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12:01 AM
@robjohn I spend too much time reading questions that no one answers.
 
@aDangerousIdea You spend too much time reading them?
 
@robjohn Yes.
 
@aDangerousIdea Poor eyesight? small print? forget to move the mouse to turn off the screen saver?
 
@robjohn I am a slow reader :(
 
@JayeshBadwaik Hello!
@WillHunting You're a star...you don't need stars =)
2
 
12:14 AM
@argon!!
 
@Melvin Hi!
 
@Argon wassup?
 
@Melvin Not much :)
 
@N3buchadnezzar The function is continuous at 0 and that is where the function has a limit.
 
Had a nice programming contest!
 
12:16 AM
@Argon hmm nice!
 
@N3buchadnezzar The function above is similar to en.wikipedia.org/wiki/Thomae%27s_function
 
user19161
@amWhy Hehe. =)
 
Hi @Will iam
 
@Argon Will he is
 
@Melvin Sam I am
 
12:24 AM
@WillHunting Hey there! Sorry I missed your greeting earlier today
 
user19161
@Argon Have you done your homework?
 
user19161
@amWhy It's OK, you are busy with important things!
 
@WillHunting Yes
 
@WillHunting I was away from my laptop; getting close to 7K, I see ;-)
 
(woops)
 
user19161
12:26 AM
@amWhy You will certainly make it to 10k this year, it will be a very merry Christmas!
 
user19161
@ada Now you look like Marvis.
 
@WillHunting Or a happy hanukkah
 
user19161
@Melvin What's that?
 
=)
 
@WillHunting you don't know?
never heard?
 
user19161
12:31 AM
@Melvin Well, I know very few things.
 
@WillHunting Marvis?
 
user19161
@aDangerousIdea Yes, there is only one infinity symbol.
 
@WillHunting Au contraire!
 
user19161
@Argon I am only a banana.
 
@WillHunting it's a jewish holiday that some compares to christmas. but it's not related
 
user19161
12:32 AM
@Melvin Is it at the same time of the year?
 
@WillHunting Are you trolling again?
 
user19161
@aDangerousIdea Always trolling.
2
 
@WillHunting this year starts dec 8
@Argon ;)
 
3 mins ago, by Will Hunting
@aDangerousIdea Always trolling.
 
user19161
@aDangerousIdea Isn't that from The Avengers?
 
user19161
12:36 AM
@aDangerousIdea There's no need to repeat something so recent.
 
38 secs ago, by Will Hunting
@aDangerousIdea There's no need to repeat something so recent.
 
Hahahaha!
 
user19161
@aDangerousIdea Isn't that Sonic?
 
@WillHunting Isn't that Will Hunting?
 
12:40 AM
@aDangerousIdea Skullpatrol is nicer than a dangerousidea
 
user19161
@Melvin He was also termed scalloptroll which sounded like skullpatrol.
 
@WillHunting That was before you got here Good Will Troll.
 
@Argon how's life?
 
@Melvin Fine, as always. You?
 
@Argon good
 
12:49 AM
I think I scared the troll away :-D
 
Hahaa
@Melvin What math do you like especially?
 
@Argon i don't have one favorite..i like math in general, like to study as much as i can , i wanna know as much as i can. it's frustrating, 'cause there's a lot of things...
 
@Melvin I ask many people this question and this is a common answer :)
 
@Argon good to know!
 
@Argon What would your answer be?
 
12:54 AM
@aDangerousIdea I can't say much about most fields of math because I don't know anything about them! However, so far, I enjoy analysis a lot.
 
Anyone seen Charlie?
 
@aDangerousIdea Hmmm... not for a few days. Odd.
 
Maybe the troll ate her.
 
I hope not! :)
 
Me too...
If he did, I Will Hunt for him :-D
 
1:00 AM
Hahahaha!!
 
I've watched a class in which the teacher said that a function is something that builds a correspondence between x and y ($f: x \rightarrow y$), and that for every x, there's one and only one y, is it possible to have functions where you can have more than 1 y for one x?
If they exist, how are they called?
 
Multivalued?
 
Dunno - if you're asking me.
 
I think this is what you mean
In mathematics, a multivalued function (shortly: multifunction, other names: many-valued function, set-valued function, set-valued map, multi-valued map, multimap, correspondence, carrier) is a left-total relation; that is, every input is associated with at least one output. Strictly speaking, a "well-defined" function associates one, and only one, output to any particular input. The term "multivalued function" is, therefore, a misnomer because functions are single-valued. Multivalued functions often arise from functions which are not injective. Such functions do not have an inverse func...
 
@Argon Thanks.
 
1:13 AM
No problem
 
odd/even are terms that make sense only when describing integers, right?
 
Generally, yes
 
There are odd/even permutations, odd/even functions.
 
