@TedShifrin is your linear algebra book supposed to be really cheap?

I usually envision buying a maths book as requiring employment

oh yeah @Ramanujan what was that about btw

@AliCaglayan hi

Greetings

What the difference between the ratio and the interval of converge

When talking about power series

for one thing, one is a number while one is an interval

I think you mean radius of convergence, not ratio of convergence

@Maks $\frac{a_{n+1}}{a_n}$ would be a sensical ratio.

@arctictern Yeah, I figured it out eventually! Thanks though!

@Maks interchange a with f(n)/f(n+1) if you like

@Ramanujan greetings

What did the thing you deleted mean

Yes sorry, I meant radius

radius is a number

you know, like radius of a disc or ball?

except the balls in one dimension are just intervals

@AliCaglayan aren't you Muslim?

@Ramanujan last time I checked I appeared not to be

If you must know I am agnostic

Ok,so it was greetings in Islam(Assalamualaikum)

do you mean arabic? or some other language?

im pretty sure islam is not a language

السلام عليكم‎‎

Islam and arabic look so similar

?

I'm being nonsense

as'salaːmu ʕalajkum

en.m.wikipedia.org/wiki/As-salamu_alaykum

@AliCaglayan why?!?

$\forall a:a=a$

yeah, you meant arabic

@meow-mix so you are studying algebraic geometry, which parts

currently studying commutative algebra in prep for algebraic geometry

what at the moment

ideals

isn't that just algebra

like abstract algebra

prime ideals, maximal ideals

yeah thats abstract algebra

but on commutative rings?

They are covered in most abstract algebra courses

i guess so. the first chapter of A-M isn't really anything new

@meow-mix Are you reading any text in particular

A-M + dummit and foote

dummit and foote is one of my fav

yeah. lot of examples

I ordered it again it should be coming soon from india

as a not for sale outside of india copy

don't tell the book police please

dude, my dad's a book policeman

I can do that... is that what it means by "simple $\mathbb{C}$-algebra"?

don't mess with me or ill tell him and get him to arrest you and put you in jail.

anyways, why do you ask @Ali ?

So, if I have $ | x | < 1 $

I can say that the converge radius is 1

and the interval (-1,1) ??

yes

@meow-mix I saw you were a middleschooler which means you have more time than everybody here

So I can talk to you easily

@user2154420 are you working through a text?

@Maks is this pertaining to some sequence in $\mathbb{R}$ or $\mathbb{C}$?

@AliCaglayan what subfields are you interested in?

@user2154420 it's a standard fact the two-sided ideals of M_n(R) are of the form M_n(I) for two-sided ideals I of R. proof is probably on wikipedia.

@meow-mix $ \mathbb{R} $

@meow-mix at the moment algebraic topology

but I am interested in lie stuff

and some differential geometry

ive only studied some basic point-set topology

@meow-mix Yeah thats all you need to start algebraic topology really

You can get up to introductory homotopy

if you know about groups and stuff

my text for topology was Munkres', by recommendation of an [analyst] friend

@meow-mix your lucky to start early

because point-set topology can easily make you bored, the best text I have found so far is Bert Mendelsons Introduction to Topology

you can do many things as middle school a lot of free time.

Munkres is nice if you solve the questions in munkres then it is not boring.

Does upto connectedness and compactness

I am also biased because its the only topology text I own that is not an appendix

I would say it is important to do until tychonoff theorem.

especially that section on quotient space they pop up everywhere.

I mean in munkres the quotient space section is very important.

$\forall a\in\mathbb{C}\backslash(\mathbb{C}\backslash\{0\}):a=-a$

@Null stop enforcing your ideology, it means nothing

@ :(

Is my proof sound and rigorous?

Well youve swapped the intersection and union symbol

oops

used bigcup

meant bigcap

I would say not so much

but other than that it looks good

because you need to pick two points inside of your intersection.

I mean these kinds of exercises are more of a list ticking

i did

points $r$ and$s$

then showed that their product $rs$ must be in the intersection because by definition, $rs$ is an element of all the $S_i$ (since they're subrings)

oh ok I didn't see it.

yeh looks good to me.

ok :]

@meow-mix even to me it is understandable.

so congratz for that alone :d

@Null how many zero puns can I make in a day?

None

@AliCaglayan i just tried to make $a=a$ more "not so boring" ;)

@null what's your favorite ring?

How many puns can you not make in none of the days?

@meow-mix The ring of fire. babdabadada ;)

@BalarkaSen If I make a pun I contradict myself, if I say a number or puns then I am simply wrong

@BalarkaSen $\infty$?

