Mathematics - 2013-02-25 (page 2 of 5)

skullpatrol

10:05

(removed...)

robjohn

@skullpatrol we could just remove you ;-)

skullpatrol

@robjohn lol

@robjohn Is it a well known fact that GCF*LCM=the product?

robjohn

@skullpatrol yep... an ASCII proof

skullpatrol

@robjohn Thanks.

robjohn

@skullpatrol Is there a question on the site about this?

JSchlather

10:22

Anyone knowledgeable at PDE on hand?

BenjaLim

@robjohn Do you know who's akon?

@JSchlather No don't think so

JSchlather

@BenjaLim :(

BenjaLim

@robjohn PDEs?

JSchlather

There's a proof in Evans where he throws in a $\sqrt{n}$ for no reason

BenjaLim

@JSchlather Do you listen to akon?

JSchlather

10:32

The rapper?

BenjaLim

yea

JSchlather

No, refer to last.fm/user/liberalkid

robjohn

@BenjaLim pardon? akon, is that a user here?

BenjaLim

@robjohn No he's a rapper

robjohn

@BenjaLim Then no, I don't, sorry.

BenjaLim

10:33

@JSchlather Guess you're not so mainstream :)

I recognise like the beach boys

robjohn

@BenjaLim I have a passing acquaintance with PDEs...

JSchlather

@robjohn I'm having an issue with the proof of Liouville's theorem for harmonic functions in Evan's PDE book.

He proves it by first giving a bound for derivatives of a harmonic function then uses this to deduce liouville's theorem, but he throws in a $sqrt{n}$ that I can't figure out.

robjohn

@JSchlather Do you have a reference that I can read online?

JSchlather

I could link you to an illegal copy of Evan's.

robjohn

@JSchlather can you transcribe the problem part into LaTeX here?

JSchlather

10:36

Yeah

So essentially the important parts are as follows

For a harmonic function u in R^n we have that $|Du(x_0)| \leq (C_1/r^{n+1}) ||u||_{L^1(B(x_0,r))}$

where C_1=2^(n+1)n/alpha(n)

and alpha(n) is the dimensionality constant

To prove that a harmonic function which is bounded is constant he uses this bound and then relates it to the l^infty norm of u.

But he introduces a sqrt(n) term into the previous bound which I don't understand the need for

so he says that |Du(x_0)| \leq sqrt(n) * other bound

robjohn

Did I correct your LaTeX about $|Du(x_0)|$ properly?

JSchlather

yes

(sorry)

So we can bound the L^1 norm by $\alpha(n) r^n ||u||_{L^\infty(\mathbb{R}^n)}$

Then putting this in to the previous bound we have that $|Du(x_0)| \leq \frac{\sqrt{n}C_1\alpha(n)}{r} ||u||_{L^\infty (\mathbb{R}^n)}$

So I don't understand where the $\sqrt{n}$ term is used at all.

@BenjaLim Also, you clearly need to expand your musical horizons ;).

robjohn

The constant's dependence on $n$ is not important. It is the fact that $|Du(x_0)|\le C/r^{n+1}\|u\|_{L^1(B(x_0,r))}$ that matters.

JSchlather

I agree. So you don't see any need for the sqrt(n) term either?

robjohn

If $u$ is bounded, then $\|u\|_{L^1(B(x_0,r))}\le Cr^n\|u\|_{L^\infty}$

JSchlather

10:47

Yes

robjohn

@JSchlather I don't know why the constant changes... I 'd have to see the book for that, but even with the $\sqrt n$, everything is okay.

user19161

@skullpatrol Works for two numbers but not more.

JSchlather

Yeah, I guess it is. Who knows why Evans added it. Thanks.

user19161

@BenjaLim akon is not a god to me; only anon is a god.

BenjaLim

@JSchlather quick question, what is the exact form of the arctan function that gives us a bijection between $(0,1)$ and $\Bbb{R}$?

user19161

10:49

@JSchlather There is a new edition of that book a couple of years ago.

BenjaLim

@AlexYoucis In australia we call a fixed gear bike a fixie

JSchlather

@BenjaLim arctan(x pi/2)

BenjaLim

@JSchlather No we don't get the negatives

Ah sorry I meant the tan function

it's $\tan ( (2x- 1)\pi/2)$

user19161

@BenjaLim Are you testing him?

