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11:00 PM
@mr.FS well, you have blackboard bold in your avatar :)
 
can't beat Springsteen 8)
 
@AlexeiAverchenko o haha i forgot about that lol, I actually prefer the sans serif computer modern to the serif one, but I like the blackboard bold letters yes
 
@Argon So so cool!
 
@Charlie I like that song
 
11:02 PM
@Argon What does it mean?
 
user19161
Hey @benja! I have been getting more rep lately!
 
@WillHunting I got my results today :D
 
user19161
@BenjaLim OK, I think you should be OK.
 
@Charlie The title means "You are beautiful"
 
@Argon hmm what about the rest?
 
user19161
11:03 PM
@BenjaLim Well done, but exams don't mean much.
 
@Charlie Some of it is in Amharic...
 
@WillHunting I know
 
it is midnight should i drink coffee?
 
@WillHunting you always make me feel better by saying these things :)
@Argon oh
 
I am the one who is always proselytizing that exams are trivial @WillHunting
 
11:04 PM
@Charlie 1 sec, I will get lyrics!
 
@Charlie depends if you are planning to sleep anytime soon :)
 
@Argon ok!
@OldJohn haha
 
@Charlie sorry - moving target again!
 
user19161
I am going for breakfast, will come back at 0000 GMT to upvote furiously.
 
that was meant for n3b
 
11:05 PM
@WillHunting Good breakfast!
 
@OldJohn i realized
@WillHunting See you!
 
@Charlie good :)))
 
@N3buchadnezzar it's 6 am, and I'm still good :D
 
@Argon WOW
 
11:07 PM
@AlexeiAverchenko Why up so late?
 
@N3buchadnezzar poor regiment :)
 
@N3buchadnezzar double espresso ... and some snus - should keep you awake :)
 
and I'm trying to wade through technical theorems about CW complexes
proving every detail from an appendix in Hatcher
 
@OldJohn snus? Kill it with fire. And I am trying to sleep, I just felt like some coffee
 
@Argon Really nice
 
11:09 PM
@N3buchadnezzar :)))
 
@Charlie I like his stuff
 
@OldJohn Do they have snus where you live?
 
@N3buchadnezzar no - but I have my sources :)
 
@OldJohn isn't snus an addictive drug? <_<
 
@AlexeiAverchenko yep
 
11:10 PM
@OldJohn you junkie :P
 
@AlexeiAverchenko :P
 
@AlexeiAverchenko No, it is unicorns and rainbows contained in a small box, which eventually gives you cancer, then you die. Horribly.
 
@Argon Seems nice thing :)
 
hello again
 
@mr.FS FS
 
11:14 PM
@mr.FS what does mean FS?
 
my initials
 
@AlexeiAverchenko вы русский?
 
Is there a way to check if a functor has left or right adjoint without trying to explicitly construct it?
@OldJohn угу
 
@mr.FS Fascinating
@Argon what other song do you like?
 
@mr.FS Francis?
 
11:15 PM
nope
 
@Charlie Hmmm....
 
you're not going to guess my first name i can promise you
 
user19161
@smack Nice red square there.
 
@Argon never listrned to this one...
 
@AlexeiAverchenko Украина?
 
11:15 PM
@Charlie Which?
 
@mr.FS smelly bottocks!
 
@mr.FS Fraser
 
@OldJohn Сибирь :)
 
lol
 
@WillHunting nice blue square :)
 
11:15 PM
@mr.FS Frumplestiltskin
 
a question here that shouldn't be difficult but i can't seem to find the answer, the problem is to find gcd(2^1120-2^1000,2^175-2^100)
 
@AlexeiAverchenko "cool" :))))
 
oh shoot
 
57 secs ago, by Argon
@Charlie Hmmm....
 
you got it
 
11:16 PM
@OldJohn indeed :)
 
@Charlie :)
 
user19161
@smackcrane Were you inspired by my blue square? =)
 
@mr.FS Frumplestiltskin?
 
yes
 
nice
 
11:16 PM
@AlexeiAverchenko :))
 
thanks
 
@WillHunting I just thought this red was a handsome color
@WillHunting haha blue is my favorite, but it was taken
 
how do I get 15 rep on mathoverflow? :D
so many cool questions and answers, and I can't upvote them
 
@Argon excuse me for a few moments :)
 
@Charlie Ok
 
11:22 PM
6 hours of math and I have almost worked through my first page
 
anyone that can help me with the question i pose a few minutes ago?
 
