Mathematics - 2017-07-16 (page 2 of 3)

Seth Taddiken

05:12

Hey I've got a question about Buddhabrots.
So wikipedia and a few other sources say that the way to calculate a Buddhabrot is to take points(c values) that aren't in the Mandlebrot set (at least up to a certain number of iterations through z^2 + c) and trace their paths to escape, since a point(c value) that escapes is by definition not in the Mandlebrot set. My question is this, how do people render Buddhabrots that show activity within the area associated with the Mandlebrot set since any value that is part of the Buddhabrot set is, again, by definition not part of the Mandlebrot set?

I'm curious because I've been trying to render a Buddhabrot from instructions on articles I find but when I try to render points that are explicitly not in the Mandlebrot set and trace their paths to escape, it produces a negative of the Mandlebrot set... which makes sense! So why do Buddhabrots produce renderings that are significantly different than the Mandlebrot set?

user400188

06:04

@Michael Hardy : A small edit to your answered here: https://math.stackexchange.com/questions/148177/what-is-the-difference-between-⊢-and-⊨?rq=1
Change "proof-checking algorithm" to "provability-checking algorithm" . Since we are after an algorithm that checks the existence of a proof rather than checking whether a proof is correct or not.
Excellent answer by the way (plus easy to understand). I bookmarked it.

Abcd

06:19

If A is acute, how do I find which is greater- sin A or cos A? I think it should be $sin A>=cos A$ because sin's value increases from 0 to 90 while that of cos decreases. Research effort: 2 textbooks don't have any answer. Googled too but couldn't find the answer to my question. Used calculator and found that sin 35 is less than cos 35. Can someone account for that also?

I think I should ask that on Math.SE

Leaky Nun

@Abcd below 45, cos A is greater; above 45, sin A is greater

desmos is a great tool

Abcd

@LeakyNun How?

Ok, let me check

@LeakyNun How to use it? desmos.com/calculator

Leaky Nun

desmos.com/calculator/591r27ro5h

Abcd

@LeakyNun 1 for degrees please

Is it not possible to draw a graph with values in degrees?

Leaky Nun

desmos.com/calculator/oamqpzk969

just choose degrees in settings

Abcd

06:27

ok, thanks

Daminark

Hey guys

Balarka Sen

06:44

hi

Araske

hello!

Daminark

Ah, you've finally rejoined us!

Araske

i exist occasionally

Daminark

Wao reax only

Balarka Sen

oh hey araske

what did i miss in the chatlog

i can't keep up with the transcript [travolta.jpg]

Daminark

06:50

.jépg

Balarka Sen

JPИSKY

I am a Russian, not a French

Alessandro Codenotti

Hi chat

Secret

Interesting. For my case, the actual dream changing is not necessary obvious to me because of how time travel can appear in my dreams and thus the narrative of the dream can mask that, but something like these do happen

Yeah, I heard people said the more you participate in dream recall, the better one remembers a dream

I also heard that the style of dreaming of each individual may be unique, thus expect that your dreams might be different form everyone else

Daminark

@BalarkaSen Vodka for you, comrade

Hey @Alessandro!

Secret

I am pretty certain I found that top secret discord group, but for some reasonI cannot join

Daminark

07:05

That would suggest that it's pretty top secret

Secret

Meanwhile somewhere in the other side of the world there's this massive chat group

Erdős Math Group

Astyx

Hi chat

Daminark

Hai @Astyx

Astyx

What's up ?

Tim The Enchanter

07:22

Hey @Astyx

Balarka Sen

@Secret It's a rather private discussion room between a few of us, sort of like the number theory room (which is gallery; nobody other than the selected few who are involved in the study group can join), except mostly not math. I very much doubt that you found it.

Daminark

@Astyx back: not too much, how about you?

Astyx

Enjoying my freedom by not doind anything

So not much either

Balarka Sen

I should be doing... Riemannian geometry!

Astyx

I should be rereading the beginning of Diff Top

Daminark

07:26

re-re-ading

jfc why are you doing stuff 3 times??

Secret

Anyway, one thing that suprised me the msot is that discord is only 2 years old

I thought it has existed for some time already

Astyx

@Daminark I like to ad things

Daminark

So it seems

Secret

07:56

Well, to be REALLY picky:

yesterday, by Astyx

I've never been redundant nor have I ever said the same thing twice in a row

-> Hi chat

but anyway, it is not important

Secret

09:37

[Random]

It amazed me how we can have a countably infinite and mostly unique decimal expansion for any real number, but there are continuumly many real numbers

So, what about trying to extend the scope of our alphabet by having a continuum decimal expansion:

Reals:
$$s=\sum_{k=-\infty}^{\infty}\frac{a_k}{10^k}$$

??? number decimal expansion

$$s=\int_{-\infty}^{\infty}\frac{a(x)}{10^x}dx$$

Richard

@AkivaWeinberger I know that "$10=(3+i)(3-i)$, so it's a multiple of $3+i$", I did write that :) Thanks

parvin

hi ! do you guys know if this book is a good one for introduction or not?

