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02:00
@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!
02:05
@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
Like $(-5,1)$ is the interval
And $| x+2 |/3 < 1$ is the 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
02:12
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
02:14
as'salaːmu ʕalajkum
@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
02:15
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
02:18
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
02:20
I keep confusing myself with the wording of "algebra" and "module"

The question at hand is "Is $\mathbb{M}_n(\mathbb{C})$ a simple $\mathbb{C}$-algebra?" So I thought it was sufficient to prove that for any nonzero $A\in\mathbb{M}_n(\mathbb{C})$, $(A)=\mathbb{M}_n(\mathbb{C})$. Then since the only ideals are the trivial ones, $\mathbb{M}_n(\mathbb{C})$ is simple
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) ??
02:22
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
02:26
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.
02:28
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
@ :(
02:31
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.
02:33
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
02:37
@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 $\infty$?
02:39
"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
02:40
@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?
02:42
@null anything bigly?
3
Q: Deriving numbers from sums

pi66Let $x_1,\dots,x_n$ be not necessarily distinct real numbers. Form a multiset $A$ by taking the $2^n-1$ sums of nonempty subsets of the numbers. For example, if we have $(x_1,x_2,x_3)=(1,2,-3)$, then by taking the seven sums, $A=\{-3,-2,-1,0,1,2,3\}$. The set $(x_1,x_2,x_3)=(-1,-2,3)$ generates t...

> us election jokes
@meow-mix well, like 4 pages
> jokes write themselves
I found an elegant solution with elementary symmetric polynomials
02:43
@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*
02:45
@meow-mix does it suck that bad on big documents or what?
but if it goes dirty, can't you just edit it after?
@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
02:47
@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
02:48
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
02:50
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
02:51
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
02:52
: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?
02:57
@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
02:59
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
03:04
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
03:07
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
03:16
I discovered I don't know how to pronounce manifold
man-if-old... right?
@meow-mix i would say: many-fold
@Sophie oh so its munni-fold i guess?
cool song hehe
03:39
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
Merhaba!
How's everyone?
03:55
@ryagami ?
04:21
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 :)
04:36
@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
04:43
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?
04:54
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)
@AliCaglayan Let $A\not=\emptyset$ and $R$ a ring.
Show that $map(A,R)$ with the following operations is a ring:

$f+g$ is defined by: $(f+g)(x):=f(x)+g(x)$ for all $x\in A$

$f\cdot g$ is defined by: $(f\cdot g)(x):=f(x)\cdot g(x)$ for all $x\in A$

Is this a simple checklist task? Or is there some ingenius trick/lemma to it?
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
04:58
@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...
05:01
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.
05:03
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
05:05
right.
@Null sorry I forgot to read it. Nothing to do with your name or anything
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,+).
05:07
@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
05:09
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?
Let $A\not=\emptyset$ and $R$ a ring.
Show that $map(A,R)$ with the following operations is a ring:

$f+g$ is defined by: $(f+g)(x):=f(x)+g(x)$ for all $x\in A$

$f\cdot g$ is defined by: $(f\cdot g)(x):=f(x)\cdot g(x)$ for all $x\in A$

Is this a simple checklist task? Or is there some ingenius trick/lemma to it?
@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
05:19
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
05:29
@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 :)
05:34
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
05:38
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
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 :)
05:45
Lets see the others
I can write then down and think about them during this week
hi
May it be reopened?
\begin{array}{cc|c@{}ccc@{}ccc@{}ccccc@{}c}
a&b&(&a&\rightarrow&b&)&\leftrightarrow&(&\lnot&b&\rightarrow&\lnot&a&)\\\hline
1&1&&1&1&1&&\mathbf{1}&&0&1&1&0&1&\\
1&0&&1&0&0&&\mathbf{1}&&1&0&0&0&1&\\
0&1&&0&1&1&&\mathbf{1}&&0&1&1&1&0&\\
0&0&&0&1&0&&\mathbf{1}&&1&0&1&1&0&
\end{array}
@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
05:54
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...
06:29
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?
06:38
you're being vague
there are many "somethings" in math
06:55
@meow still here?
 
3 hours later…
09:44
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

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