Mathematics

user19161

9:11 AM

@RajeshD Where?

user19161

@JonasTeuwen Morning! Hope you slept well!

Jonas Teuwen

Sleep? Thanks.

Rajesh D

@Jasper : Lets not continue this thing.

user19161

@RajeshD OK. I guess there are many secrets on SE. I have mine too, shh...

Rajesh D

yeah

But I don't have any.

user19161

9:13 AM

Your secret lies in that comfy chair of yours @rajesh, mwahaha!

Rajesh D

lol

user19161

That chair makes you powerful, like Spiderman.

Rajesh D

yes

sometimes, but there are already many super heros in here

silence. lol

Jonas Teuwen

11:25 AM

wiki.helsinki.fi/display/HAWorkshop2012 Plug. Nice conference. Spread the news.

Clark Kent

@anon I tried to summarize it all and posted it as an answer to someone else's closely related question. Do I have it right now?

Frank Science

12:01 PM

$\varphi(n+1)/\varphi(n)$ is dense on $\Bbb R^+$?

Mats Granvik

1:53 PM

Frank Science

What?

Mats Granvik

I used cosines in an attempt to imitate Lagarias formulation of the RH. The Harmonic numbers and sum divisors, that is. I did not know how to plot the expression with the Harmonic numbers so that the cosines would not cross it, so I cheated and shifted the expression with plus 1.

Just a plot however.

user19161

@MatsGranvik Are you hoping to solve RH? If you are using LaTeX to plot it you may ask on the TeX SE how to do the plot.

user19161

Hey @ben! School reopens next week huh?

Mats Granvik

2:27 PM

Frank Science

@JasperLoy Did you ask me on $\arg z$?

@JasperLoy Do you have any problem? I have headache now, therefore I do not want to think about tough problems.

user19161

@FrankScience No, I don't have a problem for you to solve. I am just asking if you take the argument in the clockwise or anticlockwise direction.

Frank Science

@JasperLoy aniticlockwise.

user19161

@FrankScience OK. I vaguely remember someone from your place saying that the convention there is to take it clockwise.

anon

@ClarkKent Couldn't an ascending chain in $\Bbb Z$ end as $\subset(pq)\subset(1)$? You can't say $n_k$ is prime necessarily, but your conclusion is right. Also, you mean $(1/n)\supset(1/m)\iff m\mid n$ (you've written $n\mid m$), though again your conclusion is right.

Frank Science

2:39 PM

@JasperLoy What's introduced in your country?

user19161

@FrankScience Anticlockwise as well.

user19161

If you want to laugh, please watch this video. youtube.com/watch?v=JMJXvsCLu6s

Frank Science

@JasperLoy Introduced here

@JasperLoy In my place?

user19161

@FrankScience I mean in some parts of China.

Frank Science

@JasperLoy Are they taught about $\arg$ in high school?

user19161

2:46 PM

@FrankScience It was actually a teacher I had who came from China twenty years before that time she mentioned it.

user19161

So maybe things have changed. Anyway it was just a casual remark she made.

Frank Science

@JasperLoy I cannot get it.

user19161

@FrankScience I mean I had a math teacher. She came from China when she was very young. She asked some students from China who came for exchange how the argument is done there. That is all.

Frank Science

@JasperLoy Therefore could you tell me how old the students from China are? I know that for the time being the $\arg$ was removed from most textbooks in China.

user19161

@FrankScience Well, I am not sure whether they did cover it or not in fact. It was just a question she asked because we were doing it in class that day. That was about 13 years ago.

Rajesh D

2:52 PM

$\arg$ removed! I wonder what problem the Communists had with it? any guess

Frank Science

@RajeshD It's unrelated to socialism or com...ism.

user19161

@RajeshD It's not a big deal really. No need to introduce complex numbers in high school.

Rajesh D

just kidding

we had hyperbolic functions too in high school for that matter

user19161

@RajeshD Same here.

Frank Science

@JasperLoy It's taught in that time, but I knew that it was taught as anti-clockwise. Maybe it was their pen slip.

user19161

2:56 PM

@FrankScience Well, it's just a convention that can be changed theoretically. But I guess the norm is to do it anticlockwise.

Rajesh D

and on a different note we used to make fun of the lecturer who used to teach group theory. We thought there was absolutely no need of that for anyone except him. Ofcourse we were immature both in psyche and mathematically

Frank Science

@JasperLoy It might only be their words. Did you see their figure to illustrate the $\arg$ notation?

user19161

@FrankScience No, it was just a sentence spoken.

Frank Science

@JasperLoy Well, it might be their slip.

user19161

@RajeshD Oh we did group theory in high school as well.

Rajesh D

3:00 PM

wasn't it funny

Frank Science

@JasperLoy Well, I know. I have a schoolmate in junior high school, who studies in Singapore for senior high. Now his knowledge about math is far more than mine.

user19161

@FrankScience I am inclined to think that your universities are better though.

Frank Science

@JasperLoy Why?

user19161

@FrankScience Because the ones here really suck IMHO. Of course, yours might be worse.

Frank Science

@JasperLoy But he might go to USA out of college education.

user19161

3:06 PM

@FrankScience Yeah, there is no place for math where I come from. Students just don't think. They just do endless drills all day.

Rajesh D

same case here too

Frank Science

@JasperLoy He might be familiar with abstract algebra now, which is all Greek to me.

user19161

@FrankScience Does not matter what topics are covered. I can tell you, math here sucks at all levels. Things look good on the surface but are crap beneath.

user19161

@FrankScience Indeed it is Greek. After all, they use $\alpha$ and $\beta$, lol.

