Mathematics

5:00 PM

I've heard a lot of people using 'read' in place of 'study' when they refer to the course they're doing at university, @BalarkaSen. Doesn't it seem really out of place. I can never bring myself to say 'I read xx at blahblah'.

Balarka Sen

mmhmm

"reading" is vastly different than "studying"

I have another DBZ-theory @Khallil. Interested?

Sami Ben Romdhane

I think that saying read instead study is a direct translation from the native language of the speaker. In arabic language for example we may say I read something and it's a synonym of I study something. @BalarkaSen

Khallil

Is that you in your picture, @Sami?

Not really, @Balarka.

Balarka Sen

That's a valid point @SamiBenRomdhane

@Khallil =(

Sami Ben Romdhane

Yes it's my picture :-) why? @Khallil

Khallil

5:06 PM

You're the man, right? I wanted to have a guess at where you're from, @Sami. Are you from Morocco or somewhere in the north of Africa?

Sorry, @Balarka. I'm doing a few things right now but I'd like to hear it later if you have time.

Balarka Sen

Eh, forget it @Khallil.

Sami Ben Romdhane

You're not far from in your guess, my country is the country of Hannibal and elissa;-) @Khallil

Khallil

Are you Lebanese, @Sami?

I just typed Elissa into Google and it came back with a Lebanese singer. =P

Sami Ben Romdhane

No this is her origin but she traveled to the north africa! @Khallil

Balarka Sen

@Alizter

Did you figure out that $\Bbb Q^\times$ problem?

Sami Ben Romdhane

5:11 PM

Do you know Carthage? @Khallil

Khallil

I just heard of it. My mum said it's a place with a lot of history in Tunisia, @Sami.

I know hardly anything about Africa. I've lived in the UK my whole life, but my whole family is Moroccan, @SamiBenRomdhane. That's why I asked where you're from. ^_^

Balarka Sen

That's kind of obvious from you're name @Khallil.

Khallil

It is, @Balarka?

Balarka Sen

mmhmm. Does it not sound a bit Arabian?

Sami Ben Romdhane

I thought that you have an arabic origin from you noun. Do you know its meaning. @Khallil

Khallil

5:19 PM

I think it means 'friend', @Sami. Somebody told me that my name means 'flute' in Hebrew.

Studentmath

@Khallil that would be me

Khallil

You're here!

Studentmath

And it means 'dearest friend' in Arabic

Balarka Sen

@Studentmath!

Khallil

I didn't see your avatar in the sidebar, @Studentmath. ^_^

Studentmath

5:21 PM

@Balarka !

MickLH

wow, so the word for flute in hebrew is "studentmath"

Studentmath

I just came in.

MickLH

and it's also the word for dearest friend?

Balarka Sen

hahhaha

Studentmath

@Mick indeed

Khallil

5:22 PM

Exactly that, @Mick.

MickLH

I can't play a studentmath

Balarka Sen

hello ~~pals~~ studentmaths

MickLH

hehe do I mean I can't play a flute, or do I mean I can't deceive my loved ones? (I can hijack the star board now)

3

Studentmath

I am working on some problem, and I have an issue with something. $\lim_{n\to \infty} (1-p)^n\to 0$, when $p\in (0,1]$.

MickLH

um

Studentmath

5:23 PM

But what about when $p\to 0$ as well? $n\in N$

Oh wait.

There you go.

MickLH

lol ok thank you

Studentmath

I would think it would go to $1$.

But I really can't say.

Khallil

Why?

Balarka Sen

It goes to $1$

$0 < 1 - p \leq 1$

Khallil

You're looking at a number in the interval $[0,1)$ being raised to a very high power.

MickLH

5:26 PM

I think it would go to 1 too, because actual 1 to any power would be you know...

Khallil

Hmm?

Studentmath

I think it won't go to 1. I think it's not determined.

Balarka Sen

Yes, @Khallil

That's what I meant

@Studentmath Well, prove your claim ;)

Khallil

That's the wrong inequality, @Balarka. My one is correct.

$p\neq 0 \implies 1-p \neq 1$

Balarka Sen

Ah I misread $(0, 1]$ as $[0, 1)$

Khallil

5:28 PM

Yep. So wouldn't it tend to $0$, @Balarka?

Balarka Sen

But if $p = 1$, $1 - p = 0$

in that case limit would go 0.

yes, @Khallil. typo.

Studentmath

Yeah. But if $p\to 0$.

