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12:00 AM
I'm not sure I know how to define that. I know what uniform continuity means for functions, though
like, functions whose domain is $\Bbb R$ or $\Bbb C$ or some metric space
@usukidoll I know a bit about dense sets
 
oh good. I'm having trouble with dense sets
 
Um, actually…
my explanation of indexed families was a bit off.
It's more the function than the range of the function…
Although, note:
> $\Large“$Hence, an indexed family of sets is conceptually different from a family of sets (which is just a synonym for "set of sets"), but in practice the distinction is sometimes fuzzy and the indexed family is identified with its range and treated like an ordinary family.$\Large”$
 
so if I have a function f: (0,1) -> R
and it's continuous duhhhh but
how do I apply the dense set definition to this . like there's a point in the interval...there's a neighborhood. but that's all I can get out of this.
 
I'm not sure I understand the question
 
I wanna use this definition prntscr.com/ap1f66
skip the integrable part... that's unnecessary
what is a dense set anyway?
 
12:12 AM
A set is dense if every open set intersects it.
So, for example, consider $\Bbb Q$, the set of rationals. It's dense in $\Bbb R$, because every open set contains a rational.
Your problem deals with sets that are dense in $[0,1]$. $\Bbb Q\cap[0,1]$ is an example of this.
 
so there could be a point in (0,1) ???
$Q \cap [0,1]$ is the intersection of all rational numbers and the interval (0,1) and we know that rational numbers can be expressed as p/q
so there could be fractions like 1/4, 1/3, 1/2
but what does this have to do with the neighborhood of $(x-\epsilon,x+\epsilon) \subseteq F$. There has to be an accumulation point if a neighborhood exists right? Set F of real numbers
 
@usukidoll I can't really talk right now, but you know the difference between [0,1] and (0,1), right?
 
mhm
0<x<1 open interval
0 \leq < x \leq 1 closed interval
 
Also, $(x-\epsilon,x+\epsilon)$ is an open set… but I gotta go.
 
ahhh ok
 
12:23 AM
Maybe I can explain this to you later.
 
sure
 
12:36 AM
@Mambo Looks like "cohomology of $L^1(G)$ with values in $E^\times$ for $E$ a field... or something.
 
12:58 AM
So I suppose that $G$ is a group with a measure? Perhaps a locally compact group.
 
I just started parametric equations. Functions can be parameterized to become parametric. Is there a word meaning the opposite of this?
I guess not, then. Okay.
 
Deparametrization.
Sounds like it.
 
1:19 AM
Unparameterized?
 
@usukidoll Every set of the form $(x-\epsilon,x+\epsilon)$ — indeed, any open interval — contains an element of $\Bbb Q$. For example, the interval $(-.3,.7)$ contains the rational number $.5$.
A set $X$ is dense in a set $A$ if every open interval around an element of $A$ contains an element of $X$.
"open interval around an element of $A$" will generally look like $(a-\epsilon,a+\epsilon)$. (Well, it doesn't have to be centered on $a$ like this is, but it doesn't matter if we restrict ourselves to these)
Also, if you replace "open interval" with "open set" in the definition, it's equivalent. (Assuming you've learned what an open set is)
So, yeah, the definition of "dense set" is just the first sentence of your link.
 
@AkivaWeinberger the twin prime conjecture is equivalent to the existence of an unlimited twin prime
 
1:38 AM
@usukidoll a set of numbers is said to be dense if, for any two numbers in the set we can find another number that is between those two numbers and is also a member of that set
 
@GPhys Seems right. Any theorem of the form "There are infinitely many 'nice' natural numbers" is equivalent to "There is an unlimited 'nice' natural number" where nice is any property, I believe.
 
@AkivaWeinberger Pretty much. Unfortunately, it's not as useful as it would sound
 
Yeahh
It's an interesting property of $\Bbb N$, though, which $\Bbb R$ lacks. Any infinite subset is unbounded.
Equivalently, every bounded set is finite.
This sort of thing happens in $\Bbb Z^n$. I feel like there's probably a name for it.
 
2:02 AM
\o/
 
/o\
o[
A
 
******* ME
been trying to get some matrix multiplication to work
and can't trace my error !
that is soo stupid question
 
2:17 AM
computer algebra error or pen+paper?
 