If someone has time, I'm struggling to understand steps in Mike Spivey's answer to a functional equation question.
@mr.FS Do you know the rule that gcd(ab, ac) = a*gcd(b, c)?
 
yes
 
I'm still working through it myself, but I'd start by trying to pull a term out.
 
ok
 
11:35 PM
Any luck getting anything out?
 
no i tried some things but didn't get any further
i also tried applying euclid's algorithm but hit the wall immediately
 
If it helps, I started by noticing that both $2^{1120}-2^{1000}$ and $2^{175}-2^{100}$ were even, so they should have some factors of two in common.
 
yes
i know the answer but don't know how i would get to it
 
@mr.FS is the answer anything like $2^{75}$?
 
it's $2^{115}-2^{100}$
 
11:42 PM
Then since I wanted to pull out as many factors of two as possible, the first term became $2^{1120}-2^{1000} = 2^{1000}(2^{120}-1)$, and the second term became $2^{175}-2^{100} = 2^{100}(2^{75}-1)$.
 
Is there no way to get to the main site from here?
Seems pretty poorly designed
 
@Jordan The bottom right corner has a logo, I think.
 
oh that was well hidden
Usually the bottom of the web page is where you hide the customer support links
 
@lewist I have a method, I think ...
 
@OldJohn Please feel free to step in, I'm a lousy teacher.
 
11:48 PM
$\gcd(2^{1120}-2^{1000},2^{175-2^{100}}) = \gcd(2^{1000}(2^{120}-1),2^{100}(2^{75}-1))$
$ = 2^{100}\gcd(2^{900}(2^{120}-1),2^{75}-1)$
 
sorry for lack of reactions i was watching a video
 
$ = 2^{100}\gcd(2^{120}-1,2^{75}-1)$
 
i'm sorry if this shows lack of effort
ok and that one i can do
thanks
i understand it now
 
(the $2^{900}$ term disappears since the other factor is an odd number - so cannot have a commaon factor with a power of 2
 
yes
 
11:52 PM
@mr.FS How? =)
 
Euclid
s algorithm
 
then you need to factorise $2^{120}-1$ and $2^{75}-1$
 
$2^{120}-1$=2^{45}(2^{75}-1)+2^{45}-1$
$2^{75}-1=2^{30}(2^{45}-1)+2^{30}-1$
 
Ahh, thank you folks, that last step is one I'm not good with.
 
$2^{120}-1 = (2^{15}-1)(2^{105}+\dots + 1)$
$2^{75}-1 = (2^{15}-1)(2^{60}+\dots + 1)$
so they have a common factor of $(2^{15}-1)$
 
11:55 PM
yes
 
is it finished?
 
I just got FWD:FWD:FWDed this video about math
 
Guys, look this comment. Is it plausible? I don't know what he meant with that, because it seems there are polynomials in some unusual mathematical structures. At least I still don't see these linear combination of monomials in $n$ on these Chebyshev Polynomials.
 
@GustavoBandeira I think he is actually right ...
 
I have an interesting question!
 
11:58 PM
That video is bad I think I got cancer from it
 
@OldJohn i think so, i got $\textrm{gcd}(2^{120}-1,2^{75}-1)=2^{15}-1$ by Euclid's algorithm so we just multiply this by $2^{100}$ to get $\textrm{gcd}(2^{1120}-2^{1000},2^{175}-2^{100})=2^{115}-2^{100}$
 
I don't know what he meant wit that but I can't see the linear combination of monomials on these Chebyshev Polynomials.
 
@mr.FS yep
 
You and a friend are walking along. Then a woman approaches you with 1000$, which she is willing to you. With the only condition that you both agree on how to split the money-
You want to split it fair 50/50, but your friend want to do it 90/10. And seems willing to lose all of it. What should you do?
 

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