00:00 - 18:0018:00 - 00:00

12:19 AM
quotient of groups is right inverse of direct product

12:36 AM
The point $(0,0)$ isn't an open set in $\Bbb{R}^2$ right?

no

Okay, just checking.

12:48 AM
hello all
havent been here for a while
how can i disprove this
The multiplicative inverse of a real number x, is a real number y such that xy = 1. Every real number has a multiplicative inverse.

what have you tried?

basically asking me if a * 1/a = 1
always true right? what about 0

is that the one case when that statement is false?

what is 1/0

12:51 AM
undef
thats whats holding me back..

if a=0, what is ab
b being any number

and 0 is never 1

hmm ok thank you

in particular, you can't pick b such that ab=1

12:58 AM
@gian last time i checked, it isn’t even a set

There is no largest integer.

(you meant {(0,0)}, but right singleton isn’t open in usual topology of R^2)

how can i represent that statement as a mathematical one

That seems mathematical to me

im thinking n<n+1

12:59 AM
@WillNjundong not exists n integer such that for all x integer we have n>x

If you want to be a Frenchman, say that for n in N there is m in N such that n < m.

$\neg\exists n\in\Bbb Z: \forall x\in\Bbb Z: n>x$

@LeakyNun are you a robot?

yes

1:00 AM
@WillNjundong just do what I said

cybernet has descended upon us all...

do i pass the turing test?

no
thank you again, obelo

If you use $\neg$ unironically you fail the Turing test

$\Bbb Z$ is not well-ordered under $<^{-1}$ :p

1:02 AM
I have no idea what that is supposed to mean

it has no minimal element under the order that is the reverse of the usual order
if we pretend smaller to be bigger, there is no smallest element

Yeah I got that but what possessed you to write $<^{-1}$

demon

working with integers. his statement stands...

@WillNjundong it’s a translation of “there is no largest integer”

1:04 AM
oh, I meant Z instead of N above
same thing

Every real number besides 0 has a multiplicative inverse.
so how can i spin this into a precise mathematical statement

$\Bbb R^\times =\Bbb R\setminus\{0\}$

for all x, x<0 and x>0, x*1/x=1?
hmmm thats neater, i didnt think to use that

$R^\times$ for a ring is the group of elements which have multiplicative inverses
the "unit group"

e.g. multiplicative group of units mod n

1:28 AM
Among every consecutive three integers, there is a multiple of 3.

I rewrote to:
for all integers k, a k+-1, k+-2 exists such that k%3=0
do i need to add "or"?

you don't need +-
that's pretty confusing, not good
3 does not have to divide k

but thats to show that it is a multiple of 3. I dont know other notation to represent that

k is not a multiple of 3
what if k=10?

In what class would I learn about a cobordism?

i said "a k exists"
paired with k+-1 and k+-2 does that not mean the value meeting that condition is in that range

1:35 AM
I said what you wrote is confusing
So forgive me for not getting exactly what you meant
Maybe try: For $k\in\Bbb Z$ let $T(k)$ denote the set $\{k,k+1,k+2\}$. Then...

alright, that is a lot clearer

Morning everyone

evening mate :D

1:54 AM
@BalarkaSen huh, here's something fun. in one of the appendices to his graphic novels, Alan Moore actually describes and uses the Koch snowflake as an analogy

Hi, a have a questions on paths, if anyone can help...If I have two parametrizations of the same curve in the complex plane, is it true that there is always a reparametrization between them? I wanto to prove (or disprove) that the value of the line integral is the same no matter the parametrization of the curve :s

1 hour later…
3:13 AM
We shift $k+1$ $k$ positions to the left, then we shift $k+2$ also $k$ positions to the left, etc.

So we shift each of the elements $k+1, \ldots k+\ell$ $k$ positions to the left.

At each such shift the sign becomes $(-1)^k$. We shift totally $\ell$ times. Therefore, the sign becomes $((-1)^k)^{\ell}=(-1)^{k\ell}$.

Is everything correct?

Do we apply the induction on $\ell$ ? Or on which variable?

@MaryStar Yes that is correct. You need to induct somewhere because anything with "etc." is implicitly induction.

We shift the element $k+i$ $k$ positions to the left, for each $1\leq i\leq \ell$. Do we have to induct on $i$ ? @0ßelö7

@MaryStar That seems right.
Depending on how anal the grader is you might have to justify where the $(-1)^k$ came from too. That's induction too.

3:35 AM
@0ßelö7 Ah ok! Thank you for your help!

3:50 AM
Is there anything similar to this for quadratic reciprocity ?