@TedShifrin (You are right it has to do with Burnside's lemma, by the way)
 
2:31 AM
Today's my dad's birthday!
Nothing to do with math, just wanted to say that
 
How old is he? If you don't mind me asking :)
 
2:48 AM
@Pedro: I haven't thought about it, but I'm glad my instinct was not askew.
 
Fifty-one @skillpatrol
 
Hi, DogAteMy
 
@AkivaWeinberger cool
Hello Professor @TedShifrin
 
Hi @skull
 
How's it going?
 
2:51 AM
Math is hard
 
Like a diamond
 
dogAteMy: Sounds like a discrete set in a metric space (with unbounded metric)
 
@TedShifrin You sure that's equivalent? What about the infinite-dimensional space consisting of the unit vectors along the axes along with a copy of $\Bbb N$ along one of the axes?
 
equivalent to what
 
I'm not sure of anything.
 
2:55 AM
1 hour ago, by Akiva Weinberger
It's an interesting property of $\Bbb N$, though, which $\Bbb R$ lacks. Any infinite subset is unbounded.
1 hour ago, by Akiva Weinberger
Equivalently, every bounded set is finite.
 
Clearly I'm wrong.
What's your metric in that infinite case?
 
do it in a hilbert space and use that metric
 
$\sqrt{x_0^2+\dotsb}$? @TedShifrin
It's the points $(0,\dots,0,1,0,\dots)$ and the points $(n,0,0,\dots)$
 
proper metric spaces are those with the Heine-Borel property. your property is equivalent to proper + discrete, more or less tautologically. who cares. nothing matters.
 
Ah, that makes sense.
It was all a trivial observation anyway, to be honest.
 
3:01 AM
I'm stuck on understanding the following step (from the calculation of a line integral):

$\displaystyle \int_{p_1}^{p_2} \frac{\partial \phi}{\partial x}dx +\frac{\partial \phi}{\partial y}dy+\frac{\partial \phi}{\partial z}dz = \int_{p_1}^{p_2}\,d\phi$

It's probably simple, but I can't understand how this step works. (Is context needed?).
 
@AkivaWeinberger I think I'm going to discuss sequences before I discuss standard parts in my discussion
 
Sequences/limits is one of the things that is effectively identical to the non-NSA counterpart in practice (except perhaps more intuitive? maybe?)
 
Ah. The "$\lim a_n={\rm st}(a_N)$ where $N\approx\infty$ assuming it doesn't depend on $N$" thing?
 
@AkivaWeinberger $\exists^\mathsf{s}L\in\mathbb{R},\forall^\mathsf{u}n\in\mathbb{N},x_n\simeq L$ is equivalent to a standard epsilon sequence limit for standard $\{ x_n\}$
that's "for all unlimited"
 
3:06 AM
Ah.
 
and that's equivalent to the standard part being equal to it (for ALL unlimited n) by definition of the standard part
 
Ya
Remind me how to prove that all limited reals have standard parts? I know it's equivalent to the fact that the reals are complete but I'm not sure I recall the details
 
Well, I'd like to avoid this discussion when I first talk about that definition of a limit being equivalent since it's unnecessary (and just clouds the proof), but:
you note that $^\mathsf{s}\{ y\in\mathbb{R} : y\leq x\}$ is bounded
for limited $x$
 
Ahhh. Yeah, I think I see where this is going. That's certainly one way to do it.
 
and then definite the standard part as the supremum of that set
 
3:10 AM
Yup. (I think the proof I saw had to do with decimal expansions? But that's much cleaner.)
It's like Dedekind cuts, almost.
And, since the reals are a $T_1$ space, we have $x\not\approx y$ for all standard $x$ and $y$ (right?). So that's unique.
(I think it follows from $T_1$-ness.)
Yeah, it does
 
uniqueness of the standard part?
 
you prove the absolute value of st(x)-x is infinitesimal
 
It's not unique in, say, the Sierpiński space, if you define everything in the right way
 
(then it's basically the same as saying $0$ is the only standard infinitesimal)
 
3:16 AM
That metric doesn't make sense, DogAteMy.
 
yes it does. countable direct sum, at most finitely many nonzero entries
 
@GPhys …right. We can use the fact that $\Bbb R$ is a field. But the more general the better, y'know?
 
Oh ...
 
@AkivaWeinberger I have a set of scribbled notes for a development of NSA in metric spaces somewhere
but it's not really all that much different than $\mathbb{R}$
 
@GPhys Yeah, there's a way to do it for general topological spaces, too.
@GPhys Also, idea: private chat room with just us two? Because we've been talking about this stuff a lot and nobody else ever seems to know what's going on
 
3:24 AM
okay
 
You should feel free to talk as much as you want about math that excites you in here.
 