JSchlather

Oh soryr

yeah

Yeah that works

user19161

10:55

@JSchlather I hope you don't get pinged anymore when they ping me.

BenjaLim

@JSchlather Man do I really need to write down these homeomorphisms

JSchlather

@JacobBlack Yea, no more pings.

BenjaLim

@JSchlather So annoying

user19161

@BenjaLim Yes, otherwise they might not be correct.

BenjaLim

@JSchlather In last sems algebraic topology my lecturer would have just allowed us to say it

@JacobBlack Have you seen the proof that multiplication is associative in the first homotopy group?

It's just a picture

user19161

10:57

Too much handwaving leads to errors.

BenjaLim

@JacobBlack I can tell you exactly in words how the homeomorphism works

user19161

One's hand might just fall off if you handwave too much.

JSchlather

But how can you say it exactly in words when the english language is so imprecise ;)

user19161

Why do you think we have epsilons and deltas?

user19161

If we say in words why something is continuous or not, we might get it wrong.

user19161

10:59

Some things just defy intuition.

BenjaLim

@JSchlather I know how to tell you the algorithm that gives you the homeomorphism

user19161

@benja I don't know anything about spectral sequences, but I just wanna let you know that Hatcher seems to want to include his spectral sequences book into his AT book from his website in future editions.

BenjaLim

really?

@JacobBlack I'm not that into algebraic topology

@JacobBlack I took it because I wanted to know about cohomology

that was it

user19161

Anyway, his other two AT books seem to be getting nowhere. I guess he is busy.

user19161

@BenjaLim What are you into?

Eric

11:14

My book has this problem: The group $(Z_4 \oplus Z_{12}) / \langle (2,2) \rangle$ is isomorphic to ...

What is $⟨(2,2)⟩$? I am pretty sure it is a cyclic group, but i have never seen this notation used before, and I looked in the notation index but I point to an ideal, which is about 150 pages after what i am on now...

JSchlather

It's the ideal generated by (2,2).

Eric

ok, well my book has not defined what an ideal is yet, could this possibly mean something else?

JSchlather

Oh sorry

I meant subgroup

Eric

so would this be $⟨(2,2)⟩ = \{ (2,2), (4,4), \ldots \}$

Gam Erix

Any quaternion experts here? : P

Eric

11:20

@JSchlather right?

JSchlather

Yes, eric but what is 4 in the left component?

Eric

oh yeah

so it would be $\{(2,2), (0,4), (2,6), (0,8), (2,10), (0,0), \ldots\}$, correct?

BenjaLim

@JacobBlack AG

Eric

@JSchlather am i thinking about it the right way now?

BenjaLim

@Eric Do you know how to row reduce a matrix?

use that

Eric

11:24

@BenjaLim no i dont know

BenjaLim

@Eric Your group is generated by two things $a,b$

subject to the relations'

$4a = 12b = 0$

Eric

ok

BenjaLim

2a + 2b = 0

Eric

not ok

How did you get that?

BenjaLim

when you quotient out by $\langle 2,2 \rangle$

You are basically saying that $2a +2b = 0$

@Eric Think about this: Your group is just integer linear combinations $ma + nb$ where $m,n$ run through the integers

subject to $4a = 12b = 0$

now when you quotient out by $\langle (2,2) \rangle$

you are saying now that the element $2a+2b = 0$

remember $(2,2)$ in the "basis" for your group is $2a + 2b$

Just like how we write points in $\Bbb{R}^2$ as $(a,b)$ which is also $ae_1 + be_2$ @Eric

@JSchlather Ah I think I can use spherical coordinates in $n$ - dimensions to finish the problem for me.

JSchlather

11:41

@BenjaLim What are you talking about?

Chris's sister and pals

12:32

Hi

Cortizol

i have one quick question, any hint :)
how to solve $y(y'+3)=ax^2+bx+c$, i don't have any idea
bad English, I know :(

Jonas Teuwen

Hi boys.

Chris's sister and pals

Don't you greet girls? :-)

skullpatrol

Hello ladies ;-)

Orange Harvester

What about the other people?

skullpatrol

12:41

Hi transgenders...?

Chris's sister and pals

lol :-)

Orange Harvester

Good.

user58512

hi

Chris's sister and pals

@user58512: hi

user58512

@Chris'ssisterandpals, what is your plan for learning math?

Chris's sister and pals

12:43

@OrangeHarvester: I know it's hard. For instance, if I go shopping (let's say) and I'm trying to solve some problems in my mind.

