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9:00 PM
:)
 
@Lord the latter two turn to $q/p$
I think that's right, that's was where I was heading to at least with my hints
 
@Studentmath Ok, I was expecting worse :).
 
Something bothers me. The exponent $n$ doesn't seem right.
Oh right, I keep forgetting the problem.
 
@TedShifrin $n$ from "$n$ subsets". We first pick $i$ elements as an overlap, then $n$ times $q - i$ other elements, and lastly divide over selecting $n$ times a $q$-subset from $p$.
 
The probability that each subset has specific $i$ items in it is idendepent. There are $n$ such subsets at any case
@Ted lol
 
9:04 PM
@Studentmath: I hereby abdicate my probability knowledge. It's all for you in the future. I'm sticking with differential geometry :P
 
@Lord post it as answer mayhaps? He was really there, just simplified the latter two to $q/p$ in the case of $i=1$ which got him a bit further. Sometimes simplifying isn't good, especially in inclusion-exclusion :P
@Ted too much responsibility..
 
@TedShifrin Oh, DiffGeo. That's one subject I have admitted not to be gifted for :P.
 
Well, @Lord, I won't take it personally.
 
@Ted oh, I started to read your class-notes (just the first couple of pages)
 
@TedShifrin A big relief :P.
 
9:07 PM
That doesn't mean I won't take other things personally, @Lord :P
Is that good or bad, @Studentmath? :D
 
@TedShifrin Haha, I won't get too comfy then.
 
@Ted You tell me the moment I start asking questions in that :P
 
I hope I'm slightly more competent on that, @Studentmath :D
Or else I really should have retired already ...
I guess our new mods have disappeared to learn their authorities.
 
Just like after real elections, we won't see them ever again
 
I am thinking that the site can still function without any mods.
 
9:10 PM
Well, I doubt that.
Do you recall the personal attacks on me by a certain person, @Jasper? I think not.
 
@TedShifrin Oh, Mr Rene, lol.
 
@Jasper I know many people would leave this website had it not been for some mod's work
 
Yes, I suspect his mental state is far worse than yours, @Jasper.
@Studentmath: Some claim they have left because of mods. It's not a perfect world.
hi @Henning
 
@TedShifrin Well, I don't know about that. He could just be evil and not sick. There is a difference.
 
@Ted that's also true..
 
9:11 PM
@TedShifrin Um, hi.
 
@HenningMakholm I am still here, after deleting so many accounts, lol.
 
Didn't mean to broadside you :)
 
@JasperLoy I see. Still blue.
 
He was bronzed last week, @Henning.
 
9:13 PM
Smurfs > statues.
 
You back from cavorting with the new mods, @Mike?
 
I've received so many congratulatory pings :P
 
No, there's a dog on my lap I was paying attention to.
Someone just got home so he left.
He probably thinks he's gonna get fed.
 
LOL
It's awfully early for dinnertime ... even for a dog.
 
@DanielRust Do you get pinged if they use DanielFischer?
 
9:13 PM
@DanielRust: Now you appreciate my trying to take so much care not to misping you :P
 
haha yes @Ted
 
Yes, @Ted, he's not really going to get fed. I wish I still had his naïveté.
 
Hey all. I am making progress on Problem Euler 494 :O
 
Welcome @HenningMakholm
Back :-)
 
@DonLarynx 494...! It's been so long since I've been there; I've witnessed the release of 200 :P.
 
9:16 PM
Wow. What made you stop? @Lord
I think PE will help me in my cognitive abilities
 
@DonLarynx Lack of time, mostly. Other interests were assigned higher priorities.
@DonLarynx Definitely. It aided mine.
 
Well, @Ted, Lawson's going into extreme detail about the representation theory of Clifford algebras now. I have not yet been convinced to care, even though I know I'm supposed to.
 
Interests? @Lord
 
@MikeMiller What course is this?
 
It is Winter vacation; I am not in any courses.
 
9:18 PM
@DonLarynx My studies, MSE, games. It just waned, I guess.
 
The book I'm reading is Lawson's "Spin geometry".
 
Hello!!! Could you explain me why $n$ is the nimimum element of a set $X \subset \omega$ iff $n \in X \wedge (n \cap X=\varnothing)$ ?
 
@Lord: I embarked on PE because my interests shifted to computer programming, so I understand. Games?
 
@MikeMiller I can never make the table tennis ball spin.
 
@Jasper it's in the swing
 
Huy
9:19 PM
@JasperLoy: I can teach you. I used to play professionally.
 
:O really @Huy?
 
@DonLarynx Yes?
 
@Lord: What games do you speak of?
 
@DonLarynx Computer games :).
 
Oh
been there done that :p
 
9:20 PM
Yes, @skullpatrol. Huy's real name is Pete Sampras.
 
Wrong sport, @Mike.
 
You're my hero, @Huy.
 
