantly render LaTeX? I used the bookmarks you gave me last time, but I have to click render every time someone types an equations
 
@AlexJBest prime numbers cool stuff
 
@Charlie Niiiice.
 
@AlexJBest prime numbers are so sexy
 
@anon but why would a^(p^(m-1)) have that order?
 
9:46 PM
@Charlie en.wikipedia.org/wiki/Sexy_prime intentional :D?
2
 
@AlexJBest hahah :D
 
@AlanH general group theory fact: if d|n and x has order n, then x^d has order n/d.
try to prove that
 
@robjohn nevermind, i didn't save the rendering on bookmark
 
@AlanH you can also refresh that page in your browser. That should reset the variable that blocks the rendering.
@AlanH unless you are talking about chat. Then you should use the start ChatJax bookmark
 
@robjohn yeah it works now. thanks
 
9:55 PM
Nice post, @anon (question & answer, no less!)
 
two of my wrong answers I vowed to correct and update, but haven't done so, have been upvoted recently. wonder if someone's trying to send me a message...
@amWhy thanks
 
@anon hehehe: get to work!
 
@AlanH of course, refreshing while you're editing can erase some of your edits.
the bookmark is better :-)
 
10:10 PM
Hi @peter
 
10:25 PM
shame on you @PeterTamaroff
 
10:47 PM
@charlie I was away.
 
leo
10:58 PM
@GustavoBandeira me too :-) Me and my cousins realized some kind of invariance of the head of the chicken
 
11:34 PM
@PeterTamaroff why is there shame on you?
 
11:57 PM
presumably because he did not say hello back in time
 
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