Mathematics - 2013-01-28 (page 3 of 3)

@Theorem I did not want to write it as there exists x such that there exists y such that ... but to give a slightly different way of looking at it. Probably a nutcase then.

Theorem

@JasonBourne : Its sad that very often people just like to downvote for no reason and i specially like calling them nuts :P

user19161

@Theorem They might downvote for the above reason, but that shows they have not reached that mathematical maturity to see beyond the immediate implications. =)))

Theorem

@JasonBourne : Even then i don't see a reason to downvote , if they don't understand they are free to ask or argue why its wrong . simply down voting shows that they are actually nuts

Ed Gorcenski

@JasonBourne I downvoted just to give chat something to talk about.

(kidding, of course)

Tobias

@JasonBourne I would not downvote for this reason yet. But I think with a few more links to the question here in chat, I might have

Theorem

8:27 PM

Its not that i know lots of answers , but even when i know the answer i give up writing up the answer because it takes me hours and hours to write the answer , i don't know how to improve my latex typing speed .

user19161

Hey guys you know what now? My rep is not a multiple of 5. That makes me sad. I will need to downvote some answers now... However I have run out of votes for today, so I will do it tmr.

Ed Gorcenski

Practice.

Gustavo Bandeira

As was his style, Kepler could not resist sharing his unique inter-
pretation of this mathematical observation. He assigned genders to the
solids and used duality to indicate sexual compatibility. The cube and
the dodecahedron (both dominant primaries) were male and contained the
female octahedron and icosahedron (secondaries). The tetrahedron was a
hermaphrodite because it contained itself. The faces and vertices were the
sexual characteristics because that was where the solids met.

> THE TETRAHEDRON WAS A HERMAPHRODITE!

Ed Gorcenski

Always knew that those tetrahedrons had something funny going on.

Gustavo Bandeira

@EdGorcenski =D

Tobias

8:30 PM

@JasonBourne the reason being, even though the answer is correct, I don't really find it interesting enough that it warrants advertising on chat

user19161

@Tobias Ah, so you were the one who downvoted it? It is alright if so. =)

Tobias

No, as I said, it would have taken a few more links to the answer

hhh

How can you parametrize a screw?

Tobias

and even then, I might not, depending on my general mood

Orange Harvester

@hhh conical helix?

hhh

8:32 PM

@OrangeHarvester Something like here?

user19161

@Tobias Oh well, I guess I am a rep whore then. =)

Orange Harvester

@hhh you already have a solution I gues?

hhh

@OrangeHarvester Not yet, it is not a screw

user19161

Who is the one doing all the starring?

Orange Harvester

@hhh What do you mean when you say you are looking for a screw?

Old John

8:34 PM

@JasonBourne starring went crazy last night - maybe it is the same person (cannot be me, as it was mostly my posts starred last night)

Orange Harvester

@hhh Is the image provided by nathan a close approximation?

user19161

@OldJohn Ah, I think the person who did the starring might have done the downvoting and maybe also the flagging. =)

hhh

myword.info/images/sm_screw_1g.gif <--- this is a screw that I am trying to plot

Orange Harvester

@OldJohn Jack the Starrer.

hhh

(or I believe when I can do it, I can do my original problem -- I want to get conical screw)

Old John

8:35 PM

@OrangeHarvester darned starrers - should be banned

6

hhh

@OrangeHarvester It misses the elevation of the fridge

Orange Harvester

@hhh That is a pretty tough thing to parametrize. I am assuming you are not hoping to parametrize the head itself?

user19161

@tobias I usually don't advertise my posts in chat. But sometimes I feel sad that a post is still at 0 after very long, so I do that now and then. Also, all my answers are trivial, so you don't need to bother looking at them if you want something non-trivial. =)

hhh

@OrangeHarvester err I want a cone that has a groove/fridge -- i.e. a groove spins around the surface from top to bottom where ball can roll.

If you release a magnetic ball on the top of the cone, it will go along the groove -- this is what I am trying to parametrize

Orange Harvester

@hhh If you are trying to solve the case by hand, I am not sure why would you need a protrusion for analyzing that.

hhh

8:39 PM

If I am able to find a way to get Screw-shape, then I am pretty sure how to do other shapes

Orange Harvester

The ball follows a conical helix and that is the end of it.

user19161

@OldJohn Careful, they might flag you for that. =)

Orange Harvester

@JasonBourne I consider it the responsibility of mods to prevent flag abuse, not mine.