4:23 AM
ughh

4:46 AM
Hello chat

5:24 AM
Let $A$ be some non-empty finite set and $S_A$ be its permutation group. Fix some $B\subset A$ and define $H=\{\alpha\in S_A\, | \,\forall x\in B:\alpha(x)\in B\}$.
If $B=\emptyset$, does that mean $H=S_A$? The above does not mention $B$ nonempty, and I assume every permutation on $A$ vacuously satisfies the defining property for $H$.
And likewise if $A$ is any arbitrary non-empty set. I'm not very familiar with handling $\emptyset$ or vacuous truths :/

0

Suppose M1 and M2 are two TMs such that L(M1) = L(M2). Then (A) On every input on which M1 doesn’t halt, M2 doesn’t halt too. (B) On every i/p on which M1 halts, M2 halts too. (C) On every i/p which M1 accepts, M2 halts. (D) None of above. My try : It explained here that M2 accepts string...

@DaveClarke, can you please take a look?

6:06 AM
0

Is there anything similar to this for quadratic reciprocity ? I mean, the link there explains how you can figure out the solution of cubic equation by yourself without having a suddent flash of inspiration/ genius genes. Is there some similar guide for quadratic reciprocity ?

6:45 AM
^ NE IDA IDA IDA IDA IDA IDA
THAT

7:37 AM
Could someone please tell me that in this answer how Angle $AOB = 2\pi-\theta$? math.stackexchange.com/questions/2038868/…
I have been trying to figure that out since a lot of time.
Shouldn't it be angle $AOB= 2\theta$
The thing is written after "By simple geometry of circle..."
No one?
Alright...

8:01 AM
Hello @TobiasKildetoft !! Could you take a look at my question: math.stackexchange.com/questions/2448179/… ?

8:25 AM
@BalarkaSen, this says that linear continuum can have cardinality larger than reals. Can a linear continuum be countable?

@Silent No

ok

@Silent I don't recall the precise argument

@Silent I don't know this stuff.

I think this is how Munkres proves that the reals are uncountable

8:28 AM
@TobiasKildetoft ok so i will look at that.

def vp(a, p):
i = 0
while (a % p) == 0:
i = i +1
a = int(a/p)
return i
Darn why the above programme is showing $v_{13}(27**13-1)$ to be $1$ when it's clearly $2$ ?
https://www.wolframalpha.com/input/?i=((27**13)-1)%2F1691

@AlexKChen Weird

Did you check it in python ?

because you use normal division

no, I don't have anything handy to run it

8:40 AM
replace int(a/p) by a//p

No it works perfectly for small numbers

@AlexKChen is it because Python has to change the type of the object when it becomes this large?

I just ran it in spyder
a//p works
int(a/p) has rounding errors...

@TastyRomeo Ahh, that makes sense

Ahhh yeah. Thanks @TastyRomeo
How stupid.

8:50 AM
hi chat

Hi Balarka !

Morning

Good midday lol
Hi @BalarkaSen Subho Astami BTW

i dont care

Why ?

8:55 AM
Hey
Good morning from this side

i am a hippie, i refute our half-cultural pseudo-spiritual festivals and think they are overrated
Oh hey @Brody

@BalarkaSen Ya, I perfectly agree with "pseudo-spiritual" and overrating part. Though I think the "half-cultural" part becomes prevalent after Dashami, right ?
(Though the "I am a hippie" part is surly wrong)

nah i am big on counterculture
:P
@Brody Yup. Trying to study some physics, and then math.
How are you doing?

Well actually though I live in USA, I am actually a Bengali, and this isn't my real name.

9:00 AM
Oh.

@BalarkaSen Neat. I'm pulling an almost-all-nighter, will probably sleep a couple hours before class. Ignoring poor decisions I'm pretty well lol

Poor decisions? Yolo

yolo ?

@AlexKChen generally if I saw 'Chen' I'd assume it to be a Chinese surname

@Brody What does "This isn't my real name" means, eh ? My actual title is Chakraborty.

9:04 AM
@AlexKChen let me urban-dictionary that for you

Gotcha, you only live twice.

@BalarkaSen I have a cousin that had YOLO tattooed on his wrist...

lol typical edgy kids

How do you know brody's cousin is a kid :P ?

I mean, it was cool... for a few weeks maybe, before it got uber corny and meme-ified

9:06 AM
I use it in memetic contexts

You can say anything as long as you use it "ironically" ;)

Is that a meta joke
did you just say "anything" ironically

Not really

oh well
In any case, memes aside, how is math going?

