Mathematics - 2018-08-07 (page 1 of 2)

user441848

00:06

Geometry question: How to write this $u+v+b(u^2+v^2)=0$ in the form of standar circumference (x-a)^2+(y-b)^2=c^2?

uhoh

06:02

0

Q: Algorithmic methods or techniques to find conjunctions in high N Keplerian element ensembles?

Suppose I wanted to answer the question Will Starman/Roadster pass particularly close to any asteroids in the next few years? or try to predict satellite conjunctions around Earth (e.g. Celestrak's SOCRATES), and I had simple Keplerian orbital elements, not necessarily osculating element tables n...

0

Q: Algorithmic methods or techniques to find conjunctions in high N state vector ensembles?

Suppose I wanted to answer the question Will Starman/Roadster pass particularly close to any asteroids in the next few years? or try to predict satellite conjunctions around Earth (e.g. Celestrak's SOCRATES), and I had ephemerides, TLEs, or interpolatable tables of state vectors. I could propaga...

orbital-mechanics conjunction

It's possible someone here will know something about an answer to one of these.

Tug' Tegin

06:24

Hello everyone! I am stuck in proving by induction the following formula: $\sin(x)+\sin(x+h) +...+\sin(x+nh)=\frac{\sin(x+(nh)/2)sin((nh+h)/2){\sin(h/2)}$ where n is a natural number and h doesn't equal 2πn. I addition formulas of trig finctions and related but in vain. Any hints?

Tug' Tegin

07:12

(This is not a homework question 🙂)

Rudi_Birnbaum

@rschwieb I think its in human nature that the older generation perceives the younger as "declining in culture", say. If I am not mistaken there is even Babylonian cuneiform tables from like 4000 bc where olders complain about youngers loosing culture.

On the contrary there exists something like a "real cultural decline" as it is observed e.g. in meso- and south-American cultures in the Amazonian rain forest. There are "tribes" who seem to have been completely depleted from their former "civilizational" achievements. And I am talking about pre-Columbian times. When we see decrease in alphabetization, loosing trust in scientific achievements in our societies, I think its real cultural decline.

A question on group theory for a subgroup $H$ of $G$, do we have $\{aHa^{-1}|a\in G\}=\{aH|a\in G\}$?

@Tug'Tegin maybe looking at $e^{ix}+e^{i(x+h)}+...+e^{i(x+nh)}$ would help. Your expression is just the imaginary part of that.

@Tug'Tegin Alternatively you could maybe use induction and the "trigonometric addition" $sin(a+b)=...$?

Tug' Tegin

07:39

@Rudi_Birnbaum Thank you very much! I am going to try it.

Rudi_Birnbaum

@Tug'Tegin for the first you'll need also the geometric summation $1+a+a^2+...a^n=...$

Tug' Tegin

@Rudi_Birnbaum I think the 2nd option is easier for me because I haven't started working with imaginary numbers yet :))

Rudi_Birnbaum

@Tug'Tegin check also your equation, it might be there is some "+" missing on the r.h.s.

pointguard0

any thoughts on this question? math.stackexchange.com/questions/2868922/…

MatheinBoulomenos

@Rudi_Birnbaum that never happens unless $H=G$. Note that $1 \in aHa^{-1}$ for all $a$, but $1 \not \in aH$ if $a \not \in H$

Rudi_Birnbaum

07:52

@MatheinBoulomenos Oh yes! thanks!! So the conjugate subgroups do not partition the group, since they intersect!

@pointguard0 I have linked a paper, was it helpful?

@MatheinBoulomenos: While the cosets are of course no subgroups! I slowly get that all ... ;-)

@MatheinBoulomenos Do the cosets have any other property except being sets with a binary operation?

(I guess they are semigroups, but not sure if they are closed)

mercio

a coset doesn't have a binary operation

Rudi_Birnbaum

you mean binary operation implies "closedness"?

mercio

yeah

Rudi_Birnbaum

08:08

So multiplication/addition is not necessarily a binary operation?

Since otherwise I see no use of the slogan "a group is closed under multiplication/addition".

mercio

well maybe you can say it has a binary operation but then you need to tell into what set

Rudi_Birnbaum

ok. So the cosets in general are just sets (however of a fixed order).

