That question makes no sense, @user3502615. But in the world of $\Bbb Z_n$, the equivalence class of $n$ equals the equivalence class of $0$.

how doesn't it make sense?

Because elements of $\Bbb Z_n$ are not integers.

good

They are equivalence classes of integers.

And there are $n$ equivalence classes.

hm

,,,

@0celo7 I don't think multiplying matrices wrong will get cause you to fail

in my old school such things wasn't deducted too many marks

@0celo7 That'd be impressive :P

so elements of $\mathbb{Z}_n$ are equivalence classes with the equivalence relation being congruency modulo $n$?

Right.

confusing, but it makes sense

@Danu I wouldn't make fun of that.

idk how im going to prove this now

@user3502615: Picture it as a clock with $n$ hours marked, but you can associate each hour to infinitely many (equivalent) integer hours.

@BalarkaSen I'm not---I just don't believe that 0celo7 did literally nothing right.

I don't literally believe it, either. I had a number of students over the years who would walk out from my exams muttering how they had totally bombed and messed everything up, and then would have the top score on the exam.

@Adeek oh, I did everything else wrong too

@TedShifrin is a professor?

I used to be ... for about 36+ years.

Wrong eigenvalues, wrong eigenvectors, I wrote down the wrong matrix, wrong probability, everything is wrong.

@Danu yeah very fucking impressive

wow, i want to be a professor when i grow up

@TedShifrin btw would you like to discuss something cool

So I proved yesterday that $Aut(Z_p \times Z_p ... \times Z_p) = Gl_n(Z_p)$

@0celo: I don't know how your professor will grade, and you probably don't either. I generally did not continue to take off points if the student did things correctly with their wrong matrix. But some people grade only on answers. I dunno.

@0celo7 don't worry it is gonna be ok I always stress that I got bad mark in the exam then I find out that I did really well.

@0celo7 One time in a cs exam I said I was gonna fail because I forgot to check an if condition in one of the coding excerises midterm, but then I got 95

haha

@TedShifrin I did everything wrong with the wrong matrix.

That's not so interesting to me, Karim, but OK. That's the same as why $GL_n(\Bbb R)$ is the group of invertible linear maps $\Bbb R^n\to\Bbb R^n$.

@0celo: No point continuing to stew over it. But I understand you're upset with yourself.

@0celo7 don't be like me though I used to like lock myself in a room for 2 days and don't contact people until I get over it.

it is a waste of time.

@TedShifrin Some people actually did that? That's crazy

@TedShifrin so the way I proved it is that we get $\psi \in Aut(Z_p \times Z_p ... \times Z_p)$ and we send it to the matrix where the coloumn of the matrix is achieved by acting $\psi(e_j)$ where $e_j = (0,0,...,1,0,...,0)$

Did what?

reminds me of a kid in my grade

@TedShifrin do*

who always goes ratchet when he gets anything below a 95

Grade only on final answers

@Adeek The point is (Z_p)^n is a vector space over Z_p, yes.

Oh yes, plenty of people do, @Danu.

and then we prove that this is well-defined because full rank is achieved @TedShifrin @BalarkaSen

I had my first official assignment to grade today

And group auts are linear auts under that vector space structure.

One person failed :(

Karim: You should go back and review linear algebra to understand why there's nothing new going on here.

yeah @TedShifrin yeah I understand

it is pretty cool seeing linear algebra over different fields though.

everyone talking about grading assignments makes me feel like a babby

Now, if you're going to count how many elements are in $GL_n(\Bbb Z_p)$ or $SL_n(\Bbb Z_p)$, you'll do something that's slightly different.

yeah

I counted how many elements in $GL_n(Z_p)$

pretty cool

OK, should be using the same ideas you were just discussing.

yeah

i have a stupid midterm to mark and two midterms to study for.

it sucks why isn't professor marking midterm.

professors just chill here they do nothing.

Is it a large lecture class, or just one single class?

what do you call a group of isomorphic students

an EQUIVALENCE CLASS

it is a large class @TedShifrin 160 students

and marking midterm is different than marking assignment

it takes time

Well, at most universities, the professor and the TAs all grade together, each one doing the same problem on all 160 different papers for fairness. But you cannot expect the professor to grade 160 exams by himself.

I don't know what isomorphic students are, though, @user3502615.

yeah he isn't grading anything though haha just leaving work for me.

Are you grading your own students only?

The professor should at the very least make up an answer key and give a detailed rubric on how one deducts points and assigns partial credit.

yeah he doesn't do that

he told me to pick a marking scheme by myself.