Tessa Danger Bamkin

hello

Orange Harvester

@TessaDangerBamkin Hi. Long time.

Tessa Danger Bamkin

8:44 PM

how do I know you?

user19161

@OrangeHarvester You know who Tessa is?

Orange Harvester

@TessaDangerBamkin I was previously Jayesh Badwaik.

Tessa Danger Bamkin

right i do not remember you

i apologise

anyway could i get some help

Orange Harvester

@hhh What you can do is
1. Project a function like $y=|a-x|$ on a conical helix. And the project that helix onto a cone.

@TessaDangerBamkin ask away.

Tessa Danger Bamkin

Prove the following: if |G| > 1 is not prime, then G has a proper subgroup

Tobias

8:47 PM

@TessaDangerBamkin what have you tried?

Tessa Danger Bamkin

I did a really good proof of if it is prime then it only has the identity and the group itself.

but Im not sure on this one

Tobias

right, this one is slightly more tricky

since it has size > 1 take a non-identity element

what can you say about the subgroup generated by that element?

Tessa Danger Bamkin

no idea. group theory is scaring me at this moment in time.

Ed Gorcenski

Can you use Cauchy's Theorem?

Tobias

well, what happens if this element generates a proper subgroupo?

@EdGorcenski Cauchy is way overkill for this

Tessa Danger Bamkin

8:49 PM

not heard of cauchys theorem as of yet. we were doing lagrange at the time.

Tobias

the elementary solution is instructive

user19161

It's scary huh, people starring and flagging our posts and not confessing. Could be a serial killer.

Ed Gorcenski

Lagrange's theorem is related.

Tessa Danger Bamkin

does it make a cyclic subgroup

Ed Gorcenski

But yes, you don't need Cauchy's theorem

Tobias

8:50 PM

@TessaDangerBamkin right. What can we say if this subgroup is proper (recall what it is we want to show)?

Tessa Danger Bamkin

im a girl and i do maths yaay

2

user19161

The order of the subgroup divides the order of the group.

Tessa Danger Bamkin

sorry. my husband just jumped on to bug me.

Tobias

I claim that if this subgroup is proper, we are done. Why is this?

Tessa Danger Bamkin

okay so if the subgroup is proepr then the order divides the order of the group

Tobias

8:51 PM

don't use Lagrange yet

Tessa Danger Bamkin

okay

Tobias

we wanted to show that the group had a proper (and non-trivial) subgroup

so if this subgroup we have found is proper then...?

Tessa Danger Bamkin

then it is not the identity or the group itself

Tobias

compare to what we wanted to show

user19161

Hey @tessa now I remember talking to you before, under another username...

Tessa Danger Bamkin

8:53 PM

are we all using different usernames?!

Jonas Teuwen

@robjohn Do you know something about distributions on hyperbolic measures? For instance $\partial_{x_j} \partial_{x_i} r^{-1}$ where $r$ is some hyperbolic thingie in $x_i$.

Tessa Danger Bamkin

right so I want to show that G has a proper subgroup and if I assign an element to be this subgroup then i have found it? Im really confused right now. I do heavily apologise, I feel it must be frustrating for you pros

Tobias

@TessaDangerBamkin well, if the subgroup we have found is proper, then we are done, so we can assume it is not and proceed from there

This reduces to showing the claim for cyclic groups

user19161

Whoa I find that there are more downvotes these few days compared to last year.

user19161

See how fast the questions get downvoted?

Tessa Danger Bamkin

8:56 PM

okay and what is this claim? yes i need babying

Tobias

the claim was that the group has a proper non-trivial subgroup

Tessa Danger Bamkin

tobias im getting confused slightly treat me like an idiot for a sec and explain it like you would to that person

maybe just write the proof you'd do and I'll try and understand because I don thk im going to be able to get it from my head right now

user19161

Yeah I am getting confused too, LOL

Ed Gorcenski

@TessaDangerBamkin The outline of the proof goes something like this:

Tessa Danger Bamkin

i hope thats not sarcasm Jason

user58512

9:02 PM

hi

Ed Gorcenski

Pick an element x in G that is not the identity. Make a subgroup out of it. Either that subgroup is proper, or it is not. If the subgroup is proper, then we're done! If it is not proper, then it must be the entire group. If it is the entire group, then the element has some certain given properties.

user19161

Hey @mariano I just realised you have cast so few votes, LOL

user19161

@EdGorcenski Well done.