I also talk to PVAL about garbage nobody but us parses.
 
and i ramble about stuff all the time :P
 
3:30 AM
I think the only math that should go elsewhere is math the speaker isn't actually excited about.
 
Got to go to bed now; bye
 
Night, DogAteMy
 
@TedShifrin Burnside says that $|X/G| |G| = \sum |X^g|$.
 
Yup.
 
In this case the action of $S_n$ on $[n]$ is transitive, and $|X^g| = {\rm fix}(g)$.
That's it.
 
3:33 AM
Glad I could help :)
 
3:46 AM
Hmm, I'm up early.
 
What time is it there?
 
about 9 am.
 
That's not early.
 
Well, early is relative :)
 
Sure. But there's the "sun comes up" absolute.
 
3:49 AM
I normally go to sleep when the sun comes up... so...
 
That's not good. You're still growing, Balarka.
 
time is relative ;)
 
@Pedro a generous assumption
 
4:05 AM
do you guys still go to that dj web site?
 
It's dead now, the site
 
There was a dj website?
 
yep
 
4:27 AM
@BalarkaSen Dude, I'm 99% sure you have clinical insomnia. I don't know if this is the sort of thing that you should go to a doctor for, but I think you should totally go to a doctor. Or a therapist.
 
I don't know if it should be called an insomnia, @AkivaWeinberger. I get at least 8 (most of the time 10) hours sleep a day, scrapping together all of my nap-time. I think it's a confused sleep schedule, more like.
 
…how does that even work
Also, did you say you just woke up? Don't you have school? Or is it spring vacation there
(Mine is in two or three weeks, something like that)
Still, pretty sure that that's not a normal sleep schedule. I still think you should go.
 
i would be more worried about the fact he gets the superflu once a week
 
It's vacation, yes. School starts again on August.
 
Oh, whoa. Summer vacation?!
Whaat
 
4:36 AM
@MikeMiller How do you like my proof?

http://math.stackexchange.com/questions/1731407/continuous-bijection-from-a-b-to-s1/1731524#1731524
 
long fuckin vacay
 
Post-exam vacation. We get that after final 10th grade exams.
 
looks good
 
I am so confused
OK, you have vacation April, May, June, July, and August…
Wait. When in August do you start?
 
February and March was vacation too.
 
4:37 AM
Tbh I suspect Balarka is a myth and not a man
I wouldn't think about him too much
 
lol. chat robot?
 
good!
 
My school ends on June 17 this year
 
@AkivaWeinberger maybe first or second week.
 
Whatever. I'm going to bed
Have fun vacationing
 
4:40 AM
Jelly much?
 
hashtag bitter
 
I think I just realized that all of the Jewish holidays probably makes up for it. :P
Especially the ones in Tishrei (~September, more or less)
 
he didn't choose my proof :(
I'm depressed now
 
i wrote zero details so i'm surprised mine was accepted.
 
he liked yours better @MikeMiller
I guess I left out some details too... like the fact that the continuous image of a connected space is connected, and the only connected subsets of $(0,1)$ are intervals.
 
4:48 AM
currently writing a final project for a course. the few I've had always take longer than I expect them to.
 
I wrote a lot of projects this year lol
 
5:11 AM
@BalarkaSen Four months seems like an awfully long time to be off?
 
There is nothing awful about long vacations :)
 
5:22 AM
vacation from math?
 
school
53 mins ago, by Balarka Sen
It's vacation, yes. School starts again on August.
 
@ForeverMozart there's no such thing as that
 
Once bitten by the purity of mathematics...
 
5:55 AM
@Semiclassical are you familiar with finite automata
 
sup guys
 
 
2 hours later…
7:36 AM
is there a name for the property that $f(x+y)=f(x)+f(y)$? Just "respecting addition" ?
 
@robjohn have you seen my generalizaton in db?
 
@user1618033 No, I haven't checked. I will
 
@robjohn I sent it yesterday. It's something you might like a lot.
@robjohn OK
 
7:51 AM
hi all.. anyone interested in codegolf.stackexchange.com/questions/77051/… ?
 
@user1618033 It will take a while to read it. It is quite long.
 
@robjohn hehe, don't read it now, but when you have time. Yeah, it takes some time.
 