Introduction to the theory of computation . By Michael Sipser

I checked Amazon's rating it had 4 stars :D
I'm not sure if the rating is trustworthy or not !

Secret

10:19

One thing that is obvious is that the set $$\left\{s=\int_{-\infty}^{\infty}\frac{a(x)}{10^x}dx\right\} \subset \Bbb{R}^{\Bbb{R}}$$ since the latter has nonintegrable and even uncomputable functions

however the cardinality of this set is to be found out

Leaky Nun

@Secret in other words, $2^{\Bbb N} = \mathfrak c$

Secret

Did I assumed continuum hypothesis accidentally? (though I tend to assume GCH holds for most of my analysis as I am not ready to deal with cardinalities outside GCH yet)

user84215

How can I draw a diagonal arrow in a commutative diagram with AMScd in MathJax?

Secret

I have no idea. I think you need tikz for that?

user84215

What is tikz?

Hippalectryon

10:41

A LaTeX package

26

A: How to draw a commutative diagram?

It is possible to do (somewhat primitive) commutative diagrams using \array: $$\begin{array}{ccccccccc} 0 & \xrightarrow{i} & A & \xrightarrow{f} & B & \xrightarrow{q} & C & \xrightarrow{d} & 0\\ \downarrow & \searrow & \downarrow & \nearrow & \downarrow & \searrow & \downarrow & \nearrow & \do...

SteamyRoot

If you want to make commutative diagrams, I can recommend TikZ + TikZ-cd

Doubt it works in MathJax, though.

user84215

What is TikZ?

SteamyRoot

It's drawing stuff in TeX

user84215

I think AMScd is better than {array} for creating commutative diagrams

Secret

[Infinite set] A naive dedekind cut arguement:

Consider the subset of the reals [0,1]

user84215

10:54

Infinity is an illusion.

Secret

We can perform dedekind cuts to obtain a pair of intervals of the form [0,a],(a,1]

Now $a \in [0,1]$ and we can use the bijective map $\frac{1}{\pi} \text{arctan}(x)+\frac{1}{2}$ to show that there are continuum many elements in [0,1]

We can also pick some bijective map to show that given some a, (a,1] also has continuumly many elements. Therefore the set obtained by taking dedekind cuts over [0,1] by having $a \in [0,1]$ (i.e. the set consists of all intervals of the form (a,1], [0,a] )has in total $2^{\mathfrak{c}}$ elements.

Now, because of the way the intervals are constructed, the set $\{[0,a], a \in [0,1]\}$ is one of the two continuum sized chains ($\{(a,1],a\in [0,1]\}$ being the other)

Akiva Weinberger

I don't understand how you get $2^{\frak c}$

I'm pretty sure you don't

Your elements are of the form $\langle(a,1],[0,a]\rangle$? Biject it to $a$

That forms a bijection between your set and $[0,1]$

Secret

For each $a$ we have the pair of sets $[0,a]$ and $(a,1]$. Now since $a \in [0,1]$ and $[0,1]$ is known to have continuumly many singletons then in total there are continuumly many $[0,a]$ and continuumly many $(a,1]$ so in total with have $\mathfrak{c}+\mathfrak{c}$, $=$ oopss... $2\mathfrak{c}=\mathfrak{c}$...

bah, I fail counting class again...

Akiva Weinberger

11:10

Fun fact: In ZF, it is consistent that there exists no total order on the set $2^{\Bbb R}$

and thus on any set of cardinality $2^{\frak c}$

Leaky Nun

@AkivaWeinberger total???

Akiva Weinberger

(Total order = linear order = order)

Leaky Nun

how can it not have total order...

Akiva Weinberger

(The extra word is just to distinguish it from partial orders)

@LeakyNun It's very counterintuitive.

Leaky Nun

@AkivaWeinberger what about $\Bbb R$?

Akiva Weinberger

11:12

@LeakyNun What do you mean?

Leaky Nun

is there total order on $\Bbb R$?

Akiva Weinberger

Of course there's a total order on $\Bbb R$; take the usual one

$<$

Leaky Nun

I mean, under ZF, of course.