This is true as well
 
if gcd(a,2k) = 1, does it follow that gcd(a^2,b) = 1 since a has no prime factors in common with b?
 
1:24 AM
What's b?
 
*k
Im trying to prove that 1 +sqrt(2) all over 2 is irrational lol
Just making sure im not going crazy
 
@jshin47 So?
Do you know $\sqrt 2$ is irrational?
 
Yes that one is easy ... but Im trying to do it w/o invoking closure of rationals or anything like that
Im trying to do it directly by contradiction
(1 + sqrt(2)) / 2 = a/b => b^2 = 4(a^2 - ab)
so b is even
then we can write b = 2k
 
@jshin47 Stop right there.
Let $s=\sqrt 2$. If $$m=\frac{s+1}2$$ were rational, so would be $2m-1=s$. Impossibru.
 
Hahaaha!
 
1:29 AM
Thats nice, but im trying to do it as if i dont know that sqrt 2 is irrational.
as naively as possible
and naiveity usually comes pretty naturally for me
 
Well, you need to make things swifter and swifter everyday.
 
So b = 2k => a^2 = k^2 + 2ak
 
It looks like you're on the right track, then. Go two/three steps further.
 
Then k | a
* k | a^2
But gcd(a,b) = gcd(a, 2k) = 1
So if i can show that gcd (a^2, k) should be 1, but is actually some multiple of k, im done
is actually k i mean
 
Well gcd(a,2k)=1 implies ar+k(2s)=1, so gcd(a,k)=1...
 
1:33 AM
so is it obvious enough that gcd(a^any integer power, k) = 1 as well?
 
You can easily verify that by contrapositive.
 
gotcha. I was thinking that it should be "about as easy" to show that such a number is irrational directly as showing sqrt2 itself is irrational
right.
 
You took a little longer in your proof than necessary. After establishing $k^2=a^2+ab$, we have $a|k^2$, so gcd(a,k)>1 so gcd(a,b)=gcd(a,2k)>1.
Which is what I meant by "Go two/three steps further."
 
I see. Yes that makes sense
 
user19161
@aDangerousIdea Please don't use my photo as your gravatar, that makes me very upset.
 
1:53 AM
whats that about imitation and flattery?
 
user19161
Wow, how does this site get so many questions?
 
Curious people!
 
user19161
@Argon Did I miss your integral of the day?
 
@WillHunting Didn't have one yet. I was saving it for you
Argon's Integral of the Day: $$\int_0^1 \frac{\text{artanh}\, x \,\, \log x}{x\, (1-x)\, (1+x)}\, dx=-\frac{1}{16}\Big(7\zeta(3)+2\pi^2\log 2\Big)$$
 
user19161
@Argon You must be good at series too then.
 
2:05 AM
@WillHunting I didn't come up with it :)
 
Did you find this using complex analysis?
I mean taking integral over a contour that has the segment 0 to 1 in R
 
@jshin47 I didn't prove it, but by the strange constants, I'm not sure if contour integration would work well here.
 
@jshin47 Try generalizing the irrationality problem to $\frac{\sqrt{p}+a}{b}$ where $a$ and $b$ are relatively prime.
 
The Zeta series implies to me that I this is evaluated using some type of summation, as @WillHunting suggested
 
@peoplepower ill do that now
 
user19161
2:07 AM
It is 29 Nov, this reminds me of Jer 29:11.
 
btw is there a way to make chat display tex
 
user19161
@Argon Suggested? I didn't suggest anything!
 
@WillHunting "You must be good at series too then." How is this not an implication of series, Sir [CENSORED]?
@WillHunting Favourite verse?
Nov 2 at 21:01, by Jasper Loy
Wow, I have 29 bronze and 11 silver badges now. Reminds me of Jer 29:11.
 
user19161
@Argon Nope, just happen to remember since it's a popular one. I don't really have a favourite verse and I am not Christian.
 
2:10 AM
@WillHunting Popular? Hmmm...
 
user19161
@Argon Well, it's often quoted...
 
i need some help with orthogonal compliment math.stackexchange.com/questions/246967/…
 
@WillHunting Hm... Didn't know that
 
@peoplepower does that link work in google chrome?
 
@jshin47 Yes
 
2:11 AM
Ahhh I get it
lol
oops.
 
@jshin47 $p$ is prime btw.
 
I figured
Otherwise your statement is not true :)
well actually doesnt $p$ only need to be not a perfect square
 
True, but the proof is harder.
 
I see, gotcha
 
Eventually, you just have give up and invoke the closure of rationals under addition and multiplication and their inverses. :)
 
2:17 AM
lol
Trying to follow the exact same strategy as the previous case, but unfortunately only for 2 - 1 = 1 is this so straightforward
 
 
1 hour later…
user19161
3:32 AM
@xiamx It's spelled complement and not compliment.
 