Are you doing the same problem on 160 papers?

@TedShifrin Lol :D

#thedream

it is 5 problems for 60 students and 5 similiar problem for the other 60.

Huh?

60+60=160?

60+60 is not 160

Maybe @Adeek should take the class ;)

80

yeah I should @0celo7

Wait. There are two discussion sections, each with 80?

I have been not sleeping well lately because a lot of work load

How many TAs are there?

@TedShifrin ok I got 2 a) right.

just me

Oh, geez. The professor absolutely should be helping.

I shouldn't have signed up to be TA there was an option there was an option to do help center

that would have been easier.

Do all the faculty do the same thing? No, teaching is a good experience.

Yeah @TedShifrin

But the graduate chairman who makes all the assignments needs to know that you have weeks where you have to do 15 hours of exam grading.

yeah

I mean I am proctoring TAing for two classes and marking midterms that is really a lot of work. Plus taking 3 classes.

Next semester I am just gonna do two and register in a research credit hour.

Well, no one said being a grad student is a trip to the spa.

yeah haha @TedShifrin

But they need to train grad students to become competent teachers. UGA was very good about this.

that is awesome

I mean I like my professors here they are very nice. (aside from the geometry prof). But, otherwise professors here are really awesome.

I think it's standard for grad students to take 3 graduate courses a term.

@Adeek Wow, first year grads here take 5

They don't take 5 simultaneously. That's nuts.

@0celo7 that is crazy

I don't believe it.

grad classes aren't walk in the park

@TedShifrin Algebra, Topology, Analysis, PDE, Numerical

Unless they're possibly doing remedial work and taking undergraduate courses because they don't know the material.

I mean here I am taking algebra and professor assigns like 20 questions per week.

so that alone takes a lot of time.

Algebra and Analysis are remedial

0celo, a Ph.D. student does not simultaneously take those courses at the same time.

But I'm in analysis so I see them

@TedShifrin They always have study sessions for at least 4 classes.

And I know some of them are in 5.

That should not be allowed. But I'm not going to debate here.

Bye, all.

wait @TedShifrin want to ask you something quick

@TedShifrin I want to ask you for advice. So, next semester I am planning to take representation theory of finite groups, commutative algebra, and I have the option of either taking representation theory of lie groups or take the first two classes and register in a research credit. So, I was thinking should I take the first two and I would like to learn algebraic geometry on the side ?

what do you think ?

@TedShifrin D'aw

Bye Ted

My professor sent me a bunch of books to read on algebraic geometry to read so I was thinking I shoul do that next semester?

and take 2 classes?

@ted I guess I should like take 3 and read also ? I mean if I just have a good schedule and time management then it is manageable right ?

Ok I got problem 3 completely right :D

So 2 mostly right, 3 correct, 1 completely wrong.

So I'm expecting a high F.

@shcolf Can you construct a bijection between $[0, 1]$ and $[0, 0.5] \cup [0.5, 1]$?

Or, as a slight extension, what about between $[0, 1]$ and $[0, 1] \cup [2,3]$? (A finite union as opposed to an infinite one)

for now I can only see the case when, [0,1] -> [0,0.5]. f(x) = \frac{x}{2} for [0,1] -> [2,3], maybe that would be f(x) = 2 + x. but I don't see how can I merge both answers

also I had an Idea to construct a bijection in such way. g(x) = 2n + x. where n is non negative integer, and I was hoping to find some surjective mapping between [0,1] -> N

does there exist a bijection $\phi: \mathbb{Z} \to \mathbb{Q}$?

@user3502615 Are you asking @shcolf or just anyone?

anyone :P

The answer is yes

@user3502615 math.stackexchange.com/q/7643/23353

(This is missing an endpoint, but it demonstrates the idea)

hm

then im not sure how to go about proving this

@user3502615 yes

use the usual bijection $\Bbb Z\to\Bbb N$, then $\Bbb N\to\Bbb Q$.

now I got it, let me give another try =))

Yes?

i have to prove that $\mathbb{Z} \ncong \mathbb{Q}$

They are set-isomorphic.

@user3502615 what form of isomorphism are you considering?

Are you looking at isomorphism in some other category?

group isomorphism

$(\mathbb{Z}, +) \ncong (\mathbb{Q}, +)$

They have different orders.

One is cyclic, the other isn't.

but if there exists a bijection from $\mathbb{Z}$ to $\mathbb{Q}$, then don't they have equivalent orders?

oh, only cyclic groups have orders

A set bijection.

