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00:00 - 19:0019:00 - 00:00

user19161
12:00 AM
Also, looking at the curriculum alone, there isn't really anything special about the top American universities when it comes to an undergrad math course.
 
it's 1 am i will go sleeping
good night everyone
 
user19161
But if I were to start life all over and could apply anywhere to do an undergrad course in math, I might not choose Cambridge or Oxford. I might go for Imperial College or Warwick instead.
 
user19161
@DominicMichaelis See you in your dreams.
 
@JacobBlack Over the years, more of my students went to Warwick than either Oxford or Cambridge
 
user19161
Well, because like I said before, Cambridge tries to squeeze what is covered in 4 years into 3 years and do a bit less. Then in the 4th year, they do some very exotic things which may not be so fundamental. So I rather go for the 4 years in Imperial or Warwick.
 
user19161
12:08 AM
@OldJohn Well, I did not know much about universities back then. I did not know there is so much difference. I really disliked where I studied.
 
@JacobBlack You also have to bear in mind that my experience is at least a decade ago - in the last 10 years I have not been teaching maths
 
12:20 AM
Do I detect another crank?
 
user19161
@OldJohn I wonder what exactly is written in the referenced.
 
@JacobBlack I have downloaded a copy - if you want to see :)
 
user19161
@OldJohn Nah, not too interested.
 
@JacobBlack It looks like 7 pages of confused elementary number theory manipulations - I can't be bothered to find the mistakes
 
user19161
@OldJohn But does it really claim to prove FLT?
 
12:25 AM
oh yes
 
@OldJohn I have voted to delete.
 
user19161
@anon Using your godly powers.
 
@anon Good! I have already wasted too much time following his curious logic :)
and it has gone already!
 
user19161
@OldJohn That's fast!
 
@JacobBlack Yep - I am in awe of these godlike powers!
 
user19161
12:31 AM
I am 1560 points away from these godly powers as well.
 
Just like Kaster.
 
user19161
@skullpatrol What are you talking about now?
 
Kaster came in here and passed judgement like he was a god.
 
user19161
Only anon is god, lol.
 
Time I got some sleep - thanks @anon
 
user19161
12:34 AM
The great anon!
 
G'night all
 
later
 
user19161
@OldJohn Good night!
 
:)
 
user19161
I think anon's current avatar makes him very godly.
 
12:54 AM
Hi..
 
 
1 hour later…
2:24 AM
2 days ago, by anon
I'm just going to pretend it's not there and hope my teacher doesn't have time to check it all
 
sleep eludes me :(
@anon - do I gather that your cunning ploy succeeded?
 
3:25 AM
@anon Haha
 
3:47 AM
Hi @Kaster
 
@user585104 sup
 
@Kaster Not much, you?
 
@user585104 same old
 
@Kaster I haven't seen you here before?
 
@user585104 have no idea. i definitely didn't. even if i did, i probably wouldn't recognize you, user_bunhofnumbers_ nickname, you know
 
3:57 AM
@Kaster I'm just asking if you come here often?
 
wowzers, is it just me or did it suddenly get hot in here?
 
@anon What do you mean?
 
@user585104 nah, not really often.
 
user19161
4:57 AM
@anon Maybe it got hot because of my presence. =)
 
7:06 AM
are there any visualizations of 3d turmites out there? i haven't been able to find any
grr that didn't work either. direct link to some turmites: flickr.com/photos/_robson_/7924984276/in/photostream
 
 
2 hours later…
8:49 AM
heloo everyone, i missed this room so much
 
Hi wassup
It kind of wierd , on this page chat.stackexchange.com/users/6275/skullpatrol it says your karma is 2.6K , but here math.stackexchange.com/users/22862/skullpatrol it says, its 212. @skullpatrol
 
@Tomarinator are you investigating me?
 
As a matter of fact I investigate every user that replies to me, yes I am that lame. :P
Oh its not an anomaly, its just how things are.okay i get it. <silly me>
 
 
1 hour later…
10:09 AM
Hello there.
 
user19161
Every review I have read on Chrome vs Firefox is in favour of Chrome.
 