Rajesh D

@JasperLoy Undiplomatic it sounds

user19161

3:13 PM

@RajeshD I am not afraid to voice my opinion in such matters. I am sick and tired of how silly education here is.

Rajesh D

@JasperLoy Singapore?

Its an advanced country, so I expect some better things there

Frank Science

@JasperLoy I cannot comment on Chinese education.

user19161

@RajeshD Yes. Like I told you, students only do endless drills all day. And teachers don't know their stuff very well.

user19161

@RajeshD It's only advanced in the sense that it has tall buildings...

Rajesh D

@JasperLoy So they make students do routine run in the mill things?

Frank Science

3:15 PM

@JasperLoy Do you know some Chinese students keep doing drill from 5:30 to 22:00?

user19161

@RajeshD Yes. And some of these teachers are quite arrogant. They know nothing but they think they know everything about math.

Rajesh D

@JasperLoy Clarification : What do you mean by drill ? literally?

BenjaLim

@JasperLoy What's kannappan talking about now?

user19161

@RajeshD Routine problems. Tens or hundreds of them. Students don't have to understand things to score well.

user19161

@BenjaLim Where?

BenjaLim

3:17 PM

Look at the starred comment

by BD @JasperLoy

Clark Kent

@anon But then $(pq) \subset (p)$ and $(pq) \subset (q)$. Then we can say that the maximal element is prime, no?

BenjaLim

@ClarkKent You ask a lot of CA questions!!!!!!!!!

@ClarkKent Are you studying CA at uni?

user19161

@BenjaLim I think I know what is going on. I am not the one who talks bad about him though, but I know who it is. I shan't mention it here. I tried to reconcile them though. Well, just let them settle it themselves.

Clark Kent

@BenjaLim Look, I only asked 12 questions during the last 3 days and most of them got upvoted so I don't think there is anything wrong with my questions.

BenjaLim

@ClarkKent Never once did I said there was anything wrong :D

Clark Kent

3:22 PM

I'll be back later. Just ping me anon.

user19161

@ben I used to have a problem with another user in another room, but I guess we have reconciled.

BenjaLim

@ClarkKent I am interested in CA too.

anon

@ClarkKent If you're trying to make the chain as big as possible, then (p) will be in there for some prime, otherwise you can't just say (p) will be in any ascending chain.

BenjaLim

@ClarkKent Are you self - studying CA?

user19161

@ben I did not keep track of the events between them so I cannot voice my opinion on the matter.

Clark Kent

3:29 PM

@anon Sorry, can you say that using different words? I do not understand.

@BenjaLim Yes.

user19161

@ClarkKent What book are you using, if any?

Clark Kent

Atiyah-Macdonald.

user19161

@ClarkKent Ah, a very popular text for CA.

Clark Kent

Yes, it's great.

anon

@ClarkKent You says that if $(n_1)\subset(n_2)\subset(n_2)\subset\cdots$ in $\Bbb Z$ then $n_k=p$ (a prime) for some $p$. But what about $(12)\subset(6)\subset(1)\subset(1)\subset\cdots$ or $(6)\subset(6)\subset(6)\subset\cdots$? Neither of these have $n_k=p$ for a prime $p$.

Clark Kent

3:34 PM

@anon Right. I ignored these cases because they are already stable : )

I guess that's not an ok thing to do.

So how to go about fixing the argument?

user19161

@ben With regard to the two people, I actually have no problems with either of them :-)

anon

in my opinion, I'd say the chain embeds into a finite poset (divisors of $n_1$ ordered by divisibility), and you can show the chain condition of finite posets via induction or whatever.

user19161

@JonasTeuwen I only know Mattila's name there.

Clark Kent

@anon Do I need induction if the poset is finite?

anon

The size of the poset is still indefinite. I suppose you could use something other than induction if desired, or just say it's obvious cuz it is.

Clark Kent

3:42 PM

: )

Wait, what?

The size is indefinite and finite?

What does indefinite mean?

anon

do you know the difference between indefinite versus definite integrals? means you can't say it's any specific value; the value is unknown and can be arbitrarily large.

Clark Kent

Ok. But if the poset is finite how can it be indefinite?

Jonas Teuwen

@anon Isn't it the same... Indefinite integral is like from $0$ to $x$.

anon

@JonasTeuwen What's $x$? Which real number?

Tell me which real number it is!

Jonas Teuwen

@anon Doesn't matter bro. Its all the same.

The difference is the uniqueness up to some additive constant.

anon

3:48 PM

In a definite integral you can tell me which real numbers the bounds are. In an indefinite integral, you can't, because one or both bounds are indefinite. They are unknown values.

Clark Kent

If the poset consists of $n$ elements, how can it also be indefinite in size? $n$ seems to be quite definite.

anon

@ClarkKent You say $n$ is definite? Then which natural number is it bounded by?

Is it necessarily less than 100? Necessarily less than 10^10^10?

Clark Kent

@anon It's bounded by $n+1$ and $n-1$, for example.

anon

But which naturals are those?

Clark Kent

Ok.

anon

3:50 PM

Recall the numerals for the naturals are 1,2,3,...

In other words: "it's a variable, and you can't assume anything about it's size"

Clark Kent

The way I see it is: we have a set of numbers, all divisors of $n$. Finitely many. Ordered by "divides". Certainly every chain in a finite set has a maximal and minimal element. Hence the claim follows.

My reasoning must be wrong of course, since you tell me that we need induction.

I guess I still don't get it : (

anon

well, you can use induction. but it's obvious so this is really pointless

Clark Kent

Ok. So it's good enough if I remove the claim that it end in a prime.

I'll do that.

Thanks mate!

user19161

4:08 PM

@anon The real number is $x$!