MatheinBoulomenos

$aH$ is a right $H$-set (so it has a right action from $H$)

mercio

^

MatheinBoulomenos

other than that, I don't think cosets come naturally equipped with any structure

Rudi_Birnbaum

08:13

"so it has a right action from $H$" what does it mean?

MatheinBoulomenos

you have multiplication $aH \times H \to aH$ $(ah,h') \mapsto ahh'$ this satisfies the axioms for a [right] group action

Rudi_Birnbaum

But we have also $a\in G$, so it has a left action from $G$ as well?

MatheinBoulomenos

I don't understand that

$a$ is fixed

do you mean one single coset or the set of all cosets?

there's a left action of $G$ on $G/H=\{aH \mid a \in G\}$

Rudi_Birnbaum

$G\times a H \to G (g,ah)\mapsto gah$.

MatheinBoulomenos

if it was supposed to be a action on $aH$, then you would need a map $G \times aH \to aH$

Rudi_Birnbaum

08:19

OK, sure!!

Sorry for the stupid/very basic question, but since we're already there: Is it correct that the set of sets $G/H$ has the structure of a group iff $H$ is normal in $G$?

mercio

yes pretty much

MatheinBoulomenos

yes (at least if you want a "canonical" group structure, you can always put an unrelated one on every set)

Rudi_Birnbaum

OK!

pointguard0

@Rudi_Birnbaum I've answered your comment!

Alessandro Codenotti

@MatheinBoulomenos Did you just assume AC?

mercio

08:35

:O

MatheinBoulomenos

08:49

@AlessandroCodenotti I always do

I'm not a logician/constructivist

Alessandro Codenotti

Fair, the proof that AC is equivalent to "every set admits a group structure" is quite nice in my opinion

mercio

09:06

well to choose a group structure is at least as hard as choosing the neutral element, so choosing an element

Alessandro Codenotti

Sure, but the point of AC is choosing infinitely many elements

mercio

hmm oh right i wasn't looking at it right

MatheinBoulomenos

09:57

@AlessandroCodenotti I'm confused now

Alessandro Codenotti

By what?

MatheinBoulomenos

What is wrong about the following argument that shows that "every set admits a group structure" is weaker than AC?
- "every set admits a group structure" follows from upward Löwenheim-Skolem
- Löwenheim-Skolem follows from compactness
- compactness is equivalent to the boolean prime ideal theorem which is weaker than full choice

Alessandro Codenotti

I think we talked about this already once

MatheinBoulomenos

did we?

Alessandro Codenotti

Around here

MatheinBoulomenos

10:03

ah right

you're right, but I don't see why I'm wrong

we only need upward Löwenheim-Skolem (since there's a group of cardinality $\aleph_0$) not full Löwnheim-Skolem

if we follow the standard proof for Löwenheim-Skolem via compactness, we only need for every infinite cardinal $\kappa$ an injection $\aleph_0 \hookrightarrow \kappa$ and compactness

so we need countable choice and the ultrafilter lemma

is this equivalent to AC?

I know that this isn't right, but I don't understand why

maybe there's something wrong with using Löwenheim-Skolem?

the wikipedia article has a more complicated proof for AC=> group structure

Group structure and the axiom of choice

In mathematics a group is a set together with a binary operation on the set called multiplication that obeys the group axioms. The axiom of choice is an axiom of ZFC set theory which in one form states that every set can be wellordered. In ZF set theory, i.e. ZFC without the axiom of choice, the following statements are equivalent: For every nonempty set X there exists a binary operation • such that (X, •) is a group. The axiom of choice is true. == A group structure implies the axiom of choice == In this section it is assumed that every set X can be endowed with a group structure (X, •). Let...

Alessandro Codenotti

You can just do a direct sum of copies of $\Bbb Z$ for the AC => group structure direction

@MatheinBoulomenos What exactly does "for every cardinal $\kappa$" mean here without AC?

MatheinBoulomenos

10:20

okay for every infinite set is probably better

oh wait

I didn't look closely enough at the proof for upward Löwenheim-Skolem

you actually use the downward part

nevermind then

just using the compactness theorem, you can't guarantee that your theory won't be too large

Alessandro Codenotti

Oh, of course, compactness gives you a model of cardinality at least $\kappa$

But then you need the downward LS to get a model of cardinality exactly $\kappa$

rschwieb

11:07

@MatheinBoulomenos I can't believe I haven't seen this before now, but normal domains are obviously N-1 rings (en.wikipedia.org/wiki/Nagata_ring), right? I can't help but be paranoid when something that obvious isn't mentioned in wikipedia... There could, of course, have been a forgotten hypothesis or something.