Hrm, I will look at different lines of approaching it, shouldn't be too hard

Balarka Sen

what is $\lim_{n \to \infty} 1/x^n $ for $x > 1$, @Studentmath?

Khallil

(Or more generally, $|x|>1$.)

MickLH

$\lim_{x\to i \infty } \, (1-\Re(x))^{\Im(x)} = 1$

Balarka Sen

5:30 PM

blah overcomplications

MickLH

Well at least mathematica accepts that one

Khallil

Balarka Sen

hahaha

Khallil

You lose a turn trying to figure out why you're laughing, @Balarka!

80 damage inflicted!

Sami Ben Romdhane

Yes your name in Arabic means friend and precisely the intimate friend and the kh in Kallil is pronounced like ch in Banach in the German language so this letter does not exist in english:-)@Khallil

Khallil

5:35 PM

Yep, that's why so many people here fail to say my name properly, @Sami. ^_^

nerdy

Let (X,d) be a metric space and A a subset of X. Define x to be a closure point of A iff for every open ball centered at x , it contains at least one element from A.
Define x to be a limit point of A iff for every open ball centered at x , it contains at least one element from A differently from x.
Define x to be an isolated point iff x is in A and x is not a limit point.

My question, if i know x is a closure point, does it mean that it either is a limit point or an isolated point ? Could it be something else ?

Jose Antonio

Hi. I have a question in semidirect product, someone could help me, please?

MickLH

@BalarkaSen oh, let me get your hopes up ever-so-slightly again

I've stabilized in my new job mostly, and last night I got a couple hours to work on the program :)

Studentmath

Hah. 1 it is!

@Khalil sometimes they write it as 7 (the kh, in arabic, over the net)

Jose Antonio

5:55 PM

This says the following: Let $G$ a group and $N$ a normal subgroup of $G$. If $G$ it have a subgroup $H$ s.t. $H \cap N $ is the trivial subgroup and $H$ is isomorphic to $G/ N$ then $G$ is isomorphic to $N\rtimes H$ (semidirect product). I'm not completely sure here because if we let $g$ the composition of the canonical map from $G \to G/N$ with the isomorphism, is clear that its kernel is $N$ and permute the elements in H. But from here I'm not completely sure of how to proceed

Khallil

Ah, I see. There's a double l in my name, so I didn't get the ping, @Studentmath!

Jose Antonio

$G \to G/N \to H $ so if we let $g$ be the composition, then g it has N as kernel and permute the elements in H. So I have to show that all the elements in G are of the form NH, but is not clear to me that this really happens in this case? (If so, with that and the normality of N is sufficient)

Jose Antonio

6:09 PM

someone?

at least that I have to show that g is the identity over H and that's it. But I'm not sure how to do it

Alizter

@Khallil I literally read.

@BalarkaSen Which one was that the noetherian one?

Khallil

I know what you're talking about. @Balarka and I went off on a tangent to do with something else.

Balarka Sen

say huh, @Khallil?

Khallil

Say wha, @Balarka?

Balarka Sen

@Alizter No. It was about deriving a structure theorem for $\Bbb Q^\times$.

Prove that $$\Bbb Q^\times \cong \Bbb Z_2 \times \bigoplus_{\text{primes p}} \Bbb Z$$

math101

6:32 PM

Hi guys

When is the following statement true $x^{x!}=x!^{x}$ and $x \neq 0,1,2$

Daniel Fischer

@math101 Which primes divide $x^{x!}$, and which divide $(x!)^x$?

Balarka Sen

@math101 Look at $x^y = y^x$ in general.

Whoa @DanielFischer

You knew the answer ;) That could never have occurred to me.

math101

@DanielFischer what do you mean by that?

Daniel Fischer

6:50 PM

@math101 If there is any prime dividing one but not the other, the two cannot be equal.

Alizter

@BalarkaSen You havn't showed me that before

What does $\bigoplus$ mean?

nabla blah

@DanielFischer How did you know that if there's any prime dividing one but not the other, the two cannot be equal?

GBeau

Does the division algorithm for $2/1$ proceed as $2=2\cdot 1+0$ or $2=1\cdot 1+1$

Balarka Sen

@nablablah Fundamental theorem of arithmetic.

nabla blah

Oh

Daniel Fischer

6:54 PM

@nablablah That's the unique prime factorisation of integers. Every nonzero integer has a unique representation as the product of a unit ($\pm 1$ for $\mathbb{Z}$) and finitely many primes [unique up to order of the factors].

nabla blah

I don't know about that theorem

Balarka Sen

Every number can be decomposed into primes uniquely, @nablablah

Alizter

@BalarkaSen what does $\bigoplus$ mean

Daniel Fischer

Direct sum, @Alizter.

