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12:00 AM
TENT SPACES.
 
12:13 AM
 
@skullpatrol I forgot N Sync existed.
 
What is it?
 
@Wipqozn We are all N Sync here :)
 
@JonasTeuwen A North American boy band from the 90s.
 
I was playing chess in the 90s, so probably didn't hear it ever.
 
12:16 AM
@skullpatrol Oh god.
 
@JonasTeuwen Lucky you.
 
♫ Nothing ever happened, anyone can see ♫
 
wtf?
 
♫ Nothing ever happened, nothing ever happened in heeere ♫
 
@badp I have the relevant part bookmarked :P
 
12:18 AM
bye bye bye
 
@OrigamiRobot This might've broken the bookmark
 
I demand to be suspended permanently right now!
5
 
@skullpatrol No can do
 
@skullpatrol That's @badp's favourite thing to do!
 
Please?
 
12:20 AM
@badp It did. You're so mean. :(
 
Pretty please?
 
@skullpatrol No can do. I literally have no "permaban" button.
 
What is the longest I can get?
 
I don't think the feature is even been written in the first place.
@skullpatrol It's not long enough.
 
Who damaged my intellectual property?
 
12:23 AM
Can you at least please pin this message for us?

@AlanH We only do hard ones here, sorry! Bye! - 31m ago by Jonas Teuwen
 
(By the way, it's not appropriate to think of flagging as backstabbing. Bye!)
 
What else is it?
 
@skullpatrol You probably have room owners who can do that instead, such as anon.
 
Why is it not?
It feels very much like it.
 
Flagging is a form of moderation, if you're using it as a form of revenge then you're misusing it.
 
12:24 AM
Even when you can actually say it you decide to be a coward.
 
If it succeeds it comes with a 30 minutes ban, btw.
 
Yes, that was odd once, I had like a 2,5 hour ban and it passed in minutes.
 
@JonasTeuwen Somebody probably was kind enough to wipe the suspension for you
 
Good. The right thing to do.
 
Ok I'll take the 30 minute suspension
 
12:26 AM
@JonasTeuwen Perhaps.
 
@skullpatrol Will you do that with whisky?
Make sure you also try to use a tampon and shove that up your little star.
And tell me how that went - I'm curious.
 
I'm flagging that^
1 min ago, by Jonas Teuwen
Make sure you also try to use a tampon and shove that up your little star.
 
I might approve that flag. Cmon @JonasTeuwen...
 
Oh nfity!
 
What?
 
12:28 AM
28
Q: Main Chatroom Etiquette Rules

robjohnThe previous owner of the Mathematics chat Asaf Karagila drafted some etiquette rules for the chat. They were deleted, so I am reposting them so that they might be observed. It's nice of you to drop by, after saying hello please spend a minute reading the transcript to see if there is an active...

I like that. Makes it easier to deal with any flags I see from this room.
 
I am sure Asaf would approve of that.
 
Although I rarely see flags frmo here. You guys are pretty cool.
 
Asaf who?
 
The 'previous owner'.
@skullpatrol I like that you openly admit being a backstabber. Does that make you still a backstabber? Sounds paradoxical.
 
12:32 AM
I admitted nothing.
I just want to be permanently banned from this site.
 
@skullpatrol Asaf agrees.
 
Asaf who?
 
Remove Yourself With A Push On The Button.
 
That coward is not here is he?
 
Why is he a coward?
 
12:35 AM
Because he is not here.
 
Aha.
You make no sense Sir.
You Sir, are a Tosser.
 
If you say so Sir.
 
Real men just stop coming here, instead of asking for a 'suspend', no?
2
(The answer is: yes)
 
maybe
 
Wrong! The answer is: yes.
 
12:38 AM
I have stayed here waiting for Asaf to finish what he started with me OK
 
What did he start?
 
A war.
Enough said.
 
Don't forget to remove the dust once in a while.
 
Thanks, I'll try and remember that.
I'm waiting...
9 mins ago, by skullpatrol
I have stayed here waiting for Asaf to finish what he started with me OK
 
Can we please stop flagging.
 
12:51 AM
Now what?
 
@skullpatrol Just ignore him.
 
It is a form of moderation?
 
(removed)
 
31 mins ago, by badp
Flagging is a form of moderation, if you're using it as a form of revenge then you're misusing it.
 
Can the two of you please stop flagging? You're spamming every chat user who can see the flags with your little revenge flagging...
9
 
12:59 AM
Who is flagging?
I only did that once for this mick bro, then I got the wrath of the moderators and the tosser continued after they left.
2
 
I'm not disclosing names, but who ever it is, please stop. That's all.
 