MatheinBoulomenos

@rschwieb yes I agree. Perhaps it's not mentioned because it is too obvious

rschwieb

@MatheinBoulomenos Thanks for getting back to me on it...

MatheinBoulomenos

11:22

@rschwieb one thing I find really confusing with commutative algebra (and also algebraic geometry) conventions is when some authors only define things for Noetherian rings and others do it in general

BAYMAX

11:41

Hi chat!!

How should I compute $\frac{\partial{x_{0},y_{0}}}{ \partial{x_{1},y_{1}}}$? like how is this related to Jacobian determinant?

Rudi_Birnbaum

@BAYMAX Hi

BAYMAX

Yup

was just confused about the notation!, thanks@Rudi_Birnbaum

mercio

12:02

I don't think I have seen that notation before

but I like it

Joel Sjögren

12:48

Hello

Is the tex code above supposed to look like tex code or is there something wrong with my browser?

rschwieb

13:12

@MatheinBoulomenos Yes, that certainly makes my job harder, too.

user2236

14:28

Starts at 4:30

Caucher Birkar's replacement Fields Medal ceremony

►

rschwieb

Replacement fields medal stolen seconds after being received

user2236

Mats Granvik

1

Q: Are the elements in the n-th row of the first matrix a permutation of the elements in the n-th row of the second matrix?

From my previous questions here and here the following two matrices arise for twin primes and cousin primes from Dirichlet convolution: For $h=2$ twin primes: $$T_2(n,m)=\sum\limits_{\substack{k=1 \\ k|n}}^n a(\gcd (k,m+2)) a\left(\gcd \left(\frac{n}{k},m\right)\right)\ \ \ \ \ \ \ \ \ \ (1)$$ ...

nt.number-theory analytic-number-theory prime-numbers prime-constellations

rschwieb

The only way it could be better if it was stolen in the video at the end.

user2236

he comes from such a humble beginning

Rudi_Birnbaum

14:44

@mercio did you hear rock'n roll before?

@BAYMAX welcome, was my first thought, but then in your particular form it didn't appear on the wiki, so I thought I'd better remove it before I have to explain a lot ..

@JoelSjögren you know about the mathjax parser (to be linked to browser)?

user131753

@MikeMiller: Did you get my invite?

Joel Sjögren

15:05

@Rudi_Birnbaum I use mathjax in my blog.

Rudi_Birnbaum

@JoelSjögren there is a parser which works from some browsers by clicking it. You need the link and a good browser (firefox e.g.) then it should work. I forgot all other details.

Tobias Kildetoft

Finally finishes reviewing these two papers. Now to type up the reviews.

And I need to be really careful not to mix them up, and the author of one is the editor handling the other.

Rudi_Birnbaum

@TobiasKildetoft lol!

I just got also onther one to review ...

@JoelSjögren here I think is the info I had meta.stackexchange.com/questions/97938/…

@JoelSjögren there was something else as well @TobiasKildetoft you have sent me some link about that, which was it, have you got it at hand?

Tobias Kildetoft

15:24

@Rudi_Birnbaum It is part of the room title

Rudi_Birnbaum

@JoelSjögren tinyurl.com/cfqcvpc

@TobiasKildetoft I try to understand why for some groups the sign representation doesn't make sense exactly. Last state was that for some groups there is no unique way to embedd them into $S_n$. Could we discuss that by using $A_4$ as an example?

Tobias Kildetoft

@Rudi_Birnbaum Sure

Perhaps we should start with what "the sign" of a group element should be abstractly

Rudi_Birnbaum

@TobiasKildetoft Yes.

@TobiasKildetoft Its $(-1)^{n}$ for n the number of pair permutations needed to do the group operation in a point bases.

roughly

?

Tobias Kildetoft

@Rudi_Birnbaum Now we got less abstract than I was planning

Rudi_Birnbaum

OK you go!

Tobias Kildetoft

15:37

In complete generality, it should be a homomorphism to the group with two elements

Rudi_Birnbaum

yes.