Ed Gorcenski

Then, if x has those properties, show that the order of G has properties that contradict your hypothesis (in other words, if <x> is a non-proper subgroup, the order of x relates to the order of G in such a way that means that |G| must be prime)

Rachel

@JasonBourne Naah, we're both too lazy to get married ;)

Ed Gorcenski

9:04 PM

@TessaDangerBamkin Now, using that sketch, see how Tobias's hints lead you along that path :)

user19161

@Rachel Oh. OK, as long as you are happy! Marriage is just a formality.

Tessa Danger Bamkin

thanks ed

I shall try my best

Rachel

@JasonBourne Surprised you remembered me from that :)

Tobias

@EdGorcenski no, that is not quite right

user19161

@TessaDangerBamkin Do you get it now? If not I can elaborate.

Tobias

9:05 PM

there are groups not of prime order with some element generating the entire group

Tessa Danger Bamkin

elaborate away. the more help the better! :D

user19161

@Rachel I am Jasper by the way, and I have deleted my Eng account!

Rachel

@JasonBourne Oh I remember you then :)

anon

If x generates a group of composite order and d is any proper divisor of said order not equal to 1, then x^d generates a proper nontrivial subgroup.

Ed Gorcenski

Well, then ignore my parenthetical point ;)

user19161

9:06 PM

The great anon has spoken.

Ilya

Beyond the horizon of the place we lived when we were young

Jonas Teuwen

The ocean?

user19161

@ilya What are you saying?

Theorem

@TessaDangerBamkin : This can be helpful math.stackexchange.com/questions/226365/…

and in fact its the answer

Tessa Danger Bamkin

awesome

user19161

9:08 PM

Ah, that was a nice refresher for my group theory.

Ilya

In a world of magnets and miracles

Our thoughts strayed constantly and without boundary

Theorem

@JasonBourne : which / what ?

user19161

@ilya Did you just take the wrong drugs?

Tessa Danger Bamkin

dont worry I'll be back to bug you

this course is too abstract right now for my little brain

Ilya

@JasonBourne Baby, did you forget to take your meds?

user19161

9:09 PM

@Theorem Well, how a group of composite order has a proper nontrivial subgroup.

user19161

@Ilya Haha, I call you dear and you call me baby? Your wife will be suspicious!

Theorem

@TessaDangerBamkin :most mathematicians are made , they are not born so it's same for everyone

@JasonBourne : interesting na :D

user58512

got thisproblem from another site: No group can be written as the union of two proper subgroups

Ilya

@JasonBourne you don't change

good bye, you almost asked for that

user19161

@user58512 Not quite.

user19161

9:12 PM

@Ilya Bye! See you!

Theorem

good night Volks :)

user19161

@anon I now have a feeling that you look like the person in the picture.

user58512

whats the picture from anon?

Nimza

Hi

user58512

hi

Jonas Teuwen

9:19 PM

A girl, maths, and bad spelling. Does not sound so nice 8-(.

anon

@user58512 Suppose $G=H\cup K$, pick $x\in H\setminus K$ and $y\in K\setminus H$. So where is $xy$?

user58512

I don't know, it could be in either

anon

sure about that?

$xy\in H\iff y\in x^{-1}H=H$ and $xy\in K\iff x\in Ky^{-1}=K$

user19161

@anon That also shows the flaw in the statement as given.

anon

Eh?

user19161

9:23 PM

Oh wait, I misread the statement.

user19161

Ignore my nonsense.

anon

Anyway this shows in general that $(H\setminus K)(K\setminus H)\cap(H\cup K)=\varnothing$.

robjohn

@JonasTeuwen not really. Sorry.

Theorem

@user58512 : one of the easiest arguments can be using the order of the group .

user19161

@rob I have decide not to keep my rep at a multiple of 5 anymore, it's too hard with all the downvotes these days...

Tobias

9:25 PM

@Theorem how would you use the order for that?

robjohn

@JasonBourne you've been getting a lot of downvotes?

Ed Gorcenski

@JasonBourne Everytime I see your rep not a multiple of 5 I'll make sure to downvote you so that it is.

Consider it my civic responsibility.

user19161

@robjohn Not really, but if it is not a multiple of 5 I would need to find three answers to downvote which can be quite hard, LOL.

user19161

@EdGorcenski LOL

anon

Last time I asked in chat for a precise number of downvotes (to obtain 31415) everyone and their dog jumped on the bandwagon.