Orange Harvester

@Chris'ssisterandpals 18 hours? how long do you sleep? how long do you eat? other stuff?

user58512

wow

Chris's sister and pals

Yeah, it's not that easy. When I eat I usually read something related to math.

Cortizol

i don't believe you :)

Chris's sister and pals

Of course, there are some small breaks always.

@Cortizol: I know. It's hard to believe such a thing until you know me.

Cortizol

12:48

@Chris'ssisterandpals how old are you, if I may ask you :)

Chris's sister and pals

@Cortizol: never ask a girl about her age ;)

Orange Harvester

We are asking the ages of your pals.

Ethereal

@Cortizol Ask the bro when the time is right. ;-)

Orange Harvester

@Ethereal :D cool.

Chris's sister and pals

lol

Orange Harvester

12:51

@Chris'ssisterandpals How long have you been on this schedule?

Ethereal

@OrangeHarvester HAHAHAHAHAH!

Chris's sister and pals

Orange: for some months

skullpatrol

Do you use any textbooks?

Orange Harvester

@Ethereal ?? It was not meant to evoke laughter! Dude, you might be in there for this if you studied for JEE. I have been able to sustain something similar for a year and half. I am not sure if I could have done much longer.

Chris's sister and pals

@skullpatrol: MSE is a good place to learn things. The problems in my textbooks are pretty easy.

Orange Harvester

12:54

@Ethereal Now that you are here though, I will ask you to read the first paragraph of this article.

I love it.

Also, I hate Henriques.

Chris's sister and pals

Have you seen my last question here? math.stackexchange.com/questions/313388/…

user58512

I better decide what work to do now

Chris's sister and pals

And I'm planning to open a bounty for math.stackexchange.com/questions/311801/…

Orange Harvester

@Chris'ssisterandpals About this, I think Ishan's answer gives you the class of all functions that satisfy the inequality. Do you want something more from it?

Chris's sister and pals

@OrangeHarvester: there is a limit to compute ...

user19161

13:04

Just answered 2 lhf...

skullpatrol

@JacobBlack Link please.

user19161

@skullpatrol Well, I deleted one of them. Here. math.stackexchange.com/questions/313866/…

skullpatrol

Thanks.

user19161

L. F., Paris, France

1.3k 3 15

user19161

Guys, the red square has appeared!

user58512

13:15

1

A: Prove that $a^{2^n}=1 \mod 2^{n+2}$

$$a^{2^n}-1=(a-1)(a+1)(a^{2}+1)\cdots(a^{2^{n-1}}+1)$$ If $a$ is odd then each term in the factorisation is even (i.e. divisible by $2$), there are $n+1$ such terms hence $a^{2^n}-1$ is divisible by $2^{n+1}$. Since either $a-1$ or $a+1$ is a multiple of $4$, we have an additional factor of $2$,...

nice

that factorization

it says you can write numbers in binary (if divide by a-1)

in same way (x^2 + x + 1)*(x^6 + x^3 + 1)*(x^(2*3^2) + x^(3^2)+1)*(x^(2*3^3) + x^(3^3)+1) = 9x^(3^n)-1)/(x-1)

Orange Harvester

13:39

@Ethereal "Sehwag makes a spectacle of himself
After being diagnosed with vision problems and advised to wear spectacles at the crease, Virender Sehwag proceeded to get himself out in the first innings against Australia while in the act of searching for the ball (as it fell back onto his stumps in slow motion) like a blindfolded child waving a stick at what he thinks is a piñata but is in fact kindly old Aunt Esperanca visiting from Tijuana. Said Sehwag: "The great thing about wearing spectacles is that you can attribute each successive failure with the bat to a pending adjustment in lens p…

user58512

how do I get bibtex from arxiv?

Tobias Kildetoft

hmm, I don't think arxiv does that (one usually just cites arxiv papers as Title, arxix:number where the number is the identifier on arxiv

sorry, Author in front of that of course

user58512

thanks

Tobias Kildetoft

but I guess it might still be nice to have them in ones bibtex as arxiv until they get published

Orange Harvester

You can always get bibtex from google scholar

awllower

14:11

Hello

user58512

hi

Orange Harvester

Ah I see, google scholar does not index arxiv.

awllower

I wonder if this question requires only answers by induction.
http://math.stackexchange.com/questions/313893/divisibility-problem-using-induction

Tobias Kildetoft

@awllower do you mean as in the OP only wants arguments by induction?

awllower

Yes

Because my answer completely avoids induction...