It has an annoying memory leak with me
 
user19161
It is less responsive in chat for me.
 
user19161
However it is currently the only way to get latest flash on linux.
 
user19161
@Tomarinator You misspelled "weird".
 
10:24 AM
TIL the correct spelling of word weird. thanks
 
user19161
Correcting spellings is a great way to make friends online.
 
...and enemies.
 
user19161
I see you are the black square @tom!
 
and you are the blue square.
 
user19161
10:27 AM
Hello @skull, is someone upset with me for not seeming to talk to him? =) Well, sometimes I have nothing to say...
 
...dunno...
 
user19161
@Tomarinator There are a few more coloured squares around...
 
oh really, i didnt knew that, i became the black square because a graphicon (a smiley) wasnt properly rendered by the image uploader on stackoverflow, , and I was too lazy to look up something more hippy.Thats how i came to be the black square
 
user19161
I change colours now and then.
 
@awllower i commented on that question
 
10:32 AM
What question?
oh
 
0
Q: Set of numbers pairwise relatively prime

VoyageI want to prove that $b+1,\,2b+1,\,3b+1,\,...,\,(n-1)b+1$ (given an $n$, we can find infinitely many positve $b$), where $n$ is a positive integer, are pairwise relatively prime. I assume that $b+1,\,2b+1,\,3b+1,\,...,\,(n-1)b+1$ are not r.p.. Let $1\le i \le j\le n-1$ and $ib+1,jb+1$ are not ...

 
I think your objection is invalid.
 
i don't know the right way to prove it either but i think op is on the wrong track
his false proof doesn't ever use $i \ne j$, that is why it works
 
user19161
At first I did not know what a proof is, after that I knew, now I am no longer sure.
 
indeed.
Maybe it is not for every $b$.
But, for given $n$, there infinitely such $b$.
 
10:36 AM
@JacobBlack It's sort of like the old saying about learning thermodynamics...
 
user19161
Hi @old, another day older. =)
 
Hi @OldJohn
 
@JacobBlack Thank you for reminding me of that! :)
Hi @skullpatrol
 
I was about to compare this
4 mins ago, by Jacob Black
At first I did not know what a proof is, after that I knew, now I am no longer sure.
to the old saying about learning thermodynamics...
 
Or: "I used to be indecisive, but now I am no longer sure"
 
10:42 AM
The first time you see it you don't understand it, the second time you see it you think you understand it but you really don't, the third time you see it you know you don't understand it but by then your so use to it you don't care anymore.
3
 
user19161
@skullpatrol WOW!
 
that must be understand-ception
ok that was a bad one
 
user19161
Someone should star what I said too. =)
 
Mine too.
 
user19161
10:44 AM
@DanBrumleve I don't get it.
 
user19161
Anyway @skull I won't go to the party room anymore, I will just stay here...
 
user19161
This is the site where I see the weirdest meta questions.
 
user19161
There isn't so much a problem with say homework questions on other sites.
 
user19161
I don't know why there is such a big fuss over it on math.
 
user19161
11:56 AM
@JonasTeuwen Annoying is a very annoying word bro.
 
12:29 PM
At first, I was afraid, I was petrified, ti think I could never live without math by my side...
 
12:40 PM
Nice 15 bucks to spend at the play store! Thank you google!
 
@charlie Oh please, I would rather spend 1000 dollars on a fast internet connection and get a torrent client.
 
Ok. But it's not always that I get free things
 
12:59 PM
Hello
 
Olleh
There's no life I know to compare with pure imagination, living there you'll be free if you truly wish to be
 
How do I classify simple groups of order 60?
@anon, good morning $\mathfrak{anon}$
 
blegh
 
1:16 PM
whasts wrong
 
morning in general
 
oh
you should get up earlier then morning seems like afternoon
@anon, I gained further insight into the fano plane from OldJohns suggestion
hi frog
how do I turn matrix to vector in pari gp?
 
1:32 PM
hi
 
hi
 
if I have an object consiting out of a sphere and a triangle.
How can I determine the points on which the sphere is connected to the triangle?
+ how many points can there be at least?
+ how many points can be there at most?
 