@robjohn is user1618033 me?
oh.. sorry
 
I'll be back in 30-60 min.
 
greeetings all, hey @Lembik
 
8:05 AM
Hi
 
hi skull, u sound like an acting chris'sis
 
how so?
 
3 hours ago, by skill patrol
Once bitten by the purity of mathematics...
@Lembik sounds like a nice golf-competition, unfortunately i quit from that
 
the "..." is what separates us ;-)
 
8:41 AM
What does it mean for sequences to be neighbours in a metric space? I think it might be too dumb to ask this as a question on the site.
 
@skillpatrol What really separates is the desire to beat the performance of Ramanujan ... (acting in mathematics with a different kind of spirit):-)
@Agawa001 heya
 
Don't worry about answering my question here on this chat... math.stackexchange.com/questions/1731717/…
 
8:56 AM
@skillpatrol essentially no school around the world can teach you do what I presently do. But no worry about it, my stuff is very crazy and you might not like it.
Back to finish some veryyyyy long proof.
 
bleh I'm so bad at cleverly manipulating inequalities
 
Is Introduction to Algorithms by Thomas H. Cormen worth it?
 
9:18 AM
@skillpatrol oh right !!! smdh, hey @user1618033 how is ur self-esteem right the moment?
 
@Agawa001 No matter the circumstances, I do my best to be in a perfect shape and produce amazing math.
(which I do right now)
 
9:34 AM
we all try our best
what i meant by this:
1 hour ago, by skill patrol
the "..." is what separates us ;-)
 
@skillpatrol This is slightly deceiving. Ramanujan was like us, but with a different spirit of approaching things, with a different passion, love to math, with a different willing to sacrifice himself to getting the results he got.
 
is those who will never give up ie will try forever
 
I often here the words I could never do what you do, and this is such an annoying and false statement. Many people miss the powerful cocktail of crazy passion and crazy hard word.
Just be willing to work for years, even if the result don't come, they will come at some point.
 
btw, welcome back
:-)
 
@skillpatrol hehe, thanks! :-)
Ramanujan is a limit there that has been waiting to be broken. No one should ever limit himself, herself by the performance of the others.
 
9:41 AM
Yup, in your area of math.
 
Just work extremely hard for years. It's painful to work for years and not to get the results. Keep working, you need some more, addd some more years of extremely hard work!
Where is the passion? Do you have passion? Are you burning for the stuff you do?
 
There's a movie coming out on his life story. The man who knew infinity.
 
@skillpatrol I'mmmmmm looking forward to it! :-)
 
@skillpatrol bollywood?
 
A Fields medalist was the math consultant, I think.
@Agawa001 nope
 
9:45 AM
@skillpatrol In your opinion, do you think Ramanujan had something else that made him more special than his amazing dedication, love and hard work to his math?
 
Sorry I don't know his work well enough to form such an option :-)
 
what about if they would ever think to make a movie about me, i m as assiduous as ramanujan and my life is as sad
 
@Agawa001 lol
@skillpatrol I think this that kind a spirit anyone should have to get on the highest peaks of performance. No matter his work.
 
@user1618033: Yes: Ramanujan had an almost otherworldly 'intuition'; truly unbelievable if it hadn't actually happened.
There's a lot of silliness about how he "only" had intuition and couldn't prove things until he was trained in England, this is ofc nonsense
 
@EricStucky This comes after years of passion and extremely hard work.
 
9:49 AM
No, it doesn't.
It can
and for him, it did
 
They're expecting the movie to be as popular as Slum dog millionaire, but we'll see about that @user1618033
 
The passion and the hard work were necessary
 
@skillpatrol I'll definitely be in the theatre!
 
but certainly people work much longer and at least as hard as Ramanujan, and never attain an intuition like his.
 
@EricStucky Indeed, but I suspect that it was about very deep passion, love, and extremely hard work. In a documentary his wife said about Ramanujan he worked on series day and night.
 
9:51 AM
I saw that^
 
We might have different understanding of the extremely hard work and very deep pasion.
 
I normally only watch documentaries
 
Perhaps :)
 
@EricStucky When I begin to work on my stuff I feel myself the happiest being on earth and my love is very profound it's like eating the best kind of food (but mentally).
Not sure how each person understands that love to math. Moreover I feel I can break any barrier. And most of the time I do it.
 