Balarka Sen

@Alizter Google direct sum.

Why... do I always get beat by @DanielFischer?

nabla blah

6:56 PM

lol

Daniel Fischer

@BalarkaSen You're typing too slowly today.

nabla blah

What are your wpm typing speeds

Balarka Sen

wait typing speed can be measured?

nabla blah

@BalarkaSen Try some typing tests on the internet

math101

@BalarkaSen play.typeracer.com

nabla blah

6:58 PM

Let's make a MSE room on there

Balarka Sen

whoa, @math101

nabla blah

Congrats, you just typed 122 wpm! We have to ask everyone who gets over 100 wpm in a race to take a short typing test. This is done to discourage cheaters.

lol

Chris's sis

@BalarkaSen On this site you might be suprised to find out some know things you haven't ever thought of. (for instance, some see what I was typing - before you see this message)

math101

@nablablah lmao Im like 40 WPM

Chris's sis

The same things happen if you use the latex editor of the site. They see everything you write there, no one tells you that. (I realized that due to a mistake of someone)

Balarka Sen

7:01 PM

race booked. will come back and test my speed. ;)

math101

@DanielFischer So is this just by trial and error. How would you go about this?

Chris's sis

But don't ask me about those mistakes, I won't tell anything.

Daniel Fischer

@math101 Not trial and error. Generally, which primes divide $n^k$?

Alizter

71 wpm

on this clunky keyboard

Chris's sis

And I wanna add one more thing: I also know things some would have never ever thought of.

Now, this subject is closed to me.

Let me post a nice integral ... (hmmm)

Alizter

7:07 PM

@Chris'ssis but first let me see your wpm

Balarka Sen

@Chris'ssis Oh?

Chris's sis

@BalarkaSen Please do not mention this anymore.

Balarka Sen

I don't think it's even true.

Alizter

I didnt see it damn

Balarka Sen

Why would anyone do that?

what makes you so sure?

Alizter

7:10 PM

was it an integral?

Balarka Sen

I don't believe you have details.

nabla blah

lol

Balarka Sen

I mean, it's such a far-fetched idea.

Alizter

@Chris'ssis what is it Ahhhh

anon

interesting.

Balarka Sen

7:14 PM

@anon?

darn it's 31 wpm

Alizter

@BalarkaSen :P

Daniel Fischer

@BalarkaSen You should have typed shorter words, Mr. Sesquipedalian.

Balarka Sen

throws tables at everyone

anon

catches table

Daniel Fischer

Logarithm tables?

math101

7:16 PM

@BalarkaSen Why do i feel like im in a room with a bunch of highschoolers

Ice Boy

sets table for lunch

Balarka Sen

@math101 not sure. have been baby-sitting lately?

Alizter

@Chris'ssis don't post seeecreeets because I am now dying to know :P

Chris's sis

@Alizter It was nothing important, and not about math.

nerdy

nabla blah, that theorem is really intuitive

Balarka Sen

7:19 PM

@anon throws a bench

nerdy

the decomposition of natural numbers HAD to be into numbers whose divisors are only 1 and itself ... it couldn't be otherwise

unique decomposition*

Daniel Fischer

@nerdy Why couldn't it be otherwise?

Balarka Sen

i like non-UFD rings

nerdy

it wouldn't be unique

anon

it fails in other number rings, and there's no eye-catchingly obvious reason why

Chris's sis

7:20 PM

@Alizter A very nice integral here $$\int_0^1 \frac{\log(1-x) \log^2(1+x)}{1+x} \ dx$$

anon

it's true that factorizations into primes are obviously unique, but it's not necessarily obvious when they exist or if factorizations into irreducibles are unique or when the primes are irreducibles are the same (although we know PIDness is a good culprit to pin it on)

Balarka Sen

@nerdy it fails in many other number fields. is there any particular reason you see it to be true on $\Bbb Z$?

nerdy

yes

Balarka Sen

what is it? why do you think you can't generalize that reason to $\Bbb Z[\sqrt{-5}]$?

Chris's sis

and then $$\int_0^1 \frac{\log(1-x) \log(x) \log(1+x)}{1+x} \ dx$$

nabla blah

7:22 PM

Hi @JasperLoy