Can you see them?
 
@rm-rf I came here to say this too, but I got chewed out for going to the bridge earlier and telling them to cut it out
 
The 'two of you' suggest you mean me, as there is nobody else.
And I am not that - so please stop doing that.
 
Now might be a good time to ask you gents to vote on my meta thread about chat flags :)
37
Q: Let me opt out of viewing chat flags

MDMarraI'm not a moderator. I have ~50k rep on Server Fault. I am a regular in chat. Because chat allows all 10k+ users across the network to act on chat flags, I (and many other regulars in SF chat) see some...less mature...users from other chats flagging things that aren't actually offensive. I disa...

 
1:01 AM
@MDMarra Who chewed you out? I apologize on behalf of The Bridge.
 
@MDMarra Honestly, this is a case of flag abuse. Whether we can see flags network wide or not wouldn't change that.
 
@OrigamiRobot badp. In all fairness I don't ever chat there and he was defending the people that do, so I understand it.
 
No it is not a good time Sir.
 
Hey guys... what's up with all the flags?
 
I have not seen or put any...
 
1:02 AM
@MDMarra If it was something along the lines of "we know", he wasn't chewing you out. It gets really exasperating dealing with people who continually misuse flags.
 
@OrigamiRobot Heh, yeah. It was along the lines of "we know, but you also shouldn't be slinging shit since you don't chat here"
Either way, it's amusing to see the same problem with the same users pop up in a different chat
 
@MDMarra Well we usually aren't that abrasive :P
 
Anywho, seems like the fireworks are over, back to the Comms Room for me
 
@MDMarra See you again at the next flag party!
 
@Wipqozn Glad to see you survived this one. Bye now
 
1:05 AM
sigh
 
groan
moan
 
Oli
Grumpyface Whoever is flagging nonsense, stop it please.
 
Can't you guys just suspend whoever is doing it so we can get back to normal here?
2
 
Looks like it's normal now, so let's maintain it :)
 
It is not normal to have this many of you guys here.
 
1:20 AM
@skullpatrol Yeah, so many pointy rhombi... someone's going to get hurt!
 
@robjohn Yep.
user image
3
 
Oli
The "Delete this room" link is looking pretty tempting at the moment.
7
 
@Oli Couldn't you just suspend the user(s) who are doing the flagging?
 
@Wipqozn Only if the flagger is known.
 
@GraceNote Is there no way to see the flagger? I thought a mod who is also a room owner could, and I assumed SE employees could as well (aka you).
 
1:33 AM
@Wipqozn Only if the flag is live.
 
@GraceNote Ah, I see. Well that's annoying.
 
1:45 AM
Guys... please don't get our room deleted before I get back from walking Lilly.
BBL
 
What the hell is going on?
3
 
1:59 AM
in the one dimensional case of the inverse function theorem, are the conditions (continuity, monotonicity) also necessary conditions for a function to have an inverse?
 
2:48 AM
@robjohn Okay, I'll wait till you get back before getting the room deleted ;-)
 
wtf?
 
@skullpatrol I just saw that you asked if I could pin the msg for you guys...What does that mean?
 
@math101 Hi! Wassup?
 
3:03 AM
@OrangeHarvester I am great. My brain is quite fried though
 
@math101 Ehh. No healthy mind in healthy body for you? :P
Why is your brain fried?
 
Studied too much of Real Analysis this evening
 
Ahh. I see. With me, it is the opposite, my brain fries when I do not study.
 
@AlanH it was sarcasm because someone flagged it as "offensive."
 
@skullpatrol Do you watch "How I Met Your Mother" ?
 
3:07 AM
no
i really don't like being threatened, but out of respect for robjohn i will wait
 
hahahahaha
 
what's so funny?
 
Your comment
This morning you made this funny comment and I was in the middle of some meeting and noticed. I cldnt stop laughing. The guy thought I snapped
 
@math101 You access MSE chat from a meeting!!
 
do you know what the threat is?
 
3:14 AM
@skullpatrol no.
 
Is it regarding deleting this room?
 
2 hours ago, by Oli
The "Delete this room" link is looking pretty tempting at the moment.
 
Seee I am brilliant.
 
2 hours ago, by robjohn
Guys... please don't get our room deleted before I get back from walking Lilly.
29 mins ago, by skullpatrol
@robjohn Okay, I'll wait till you get back before getting the room deleted ;-)
 
@skullpatrol Was something pedestrian was flagged and then a series of revenge flagging took place?
 
