10:00 PM
Last night I had a nightmare that I was in a bio class.

Worst dream.
@JessyCat Do you know how to fast put the bio chapters in memory?

No. If I had, I'd probably be a doctor. LOL

@JessyCat Need not be, you must have developed techniques to memorize, back in school.

You need to make it meaningful.
Something personal to you.

I need to make bio meaningful?

10:11 PM

@robjohn Thanks, but I am in serious time crisis now. Will check tomorrow :)
@robjohn Took a fast look. I wanted it to be in terms of $ABCD$. See the comment on question?

10:33 PM
Guys, how long does it usually take on average for a comment to appear on a question?
?

@user129967 Instantly, really.

i Have asked a question and over 30 people have read it but not a single person has responded
the wierd thing is it hasnt even been downvoted
its just as if people just don't give a damn

@user129967 Maybe they don't. You shouldn't take it personally.
What's the question?

haha I will

Wait.

10:37 PM
0

After some research, I have discovered that proving the statement; There exist an infinite number of positive integers K such that; $K \neq 6ab \pm a \pm b$ and $K \neq 6ab \mp a \pm b$ is equivalent to proving the twin prime conjecture. I have naively attempted to think of ways in which on...

I know its sort of, of the wrong format, but people are not downvoting.

Also, you asked your question 1hr ago.
And it is not an easy question, I think. Also long.
People are usually spooked by long questions.
I didn't read it fully, really.

Haha I can tell
everyone is probably just shuddering as they see the size of the question

@PedroTamaroff can you see this question? There is a link in it that might interest @user129967.

@IanMateus Looks kinda cranky. =P

I know I just want to know what is wrong with it because it is most likely wrong.

10:44 PM
@Pedro You never gave me a straight answer.

nahahahahahahaa

@Mike What to?

Is it racist to dance the robot in front of a robot?

@PedroTamaroff caution: Microsoft Word

@Mike Interesting...

10:47 PM
@Mike: Is it racist to rap in front of a black man? Is it racist to each chilli in front of an Indian?
Such are the questions of social life.

@Nick What are you doing at this time?

@Nick There are many people to do that. But why are you awake at this time? 4 am?

@user129967 this link might interest you, this is probably the source of your statement

@Nick There's a huge difference.
A robot can move in no eh except the robot.
so you're essentially mocking his very existence
Black men can make more music than rap

10:53 PM
@Sawarnik: Long story short, I woke up at 5 am yesterday to catch the Oscars live. To compensate the lack of sleep, I slept the whole day and now I'm not in the mood to sleep. Does that satisfy your curiosity of my personal life?

@Nick Hmm..fine. You got angry a bit. Sorry.

@Mike: Yes but the question was whether if it were racist, not the degree of racism involved.

@Mike what about walking in front of someone who can't?

@Sawarnik: $\ddot\smile$ No I didn't, silly. So, what's up with you? Early morning studying or something?

@Nick I'm saying neither of our situations are comparable.

10:57 PM
Haha this is such a strange conversation

@Ian But for those of us who can, walking isn't done specifically because of the crippled man
Maybe I should say this isn't at a club

@Nick $\ddot\cry$ [Didnt work] Studying for my bio test in 3 hours. 1 more chapter to deal with.

this is in an office environment, where people dance in front of the robot intentionally

@Sawarnik: 1) *\frown , 2) Swallow as much as you can and run.

@Nick Exactly. The next one to deal is reproduction in just half an hour !

11:00 PM
@Sawarnik: ... Wow, I think you've predicted my future if I go into Math

@Nick I don't think so. Which class are you in, considering our same level, upto basic calculus?

@Sawarnik: (1) 11th, (2) I'm glad you didn't get the joke.
@Mike: That's just wrong? Is that like your job or something?

rawr

@Nick Hee. I m in 8th. 2) Wat was the joke really?

@Sawarnik: I'll tell you in 3 years.

11:05 PM
@Nick I would remember it.

howdy @Nick, @Mike, @Sawarnik

@Sawarnik: Concentrate on bio!

@Ted: Reach for the sky! You're my favorite deputy.

Huh?!

11:06 PM
@Nick Good advice. Bye for now.

@usukidoll: And a fine rawr to you too.
@Sawarnik: Catch you later, alligator.

:/ 8 cases of this...how the? http://math.stackexchange.com/questions/697399/prove-or-disprove-a-b-cap-c-a-cap-c-b-cap-c

suppose I want to disprove the statement...does this mean that I have to do it 8 times as well?!