Ed Gorcenski

9:27 PM

@anon LOL

user19161

@anon What is special about that number?

Ed Gorcenski

It's exactly 10000 times pi.

anon

exactly?

Ed Gorcenski

(that feeling was everyone in the room twitching)

user19161

Truncated.

user19161

9:28 PM

Not even rounded. or it will be 6 and not 5.

anon

Indeed.

user19161

I wonder if I will ever know the identity of the great anon...

anon

Consult your nearest 8-ball.

user19161

Wow, interesting that crystal = glass = glasses = 8.

Theorem

9:44 PM

@Tobias : looks like i have to spend some more time thinking about it .

@Tobias : I am not sure if it works , but lets see

user19161

I am kind of sick of the answers that go as XXX pointed out or a YYY pointed out. It's just noise to me. Yes, I know about the acknowledging others part but that's really not necessary to me. We are not stealing people's research here, LOL

Jonas Teuwen

10:02 PM

Give credit where credit is due.

2

user19161

Even for 1+1=2?

user19161

So now my answer goes like this: As Jonas pointed out, 1+1=2, but he only commented so I am typing this as an answer.

user19161

I would just prefer to see an answer that says: 1+1=2. QED.

Jonas Teuwen

Precisely, that is what 'is due' is supposed to say.

It means 'someone in his right conscious mind should do the right thing'.

user19161

10:22 PM

Hey guys, I actually think that some downvotes could be accidental.

user19161

Many times, I wanted to click on upvote but because the page moved I clicked on the downvote instead.

user19161

This happens especially on math when the tex takes long to get converted.

user58512

neverendingbooks.org/monsters-and-moonshine-a-booklet

I wanna download this book anyiidea how?

Old John

10:50 PM

Hi @WillJagy

draks...

Howdy

user58512

hi

draks...

I was about to post the following, but it feels not enough for a full question.

So I'd like to post it here:

How does the concept of factoring natural numbers carry over to complex natural numbers?

with regard to arithmetic functions I mean...

user58512

@draks..., prime numbers equal to 1 mod 4 split and ones equal to 3 remain prime

draks...

Yes I know

user58512

10:54 PM

what are you asking?

draks...

I thought about ..wait ia second

David Wheeler

who is wolfgang muckenheim, and why does he tick ppl off so?

user19161

@draks... Do you mean the Gaussian integers?

draks...

something like this $ g(n) = \sum_{d|n}f(d).\; $
So $d$ could be a compex number then right?

David Wheeler

i would think it would have to be as $\Bbb N[i]$ is not closed under multiplication

user58512

10:56 PM

@draks..., have you tried defining a mobius function for it?

draks...

@user58512 stop reading my thoughts

how did you guess that?

user19161

Hey @math

math101

Hey @JasonBourne

user19161

@math101 So have you thought about the exercise I gave you?

Old John

@DavidWheeler probably because he refuses to listen to logical arguments and just bangs on and on about his own crazy theories?

math101

10:59 PM

@JasonBourne Not at all. I just came back from work :)

user19161

OK guys, now that I have cast exactly 200 downvotes on main and 20 downvotes on meta I declare that I won't cast any more downvotes on SE.

draks...

@DavidWheeler who or what is ppl?

user19161

@math101 Oh, alright.

math101

Ya gonna catch something to eat and head straight first into math

David Wheeler

@draks... ppl = abbreviation for "people"

draks...

11:00 PM

@JasonBourne should we star this to have some evidence when you ever do it again?

user19161

@draks... Haha, I always change my mind!

dwarandae

@DavidWheeler, Wolfgang Mueckenheim: math.stackexchange.com/users/29503/wolfgang-mueckenheim

David Wheeler

@OldJohn what kind of crazy theories?

user19161

Erm why is everyone talking about this guys? Who is he?

Old John

@DavidWheeler he has some very non-standard ideas about real numbers and countability - I think he refuses to accept Cantor's diagonal argument proving the reals are uncountable, for instance

I followed some of his "arguments" out of amusement - before they were all deleted

user19161

11:03 PM

@OldJohn Is he a genius or a crank?

draks...

@DavidWheeler how would this affect a complex variant of the möbius function?

user58512

why is "cantor was wrong" so common?