I am wondering if there is a way to make this analogy of representation theory to physics more fitting: The simple modules are the atoms, the standard or costandard modules are the [something] molecules, and the tilting modules are the [something bigger] molecules
then restriction to a maximal torus (which is essentially what one does when considering characters) is looking at electrons and so on

9:13 AM
@BalarkaSen Just did my algebra homework. As usual, not fully confident anything is right

Just not sure what sort of molecules would make sense for the costandard and tilting modules, since tilting modules are build from costandard modules

@Brody I see

I failed to solve couple of trivial NT problems :/

@Tobias my favourite analogy is that fundamental particles are positive energy representations of the universal cover of the Poincaré group ;)

@s.harp I am looking for one in particular for reps of algebraic groups

9:15 AM
On a related note, psychology class last spring required taking a Myers-Briggs typology test @Balarka

Also, I failed to find out why 2+i has no inverse modulo 5
Leaky told me after that, though.

My result was INTP. The description was surprisingly specific and accurate despite the mundane questionnaire

@Brody Haha nice

@Tobias simple modules are irreducible representations?

@s.harp Yeah

9:18 AM
What role does the group play in your analogy?

@s.harp None at all, apart from determing what the simple modules are (basically, it tells us which category to work in)
I am looking for an analogy that will sound familiar to people with a basic understanding of science, but no idea about any algebra

but then your analogy is no different than for vector spaces or any other abelian category rihgt?
where its basically just there are elmentary building blocks and you can mash em together to get anything else

@s.harp Except that there are certain distinguished modules other than the simples
also, the analogy is actually better here, as you can get different modules using the same building blocks, by putting them together in other ways

@BalarkaSen apparently INTPs tend to lack confidence in their conclusions because they feel they overlooked some data

Aha

9:21 AM
@TobiasKildetoft I don't understand here sorry, up to isomorphism direct sum is commutative and associative right?

@s.harp Sure, but we have more ways to put together the modules since the category is not semisimple

there's confirmation bias and such oc, it's just a very refined horosope after all

Doesn't that rather mean there is no way to get some of the objects from the "atoms"?

@s.harp So I am looking for two classes of molecules such that those in one class can be build from those in the other one

what are you doing in physics @Balarka?

9:23 AM
@s.harp No, we can still get them from the atoms, it just matters how we put together those atoms (just like it does in real chemistry)

@Tobias Amino acids are molecules, chains of amino acids form proteins

@s.harp Awesome. Great idea

@TobiasKildetoft Can you give me an example because I am really confused

@s.harp Consider the two-dimensional representation of the integers given by mapping an integer $n$ to the matrix $1 n 0 1$
(that was less readable than was meant, but I forgot how to make nice matrices in MathJax)

ok, i got the representation

9:26 AM
this is two-dimensional and build from two copies of the trivial representation, but it is not the direct sum of those. One is a submodule and the other is a quotient

ah

@Brody Electromagnetism mostly. Induction, etc

(also, as you may have noticed, I use representation and module interchangeably which is probably not a great idea)

a representation is just a module of the group algebra so I dont mind

@s.harp Well, for algebraic groups that is no longer really the case, but the term $G$-module is still used

9:29 AM
because continuity considerations are not reflected in $\Bbb K[G]$?

what precisely is meant then tends to vary (I usually specify at the beginning of my papers that I mean only rational ones, and sometimes only finite-dimensional ones)
Right. One wants comodules of the coordinate algebra instead
but hardly anyone thinks of them like that

@BalarkaSen Oh very fun concepts (exercises are hit or miss). You guys do special relativity or QM?

@Brody Nope, I don't think so, unfortunately.

In school ?

right

9:38 AM
AFAIK, in WB they don't have SR/QM in school. (Though they're required for that national physics Olympiad similar to USAPhO)

that's fine Balarka. hope you still enjoy the content :)

I like EM so far.
I want to learn gauge theory from the physical point of view at some point

@AlexKChen here in the States, AP Physics C (but not B?) included some very basic SR and QM, but iirc at the instructor's discretion if time allowed...
@BalarkaSen ah so you're acquainted with the mathematics of it

Oh no not really. I know the basic definitions but that hardly counts as being acquainted to it.
Mostly I learnt whatever I did from a talk in a differential geometry workshop I went to

Hmm... I have solved the P vs NP conjecture !!!
It's true iff P=0 or N=1

9:45 AM
@BalarkaSen haha I was hoping that finally learning some algebra would help me some. guess not

Learning algebra is useful. What have you learnt so far?

brb

Jeez, they don't mock about in this grant application. As part of filling out personal information it has "Give an account of your most significant contributions to science"

Ack

Back

9:52 AM
Hack

@BalarkaSen just very basic group stuff so far

kek

@BalarkaSen But that would have completely broken the chain

@Brody What book are you following?
@Tobias Not the rythm though

9:53 AM
true

I can't remember....
wait
Pinter, 2nd ed

ah... I have heard of that book

We're now up to isomorphism
No pun intended

Cool!
Keep learning

@Brody "A book of abstract algebra"?