Oh, is that all?

Tobias Kildetoft

Now, this is the same as a subgroup of index $2$

(unless we want the trivial sign, which gives all elements the same sign)

Rudi_Birnbaum

You mean then we look at $G/$ that subgroup?

Tobias Kildetoft

right

the sign is the canonical map to the quotient, and it is uniquely determined by the subgroup

Rudi_Birnbaum

Ahh, so I only works (canonically) if there is a(=the) subgroup of order 2?

Tobias Kildetoft

15:41

right. If there is more than one such subgroup, we get different "sign"s, and if there are none, we only have the trivial and uninteresting one

Rudi_Birnbaum

:-)

OK then forget $A_4$ Then I want to know about what we call $T_d$ order=$24$.

Tobias Kildetoft

How is that defined?

Rudi_Birnbaum

Its the group of a non-chiral tetrahedron.

$A_4$ is the chiral tetrahedron.

$T_d$ has mirror planes.

at times called "full tetrahedral symmetry"

Tobias Kildetoft

Hmm, isn't that the full symmetric group on $4$ objects?

Rudi_Birnbaum

It has two-1, one-2, and two-3 dim. irreps.

(over $\Bbb C$).

Tobias Kildetoft

15:49

right, it is what I refer to as $S_4$

Note that you can permute the corners of the tetrahedron any way you want

Rudi_Birnbaum

Ah, OK (interesting way to look at it like that ...)! Whats the sign(s) here?

Tobias Kildetoft

Since there is just one subgroup of index $2$ ($A_4$), we just get the regular sign of a permutation

Rudi_Birnbaum

I'd call it $C_2$ :the rotation around 180°.

But wait, isn't $A_4$ what we call $T$ (order $12$)?

Tobias Kildetoft

certainly order $12$ (I don't call it $T$ :) )

Rudi_Birnbaum

Ah you talk about the quotient group now!

Tobias Kildetoft

15:54

Right, it has order $12$ and hence index $2$

Rudi_Birnbaum

$S_4/C_2=A_4$

Tobias Kildetoft

Other way around

$C_2$ is not a normal subgroup

Rudi_Birnbaum

Oh, OK right!

You mean "index $2$ in $S_4$."?

Tobias Kildetoft

right

Rudi_Birnbaum

Now we still didn't mention representations

Tobias Kildetoft

15:59

Right, so the connection between the sign and the sign representation is that the sign representation is usually defined to be a $1$-dimensional vector space on which group elements act via their sign (i.e. either as $1$ or $-1$).

Rudi_Birnbaum

yes.

oh wait

It must then be like the a single vector alinged perpendicular to some axis.

makes sense?

Tobias Kildetoft

not really, no

Rudi_Birnbaum

OK then lets forget it ...

Leaky Nun

is there no analytic number theorist among the regulars here?

Rudi_Birnbaum

But a one-dim vector space is for example $\Bbb R$, isn't it?

Tobias Kildetoft

16:05

sure, if we work over the reals

Rudi_Birnbaum

So isn't the only question then how to embed $\Bbb R$ in our space?

Tobias Kildetoft

@Rudi_Birnbaum Ahh, I see what you mean (I keep forgetting that you have the group given as some set of symmetries of space)

Note that since we want the subgroup $A_4$ to act trivially, we can start by looking for subspaces on which this is the case

Rudi_Birnbaum

Sounds complicated, but I'd say thats a (certain) tetrahedron with arrows as edges.

Now looking at my character table I'd say its one of the two three-dimensional ones. Its the one which changes sign with $S_4$ (which is your $S_2$ group operation I guess).

What do you think?

Tobias Kildetoft

For this group, the sign representation is the non-trivial $1$-dimensional one

Rudi_Birnbaum

16:21

Oh sorry I was stupid

It must be 1-dim

,right

?

Tobias Kildetoft

yeah

Rudi_Birnbaum

OK. That changes things for me ... Then still what is it for $A_4$?

Tobias Kildetoft

For $A_4$ the only $1$-dimensional irreps on which all elements act as either $1$ or $-1$ is the trivial one

(on the other ones, they can also act as third roots of unity).

Rudi_Birnbaum

Hm yeah.

OK, maybe now once again. I am after a representation that gives the rotation around the main axis (in 3-D space), and as I suspect its the same as a simple $C_2$ operation around that very axis.