Old John

@JasonBourne I think he is a well-known crank - although he is a professor of something like physics at some university

user58512

I understand "godel was wrong".. but not that

user19161

@OldJohn Ah, I don't really trust someone with the title of "professor" even in math. I have met enough professors who talk crap in classes.

Old John

11:04 PM

I really don't know - I read Cantors's book on the founding of the theory of transfinite numbers when I was at school - and thought it was wonderful

David Wheeler

well i've read some very interesting articles about what cantor's argument might mean if we adopt a non-standard model of ZF

Old John

@JasonBourne me too

user19161

When I attend these prof's lectures, I wonder how they got their PhDs.

draks...

are there any prof around yet?

user58512

@DavidWheeler, what's that about?

user19161

11:05 PM

Sure, teaching is not the same as research, but some profs really don't care about teaching at all!

Orange Harvester

@math101 work?

draks...

@user58512 Hey com' on. we had a good start...

Old John

this is the guy

user58512

@draks..., hi

dwarandae

:O

user19161

11:06 PM

@OldJohn From my dreamland too.

draks...

@user58512 yeah hi

David Wheeler

from what little i've seen, he ruffles people up by being argumentative

sometimes to the point that peoples' replies seem to be more ad hominem than "on-topic"

draks...

@OldJohn as long as we don't get another M....o K..o, who cares?

Brian M. Scott

@OldJohn One of the more notable mathematical cranks. It’s appalling that he holds an academic position.

Old John

@BrianM.Scott Yes -I thought he was quite good for entertainment for a while - once you get past his argumentative style.

But I got bored of his stuff after a few days :)

user19161

11:10 PM

@DavidWheeler I never met him on main, but he does not really post that much either.

Old John

@draks... MK seems to have gone rather quiet recently, I think

user19161

Wolfgang Mueckenheim

1 7

Brian M. Scott

@OldJohn I’ve seen it all over: here, de.sci.mathematik, sci.math, at least one other web-based forum, and at least one paper somewhere. And yes, by now it’s pretty boring! (And I get a kick out of crackpots if they’re at all interesting.)

user19161

What was he suspended for?

draks...

@OldJohn maybe he's got what he was looking for...I hope!

David Wheeler

11:12 PM

i'm interested in "non-standard mathematics"...i'm not entirely prepared to embrace ZF set theory as true...just "convenient".

Old John

@BrianM.Scott I am glad he didn't discover this chatroom, then!

dwarandae

I wonder the same thing

user58512

@DavidWheeler, can you tell me about it?

robjohn

@JasonBourne He does not believe in infinity.

user19161

@robjohn There can be infinity when you believe.

draks...

11:15 PM

believing in infinity is the only way. You will never encounter it in real life...
despite in [human's stupidity](http://quoteinvestigator.com/2010/05/04/universe-einstein/)

Orange Harvester

@robjohn I have read about this. Are they the same things in relation to infinity? Or this

robjohn

@OrangeHarvester I think it is close to what he believes, if not exactly. It is hard to tell.

David Wheeler

@user58512 about what?

user58512

non-standard

Orange Harvester

@robjohn okay.

David Wheeler

11:18 PM

oh...well one can insist that all sets be finite, or countable, or hereditary, for example

draks...

@JasonBourne How to do these broad ads?

user58512

oh right

robjohn

@OrangeHarvester to the ultrafinite people, even big does not exist.

user58512

is that the cumulative heierarchy

David Wheeler

that's one kind of hereditary set list, yes

user19161

11:19 PM

@draks... Well, pasting the link of a user profile, question or answer, youtube or wikipedia autoexpands in chat if there is nothing else on the same line, but I do this only sparingly sometimes to exaggerate something as some people don't like too much expansion in this chat, so to avoid it just type something else on the line.

draks...

ok here it goes:
Is there a complex variant of Möbius' function?
http://math.stackexchange.com/q/289344/19341

Orange Harvester

@robjohn yes, I am familiar with finitism and such. I must confess even I have wondered some years ago as to how can be talk about numbers which may not even exist.

Jonas Teuwen

As of tomorrow there is 'Professor' on my office door... but only temporarily as the occupant is not occupying it 8-).

Orange Harvester

You moved back?

Jonas Teuwen

This is even a bigger one than my previous one.

With a sofa, (...)...

Orange Harvester

11:28 PM

Lucky profanity.

robjohn

@JonasTeuwen ooh... a coffee machine