Secret

What's the cardinality of $\omega_2$, is it $\aleph_2$?

Akiva Weinberger

Yeah @Secret

@LeakyNun Yeah

Leaky Nun

11:19

@AkivaWeinberger could you expand on that?

Akiva Weinberger

I think it's consistent with ZFC, even

Not sure

Let me find out

Oh wait duh

user84215

To be Well-ordered needs the axiom of choice not for having a total order.

Akiva Weinberger

Of course it's total orderable with ZFC; it's even well-orderable

But yeah it's consistent with ZF that $\mathcal P(\Bbb R)$ has no total order

I found this answer on the subject

user84215

I mean for R not for 2^R.

Akiva Weinberger

It is consistent with ZF that $\Bbb R$ has no well-order, yes

but it definitely has a total order (the usual one)

Secret

11:27

Ah, $\omega_1$ is different from $\Bbb{R}$. You can have subsets in $\Bbb{R}$ that has an infinitely decreasing sequence, but not in $\omega_1$

Akiva Weinberger

The statement "Every set can be well-ordered" is famously equivalent to the axiom of choice. The statement "Every set can be totally-ordered" is strictly weaker than the axiom of choice, but still stronger than ZF.

@Secret They could be equinumerous (that's the continuum hypothesis)

Secret

yup

But I do need to be careful of the ordering properties of infinite sets if I want to understand the higher cardinals

user84215

I have a problem with the following formula. I want the parentheses (left and right) to have the same size.
$$\begin{aligned} a= & \Biggl (1+2+3+ \cdots \\ & \left. \cdots + \infty -\frac{(x+2)^7}{\frac{\frac{\frac{x^2}{(x+3)^2}}{(x+2)^5}}{(x+y)^3}}+\infty -1+\infty \right ) \end{aligned}$$

SteamyRoot

If you split it, then you have to manually change the bracket sizes

Akiva Weinberger

@aminliverpool \vphantom might be helpful

user84215

11:31

What is that?

Akiva Weinberger

\vphantom{stuff} inserts a zero-width invisible thing with the height of the "stuff"

SteamyRoot

It puts some invisible stuff there

Akiva Weinberger

Stands for "vertical phantom"

Without the v you get a space with the width of "stuff"

user84215

Thanks, good idea.

Akiva Weinberger

$\left(\vphantom{\frac{(x+2)^7}{\frac{\frac{\frac{x^2}{(x+3)^2}}{( x+2)^5}}{(x+y)^3}}}x\right)$

^Example

Balarka Sen

11:33

why are we creating LaTeX Frankensteins

Akiva Weinberger

Why aren't you creating LaTeX Frankensteins

Secret

Hmmm...
$S_0 = \{0\} = \{|\}$
$S_1 = \{-1,0,1\} = \{|0\},\{|\},\{0|\}$
$S_2 = \{-2,-1,-0.5,0,0.5,1,2\} = \{|-1,0,1\},\{|,0,1\},\{-1|0,1\},\{-1|1\},\{-1,0|1\},\{-1,0|\},\{-1,0,1|\}$
$S_3 = \{-3,-2,-1.5,-1,-0.75,-0.5,-0.25,0,0.25,0.5,0.75,1,1.5,2,3\}$
...
$S_{\omega} =\Bbb{N} \cup \{\frac{1}{2^n},n\in \Bbb{N}\}$
ok this is getting too messy to follow...

user84215

What is LaTeX Frankensteins?

SteamyRoot

Mashing disgusting looking code together to get a working solution, I guess.

Secret

Another example of a latex frankenstein

$$\overset{+}{\left.\begin{matrix}
\\
\\
\\
\end{matrix} \right|} \begin{matrix} * & * & * \\0 & 0 & * \\0 & 0 & *\end{matrix} \overset{+}{\left.\begin{matrix}
\\
\\
\\
\end{matrix} \right|}=0$$

Balarka Sen

12:01

Hi @MikeMiller

Leaky Nun

12:26

How many ways are there to make 200 from 1,2,5,10,20,50,100,200 up to permutation?

@Secret obviously it isn't just $\frac1{2^n}$

Secret

Yeah, it in fact includes all reals and also the first generation infinite and infintesimal numbers

and STILL not bigger than $\Bbb{R}$

Leaky Nun

I mean, if anything, it is $\frac{k}{2^n}$

GFauxPas

good morning frands

Leaky Nun

12:54

@GFauxPas good morning

Soham Chowdhury

13:37

Consider an integral domain $A$ integrally closed in $K := K(A)$, $L|K$ a finite extension, and $B$ the integral closure of $A$ in $L$.