I have seen from my tables its for all space-groups except tetrahedral and octahedral ones the same as the sign representation.

For those I was not sure what the sign rep. was there, now I know its not the same. because that rotation (real 3D space one) is in these groups (one of) the lowest non-one dimensional ones.

Well its actually always a three-dimensional one in the isometric groups ...

Tobias Kildetoft

Ahh, so for $S_4$ and the octahedral group there is a "distinguished" $3$-dimensional representation, called the reflection representation (or the fundamental representation for $S_4$)

Rudi_Birnbaum

16:34

Oh! Whats that exactly?

Tobias Kildetoft

It comes from the group being the Weyl group of a root system, which in both cases is a $3$-dimensional real vector spaces

ÍgjøgnumMeg

@Mathein ja kann sein, wenn ich mich für Heidelberg entscheide hahah

Semiclassical

it's sorta interesting to me that the smallest nonabelian group of odd order is of order 21

it feels larger than I'd have expected it to be

Rudi_Birnbaum

@TobiasKildetoft would that somehow have an analogue in the non-isometric groups?

Tobias Kildetoft

@Semiclassical Well, $7$ is the smallest prime $p$ such that $\varphi(p)$ is not a power of $2$.

Semiclassical

16:37

huh

Tobias Kildetoft

@Rudi_Birnbaum Not sure actually.

ÍgjøgnumMeg

@Semiclassical that clears that then

lol

Semiclassical

and Z3 is the smallest nontrivial odd order group

Rudi_Birnbaum

@TobiasKildetoft Maybe could we first check if its the same one in the 7 isometric groups as my Rot z?

Semiclassical

so might as well take a semidirect product to get a nonabelian group

Tobias Kildetoft

16:38

@Rudi_Birnbaum Unfortunately I need to go soon.

Rudi_Birnbaum

@Tobias Oh ok. But maybe we can continue that?

Tobias Kildetoft

sure

Rudi_Birnbaum

@TobiasKildetoft so thanks a lot for the moment that was good progress for me!!!

Shobhit

16:57

Helloguys

MatheinBoulomenos

@ÍgjøgnumMeg bist du dir unsicher? Du wolltest du doch unbedingt nach Deutschland

ÍgjøgnumMeg

@Mathein ja stimmt schon, aber Nottingham wäre viel einfacher hahaha

naja ich schau noch, hab bis zur Immatrikulationsfrist

(und ich muss trotzdem noch eine Sprachprüfung bestehen lol)#

MatheinBoulomenos

einfacher ist es sicher nicht finanziell

ja, du hast bestimmt große Probleme mit der Sprachprüfung, lol

ÍgjøgnumMeg

hahaha

Ja solange ich mindestens so.. ca. 600€ im monat verdienen kann solls schon gehen

MatheinBoulomenos

ich muss dich jetzt von Heidelberg überzeugen ^^

wir haben eine Bäckerei ~50m vom Mathegebäude, bei der es Fleischkäsebrötchen gibt

2

ÍgjøgnumMeg

17:08

Pff ich bin schon da

tschau england

hahah aber ja ich bin schon überzeugt, ich hab nur angst dass ich nicht genug geld habe

lol

MatheinBoulomenos

achso ja hmm

bist du Bafög-berechtigt als EU-Bürger?

glaub schon

Alessandro Codenotti

@ÍgjøgnumMeg Das werde kein Problem für dich sein (I don't remember how Futur I works)

ÍgjøgnumMeg

Das weiß ich nicht, ich glaub ich hab mal gelesen dass man als EU-bürger nicht automatisch Bafög-berechtigt ist

@Alessandro wird*

sonst perfekt

haha

MatheinBoulomenos

ah nein stimmt

man muss irgendwelche Voraussetzungen erfüllen

bafoeg-rechner.de/FAQ/bafoeg-fuer-auslaenderinnen.php

Alessandro Codenotti

Damn I knew it looked wonky

MatheinBoulomenos

17:12

aber 600€ im Monat ist machbar (auch wenn ich nicht so die Ahnung von Arbeiten außerhalb der Uni hab)