@Ted: I thought we were quoting Woody from Toy Story.

I have no earthly or unearthly idea, @Nick.

I'm hearing crickets

11:10 PM
@Ted: Get idea, sirji

If you want to disprove an equality, @usukidoll, give an explicit counterexample. Otherwise, prove it without relentless case analysis.

@usukidoll: Oh, I stopped listening to it the moment Sachin Tendulkar left the game.

like maybe if A B or C was an empty set and then we try to disprove it that way? no wait I've tried it last night and I got empty set all the way through

Why are you so convinced it's false?

the first time I did it....I used complement def on $A+B$

11:12 PM
Can you use the definition of $+$ to write everything in terms of $\cup$ and $\cap$ and use known laws?

and then I distributed the $\cap C$ I've noticed that the middle part wasn't $+$ so I thought that they weren't equivalent
but apparently that's not how it works

So you're looking at $(A\cup B \backslash A\cap B)\cap C$?

yeah ... the question was prove or disprove $(A +B) \cap C = (A \cap C) + (B \cap C)$
since $A+B$ was $A \cup B) \backslash (A \cap B)$
I used the complement definition.. to make it just $A \cup B$ because by complement def we have $[x: x \in A \cup B \land x \notin A \cap B$
so I was left with $(A \cup B) \cap C$ on the left hand side

Right. Can you decide if $(X\backslash Y)\cap C = (X\cap C)\backslash (Y\cap C)$?

hmmmmm... I'll start on the right... $(X \cap C) \backslash (Y \cap C)$ so by complement def I would have $[x: x \in X \cap C \land x \notin Y \cap C$ so that leaves me with $X \cap C$

11:18 PM
typo ... big $X$ should be $x$ at the end
no, it doesn't leave you with $X\cap C$ :)

:/ ok and why
?

because you can't have any $x$ that live in $Y$ (and $C$).

yeah isn't that complement definition though?

Hey everyone :)

hi Mick

11:20 PM
:/

@usukidoll, perhaps a Venn diagram picture would help you see what's going on?

sure
en.wikipedia.org/wiki/File:Venn0110.svg that's the venn diagram for $A +B$

I've got a very simple question, sorry it includes no amazing integral, but I'd still hope for some answers and/or speculation from any lurking mathematicians. @robjohn @Chris'ssis everyone...

Are there actually practicing mathematicians who reject the law of excluded middle or any non-constructive proof?

fanatics, perhaps, @Mick. Mostly among the ranks of undergraduates and graduate students. None that I've ever heard of among serious researchers.

Oh thank physics!

11:24 PM
Most of us agree that $\sqrt2$ is irrational, @Mick, I do believe.

@TedShifrin hey, but that one has a constructive proof.

So, @usukidoll, write a proof that what I told you to do is correct.
Oh? @Ian ... What's that?

how the heck does it not leave me with $X \cap C$!?!?
I mean if I start from the left $(X \backslash Y) \cap C$
complement def of $X \backslash Y$ is just X
$[x: x \in X \land x \notin Y$
and I am left with $X \cap C$

NOOOOO ... why do you think the complement of $Y$ is all of $X$?
You need to learn some serious definitions.

Haha thanks, I was nervous about that having had someone bring the non-constructiveness of a proof I made into question

11:26 PM
However, @Mick, I do not like proofs by pleonastic (unnecessary) contradiction.

I prefer to avoid ridiculous logic if possible

"Ridiculous" be in the eye of the beholder.

Of course

what! why is the complement $Y$ all of $X$?

it is NOT. You said it was.

11:27 PM
@TedShifrin we can prove that $|\sqrt 2 - a/b|\geq 1/3b^2$ for every coprime $a$ and $b$. See here

oh, you're typing stuff I'm misinterpreting.

o-o

I just hope that it's a common view that rejecting pure logic only on the basis that there is no tangible object produced can be called "ridiculous"

But the criterion for irrationality, @Ian, are you sure there's no proof by contradiction in that? I think there is. Certainly the proof of Liouville's Theorem I know has one firmly there.

@Ted Can I ask you a serious question?
BTW, I think the word pleonastic is itself pleonastic

11:29 PM
@usukidoll: $(X\backslash Y)\cap C = \{x: x\in X, x\notin Y, x\in C\} = \{x\in X, x\in C\} \backslash \{x\in Y, x\in C\}$.
smacks @Mike
haven't seen you in days .. what's the serious question before I run off to dinner?

we don't have an x that belongs to Y on the left...
OHHHHHHH MY GAWD ETHAN! faints

right ... I took those out ... and nor do we on the right @usukidoll.

yes that's true
so aren't I left with... $X \cap C$?! or is it something else

You can replace my commas with $\wedge$, @usukidoll

@Ted OK. Is it racist if people are intentionally and spitefully dancing 'the robot' in front of a robot
that can move no other way?