Tobias Kildetoft

14:14

give that answer and see

awllower

Well, maybe a little, but the principle is to avoid induction.

Tobias Kildetoft

personally, I tend to prefer arguments not by induction, as those show more precisely why some statement is true

awllower

Oh, My answer is in the bottom.

Tobias Kildetoft

(of course, sometimes, induction is just so much easier)

awllower

Indeed.

Tobias Kildetoft

14:16

one of my favorite examples is the difference between showing that $q^n - 1$ is divisible by $(q-1)$ and that $\frac{q^n - 1}{q-1} - n$ is divisible by $(q-1)$

the first is easily done by writing up what the quotient is

one couls also do that for the second, but the quotient is really ugly (I don't even remember exactly what it is)

awllower

I agree.

In retrospect, I find my answer to be quite lengthy and annoyingto soe extent, as in most cases. Hope it is not too boring...

awllower

14:33

@TobiasKildetoft Do you have any idea, what a group of order $n^2$ while having no subgroup of order $n$ can look like?

Tobias Kildetoft

@awllower no, not really

I talked abit about this with m.k on IRC

awllower

A nice question to ask indeed.

Oh

Tobias Kildetoft

a first place to look is if the order is $(pq)^2$ for distinct primes $p$ and $q$

in this case, one can see that if there is no subgroup of order $pq$ then the multiplicative order of $p$ mod $q$ is $2$ (or the other way around)

so there are some limits to what primes might be possible at least

awllower

Yes

Tobias Kildetoft

I never got any further than that

awllower

14:36

It is a start at least!

Tobias Kildetoft

I guess one should take the smallest pair of such primes and check in GAP if there is a group of that order with no subgroup of order $pq$

BenjaLim

@anon I just showed that the unit ball is homeomorphic to $\Bbb{R}^n$ using the tangent function

Tobias Kildetoft

given such primes, there is a group of order $pq^2$ (or $p^2q$) with no subgroup of order $pq$, and the question is then whether this can be embedded in a group of order $(pq)^2$

awllower

I am npt familiar with GAP, though.

Tobias Kildetoft

@awllower it is a great tool for this sort of thing

Sage can do the same things, but I think usually by using GAP anyway

awllower

14:39

I guess so.I heard of it severaltimes before.

anon

@BenjaLim my first thought is $x\mapsto \frac{x}{1-\|x\|}$

BenjaLim

@anon I have $(x,y) \mapsto (\frac{x}{\sqrt{x^2 + y^2}} \tan (\sqrt{x^2 + y^2}\frac{\pi}{2}), )$

similarly for the $y$ coordinate

@anon that's continuous and it works

Tobias Kildetoft

@awllower and most of the ways to use GAP are fairly intuitive

awllower

Mh.
I guess I will take a look at it, when I have a chance.
As now I am very sleepy...

BenjaLim

@anon Hmmm I think at the origin we need to define it to be zero

anon

14:48

the tan one, yes

BenjaLim

@anon But do we still have continuity there?

@anon hmmm is $x/\sqrt{x^2 +y^2}$ cts ?

anon

yes because the thing in front of tan is bounded and tan0=0

yes it is continuous except at the origin

BenjaLim

at the origin if we define it to be zero

devWaleed

Hi

awllower

hi

BenjaLim

14:49

@anon I think we still should have continuity there

Ilya

Does anybody know, whether it is possible to change a title of the arXiv paper?

Tobias Kildetoft

@Ilya you mean you have a paper on arxiv, and you want to change the title?

devWaleed

Does a/b/c = ac/b or a/bc ?

Ilya

@Tobias aha

anon

@devWaleed depends on if you mean (a/b)/c or a/(b/c).

BenjaLim

14:50

@anon I'm not sure if we still have continuity at the origin even if we define it to be zero there.

devWaleed

@anon tell me both conditions please

Orange Harvester

@Ilya I think you can do so using the replace function.

Ilya

@OrangeHarvester perhaps, yes

anon

@BenjaLim your map is $\vec{x}\mapsto\frac{\tan \frac{\pi}{2}r}{r}\vec{x}$, and $\frac{\tan \frac{\pi}{2}r}{r}$ is continuous on $(-1,1)$ even at $r=0$

BenjaLim

@anon ah that's right I remember that from calc

@anon Man it's been a while since I dealt with all these trig functions

anon

14:53

@devWaleed (a/b)/c=a/(bc) and a/(b/c)=ac/b.