I don't understand
 
1:57 PM
can you help me with grroup theory please :(
let G be simple of order 60, then why can't it have a subgroup of index 4?
 
4! <60
Hence the core of any subgroup of index 4 would be nontrivial.
 
Hi all
@user58512 Did you solve the Fano plane thing?
 
@OldJohn, yeah that was wonderful!
 
I thought about it again this morning
 
I couldn't sleep so I did that and it came out to 168!
 
2:02 PM
I think I have a simple argument now
 
Hi @oldjohn !!!!
 
choose 3 points a, b, c which are not all collinear and look at possible images - 7 choices for $f(a)$, then 6 choices for $f(b)$ and then 4 choices for $f(c)$ (since it cannot be the third point on the line through $f(a)$ and $f(b)$) - then 7.6.4 = 168
 
@peoplepower, I don't understand that at all :(
 
Hi @Charlie!!
 
@OldJohn how are you?
 
2:05 PM
@Charlie Tired - from a cold which is keeping me awake at night - but it is getting better gradually. And you?
 
@OldJohn, I did rougly the same: Take one point anywhere 7 choices, then another point anywhere 6 choices, that determines a line - without loss of generality we can take that line to be the base of the triangle: Now there are 2x2 symmetries of the top part of the triangle. -- hardest part for me was realizing all the lines of the fano plane are actually circles
it was only matching up the triangle diagram and the C_7 diagram that I realized this
 
@OldJohn I'm fine, here is hot
 
@user58512 Yep - pretty much identical argument. It takes a while to get the hang of lines not being like normal lines :)
@Charlie how hot? (about 8 or 9 here)
 
@OldJohn, so your comment solved my question here ! thank you very much :D .. if you want to just copy & paste it I'll accept it (if you can't be bothered no problem)
 
@OldJohn ow, yeaterday it was 31, today may get 32...it's currently 27 celsius
 
2:08 PM
@peoplepower, oh! "The normal core of any normal subgroup is the subgroup itself"
wait you said nontrivial.. I don'tunderstand
 
@user58512 Not sure I have the time to do it today - but if you want to post your own solution (and maybe mention you had a hint :) ) - that would be fine
 
ok
 
@Charlie 32 is starting to feel too warm for me :)
 
@OldJohn i don't feel good it this heat... I'm not fond of it...
 
@Charlie I prefer to be a bit to cold rather than a bit too warm - but 8 or 9 is just a bit too low for comfort :)
 
2:11 PM
@user58512 Here's the result I was using. Let $G$ be any group; if $H<G$ has index $n$, then the core of $H$ has index dividing $n!$. You can prove this by considering the kernel of the action of $G$ on the $n$ cosets of $H$.
 
time this old guy had a nap, I think - later
 
@OldJohn hehe I prefer a bit cold too :P
 
@Charlie :)
Later folks
 
Later, John!
 
bye, thank you!
 
2:16 PM
@michael Hi!
 
@Charlie Hello
 
Wassup?
 
the sky :P
 
@MichaelCorleone that's the best answer ever! Thanks!
 
Really?
haha
I never know what to say when someone asks what's up
 
2:18 PM
Yeah it's funny and logical
 
:)
 
@peoplepower, I still don't see it :(
 
How are you @michael ?
 
how would it be nontrivial? 1 divides n!
 
2:21 PM
@Charlie I'm good, and you?
 
oh wait I got it
the index divides is something.. so the order of the group is something else
 
@MichaelCorleone I'm fine
 
Good to hear that :)
 
is it |H : core H| | n! or | G : core H | | n!?
I guess the second
 
The second.
 
user19161
2:24 PM
@people You seem really interested in groups?
 
so the size of the core is a nonzero multiple of 5!
 
@JacobBlack It's just the subject I have studied the most. I actually do not have much interest in the theory of groups by themselves.
 
@MichaelCorleone have you been studying a lot?
 
@user58512 Yes, if that was an exclamation point.
 
hehe
 
user19161
2:25 PM
@peoplepower So what are you most interested in currently?
 
it was
and for the orders 2,3,5 I used a lemma that if G/N is cyclic then G is abelian
 
@Charlie Not enough :P. I have learned quite a lot the past while so I guess I have been studying a lot
 
so I'm proving nonabelian simple group of order 60 - the abelian are easy with the structure theorem
 
@MichaelCorleone good, good
 
and you?
You had a long break at winter didn't you? Are you back at school again?
 