3:18 AM
yep
 
pedestrian?
 
@math101 not anything out of ordinary.
@skullpatrol Why the revenge flagging?
 
I have trouble comprehending why four messages on this topic were starred, or even why the conversation is ongoing.
 
@peoplepower right you are.
 
@OrangeHarvester ...because of a grudge...
 
3:24 AM
@peoplepower Same.
@OrangeHarvester Yoda!
 
@skullpatrol that is only going to affect you, no one else.
@GustavoBandeira boa tarde da noite
 
@GustavoBandeira Did I get it right?
 
@OrangeHarvester Brasileiro também?
 
@GustavoBandeira Dude, I used Google Translate.
 
3:29 AM
@OrangeHarvester Oh, you killed my hope. But that's not right.
 
@GustavoBandeira Okay!
 
I can understand/speculate what you meant with that though.
 
I meant, Good Late Evening.
 
Good to see the room is still here.
 
3:35 AM
@robjohn :-)
 
@math101 do you like this?
 
Who's flaggin this room for deletition? Who wants this room (removed)?
 
4:03 AM
Is there any good sufficient and necessary condition for a function having primitive (anti-derivative)?
 
@FrankScience If you mean in sense of Riemann Integration, Riemann Integrable functions, that is precisely those with countable discontinuities on the domain.
 
@OrangeHarvester It's unrelated to Riemann integration, I think. $f(x)=F'(x)$. $f(x)$ need not be Riemann-integrable.
 
hi all!
i done uni, worked in industry, now is time to think about what exp(ix) really means!
 
@OrangeHarvester Incidentally, your proposition of Riemann integrability is not right enough. It's exactly Lebesgue-measure-zero discontinuities on the domain of a bounded function.
i.e, null set.
 
@FrankScience It is necessary that if $\lim\limits_{x\to a}f(x)=b$ then $f(a)=b$. That is a necessary condition.
That is required by the mean value theorem
 
4:21 AM
I know some results part of which I can't prove.
For example, the continuities of $f(x)$ is dense.
 
4:34 AM
Hey @Oli!
 
@FrankScience, a necessary and sufficient condition is that the set of points of discontinuity of your function be of measure zero.
 
4:58 AM
@skullpatrol Thats totally not my style. lol
 
@WillJagy I have not learnt Lebesgue measures yet, but does not measure zero mean continuous almost everywhere? That is the impression I got from Principles of Mathematical Analysis Rudin ( I have yet to do the last chapter on measures there.)
 
5:11 AM
I want context for each and every star right there. Even the transcript is not able to help.
 
@Novice leave it dude.
It is apparent that something has been flagged and there was a quarrel.
 
I see.
 
@FrankScience I guess you are correct. I have not understood the lebesgue thing well enough as of now.
 
Hi guys
Anyone here?
 
no
 
5:18 AM
Oh no
Mind to help me with my confusion regards to a proof?
 
HI @Ethan
 
I don't think it's post-worth so I'd rather "chat" it out here.
*post-worthy
Hi Eathan and math101 btw :)
 
Hey @Sean!
 
Hi Novice
Can anyone familiar with set theory explain this to me?
 
Hey @Sean. I dont think I will be of much help :(
What is ur question?
 
5:22 AM
How can there be "x belong to A but not belongs to f(x)"
lol I would still appreciate your help
I want to prove that "there is no surjection from a set to its powerset"
I want to induction on the cardinality of powerset: Assume true from 1 to n, using n<2^n, we know n+1 < (2^n)+1, so (2^n)+1 < (2^n)+(2^n) = 2^(n+1)
but the set may be countable, so I don't know if i can assume n<2^n without using more propositions, corollary, theorems,...etc
 
5:38 AM
I have my first Linear Algebra quiz tomorrow morning >.<
It's just basic vector stuff: a little silly
 
Good luck ;) Genintendo
 
Thanks :P
 
@Sean $n < 2^n$ even for $n=\aleph_0$. In terms of cardinality of course.
 
But the set may be countable
So I dont want to assume n<2^n
(I understand that there is a proposition says exactly that....but will my proof still work without involkling the theorem/theorem)? (sounds silly I know ;)
 
I either love matrices and think they're the greatest thing ever, or I absolutely abhor them.
I haven't decided which it is yet.
 
5:50 AM
I abhor them
I always tend to make a mistake somewhere down the line
and by the time I realize my lil error its kind of too late and I gott redo the whole thing
 
I agree with you guys...
 
Its a form of tedious arithmetic
 

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