BenjaLim

@anon Thanks anon.

devWaleed

@anon Lets say we have a/b and b=b/c so after substituting, a/(b/c) is it correct?

@anon further it becomes ac/b ?

anon

@devWaleed try saying this again, but making sense

:)

devWaleed

@anon :|

Tobias Kildetoft

@devWaleed be careful not to reuse letters already used once (as variables)

BenjaLim

14:56

@anon Thanks.

anon

I suppose you mean a/x and x=b/c, so that a/x=a/(b/c)=ac/b, yes

BenjaLim

@anon Nite.

@TobiasKildetoft Nite.

anon

morning here

but night

devWaleed

@anon ok :)

Tobias Kildetoft

it's morning here too

(well, late morning I guess)

devWaleed

14:57

but If there's x/c and x=a/b then x/c = a/b/c = ac/b ?

anon

if x=a/b then x/c=(a/b)/c=a/(bc), not ac/b

devWaleed

ok, It means (a/b)/c/1 , a/b x 1/c = a/bc

anon

you mean (a/b)/c=(a/b)/(c/1)=(a/b)*(1/c)=a/(bc), yes

writing things without parentheses is not recommended by any doctors I know of

devWaleed

no, I mean (a/b)/c has 1 as a denominator for c?

so, a/b x 1/c

anon

c might not be in the form of a fraction so might not have a denominator technically, but if you want to imagine it has one using c=c/1 for the purposes of helping you, then sure

devWaleed

15:03

:/ a/b/c has double denominators and should be ac/b where is If 1 is taken denominator for c then it becomes a/bc, we can't neglect 1, right?

Tobias Kildetoft

@devWaleed since / is not associative, one should never write a/b/c

peoplepower

We can't neglect parentheses.

anon

a/b/c is not a mathematical expression, it is a mental weapon against mathematicians

user58512

hello

do you believe in axiom of choice?

devWaleed

what do you mean?

(a/b)/c != a/b/c ?

user58512

15:05

what will we do without choice?

peoplepower

You can choose / to be left-associative, but you will not necessarily find agreement.

devWaleed

>(

peoplepower

This might be because of the existence of simpler forms (ab)/c or a/(bc).

user58512

hello :(

devWaleed

@user58512 Hi

Tobias Kildetoft

15:08

@user58512 I use AC when I cannot avoid it, but I always prefer an argument that does not use it, if such an argument exists

user58512

I have some sugar but there are weird brown bits in it, I think they might be from instant hot chocolate - how can I test this?

Tobias Kildetoft

try tasting it

user58512

I wanna find out if I can use this sugar or not, it was left behind from the previous room mate

Orange Harvester

@user58512 send it to forensic laboratory.

user58512

I think it's agood not to waste it, but it might be bad..

Orange Harvester

15:09

@user58512 how much is it?

devWaleed

Ok, I was solving an equation using Binomial Expension...

so, at one step I get:

:/

12 - x /6 / 4 + 3x / 2

user58512

it doesnt taste of anything

peoplepower

Do you want $\frac{12-x/6}{4+3x/2}$?

devWaleed

yes :)

peoplepower

Assuming you want to simplify, multiply by lcm(2,6)=6 on top and bottom to cancel those inner fractions.

Orange Harvester

15:33

HAHAHAHAHA

Joey Morani

Hi. Anyone know if I've worked out this algebra correctly? I need to make k the subject in this equation: 4t = 3k/7t - 5k

I worked it out to be k = 11t/-2

Please don't tell me the answer if it's wrong :P

Arkamis

just factor out k

4t = k*((3/7)*t - 5)

Joey Morani

oh right

Charlie

16:01

Hello

Joey Morani

Hm, I don't get it. What I did was remove the division by multiplying it with 4t. So it becomes 11t = 3k - 5k

Is that right?

Ethereal

Whut

Arkamis

That doesn't make any sense, @JoeyMorani

First, you need to edit your question by putting parenthesis. Second, you have k as a common factor of both right-hand side terms

so you just factor out k, then divide by what's left.

Joey Morani

Can't edit it now. Won't let me

Arkamis

Re-type it

Joey Morani

16:07

4t = (3k/7t) - 5k

Arkamis

sigh

put parenthesis after your division

in other words are you dividing by 7, or 7t?