Lol
 
Huy
@MikeMiller: I wish I could tell you the same.
 
Well, I just crossed another K threshold, @Mike.
 
9:21 PM
I'm still down in the 7s... and I'm going to lose 100 on a bounty I give out later.
 
Well, I made someone very happy with today's answer :)
 
I need to go answer a hundred questions and make it past the big 10.
 
@DonLarynx :)
 
No, I think you're better off being a graduate student, @Mike.
 
@Mike do you have the Hannukah hat? Since it ends in... well, supposedly today
 
9:22 PM
@Ted Convince me to care about the representation theory of Clifford algebras.
 
Could I ask someone something about set theory?
 
I've everything I could have by now, @Studentmath
 
@Mike: Not I. I've never learned enough representation theory.
 
And I dare asking
 
OK, then I won't.
 
9:23 PM
@evinda Do you remember each natural number is the set of all smaller naturals?
 
I have to admit it's one of those taste barriers I never got over.
 
Although I did ask someone else the question I gave you months ago ... about giving $T(G/H)$ as an associated bundle.
 
That was nice.
 
@HenningMakholm Yes...
@HenningMakholm What do we get from that?
 
I wonder if @Pedro and @DanielF will be too good for us, now :)
 
9:24 PM
@evinda So $n\in X$ means that, um, $n$ is in $X$, and $n\cap X=\emptyset$ means that $X$ doesn't contain anything smaller than $n$.
 
@Studentmath @TedShifrin @A.E My attempt. 't Feels good to write MSE answers once more :).
 
Pedro's already too good for me. I said hi on facebook post-election and he told me he was leaving.
 
I will have to go now, see you all around!
 
We're glad to have you back, @Lord. Take care.
 
Later pal @Lord_Farin
 
9:25 PM
@TedShifrin Thanks for the warm welcome, bye :).
 
@TedShifrin At the moment, we're busy learning new stuff. So I think we'll be around here a little less in the next days.
 
Should I cook you dinner and send it over, @DanielF? :D
 
@DanielFischer I see that your name is in blue.
 
Yup, he's now officially officious :)
 
@HenningMakholm Could you explain me why this is the condition for a natural number to be minimum?
 
9:26 PM
To the new mods, it's not too late to retire, lol.
 
Did you learn any nasty secrets about me, @DanielF?
 
@TedShifrin With the speed of mail, I think that's a bit much of a gamble.
 
@evinda What else would you want "minimum" to mean? It is a member of X, and nothing smaller is a member of X?
 
Especially around Christmastime, @DanielF :)
 
Did they take away your picture @DanielFischer?
 
9:27 PM
@MikeMiller Yes. You can't cook. Ted told me.
 
@DanielFischer I guess he can cook, it's just the result which could be better
 
@skullpatrol I had to change my chat-parent from SO to Mathematics to get into the mod chat room.
 
Icic
 
I have never cooked anything in my life.
 
@Jasper: The first time I tasted bacon was at 18, in college
 
9:28 PM
@DanielFischer Is there a mod chatroom? Do tell dirty jokes and plan trips to las vegas?
 
@HenningMakholm Could you maybe explain it further to me? I am confused now... :/
 
@DonLarynx The first time I ate an oyster, I spit it out.
 
That's not true, @DanielF. I know how to make chicken, brussel sprouts, and pasta sauce.
 
@Jasper: I had to hold (the oyster) in to not look doofy in front of the chess team.
 
Huy
Which pasta sauce, @MikeMiller?
 
9:29 PM
@N3buchadnezzar I haven't been there much yet, so far, it was all "hello" and tips for what moderators should know. The dirty jokes will surely come later.
 
$$ \int_0^\infty 1 - \left( 1 - e^{-a w} \right)^n \,\mathrm{d}w = \frac{1}{a} \sum_{i = 1}^n \frac{1}{i } = \frac{H_n}{a} $$
 
Meat sauce. I can't make pesto.
 
Brussels sprouts are easy to ruin ...
 
I'm liking the blue name, @DanielFischer! ^_^
 
I had the best pesto of my life two days ago.
 
9:29 PM
How many members are in the smallest set @evinda?
 
Pesto is easy, @Mike. I see we'll need lessons.
 
@DonLarynx I used to play chess too, but stopped after elementary school, lol.
 
But is the best pesto of my life easy, @Ted?
I won't settle for less.
 
It's good to set the bar high.
 
I see grades are still not submitted yet. Hmph.
 
9:30 PM
I think the easiest dish to cook is salmon. Just add hot water and boil it and you are done.
 
@skullpatrol In the smallest set generally?
 
Huy
Good pesto is rather difficult to make, imo.
 
@TedShifrin That excludes the dwarves.
 
@evinda yes
 
@skullpatrol $\varnothing$ contains the less elements...
 
9:31 PM
@Jasp: I played chess professionally.
 