Gourmet food for thought
 
9:54 AM
Does someone of you have an idea about my question:
0
Q: Nilpotent - How can we find a contradiction?

Mary StarI want to show that the dihedral group $D_n$ is nilpotent if ane only if $n=2^i$ for some $i$. I have shown the direction $\Leftarrow$. Could you give me some hints for the direction $\Rightarrow$ ? We suppose that $D_n$ is nipotent and $n=2^im$, where $2\not\mid m$, or not? How can we f...

?
 
recently my interest to maths and computer science reduced because i got lil bit elder and my engagements with life took a lot of my attention, i think this is a sad thing and ramanujan hadnt known this lil stumble in his living
 
Ya, not just petrol for the skull.
 
@Mary: Do you know the commutator subgroup of, say, $D_5$?
 
I have shown that the commutator subgroup of $D_n$ is isomorphic to $Z_n$ when $n$ is odd and it is isomorphic to $Z_{\frac{n}{2}}$ when $n$ is even$, and that $D_n'=\langle s^2\rangle$. Does that help us?@EricStucky
 
Are all of those elements in the cyclic subgroups rotations, then?
I don't know what your second comment means: $s^2=1$, right?
 
10:05 AM
@EricStucky Which second comment do you mean? That $D_n'=\langle s^2\rangle$ ? @EricStucky
 
yes
please stop tagging me
2
 
Ah ok, sorry...
I have shown that each element that belong to D_n' is if the form $s^{i}$.
 
I am not sure what $G'$ means, but okay.
But yes, the commutator subgroup calculation should be most of the work.
 
It is the commutator group of G, G'=\{xyx^{-1}y^{-1} \mid x,y\in G\}
 
You use $s$ for rotation?
 
10:11 AM
Yes
 
Now the commutator $[D_n, D_n']$ should be $D_n'$ again, when $n$ is odd, and so inductively the lower central series for $n$ odd does not go to the trivial group
For even $n$ you hope to apply some inductive argument
 
Why do we look at the commutator $[D_n, D_n']$ ? I got stuck right now...
 
I believe the plan is to show that the lower central series sloughs off a factor of 2 at each step if it can, and once there are no more factors of $2$ it stabilizes.
This would stabilize $D_{2^k}$ at $Z_1$, which makes it nilpotent. And also it stabilizes $D_n$, for $n$ anything else, elsewhere, which makes it not.
 
Mary, actually a more sensible thing to do is look at $[D_n',D_n']$ if you are only interested in the not-power-of-2 case. Because hopefully this will do the same thing, which shows the group is not solvable (and thus not nilpotent). And this lends itself better to an inductive argument.
 
10:25 AM
Ok... I will think about it... Thank you very much for your help!! :-)
 
npnp
 
10:37 AM
Could anyone please confirm if my logic is sound? I have been told that if I formalize the sentence: $If$ $ x<x'$ $ then$ $ x''=x'$ using Peano Axioms as $\bf [\forall x''' \sim x=(x'+x''')]x''=x'$, then the case where $x=x'$ will be undefined. Therefore I have rewritten my sentence as $\bf [\forall x''' \sim x=(x'+x''')][\sim x=x']x''=x'$. It this better?
 
10:58 AM
why does 'please stop tagging me' have 2 stars :/ :/ :/
2
 
11:19 AM
@AkivaWeinberger and @Semiclassical, would you please be able to quickly check my post above?
 
11:45 AM
No idea @EricStucky :P
 
12:17 PM
What is the reason for the 1/2 in the definition of the Riemann Xi function?
I mean, without the 1/2 it still has all those nice properties, so maybe it's just there for historical reasons?
 
12:39 PM
"you can show the last part by double contenence" - I suppose the poster means "both inclusions" by "double contenence". Is it propoer English term for this? (I am not a native English speaker.)
 
I don't think so.
It's not in common use, at least.
 
@EricStucky About terminology here: You probably mean pinging rather than tagging.
(Sorry for pinging you.)
Pings in chat are described here or in chat faq.
is something different. (At least in the context of this site.)
(And sorry for nitpicking.)
The phrase subset "double containment" returns a few hits in Google Books.
 
Yeah that's a thing
 
Searching for subset "double contenence" returns no results in Google Books.
 
12:58 PM
@barto it makes for a tidy-looking reflection formula.
 
@J.M. But $f=2\xi$ also satisfies $f(s)=f(1-s)$
 
1:10 PM
@barto Oh, I thought you meant the factor inside the gamma function. You're right, the reflection formula still applies.
 