"On Algebraic Structures Consisting of Fire" by Null O. Eye-dent-it-ee

So I will say I can make no none puns none of the days

I need some help with symbolic computation. $a_1=x_1$, $a_2=x_2$, $a_3=x_3$, $a_4=x_1+x_2$, $a_5=x_1+x_3$, $a_6=x_2+x_3$, $a_7=x_1+x_2+x_3$ and I want to know the sum of the three by three products $a_1a_2a_3+a_1a_2a_4+a_1a_2a_5\dotsb a_5a_6a_7$, is there a good way to automate this?

@Sophie Stick it into a CAS?

@meow-mix you latex or how you wrote your proof?

(i mean you could have just posted it in here..)

yeah i use texmaker

@AliCaglayan I tried putting it on wolfram alpha, but it just didn't give me anything back

i used to use gummi, one of those fancy live pdf updater things, but i felt it was just useless and a waste of computational power

I did it by hand but I think I made an error in computation

oh and i used to use MS word, that was a disaster :D

don't worry, those dark days are long gone

@meow-mix i was amazed by its wysywig factor, but you say it's trash for anything closely big?

@Sophie Where did it come from?

@null anything bigly?

> us election jokes

@meow-mix well, like 4 pages

> jokes write themselves

I found an elegant solution with elementary symmetric polynomials

@Null idk i find latex's typesetting to be amazing

So you have 7 roots?

also, its much easier than inputting symbols by hand

which i'm pretty sure you need to do in the case of ms word.

Why not type into WA, (x-a1)(x-a2)...(x-a7) then see what it simplifies it to

You only need to check one of the coefficients

@meow-mix ah i meant, specificly Gummi. You say its bad because its a waste of computational power. But on the other hand i find it nice, because you see quick if it goes dirty

@Null OH. i aplogize

apologize*

@meow-mix does it suck that bad on big documents or what?

but if it goes dirty, can't you just edit it after?

@Sophie for example wolframalpha.com/input/?i=prod+from+n%3D1+to+7+(x-a_n)

@Null yeah, past like 6 pages youll notice a considerable length of time before it updates

@meow-mix true that, i guess im just spoonfed by ms office lol. (or libre office, which in terms of math, is the same)

my teachers always yell at me because i forget to save as docx instead of odt

@AliCaglayan A modified version of this worked. Thanks

@meow-mix omg..

@Sophie yup, no problem.

@Null does anyone use that math program with libre office?

like, anyone?

it's called "libreoffice calc"

I used that once, but latex is much better

@meow-mix i used it only 10 minutes, then i got how you insert an equation there. after that byebye xd

Other than latex why would you write math on a computer

someone needs to invent a computer screen that you can use as a whiteboard

and no, i don't mean a projector onto a whiteboard

i mean an actual screen

you mean a touch screen stylus computer

@meow-mix scratchpads can do that, but no ocr of course

its just not the same

paper triumphs above all

if i were to lecture

i'd prefer whiteboard over blackboard

blackboard being, chalkboard of course

I prefer blackboard because it can be seen clearer

@meow-mix or a black whiteboard :O (puns)

Whiteboard pens tend to be wimpy in thickness

@Null would white marker even work on a black whiteboard?

I have seen a black whiteboard before

They used charcoal

haha

and then sponged it down after the lecture

well the thing is

chalkboard is easy to use

and easy to maintain

Whiteboard is good in close proximity

i want to be a professor as a...

wait for it...

but for a lecture chalk is better

profession

:D

> ba dum tss

LMAO

i need a magic 8 ball

you can have PhDs for the wierest things. like "Life management"

Hi I'm Robert Miller, Ph.D in Children's Board Games

I wrote my thesis "On Lands of Candy, and Chutes and Ladders" in 2009

@AliCaglayan are you in university right now?

@meow-mix yes

what year?

$\approx 10^0$ years i guess. (sooorry xd)

@meow-mix 1st

Wouldn't $0^0$ = 1

but i have been reading math for a lot longer

everyone says it's undefined, but the empty product is always 1...?

at least my 6th grade math teacher

said its undefined

it is undefined

@meow-mix holy cow, do you have still contact with your 6th grade teacher?

yeah, she's in her classroom everyday lol

ok so i don't feel like typing this up in latex

but as a matter of definition, $0^0=1$ has no adverse side effects that I am aware of

and i dont even know if this is right

but let me just type away

"Show that every boolean ring is commutative"

Suppose elements x,y in R (R is the boolean ring)

we wish to show that xy=yx

wait hm, i was right, im too tired for this shit

let me ponder it

oh wait

What is the definition of a boolean ring

multiply each side by x. I can do that with rings, right?

@meow-mix i dont know if this will help you, but proofwiki.org/wiki/Definition:Boolean_Ring

@Null no spoilers, im not gonna click that link :[

@AliCaglayan a ring such that $a^2 = a$ for $a \in R$.