2:28 PM
@JacobBlack Hm.. Category theory has interested me most recently.
 
no im really confused
 
@MichaelCorleone summer, actually. Yes, I'm back !
 
user19161
@peoplepower OK, I hope to see your papers soon!
 
@user58512 Presumably you know the Sylow theorems.
 
in a comment someone told me that I will find a subgroup of index 5: But if I have a group G of order 60 with a normal subgroup N of index 5 then because G/N is cyclic G must be abelian
 
2:29 PM
@Charlie O I thought I had read that you had a month off around christmas
 
@user58512 It will not be normal. A4 is a subgroup of A5.
 
hmm ok... so in fact it has no subgroups of index < 5 at all
 
@MichaelCorleone yes, I did, but it is summer here :) not winter, during winter, is less time
 
2:31 PM
@JacobBlack :) I probably have some mixture of ADD and an addictive personality.
 
@MichaelCorleone :P :)
 
oh ugh
G/H means the cosets
I assumed it meant H was normal
I write G:H for cosets
 
user19161
@peoplepower Oh OK. Hmm, perhaps I might email you some day...
 
@user58512 Hello!
 
hey @awllower, how ar eyou
 
2:36 PM
I am recently troubled by the verisimilar proof that I came up with in my post.
 
user19161
@awllower Where?
 
I thought that this isomorphism must be simple, but, when I tried to construct one, it took some considerable time indeed.
 
What are you studying, @michael ?
 
@Charlie Maths and physics
 
@MichaelCorleone hmmm physics too, nice :)
 
user19161
2:41 PM
I lost interest in physics as there was too much math to learn already.
 
how long would it take to learn all of math?
 
user19161
Never.
 
@Charlie It should be, but I'm not enjoying the practicals too much, so I'm considering stopping physics next year
 
@MichaelCorleone oh!
 
user19161
@MichaelCorleone My practicals in high school were 3 hours.
 
2:43 PM
what's wrong with having a homomorphism from G into the symmetric group $S_{|G:H|}$?
 
@JacobBlack My practicals are a full day. I'm not in high school, if that was what you were implying
 
because $|G|=2^3 \cdot 3 \cdot 5$ and $|G:H| < 5$ the order 5 elements get mapped to 1, but that's ok?
 
2:54 PM
oh the kernel would have size 5
so a normal subgroup of G
 
3:07 PM
 
user19161
@Ilya Congrats!
 
if G has order 60 and suppose (for contradiction) that it has 4 3-sylow subgroups, then let it act on the set of those subgroups
This gives a map G -> S_4, and since |G| is bigger than |S_4| its got nontrivial kernel
but how do I show the kernel is not the whole group?
for all sylow subgroups P, P' there exists g in G with P^g = P' ? so the map is surjective?
 
Recall that the Sylow groups form a single conjugacy class, so the map is nontrival. I.e. the kernel is not $G$.
 
how does the fact they are a single conjugacy class imply that?
oh you can conjugate one to another
I see
this is really interesting
 
user19161
@peoplepower Is your email still the same one that you mentioned?
 
3:15 PM
@JacobBlack Yes.
 
user19161
@peoplepower OK, I just sent you an email so that we can keep in touch, and also to share some things.
 
@JacobBlack Thanks. :)
 
:)
 
user19161
@peoplepower Did you get it? If so, I expect a reply of some sort. =)
 
@JacobBlack Yes, you shall get a reply.
 
3:26 PM
<=>
 
I wrote a script that prints out the possible number of Sylow subgroups for groups and simple groups
 
@michael what is your favorite The Godfather?
 
hi @AndrewSalmon are you interested in simple groups
I would really like a nicer way to show that for a simple group of order 60: having 15 2-sylow groups is impossible than the one given here
any suggestions?
 
user19161
@user58512 To be honest, I kind of hate groups, lol.
 