A set bijection need not be a group homomorphism.

rip

Indeed, the one I was suggesting is not a homo.

yeah

the order of a cyclic group is the minimal natural number $k$ such that it's generator $g$ satisfies $g^k = e$

@user3502615 It should be clear that if a group is iso to a cyclic group, the first group is cyclic.

Check that to make sure.

Hey everyone, just a quick question, between Lang, Artin and Dummit and Foote, which is the best to tackle Abstract Algebra?

I, myself, don't know, but I suspect the first question anyone is going to ask you is if you have prior experience with algebra or not.

I'm currently going through Axler's Linear Algebra Done Right, and I figured one of these would be the next best book to follow on

I would really like to see an answer to this question. I had the same question. math.stackexchange.com/q/945174/371514

@user3502615 did you show the cyclic thing?

what cyclic thing?

oh, that a cyclic and non-cyclic group are never isomorphic?

@Semiclassical What? Come on, only in the thermodynamic limit

@user3502615 yes

easier to prove the converse/contrapositive

$G$ cyclic, $\phi:G\to H$ iso $\implies H$ cyclic.

@Perturbative Artin is great

Hi @Tobias

@BalarkaSen Hi

How's things?

@apnorton, could you please check this, I've got something like this, but not sure: f(x) = n*x + i, where x \in [\frac{i}{n}, \frac{i+1}{n}]

Can someone give me some hints for b and c please? Not sure how to use the definition of continuity.

Hello.

What kind of horrible notation is that @GridleyQuayle

@BalarkaSen Good. Not been doing much math recently, but a bit and I am starting to prepare for my visit to Aarhus where I will probably give two seminars

$\underline{x}$

@shcolf That looks good, but that will only work for a finite union, not an infinite one.

Cool.

@0celo7 You just have to deal with what the lecturer throws at you haha

ahh, I see. but how can we "generalize" it for an infinite case?

I'm assuming I need to try and show that grad f is 0, but I'm not sure how I can use continuity to help with that.

@GridleyQuayle You're gonna use (a).

it's a little misleading to say use (1)

You might also need a connectedness argument, do you know what that means?

Guys any recommendations for Set Theory please ?

(there might be a better way of doing it, but that's how I would do it)

@0celo7 Nope.

@0celo7 Tell me an integer from 1 to 5 (1, 5 included).

3, always.

Excellent.

Thanks

What for?

@GridleyQuayle You can show that $f$ is locally constant using (a).

For helping me decide what to do.

I made a list and chose the 3rd item.

which is?

Physics.

Guys.

@apnorton, now I'm really stuck, couldn't we just let n to approach infinity?

@shcolf I'm honestly trying to think of it myself--I haven't taken analysis yet, which is when this sort of thing would be covered in my college's curriculum

I believe we can let n go to infinity, but I'd prefer to have someone else confirm

Never Mind

@GridleyQuayle Do you understand it yet?

You know that $f$ is constant along each ray of the form $\Bbb R_{>0}x$, $x\in\Bbb R^n-\{0\}$, right?

That's what (a) is telling you.

You use continuity to show that "constant along each ray" implies "constant on $\Bbb R^n-\{0\}$".

Sigh... I've done all of my side-tasks... Time to get back to my own work :(

@0celo7 I understand the intuition I think. In crude terms f is constant along vectors parallel to a basis of R^n, and if its continuous then it can't 'jump' between rays and so has to be constant everywhere. I'm not sure what to use to prove it though. Do I do it through manipulating $ \left | f(\mathbf{x}) - f(\mathbf{y}) \right | <\varepsilon $

Yeah, the "jump" is the idea.

Do I do it through contradiction then?

I think that works.

But since continuity is local, you'll end up with local constancy.

So you need a connectedness argument to conclude it's constant everywhere.

is $1$ always a generator for $\mathbb{Z}_n$?

I've not come across connectedness yet. This is a vector analysis module, so far we have covered conservative vector fields, line integrals, flux, FTC and differentials

@GridleyQuayle Ok, I have a better approach.

Thanks

Let $h:\Bbb R_{>0}\to\Bbb R^n$ be defined by $h(t)=t\mathbf x$, for fixed $\mathbf x$.

Clearly this is differentiable.

that's how I did part a

Ok, well you can differentiate $f(h(t))$, right?

@user3502615 Yes

And that gives you 0 because $f(h(t))$ is constant by the result from (a)

@GridleyQuayle Actually you only need continuity at one point.

Yes, so gradient [f(tX)] dot X =0

What you need is $\lim_{\mathbf x\to 0}f(\mathbf x)=f(0)$.