ÍgjøgnumMeg

Genau :(

Ja ich glaub teilzeit kann man so 600€ verdienen

kann dann nebenbei Nachhilfe geben oder so

MatheinBoulomenos

mit Nachhilfe kann man gut Geld verdienen

ich hatte mal eine Austauschschülerin aus Wien, die hat 15€ pro Stunde gezahlt

ÍgjøgnumMeg

ja schon nice, man muss halt schüler finden die nachhilfe brauchen

hahaha

MatheinBoulomenos

du bist Englischer Muttersprachler

damit findest du garantiert was

Alessandro Codenotti

Brauchst du hilfe mit Klassenkörpertheorie? Rufe jetzt an! Nur 15€ pro Stunde

3

ÍgjøgnumMeg

17:16

hahaha

@Mathein ja stimmt schon

hab vergessen dass ich ziemlich gut Englisch kann

MatheinBoulomenos

@AlessandroCodenotti das wäre der Traumjob

@ÍgjøgnumMeg also Englisch und Mathe ist bei Nachhilfe sowieso gefragt und du bist Muttersprachler in Englisch und hast schon einen Abschluss in Mathe

da wirst du überhaupt keine Probleme haben, was zu finden

ÍgjøgnumMeg

:) Hoffe ich, ist halt nicht so stabil

Rudi_Birnbaum

Servus Leute, ist heute German day?

ÍgjøgnumMeg

kann schon sein

Alessandro Codenotti

Es sieht so aus

MatheinBoulomenos

17:20

@ÍgjøgnumMeg meine Mutter hat sich ihr Studium auch komplett selbst finanziert

also das geht schon

Rudi_Birnbaum

Ich hätte mein Studium nie geschafft wenn ich nebenher arbeiten müssen, zumindest nicht in der Regelzeit.

ÍgjøgnumMeg

:) meine Mutter hat auch gesagt sie kann mir so 100€ im Monat geben oder so

was auch hilft

@Mathein wie viele Kurse nimmt man pro Semester?

Rudi_Birnbaum

Aber das sollte über Bafög/Eltern auch kein Thema sein.

ÍgjøgnumMeg

Bafög krieg ich nicht :(

Rudi_Birnbaum

Dann verdienen die Eltern so viel dass sie dich unterstützen können müssen.

MatheinBoulomenos

17:23

@Rudi_Birnbaum er kommt nicht aus Deutschland, das ist das Problem mit Bafög

ÍgjøgnumMeg

Nein ich kriegs nicht weil ich kein deutscher Staatsbürger bin

Rudi_Birnbaum

Ah OK. AT oder ?

MatheinBoulomenos

@ÍgjøgnumMeg 3-4

ÍgjøgnumMeg

@Rudi na Großbritannien

hahaha

MatheinBoulomenos

also 3-4 ist Standard und damit wirst du in Regelstudienzeit fertig

Rudi_Birnbaum

17:24

Uuups

MatheinBoulomenos

aber du kannst natürlich auch mehr machen wenn du willst

Alessandro Codenotti

@ÍgjøgnumMeg Genau so viel wie di angebotene Abstrakte Algebra Kurse (I strongly doubt this sentence works)

ÍgjøgnumMeg

@Mathein aso dann kann ich knapp 20std in der Woche arbeiten oder? lol

Rudi_Birnbaum

Ja das ist blöd.

MatheinBoulomenos

@ÍgjøgnumMeg ja

ÍgjøgnumMeg

17:26

@Alessandro Nicht so schlimm, man versteht schon was du damit meinst hahah

MatheinBoulomenos

und mehr als 20h darfst du als Student eh nicht arbeiten

ÍgjøgnumMeg

aso darf man nicht oder wie

Willst du mein Buddy sein? hahahaha

Shobhit

Hello

I used to think that $ Q[\sqrt{2}] = \{ a + b (\sqrt{2}) ;$ where a and b were rational$ \} $, but in my book its written that $Q[3\sqrt{2}] \ne \{a + b(3\sqrt{2}) ;$ where a and b are rational$ \}$, and they wrote that it is infact $ Q[3\sqrt{2}] = \{ a + b(3\sqrt{2}) + c(3\sqrt{2})^2 ;$ where a,b,c are rational$ \}$, but $c (3\sqrt2)^2=18$ which is rational and would join with $a$ so why have they done that.Please help.