9:58 AM

Seems like I only have the 1st edition here

It was by far my cheapest textbook this semester
so it is very appreciated in at least one way

does it reflect badly on me if i think this is fantastic music lol

@Brody Wow, it certainly isn't short compared to the content, which is probably a good thing for an introductory textbook

@TobiasKildetoft yeah he really likes to expound and teach through prose
@Balarka no but to each his own

10:09 AM
the whole album is full of sonically and lyrically horrid stuff

Pick a number in [0,10]
With 100% accuracy I ccan guess it

@BalarkaSen the music's a little too atonal to enjoy
but great video on epizootics. it's oddly satisfying

In terms of music, did you hear the songs of Luar na Lubre @BalarkaSen

@Brody True, Bish Bosch is intended to be sonically hard to listen to.
Oh, Epizootics is a lyrical masterpiece in my opinion

10:17 AM
I like "Zercon, A Flagpole Sitter" too
21 minutes of full nightmarish masterpiece

I recommend listening O son do ar, Sereas, Cantiexre to name a few (of Luar Na lubre)
@Brody Done ?

@AlexKChen mm-hm

It's an integer muahahaha... Am I right ?

no

Mm... pi ? e? pi+e ? sqrt(2) ?

10:19 AM
nope, nope, nope... nope

Oh, a fun fact:
pi+e = pi

"u silly"

Is it true that there are real numbers that are absolutely unutterable?

@Brody I... err... I guess .... erm... my ... er.... my soothsaying powers ... yeah..yeah.. are stolen by the wizard @BalarkaSen.

10:21 AM
Depends on your definition of "unutterable"
There are real numbers which are algebraic and yet cannot be written as surds

@Brody OK, what was the number ?

That's what amazes me most
Of course, that's the main theorem of Galois theory
Take a root of x^5 - x - 1 = 0 say

i.e. they cannot be mathematically described
I know this sounds ill-posed
@AlexKChen 10 minus euler-mascheroni

lol

but I did think about unutterably incomputable, unrepresentable numbers

10:25 AM
Oh, hm, there are things called uncomputable numbers aren't there
Chaitin's $\Omega$ something something

there are also hyperreal and surreal numbers, come to think @Alex
yeah computability theory is freaky-deaky (and computable analysis is a thing)

yeah iunno this stuff

New word: iunno

10:43 AM
@Brody There is also something like "the smallest number which cannot be described with less than 13 words"

You mean 12 words right ? (Or IDK how to count)

@AlexKChen possibly

@TobiasKildetoft well surely there is no such least number!

@Brody But there are numbers that can be described like so. So by well-orderedness there is a least one.

If your description comprises thirteen or more words, can one not add "reduced one" to make a lesser number?

10:50 AM
[Random]
The following topological (possibly metric) space will be elaborated later. It is constructed using the ordinals in $\omega +1$

@Brody Sure, but that number might be describable using fewer words also
(this is all fairly silly of course)

The space in words:
Each point has a neighbourhood given by the radial direction being the ordinal $\omega +1$ smearing uniformly in the angular direction
such that one can go nowhere until they walked countable number of steps

11:09 AM
I have a bit of trouble understanding that
with words you obviously wnat to restrict to a finite pool, say whatever you find in a dictionary
then a person needs to be able to understand the description
but what does understand mean in this context?
does it mean be able to generate an algorithm in order to find hte number?

11:59 AM
Can there exist a non diagonal matrix $A \in M_{2}(\Bbb{R})$ such that $A^3 = I$

We have the permutation $\pi: (1, \ldots , k, k+1, \ldots , k+\ell) \mapsto (k+1, \ldots , k+\ell, 1, \ldots , k)$.