@meow-mix it is only the definition of a boolean ring, no proof ;)

so we wish to show that xy=yx

then xxy = xyx

thus, xy = xyx

wait

where am i going wit hthis

ok ignore all that, im going to stop typing until i get a solution

You could show that xy-yx has the properties of 0

in rings, am i allowed to do the whole "operation to both sides" thing?

like "left multiplication to both sides"?

@meow-mix if the operation itself is fine then yes

Yes, you can prove this

if a = b and a + x =/= b + x then we have no ring

same for mult

so adding and mult to both sides is fine

You can extend this to subtraction

but unless you have no zero divisors, the same doesn't hold for division

im so bad at math

ok so

wow this problem sucks

I discovered I don't know how to pronounce manifold

man-if-old... right?

@meow-mix i would say: many-fold

oh this reminds me of something

@Sophie oh so its munni-fold i guess?

cool song hehe

Could someone please help me understand the answer here? chat.stackexchange.com/…

More specifically, I do not see how the first line is true.

@Sophie I'm mad I didn't know about this.

@Raptor your link is prolly not what you wanted to link

@Null sorry math.stackexchange.com/questions/76478/…

Merhaba!

How's everyone?

@ryagami ?

?

What do you mean by ''merhaba"

Google is your friend

it is hi in arabic

Man-I-fold, all

@AliCaglayan: I wish.

@Null: I see you have your new name. I've already forgotten who you are :)

@TedShifrin I solved many differential geometry questions

@TedShifrin do you have any interesting problems that you can think of at the moment?

I am happy so far I think I will do well in tomorrow's exam

@Adeek good luck

thank you @AliCaglayan

@TedShifrin ^^ i got a tough nut to crack

time to get the sledgehammer out then?

@Null what is the nut

Question about isomorphisms: If I'm trying to disprove that a set (under an operation) is not isomorphic to R^2 (direct product of R + R) and the set is not closed is this enough? The set R^2 is both open and closed, so I don't feel that this follows.

@Pythonista The image of the direct product isn't exactly 2 dimensional?

Do you mean the direct sum

I thought you meant dot product sorry my bad

@Pythonista However in this case we have $\Bbb R \oplus \Bbb R \cong \Bbb R^2$ so I fail to see thie question

^ And edit. R^2 is a group under addition since R is an abelian group under addition I was think about being topologically closed and open. I guess my real question is this: If we have a set (with some operation) and we want to show that it is not isomorphic to R^2 it's simply enough to state that the set isn't closed and hence cannot be isomorphic to R^2?

The set in consideration isn't closed *

closed wrt what?

Is it a subset of R^2?

What is the set/operation under consideration?

Under the operation of the set. No it's a set of matrices (3x3). With real entries along the diagonal and the middle entry being 1. The operation is addition.

Seems like that set is too big (compared with R^2) to be isomorphic to it

You've got a total of 2+2*6=14 real parameters (two real values along the diagonal, six complex values off the diagonal)

Have you done this one, @Ali? If the minimal polynomial of $T$ has distinct roots, prove $T$ is diagonalizable.

@Pythonista you have no zero in the group, i.e $\operatorname{Diag}(a, 1, b)$ has no $a,b$ to give $0$ hence no identity

@AliCaglayan Right which was my first thought. If there was some identity it would have to be with a=0, b=0 (supposing an isomorphism) since (0,0) is the identity in R^2 by properties of isomorphisms. Since the identity would be carried. But, the question doesn't explicitly state a group isomorphism. The question just asks about the sets. Although, a group is jsut a set under an operation. So, the wording is a bit confusing since we can talk about set isomorphisms etc.

What is a set isomorphism?

That's a pretty bad question, then.

We don't talk about set isomorphisms.

We call them bijections.

@TedShifrin I haven't done that one, I shall give it some thought. Thanks for the question.

No idea, read a question here on stack-exchange a few moments ago that said more or less the two sets being equal (bijective mapping) without operation preserving being considered as it would for a group isomorphism. Although, I've never heard of that before.

@Semiclassical Agreed, what's even more confusing is that the matrix set isn't even a group since it's not closed so...

I mean in the category of sets an isomorphism is a bijection

When I get home, I can send more, @Ali. That One's pretty nifty.

What's the precise statement of the question?

@Pythonista That would mean that you have to show a bijection between $\Diag(a, 1, b)$ and $(a, b)$ which is trivial. However I fail to see how the operation comes into this

Is H under addition isomorphic to R^2. And H is just the set of matrix defined earlier. H - {(3x3 matrix) | a,b in R} where the 3x3 matrix has real entries a,b across the upper and lower portion of the diagonal. The middle of the diagonal is 1.

@Semiclassical ^ That's the exact wording.