I kind hate "lol"
 
user19161
3:35 PM
@Charlie Really?
 
pls
 
Yes, I hate
 
how to show a group doesn't have too many sylow subgroups?
 
Time to sleep.
Bye Ladies and gentlemen.
 
good bye
 
3:42 PM
Bye
 
hows this for a proof: Let G simple of order 60, by sylow theory it has 5 or 15 2-sylow groups, 10 sylow 3 (= 3 cycles), 6 5 cyles.. and assuming it doesn't have 151 2-sylows then this is all the same as A_5 so it's almost certainly A_5. QED/2.
 
user19161
@awllower See you in your dreams.
 
Hello :)
I have a basic question
if
 
@AmithKK hello
 
user19161
3:48 PM
Once you know tan, you can work out sin and cos.
 
user19161
Simply draw a right triangle.
 
Yeah, I just wanna check if I'm doing it right :)
 
user19161
Yes, right.
 
-15/11?
is that right?
 
user19161
Well, you can work that out yourself I think.
 
3:51 PM
that seems wrong
aren't you supposed to get a positive ans?
oh got it :P
 
user19161
First work out the denominator.
 
this is great, any subgroup H of a simple group G - normal or not - has index [G:H]! >= |G|
so for a Sylow subgroup to exist we require |G| <= (|G|/p^r)! ... but Sylow subgroups always exist
 
Hmm
 
so I conclude that 168 <= 51090942171709440000
 
Fascinating
 
user19161
4:06 PM
How is that inequality fascinating?
 
user19161
168 is of course not greater than 51090942171709440000.
 
@Charlie I think number 2
 
@MichaelCorleone I like the first. Incredible.
 
I like the first as well, the third one wasn't very good in my opinion
or at least not nearly as good as the other two
 
@MichaelCorleone yes, the third is not that good
@user585104 hi
:-/
 
4:21 PM
hi
 
@user585104 wassup?
 
Hi @Charlie how are you?
 
hi
 
@skullpatrol I'm fine, and you Skull?
 
@Charlie Still alive ;-D
 
4:25 PM
@skullpatrol that makes me glad
 
@Charlie I'm glad you're glad.
 
@skullpatrol we're both glad ' cause you're alive ;)
 
@Charlie Yup, I was born to be alive.
 
> This completes the proof and thereby completes the classification of all simple groups of order 60.
yeah, he classified all one of them...
 
@skullpatrol OH YEAHH!!!!
 
4:41 PM
Do you^ mind keeping your voice down? :p
 
@user585104 NO ;)
 
can you help me understand thisplease
the second part
how do they find out the number of conjugates?
 
5:06 PM
@user585104 DOES MY CAPS BOTHER YOU?
 
Can't tell if this joke is in base 10... or base 10. O_O
 
@Charlie Are you trying to piss me off?
 
@user585104 HMM. I THINK YOU HAVE SOME VISUAL PROBLEMS; NOBODY IS USING CAPS.
@user585104 WHY ARE YOU ON CAPS LOCK?
 
Hmm. I think you have an attitude problem.
 
@user585104 DO NOT SAY THAT IN CAPS!!!! YOU LOOK REALLY DESPERATE!!!!!!
 
5:15 PM
meh
 
Well anyway, anybody here who has appeared in IMO?
 
@user585104 JUST A BIT, IS IT WORKING?
 
BBT
Good Night
 
morning all
 
5:24 PM
Morning
 
@Amith Hey, I don't see you here.
Aren't you seen in Ubuntu chat?
Welcome!
 
5:39 PM
hey @Arkamis
 
5:56 PM
im bored....
how do I find the normalizer of a group generated by permutations?
 
@user58512 Is that you?
 
6:12 PM
:P
 
6:28 PM
@user585104 welcome ;-)
@skullpatrol welcome ;-)
 
@robjohn About what?
 
@skullpatrol Just welcoming you back...
 
@robjohn Thanks (@)8<
 
my hands are cold :( :(
 
@skullpatrol (@)8< looks like a pineapple
 
6:35 PM
(||)8<
)))8< $(mean)^3$
 
it seems like the uniqueness proofs can get a bit bad
for simple groups
 
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