1:23 PM
Hello!!! If we have the following: Find where the tangent plane of $z=e^{x-y}$ at $(1,1,1)$ intersects the z-axis.
we can find the tangent plane using the formula $z=f(x_0, y_0)+ \frac{\partial{f}}{\partial{x}} (x_0, y_0) (x-x_0)+\frac{\partial{f}}{\partial{y}} (x_0, y_0) (y-y_0) $

and then set $x=y=0$.

If we want to find where the tangent plane of $x^2+2y^2+3xz=10$ at $(1,2,\frac{1}{3})$ intersects the z-axis could we divide by 3x and solve for z and use the above formula?

I think not, because we would have to set x=0, but $z=\frac{10}{3x}-\frac{x}{3}-\frac{2y^2}{3x}$ does not hold fo
 
1:33 PM
@DanielFischer Do you have an idea?
 
@robjohn then also the version with $1+1/2^3+\cdots 1/n^3$. As far as I know no one did it so far, and I don't even mention I finish it by elementary manipulations of series only (I have 3 ways to put it down).
 
Just plug in the $x$ and $y$ coordinates of the point you're interested in. You have a formula to compute the $z$-coordinate then.
 
So we can compute the tangent plane using the formula $z=f(x_0, y_0)+ \frac{\partial{f}}{\partial{x}} (x_0, y_0) (x-x_0)+\frac{\partial{f}}{\partial{y}} (x_0, y_0) (y-y_0) $ , right? @DanielFischer
@DanielFischer Also if we want to find the tangent plane of the surface $x^2+2y^2+3xz=10$ at $(1,2,\frac{1}{3})$ we get $3x+8y+3z=20$.

Do we consider that $3x+8y+3z=20$ is the tangent plane or $z= \frac{20}{3}-x-\frac{8}{3}y$ ?
 
@robjohn maybe in the past I would have been very happy for having stuff no one did so far, but it seems these days this fact gives me no more satisfaction.
I have far higher objectives: my (master) theorems will conquer the world.
Just patience (a matter of patience all this).
 
1:49 PM
If we consider this : $3x+8y+3z=20$ the plane will be in $\mathbb{R}^3$ but if we consider this: $z= \frac{20}{3}-x-\frac{8}{3}y$ it will be in $\mathbb{R}^2$.

Which is right? @DanielFischer
 
Both equations describe a plane in $\mathbb{R}^3$.
 
Ah I see... Thank you :) @DanielFischer
 
2:06 PM
@skillpatrol The bad part of the movie is that many people will forget the next day the movie, but the movie is about the most beautiful part of life which is going to be missed: the math of Ramanujan which also includes what I do. Such a loss, not to have the opportunity to taste it. Nothing can be compared to it in terms of beauty.
 
DLN
I asked this question earlier today. math.stackexchange.com/questions/1731818/… but still no answer, can anyone please help?
 
@DLN 3 hours is a relatively short period of time; have patience.
 
And I would only accept to be contradicted by people that got results like Ramanujan (however, there is a paradox, they would never do that).
Too much talk, back to my work.
 
have fun :)
 
@litmus i'm afraid that question is outside the realm of topics about which i can say semi-intelligible things :)
 
2:16 PM
now that^ is a rarity :P
 
@Adeek nope
 
now i know why you @Semiclassical don't hang out at the h Bar
 
one over dominant user can ruin a chat room
 
2:22 PM
i thought he was cool until he ditched ted >8(
 
i don't actually know who you're talking about, but i think i'm okay with that
 
kids these days...
 
@skillpatrol thanks. I really have fun, it's one of the critical parts of mathematics, to have fun and laugh a lot. ;)
But at the same time to produce very top mathematical stuff.
 
true dat
 
Huy
is user 1618033 chriss'sis?
 
2:35 PM
@Huy No, I bought the account of chriss'sis (for 1 buck).
3
 
are you Alec?
 
Huy
@skillpatrol: nice observation
 
@skillpatrol I'm Ramanujan (the more powerful version).
 
2:38 PM
:D
Back to the serious work. Later.
 
cya later
 
morning
 
hi
@0celo7 you wanna come in here and talk?
 
@skillpatrol No.
 
good
 
2:46 PM
What?
 
ignore this user (everywhere)
 
Lol
 
and take off those raider colors you back stabber
band wagon jumpers need not apply :P
 

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