Which is what continuity at $0$ means.

So you know that $f(t\mathbf x)=c$, right?

So take a sequence $t_n\to 0$, but with each $t_n>0$.

Then $t_n\mathbf x\to 0$, agreed?

Does the definition of a kernel of a group action coincide with the definition of the kernel of a homomorphism? If $\phi : G \to H$ is a homomorphism of groups the kernel is defined as $\operatorname{ker}\phi = \lbrace g \in G: \phi(g) = I_H\rbrace$ where $I_H$ is the identity in $H$. However if I have an action $\pi : G \times A \to A$ how can I define the kernel, since $A$ is a set?

@GridleyQuayle How about this. I'll write the proof so you don't need to hear me ramble.

Be right back.

Or maybe I'll ramble, I dunno

@GridleyQuayle Just note that continuity means $\lim_n f(t_n\mathbf x)=f(0)$.

But every term in $f(t_n\mathbf x)$ is constant

So each term equals $f(0)$.

This works for any $\mathbf x$, so $f(\mathbf x)=f(0)$ for any $\mathbf x$.

Okay, if f is constant we should find that for every sequence $y_n$, the limit of $f(y_n)$ is f(0).

Not quite.

@Ed_4434 an action on the set $A$ corresponds to a homomorphism from the group to $Sym(A)$

One sec.

Can anyone please explain what the $$\text{Axiom Of Extension}$$ Is ?

where $Sym(A)$ denotes the group of bijections from $A$ to itself

as long as $y_n$ has a limit?

@Ed_4434 And the kernel is the kernel of this homomorphism

@GridleyQuayle So you want to show that $f(x)=f(0)$, right?

I'm dispensing with vector notation.

yep

@apnorton, I've posted a question here if you're still interested :) thanks for your help. math.stackexchange.com/questions/1984843/…

So pick the sequence $(x,\frac{1}{2}x,\frac{1}{3}x,\dotsc)$.

What is the limit of this sequence?

For fixed x, 0

@TobiasKildetoft I realise this but are they identically the same or is there another notion of kernel for a group action? I'm asking because I'm confused at an exercise in Dummit and Foote which asks one to prove that the kernel of an action of $G$ on $A$ is the same as the kernel of the homomorphism from $G$ to $\operatorname{Sym}(A)$

Ok, so continuity means $\lim_n f(\frac{1}{n}x)=f(\lim_n \frac{1}{n}x)=f(0)$.

But $f(\frac{1}{n}x)$ is constant as we vary $n$ by (a).

@Ed_4434 The kernel is defined as a certain subset of $G$ (the set of those $g$ such that $g.x = x$ for all $x\in A$).

The limit of a constant sequence is that constant, so $f(\frac{1}{n}x)=f(0)$ for all $n$.

In particular, for $n=1$ we have $f(x)=f(0)$. Done.

Never Mind Again

@TobiasKildetoft Okay thanks for your help! :)

Okay, that makes sense.

@GridleyQuayle I suspect that (a) implies continuity on $\Bbb R^n-\{0\}$ already.

You'd need a bit of an argument to show that.

You mean that a is sufficient for f to be continuous?

On $\Bbb R^n-\{0\}$.

It tells you nothing about continuity at $\{0\}$.

G'Bye.

So for c) we can find a function cts every but 0?

right

it also has to have a gradient at 0

@0celo7 Can a function not be defined such that it is 1 when (x,y)>0 and 0 else?

What does >0 mean there?

so both x and y are bigger than 0

sure you can define that

you need to be careful that $\nabla f$ exists.

@0celo7 how old are you? if you dont mind me asking

19

oh youre not that old :P

why would I be old?

because i feel like everyone here is twenty thousand times older than me

@BalarkaSen is fresh off the press

Hello could someone help me with lim x->-inf of (ln(2^x + 1))/ln(3^x +1)

Did you try el Hospital @user379685 ?

@user379685 looks like l'hopitals rule could apply here

Im supposed to do it without

...then do it graphically?

I cant sketch this function by myself

then use a graphing calculator

Besides thats not a valid method

The limit is infinity

And how can i get to this anwser

are you in calculus?

I don't remember these kinds of problems being hard

Analysis

First month of the first year

College

College? You're American?

Polish

@0celo7

how did you get that answer?

oh wait

I graphed it. Don't tell me it's wrong :P

lim->-inf of 2^x + 1 = 1, right

whoa

this function is freaky

yeah

it's like squarish

I'm trying the "usual" tricks

like squeeze theorem or whatever

but if you drop the $+1$ anywhere the function goes crazy