@Huy I think imo is not needed, as everything one says is his opinion.
 
No, @Huy, like most good Italian food, it's about the quality of the ingredients. Very few needed, but one needs good fresh basil and top quality olive oil. Garlic, parmeggiano reggiano, pignoli nuts. Done.
Salt and pepper, of course.
I might put in a bit of cilantro or arugula to be different sometimes.
 
Huy
@TedShifrin: Then, my ingredients were mostly of bad quality, because my pesto is never as good as I imagine it to be like.
 
My favourite Italian food is spaghetti carbonara.
 
That and being willing to start from scratch... the way most people ruin their meat sauce is by starting from pre-made stuff.
 
9:32 PM
@evinda Okay, what is your definition of "mimimum element"?
 
@Jasper: It's not good for me, but I love it. I make it occasionally.
@Huy. Really good olive oil is expensive.
 
Mine is pizza
 
@TedShifrin Why is it not good for you, too fatty?
 
We're about to be flooded and have tornados here. So if I disappear, send a boat for me.
 
Huy
@TedShifrin: What does expensive mean, in terms of olive oil?
 
9:33 PM
Jasper called you a fatty, @Ted
 
@JasperLoy that sentence becomes much funnier if you remove too-
 
Yes, @Jasper ... bacon and butter and eggs.
 
@HenningMakholm Let $(A, \leq)$ be an ordered set.
We say that $a \in A$ is:
mimimum when $(\forall x \in A) a \leq x$
 
smacks @N3B and puts him on ignore
 
I said it first. Shouldn't it be me?
 
9:34 PM
:O
 
@evinda Right, and if you say $n$ and $X$ instead of $a$ and $A$, you get ...
 
You, too, @Mike.
 
Finally!
 
@evinda ... $n\in X$ which you already have, and $\forall y\in X: n\leq y$.
 
9:36 PM
Who here is a fan of Tycho?
 
But the ordinals are totally ordered so $n\leq y$ is the same as $y\not < n$. And $<$ on ordinals is the same as $\in$, so $n\leq y$ is the same as $y\notin n$.
So for the second condition we have $\forall y\in X: y\notin n$, which says exactly that $X\cap n=\varnothing$.
 
Got it^ @evinda?
Take your time and think about it :-)
 
@evinda $a$ is a minimum of the set $A$
induction proofs are so awesome
and so are RAA's
 
What's an RAA, @DonLarynx?
 
Reductio ad absorptum, also known as contradiction
 
9:47 PM
Let $(a_n)_{n\ge 1}$ a strictly increasing sequence in natural numbers. Prove that

$$\lim_{n\to\infty} \left(\frac{1}{a_1} +\frac{1}{a_1\cdot a_2}+\cdots + \frac{1}{a_1\cdot a_2\cdots a_n}\right)\in \mathbb R \setminus \mathbb Q$$
 
@Chris'ssis HINT: $\sum_1^{\infty}\frac{1}{n!}$
 
I thought it was reductio ad absurdum, @DonLarynx.
 
@Chris'ssis the sum is $e-1$, which is in $\Bbb{R}$ and not in $\Bbb{Q}$
I think you should stick to your integrals from level 9 in Dante :)
 
@DonLarynx I don't think that is relevant for proving this fact, maybe for proving the fact the limit exists. And yes, as @HenningMakholm said, you only proved a particular case.
 
@DonLarynx Only if $a_n=n+1$.
@Chris'ssis Suppose the sum is p/q. What happens the first time a_n is larger than q?
 
9:52 PM
Let $q \in \Bbb{Q}$. then the sequence can be written in the form $\sum_1^{\infty}\frac{1}{{q_n}{a_1}}$
 
@HenningMakholm I know how to do it. I posted it here for you (it's just for fun).
 
Then we get $\frac{1}{a_1}\sum_1^{\infty}\frac{1}{q_n}$
 
@Chris'ssis Oh, sorry.
 
And since $q_n$ is increasing, then we get that this is a monotonic decreasing sequence. What remains is to prove this sequence............
 
Yeah! Someone downvoted one of my posts so now my rep is $\equiv 0 \pmod{10}$.
 
9:57 PM
@KajHansen wop wop
 
That didn't last long, @Kaj
 
Huy
#rekt
 
@Chris'ssis Using the fact that "every real number is the sum of countably many rational numbers", we finish.
:)
QED
 
@KajHansen They downvoted your integrable answer because you did not point out the mistake in the asker's reasoning.
 
@KajHansen I downvoted you because I wanted to see you suffer.
 
9:59 PM
I figured @JasperLoy
 
@AlexanderGruber something really weird happened
 
@DonLarynx It was given in a high school contest.
 
But then I undownvoted you because I didn't want you to suffer for long.
 
haha @Mike
 
@Chris'ssis: I shall put more effort into solving your question later.
 

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