11:32 PM
No, you're left with $(X\cap C)\backslash (Y\cap C)$ !!

Damnit I tabbed away for like 2 seconds and I come back to a screen full of fucking sigma zeta beta yayta bleata bass drop WUBWUBWUB

ROFL @Mike slaps harder

AND WHY IS THAT!!!! D:

Hey, guys. Is there a... like, notebook or something (equipped w/ MathJax) that I can use for my own referencing?

Nobody will give me a straight answer!

11:32 PM
like OH FFFFFFF******* I think the lightbulb just ignited

@Mike Yes

@Mike: I rarely do anything straight.

Like a MathStack Exchange personal notepad?

As long as you don't restrict "race" to mean human races

@TedShifrin Liouville is way stronger than this one, I'd blindly bet it is nonconstructive :P I have never studied that. I think this is a cute constructive proof though: $\sqrt 2$ is not any rational you can think of.

11:33 PM
Or do any of you know of some kind of application equipped w/ MathJax that I can use as such?

that's not a constructive proof in the least, @Ian.

It is if you're an alien with infinite thinking bandwidth

wait hold on let me try this again OH!!!

$(X \backslash Y) \cap C$
$(X \cap C) \backslash (Y \cap C)$

@Ted I'm going to cap off my RS course with the proof that complex curves are algebraic.
It's a satisfying ending.

I really think you should learn Riemann-Roch, @Mike, but OK.

11:34 PM
-.- I'm just using distributive law for $\cap C$

It's such gorgeous mathematics.

@Ted I spent last quarter doing that...
That was the proof I presented during my final.

But you don't know it, @Mike.

I don't understand your meaning.

Oh, did you really? In terms of linear systems and meromorphic functions?

11:36 PM
@Ted yeah I am left with $(X \cap C) \backslash (Y \cap C)$

because we had $(X \backslash Y) \cap C$
by distributive law
$(X \cap C) \backslash (Y \cap C)$

OK, dinnertime for this Bonzo. You all have fun without me.

In terms of sheaves and cohomology :P

Oh, right, that abstract crap in Gunning. Sigh.
OK, @usukidoll. Now proceed.

Jeez, now I'm offended! :D

You want an excuse to hate me, anyhow, @Mike. :) When do you go to LA?

11:37 PM
maybe the fact that I like that abstract crap says I'm more of an algebraist.
8 days from now.

indeed @Mike, indeed :(
very cool ... Give Jacob a hard time for me :)

@Ted I find an incredible value in building my geometric intuition and not living in abstraction. So I appreciate your POV. :)

so if I want to prove $(A +B) \cap C = (A \cap C) + ( B \cap C)$
starting from the left
$(A +B) \cap C =$
using distributive law
$(A \cap C) +(B \cap C) =$
and that's the same as the right
$(A \cap C) +(B \cap C) =(A \cap C) +(B \cap C)$ @Ted

Is anyone willing to suggest what area of mathematics I should pursue next? My skill level is basic calculus with a good number of special functions available

Do you think there's a distributive law for everything? @usukidoll
I know it only for $\cup$.
LOL @Mike ... No fair patronizing me now. :D

11:41 PM
Hey, I get it from you. I get to take my potshots once in a whole.

But, yes, @usukidoll, it is true and you need a basic proof based on what I told you and then using union.

using union?!

Take it once in a half instead @Mike :P
the definition of $+$, @usukidoll.
OK, I'm outta here.

OH THAT ONE!

Me too.

11:42 PM
$(A+B) = (A \cup B) \backslash (A \cap B)$
$(A+B) \cap C) =$

$[(A \cup B) \backslash (A \cap B)] \cap C$

so distributing $\cap C$ I would get
$[(A \cup B) \cap C) \backslash (A \cap B) \cap C$

Specifically, I'm stuck on what seems like a simple problem, but I am painfully unaware of where to begin. I've built a relatively simple function, but no closed form is available. Is it possible for me to take the integral of this function which I can't even express directly?

and then $(A \cap C) \cup (B \cap C) \backslash (A \cap C) \cap (B \cap C)$