:/

There's a very interesting variant of that question which a bunch of us have discussed. Can two non-isomorphic vector bundles have diffeo(homeo)morphic total spaces?

one thing i'll note is that those matrices aren't assumed to be diagonal, i.e. the question says nothing about the off-diagonal entries

if it were, then those matrices would form a group under matrix multiplication

@AliCaglayan just to be sure, did you read my thing? just wont to know what kind of task this is, because at the moment im not thrilled to ride definitions xd

Well, my subject is abstract algebra so I had assumed showing group isomorphisms, but the wording is very vague. Add in that the set wouldn't even be a group under the operation... and yeah no idea what's being asked.

I think the intent is diagonal ...

Right, the previous portion was showing exactly that. That under multiplication the set of matrices was isomorphic to R*^2

right.

@Null sorry I forgot to read it. Nothing to do with your name or anything

LOL

i dunno. i'd talk with the person who posed the question if possible.

The name is my fault, too :D

because at the moment the simple response is: H isn't a group under addition, so there's no way it can be isomorphic to (R^2,+).

@AliCaglayan haha, baddumts ;)

@Semiclassical yes I agree. When you say set, it doesn't really make sense to talk about isomorphisms unless you defined them

Think that's what's being asked, but I'll definitely ask for clarity. Thanks for the help I was staring at that for awhile.

@Null is map(A, R) just the set of all maps from A to R

yes @AliCaglayan is there a better notation?

no, just checking there arent any further restrictions imposed on these maps

i mean, as sets, they'd both have the cardinality of R. so in that sense there's a bijection. but there's not one that can preserve an algebraic structure that doesn't exist.

@Null: What are $A$ and $R$?

@TedShifrin A is an arbitary nonempty set and R is a ring

And what was the question?

@Null you need to show that map(A, R) is an additive group, which is simple as it inherits addition from R. You need to show there is identity and mult assoc. Then finally distributive. These are all inherited properties from R so you should have a ring. I think the confusing part is that it is utterly useless

Nothing ingenious, nor anything clever.

Not useless, but normally something people would say is clear.

Using the structure on $Y$ To get the structure on maps $X\to Y$ shows up a lot.

i think i overthought it tbh

but is + the same + as in the ring R?

Point by point, yes. But you're defining addition of functions (just like in calculus).

@TedShifrin Is saying, if T is not diagonisable then its minimal polynomial does not have distinct roots sufficient ?

Noooooooo

its 5:30 am but I am confused to why that is wrong

A => B

¬B => ¬A

@AliCaglayan if i am chatting, i am logged on HERE

Oh, wait, you're using Jordan form :(

I wanted a more elementary argument. Damn.

Where is the Jordon form?

Oh, I thought you were going to deduce the contrapositive from Jordan form.

Well ... of course i was ;)

I'll do it directly

There's a beautiful direct argument. It should even work in infinite dimensions, I think.

Get a nap :)

Well

If there are repeated roots

then eigenvalues repeat

hence non-diag

@AliCaglayan mmh, what kind of argument you expected?

Preferably one where I am correct

haha

@AliCaglayan far From true.

diagonal matrices can have repeated diagonal entries bro

hmm ok

But we're talking minimal poly ... hi tern :)

@Ali; The hint is that you have to show the eigenspaces span.

Is this because the eigenspace will be of dimension n-k where k is the number of repeated roots?

I am gonna actually read the chapters I need for this so I can solve it properly without talking nonsense

Huh?

I am close to the chapters so should be qualified to answer question in about 2 days

Oh, I had no Idea what you know ..,

I have others for you

if you want em .,,

This is a relatively deep question.

Night :)

Lets see the others

I can write then down and think about them during this week

hi

math.stackexchange.com/questions/2008976/… Is my question now clear?

May it be reopened?

@AliCaglayan there you have it

Thanks @Null its good to see I still have part of my sanity

I'll have to email when I get home, Ali.

@TedShifrin Thats fine, no rush. Thank you for this problem anyway :) I hope you have a good rest of your day

Long drive after sleep ... G'nght!

@AliCaglayan is there some illustration why map(A,R) is a ring if R is a ring?

i mean i could prove it, but i wont have any insight...

I don't have an intuitive explanation, but sets of function usually inherit their structure from the codomain

@Null in algebra and calculus we add and multiply functions all the time

@Alessandro this does help me still :)

the reason we can is because the functions take values in a ring with addition and multiplication

Similarly map(A,K) is a vector space if K is a field, regardless of A

ah so map(A,B) is something if B is something? (and A is not empty..)

or does this not work at some point?

you're being vague

there are many "somethings" in math

@meow still here?

hi everyone

Hi @DHMO

Has anyone come across the notion of "group action with finite covolume"? Some sources say it means that there is a fundamental domain of finite volume while others say that the volume of the quotient space is finite. Is this the same?

Hi @DHMO