@MattN. My eventual goal was to stay awake. Caffeine did not help after all. :P
@WillHunting @aDangerousIdea used your pic!!
 
user19161
@JayeshBadwaik I am very pissed with him now.
 
user19161
The last time he used Jonas's picture without asking him too, and I already mentioned it to him.
 
@WillHunting Hmm.
 
user19161
@JayeshBadwaik Maybe he will use yours next!
 
user19161
3:42 AM
@peter What are your plans this December? Are you going anywhere overseas?
 
@WillHunting I don't think so.
 
user19161
@PeterTamaroff Hmm OK, you can just study more math!
 
@WillHunting Maybe I will. I'll take some vacations on mid February.
 
@WillHunting Hmm, if he uses my personal picture, I will flag him for sure. @aDangerousIdea Let this be a warning to you. :-)
@PeterTamaroff Do you guys have three month vacations in around dec-feb?
 
@JayeshBadwaik I am already on vacation since Nov-22nd
 
user19161
3:45 AM
@JayeshBadwaik I think the smile should not be there. I am seriously pissed. This is not a joke.
 
user19161
@PeterTamaroff It should be vacation and not vacations.
 
@WillHunting OK.
 
@WillHunting Its the smile of impending evil. Just like one of my professors who had this sign on his door "Never trust a smiling professor."
@PeterTamaroff Ahh, nice.
 
@JayeshBadwaik ...and that's because I studied. Hehehehhe
 
@PeterTamaroff I don't get it? What happens if you don't study?
 
3:47 AM
@JayeshBadwaik I have 2 exams before the final exam in each course. If I average 70% or more, I don't have to take the final exam, which is around december.
 
user19161
December is coming soon. It will be beautiful.
 
@WillHunting That is questionable,
 
user19161
@PeterTamaroff Well, certainly not as pretty as you!
 
@WillHunting ...
 
@PeterTamaroff Ohh. That's great. I always like such things. There were one or two courses where I had passed before the final exams (I had to take it anyway due to the rules, but it was just like a vacation.)
 
user19161
3:49 AM
@PeterTamaroff I expected bleh instead.
 
user19161
I intend to have higher standards of grammar even in chat from now.
 
user19161
So I will try to avoid comma splices here.
 
TIL cassandra0= jdoe. Why does he keep changing his name so much?
 
@WillHunting You get nothing.
 
user19161
Over and out, see all of you in your dreams.
 
3:54 AM
@PeterTamaroff How many courses did you have this semester?
@WillHunting Good Sleep.
 
@JayeshBadwaik 3
Quadmester?
 
@PeterTamaroff Okay. You have Quarters?
 
@JayeshBadwaik Yes, that.
 
Linear Algebra, Calculus and ?
 
@JayeshBadwaik Linear Algebra, Chemistry and "State and Society" (basically, Argentina's history)
 
4:03 AM
@PeterTamaroff Ohh, then what was with all the Spivak excursions?
 
@JayeshBadwaik Personal excursions
I am almost finishing it, finally.
@anon Yo
 
@PeterTamaroff Good.
 
user19161
I am still here spying on all of you.
 
@PeterTamaroff Okay.... I weasled out of my humanities courses after I failed in first one of it. I preferred to replace it with departmental electives. I similarly weasled out of Engineering Graphics Lab since my hand drawings were always very dirty and I was always told to redraw.
 
@JayeshBadwaik Oh, well. I averaged 65% on S&S, but it rounds up to 70% so I was good.
Heheheheh
 
4:06 AM
yo
 
@PeterTamaroff Sweet.
 
user19161
@anon Yo yo is something I can't do well.
 
cool story bro
 
user19161
@JayeshBadwaik Sweet sounds so ... sweet.
 
user19161
I try to avoid using the word because it is too sweet.
 
4:09 AM
@anon Man, I just proved that if $f$ is a projection and we set $g=\text{id}-f$, then $g$ is a projection and $Im(g)=Ker(f)$; $Ker(g)=Im(f)$
 
user19161
I have a question to ask anon which I will save till a few years later maybe...
 
I recall you told me something about that a while ago.
 
user19161
@PeterTamaroff You love to call him man.
 
@WillHunting I take it he is one. If this is not the case, please do tell, @anon
 
user19161
@PeterTamaroff I take it he is a man in both senses of the word.
 
user19161
4:12 AM
But maybe I am a unicorn from unicorn land.
 
I. Am a man.
 
user19161
I come to this chat to select another unicorn to take with me back to unicorn land.
 
@anon Yes, sir!
 
user19161
@anon So cute!
 
user19161
4:16 AM
Anyway, I have a feeling that something big is going to happen soon in my life...
 