Is the following correct? Or should I use in an other way the permutation?
$$\psi_1\land \ldots \land \psi_k\land n_1\land \ldots \land n_{\ell}=n_{\pi(k+1)}\land \ldots \land n_{\pi(k+\ell)}\land \psi_{\pi(1)}\land \ldots \land \psi_{\pi(\ell)} \\ =\text{sign}(\pi)\cdot n_{k+1}\land \ldots \land n_{k}\land \psi_{1}\land \ldots \land \psi_{\ell}$$

12:22 PM
Should it maybe be:
$$\psi_1\land \ldots \land \psi_k\land n_{k+1}\land \ldots \land n_{k+\ell}=\text{sign}(\pi)\psi_{\pi(1)}\land \ldots \land \psi_{\pi(k)}\land n_{\pi(k+1)}\land \ldots \land n_{\pi(k+\ell)} \\ =\text{sign}(\pi)\cdot n_{k+1}\land \ldots \land n_{k+\ell}\land \psi_{1}\land \ldots \land \psi_{k}$$

oh god
what demonic entity is forcing you to do this

Geometers, probably

I'm a geometer and we just take it on faith
some algebraist has done it correctly
@MaryStar at least use \cdots

hey ur pic changed,I saw in Physics SE right@0ßelö7

what?

12:33 PM
I remember ur nick similar to some1 in Physics SE
sorry perhaps it's not you
sorry for ping.

it was me

nice!

1:02 PM
@MaryStar first switch the $\psi$s past the $\eta$s
$\psi_1\wedge\cdots\wedge\psi_k\wedge\eta_1\cdots\wedge\eta_l=(-1)^{k\cdot l} \eta_1\wedge\cdots\wedge\psi_k$

@BalarkaSen from me

possibly

1:22 PM
morning
Semi

morning

1 hour later…
2:39 PM
Hi. Suppose that $E[X_K]=O(K^{-\alpha})$ for some $\alpha>0$, and that $W_K$ is such random variable that $0\leq W_K \leq 1$. Is it possible to say something about the rate of convergence of $E[X_K W_K]$ in terms of $\alpha$ and $W_K$ (suppose that everything about $W_K$ is known)? Clearly $E[X_K W_K]=O(K^{-\alpha})$, but maybe something more can be said

1 hour later…
3:49 PM
Hello
I have a question...
Does $p_1^{a_1} \cdots p_r^{a_r}$ have $a_1+ \dots+a_r$ divisors?

That number sounds more like the number of divisors that are prime powers.

It has $(a_1+1)(a_2 +1) \cdots (a_r + 1)$ divisors I think

@TastyRomeo Oh yes, right
@BalarkaSenHow can we show this?

A divisor will be of the same form, but with different powers for each prime factor
that power can be anything between $0$ and $a_i$

Hmm. $\int \frac{\int}{}$

4:03 PM
each divisor will be of the form $p_1^{b_1} \cdots p_r^{b_r}$, where $b_i \in \{0, \dots, a_i \}$ @TastyRomeo

There are $(a_1+1)$ possible values for $b_1$,.., (a_r+1) possible values for $b_r$ and so totally there are $(a_1+1) \cdots (a_r+1)$ possible divisors of $\prod_{i=1}^r p_i^{a_i}$, right?

I guess so :P

nice, thanks :)

1 hour later…
5:17 PM
library(MASS)
y <- quine$Days n <- length(y) ysum <- sum(y) g <- function(r) sapply(r, function(x) sum(lgamma(y+x))) a <- function(r) g(r) - n*lgamma(r) + + lgamma(ysum+0.5) + lgamma(n*r+0.5) - lgamma(ysum+n*r+1) - 5080 f <- function(r) exp(a(r) - 2/3*r) C <- integrate(f, 0, 3, rel.tol=1e-10)$value
curve(f(x)/C, 0.001, 3)

Anyone know how to find the maximum value of that curve and the x value for which it occurs? (in R)
seems to be 3.1-3.2 but I don't know how to find it numerically

5:32 PM
I was wondering around
if a point is the middle of a line segment
and a segment is the midst of a square
and a square is a symmetrical shape of a cube
what would a cube be symmetrical of ?

if by symmetrical you mean a section, then tesseract

impressive
i thought it to be a cube of cubes

yeah a tesseract is basically a 4 dimensional analogue of a cube

Great... I inherited all files of the previous TA of this course.
Too bad all of them are unordered, in a single folder, with not very descriptive filenames

5:38 PM
so this applies to Space Time Continuum
since there is no human conceivable shape of a superior dimension than a sphere

[citation needed]

it was a question

[Random]
$(r,\theta), r \in \omega + 1, \theta \in [0,2\pi)$

5:52 PM
4th dimension is the freakiest thing to understand
and half of youtube is entitled "4th dimension explained"

00:00 - 18:0018:00 - 00:00