Joey Morani

7t. It's 3k over 7t

Arkamis

Is it (3k)/(7t)?

Ok

Joey Morani

Yes

Arkamis

Then 4t = (3k)/(7t)-5k

4t = k*(3/(7t)-5)

k = 4t/(3/(7t)-5)

Alternatively

You can multiply both sides by 7t

4t*7t = (3k)*(7t)/(7t) - (5k)*(7t)

28t^2 = 3k-35kt

28t^2 = k*(3-35t)

k = 28t^2/(3-35t)

Joey Morani

16:12

Ohh, I see where I went wrong

I assumed 4t*7t would be 11t because they are both t. Didn't think I had to multiply them.

Got it. Thanks @Arkamis

Peter Tamaroff

@BenW. PEW PEW PEW

Ethereal

16:28

Guys, how will you show that $\pm 1, \pm i$ are the only Gaussian units?

user58512

a+ib is a unit requires a^2+b^2 = 1

Ethereal

Ah.

Charlie

@peter likes to shoot

Tobias Kildetoft

@user58512 well, technically, it would also work if it was equal to $-1$

anon

vacuously

user58512

16:30

@TobiasKildetoft, I'm going to try to solve my group theory problems

but im really bad at solving problems

Peter Tamaroff

@Charlie Hey.

Arkamis

This complex analysis homework is kicking m ass

Peter Tamaroff

I just signed up for my courses! Yay

Tobias Kildetoft

@user58512 feel free to ask here if you need help with any of them

Peter Tamaroff

@Arkamis What is it? Just curious, don't think I'll solve it.

Tobias Kildetoft

16:31

I always like to discuss group theory problems

user58512

@TobiasKildetoft, thanks! I probably will ask some things - also spend a lot of time digging around in my notes

Charlie

@PeterTamaroff hola, Pedro

user58512

the course really goes way too fast for me, but I am learning and it is good

Arkamis

@PeterTamaroff I've had a few problems

Um, let me send you a link to the text

filetosi.files.wordpress.com/2010/12/1441972870complex.pdf

Chapter 8, problem 9 is what I'm stuck on right now.

f analytic in the plane minus the non-positive real axis, let f = x^x on the positive real axis. Find f(i) and f(-i), and show that f(conj(z)) = conj(f(z)) for all z

I need to use Schwarz reflection there somehow

As far as I can tell, the material in the chapter has nothing to do with this problem.

This book is annoying like that.

Peter Tamaroff

@Arkamis I'll look at it.

Arkamis

16:41

i'm heading to office hours in a few hours anyhow

Peter Tamaroff

@Arkamis I'm clueless about complex analysis, but I'm pretty sure you should be looking at $x^x=e^{x\log }$ for starters.

user58512

math.stackexchange.com/questions/314001/qwertyuijhgfdsxcv lol

Tobias Kildetoft

looks like someone trying to obscure homework

Orange Harvester

What actually impresses me is the "replacement text"!!

This time the person has chosed spanish for the text I think, which google translate translates to
"There was once a country where there was nobody. Ah, yes ... There were some small and tiny inhabitants that hardly anyone saw. Not very well appreciated. It was a ... Pulguitas."

Arkamis

@PeterTamaroff Aye, that looks like the right approach

Peter Tamaroff

16:55

@OrangeHarvester HAHHAHA

@OrangeHarvester Pulguitas are Fleas!

(Tiny) Fleas

Orange Harvester

@PeterTamaroff HAHAHAHAHAHAHAHAHA.

user58512

what does this comment mean? math.stackexchange.com/questions/314001/…

and why is it upvoted..

what a jerk

Peter Tamaroff

@user58512 Well, questions here are hardly ever deleted.

Orange Harvester

@user58512 He is egging on the OP to confess! C'mon, one can understand that much sarcasm I suppose! Also, if there is a very genuine reason, then a question can be deleted. (Though there is none here, but I think it should still be okay to ask, something like "You have the right to remain silent etc etc".)

Peter Tamaroff

@user58512 Don't make it a comment war.

He's right.

You've been using the site for a little more than a month, so try to understand how it works @user58512 before calling someone a "jerk".

Tobias Kildetoft

16:59

I really wish they had not made it possible to edit posts in chat. It completely breaks the timeline

user58512

it's only possible to edit for a short time

Transcript for

$Mathematics$ Mathematics