@WillHunting Why?
 
user19161
@PeterTamaroff I don't know. I guess I just need some "miracles".
 
user19161
By the way did I tell you that I have been called Chicken Little and also Spongebob before? Hehe.
 
@anon Hahaha.
 
@WillHunting Make them yourself.
 
4:56 AM
@anon Hey
 
zup
 
@anon I just had a blackout
suppose we have an element $\sigma,\tau$ in the Galois group
 
too much alcohol
 
of some extension doesn't matter
if we do $(\sigma \tau)$ applied to an element
is that the same as applying $\sigma$ to $\tau$ of that element?
I.e. if $u$ is that element
is $(\sigma\tau)(u) = \sigma( \tau(u))$?
 
yes, the automorphisms are maps on the underlying sets
 
4:58 AM
phewww
All of a sudden my mind went blank
@anon thankx
 
 
3 hours later…
8:28 AM
2.999... hours later
 
Not much
Trying to understand this: If $p$ is a prime, then $p$ divides $a^{p-1}-1$ for every integer $a$ that is relatively prime to $p$.
 
(Z/pZ)^x is a group of order p-1, so of course its elements have exponent dividing p-1
number theory is easier with abstract algebra
 
@anon I believe in you, but I'm still too noob for that.
Actually, I've seen a video: "An informal introduction to abstract algeba"
The concepts were kinda simple to understand, groups, rings, etc.
BUT, he gave me only a shallow description on what they are.
A ring is a set + one operation (I guess)
In the form {*,A}
 
8:36 AM
I recommend
http://people.brandeis.edu/~igusa/Courses.html
http://www.math.uconn.edu/~kconrad/blurbs/
http://www.mimuw.edu.pl/~jarekw/pdf/Algebra0TextboookAluffi.pdf
4
anyway, I'm off to bed
 
Ok
Thanks for the recomendations.
 
Hi @OldJohn how are you?
 
@aDangerousIdea Hi - not so bad, thanks - and you?
 
@OldJohn Fine, thanks.
 
Good grief - max temperature here today is to be +3 (but at least it is a lovely morning)
 
8:51 AM
What's the min?
 
@aDangerousIdea I swear I've read: Hi, how old are you.
I really need to sleep. =/
 
@aDangerousIdea -1
 
@GustavoBandeira Sleep well.
@OldJohn Brrr...
 
I'm not going now, need to read some things.
 
@aDangerousIdea Yeah - might be cool walking back from the gym this morning
time I did stuff - back later, folks
 
8:56 AM
later
 
@aDangerousIdea Age?
 
 
1 hour later…
10:05 AM
@GustavoBandeira 0.999... hour older
 
10:22 AM
Hi
 
:-| :-P
 
10:28 AM
<B)
 
Yesterday I posted a question that was closed for the reason that there was a general case of the problem already discussed. But some particular cases of some problems do not apply to the general case. I think this behaviour may be rather risky for the future. Here math.stackexchange.com/questions/246739/…
 
:/
hmm..
 
@Chris'ssister There have been one or two threads on meta about such situations, from looks of it, I am sure people have thought well enough about it. So, its all good I might say!.
 
@Jayesh: It's not healthy at all for Math SE. Just think about this fact. The link with the general solution require knowledge that overpass the staff you learn in highschool. But note that the particular case that I posted may be evaluated with the highschool knowledge.
 
10:35 AM
Very good point^
 
good day!
 
@JayeshBadwaik: Math SE is meant to help all people interested in mathematics. I have no problem with the general case because I understand all is written down there, but I'm convinced that there are people that do not have advanced knowledge of calculus. It's fair to take into account this case.
 
hi @JayeshBadwaik!
 
@Chris'ssister Ahh, right. I completely agree. In which case, post such a comment to the question and contact one or two moderators. Note: Closing is not the same as deleting. (P.S. I am not able to find those meta questions right now, this is a nice policy I feel.)
@Nimza Hi! Wassup?
 
@JayeshBadwaik heh, answer "nothing" is clear? I think it is traditional russian answer, but I may be mistaken
 
10:40 AM
@Nimza: hi
 
@Nimza Its perfectly correct. :D :D
 
hi @Chris'ssister!
 
hello
 
@Nimza, I linked a PDF to you when I wasn't here did you see it? about the inverse mellin
 
10:42 AM
@JayeshBadwaik historically it comes in Russia from fear of evil eye:)))
@cassandra0 ah! Thank you, yes, I've seen it!
 
@Nimza Ahh. Interesting.
I wonder how relevant the history is in current times. ;-)
 
I'm a bit confused about that algebra 0 link
 
be back in a few
 
I still find it so weird that both abgrp and grp have coproducts, but different ones..
 

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