Alessandro Codenotti

@ÍgjøgnumMeg How would you say? I'm trying to review some German before moving to Bonn, but I forgot a lot of what I knew

MatheinBoulomenos

@ÍgjøgnumMeg ja komm das machen wir

aber ich weiß nicht, kann man sich das beim Buddy-Programm aussuchen?

ÍgjøgnumMeg

17:28

@Mathein keine ahnung, da steht bloß dass man ein Buddy bekommt

MatheinBoulomenos

@AlessandroCodenotti don't worry, you know the word Klassenkörpertheorie, that's all that matters

@ÍgjøgnumMeg ich habe eine Mail bekommen, ob ich als Buddy für jemanden fungieren will

Alessandro Codenotti

Not every university in Germany is an algebraists's den like Heidelberg!

ÍgjøgnumMeg

@Alessandro I dunno how I would say that actually, "genau so viel wie's Algebra kurse gibt" oder so

@Mathein Was kriegst du eigentlich wenn du als Buddy arbeitest? Oder tut man's einfach freiwillig

MatheinBoulomenos

@ÍgjøgnumMeg freiwillig

ÍgjøgnumMeg

@Mathein gut dann kriegst du einen Leberkässemmel von mir

hahaha

MatheinBoulomenos

17:31

@ÍgjøgnumMeg okay das ist ein Deal

ÍgjøgnumMeg

@Shobhit I expect they mean $\Bbb Q(\sqrt[3]{2})$

Rudi_Birnbaum

@ÍgjøgnumMeg hast du schonmal nach Stipendien für GB-ler in DE geschaut?

ÍgjøgnumMeg

@Rudi ja ich und mein Betreuer haben vor ein paar monaten gesucht

MatheinBoulomenos

ich geh morgen zum Studierendensekretarität für Internationale Beziehungen und frage nach, wie das mit dem Buddyprogramm ist, wenn ich schon jemanden kenne

Rudi_Birnbaum

war nix dabei?

Shobhit

17:33

@ÍgjøgnumMeg no it is $3$ times $\sqrt2$

ÍgjøgnumMeg

o.O

Shobhit

$3\sqrt2$

ÍgjøgnumMeg

@Rudi glaub nicht, ich kann nochmal schauen aber die fristen sind wahrscheinlich schon längst vorbei hahaha

@Mathein cool :)

Rudi_Birnbaum

Naja es gibt immer alles mögliche, man sollte auch religiöses oder politisches nicht ausschliessen (auch wenns eklig ist, aber geld stinkt ja nicht).

und oft gibts auch keine fristen

Wenn ich in Heidelberg wär, würd ich dich als Hiwi anstellen, dann müsstest Du aber meine Gruppentheorie Probleme bearbeiten :P

ÍgjøgnumMeg

Daaamn wär nice

lol

Rudi_Birnbaum

17:37

(Ja für mich auch - wissenschaftlich zumindest ....)

Shobhit

@ÍgjøgnumMeg help

ÍgjøgnumMeg

@Shobhit what you said was correct, $\Bbb Q(3\sqrt{2}) \neq \lbrace a + b(3\sqrt{2}) + c(3\sqrt{2})^2 : a, b, c \in \Bbb Q\rbrace$

so I expect that there is a typo in the source you're using

because this looks a lot like the description for $\Bbb Q(\sqrt[3]{2})$

Shobhit

@ÍgjøgnumMeg so $Q[3\sqrt2] = \{ a + b(3\sqrt2) ; a,b \in Q \}$?

ÍgjøgnumMeg

@Shobhit and the reason you gave was good, because $(3\sqrt{2})^2 \in \Bbb Q$ so $\lbrace 1, 3\sqrt{2}, (3\sqrt{2})^2 \rbrace$ is not a $\Bbb Q$-linearly independent set

Alessandro Codenotti

@Shobhit Equivalently $\Bbb Q[3\sqrt{2}]=\Bbb Q[\sqrt{2}]$

ÍgjøgnumMeg

17:40

exactly

Shobhit

oh, thank you both :)

ÍgjøgnumMeg

@Shobhit because $3, b \in \Bbb Q$ also $3b \in \Bbb Q$ so $a + b(3\sqrt{2}) = a + b^\prime \sqrt{2}$ with $a, b^\prime \in \Bbb Q$

I mean

"$=$"

lol

Shobhit

o.O

Transcript for

$Mathematics$ Mathematics