I can prove that every element of $L$ is $b/a$ for some $b \in B$ and $a \in A$, so $K \subset K(B) \subset L$.

Is $L = K(B)$?

(I'm probably just being dense)

s/is/isn't/

Balarka Sen

13:54

Hi @Soham

Soham Chowdhury

hello

Balarka Sen

Z is integrally closed in Q; take the extension Q(sqrt(5))/Q.

Hm

maybe that doesn't work

Dattier

Hi, I think I have solve sat, how speak in french, here the solution :

forums.futura-sciences.com/mathematiques-superieur/…

NP=complexity of convexity optimisation

That's correct ?

GFauxPas

what is "sat"?

Dattier

Problème SAT

En informatique théorique, le problème SAT ou problème de satisfaisabilité booléenne est le problème de décision, qui, étant donné une formule de logique propositionnelle, détermine s'il existe une assignation des variables propositionnelles qui rend la formule vraie. Ce problème est très important en théorie de la complexité. Il a été mis en lumière par le théorème de Cook,, qui est à la base de la théorie de la NP-complétude et du problème P = NP. Le problème SAT a aussi de nombreuses applications notamment en satisfaction de contraintes, planification classique, model checking, diagnostic, et...

GFauxPas

14:03

Oh

Dattier

do you speak in french @GFauxPas ?

GFauxPas

no but I have Google translate

what are the variables, and what are the subscripts? I need context

are they Boolean variables, 1 = TRUE, 0 = FALSE?

Dattier

Boolean satisfiability problem

In computer science, the Boolean satisfiability problem (sometimes called Propositional Satisfiability Problem and abbreviated as SATISFIABILITY or SAT) is the problem of determining if there exists an interpretation that satisfies a given Boolean formula. In other words, it asks whether the variables of a given Boolean formula can be consistently replaced by the values TRUE or FALSE in such a way that the formula evaluates to TRUE. If this is the case, the formula is called satisfiable. On the other hand, if no such assignment exists, the function expressed by the formula is FALSE for all possible...

GFauxPas

I'm looking at that, but that is a page about statements. Your variables seem to be numbers

Dattier

yes in {0,1}

Balarka Sen

14:06

@SohamChowdhury Ok, I'm a little confused. Your theorem says every element of $L$ is $b/a$ for some $b\in B, a \in A$. So that automatically says every element of $L$ is of the form $b_1/b_2$ for $b_i \in B$, hence $L \subset K(B)$ (because elements of the fraction field $K(B)$ are precisely of that form?)

GFauxPas

right, I was calling those Booleans

Balarka Sen

Am I dumb?

Dattier

but the trick is to work on real

GFauxPas

and what is the relationship between $x_1$ and $x_{-1}$ Dattier?

Dattier

with convex function

GFauxPas

14:07

what's the site's convention/policy for asking questions in another language, anyone know?

Dattier

t

Soham Chowdhury

@BalarkaSen that makes sense, I think I have the containments wrong.

how do I know that $K(B) \subset L$?

Dattier

Thanks it's difficult for me to be understanding in english, I go speak in french forum @GFauxPas

Balarka Sen

every element of $K(B)$ is $b_1/b_2$

ah, hm

GFauxPas

Dattier ask your question in French and request a translation

Soham Chowdhury

14:09

we need to use the finiteness/integrality conditions, ofc

Dattier

Ok.

GFauxPas

say at the beginning

Dattier

J'aurais résolu le probléme sat

GFauxPas

no, on the website

on the main website, say "I don't speak English, please translate my question" or similar, and then ask

Balarka Sen

@Soham Well, $b_i$ are elements of $L$ because $B \subset L$, right?

Dattier

14:10

j'ai donné un lien de ce que c'était, il permet de résoudre les problèmes de résolution de problémes de maths (par exemple et tout les problème de la classe NP)

Balarka Sen

fractions are closed in fields, so $b_1/b_2 \in L$

Soham Chowdhury

@BalarkaSen oh duh

Balarka Sen

right, so it was more of a symbol pushing thing

Soham Chowdhury

I have not done any math since we last spoke in chat / I went to MM, and I get a little dumb when I'm away

Balarka Sen

oh, haha, it's fine

i am dumb too

Soham Chowdhury

14:12

check Hangouts if you can

yep, star-board confirms

Dattier

je ne sais pas si NP=P, mais cette algo permet de résoudre sous la condition que l'on sache résoudre les problémes d'optimisation convexe

Balarka Sen

oh ****

Dattier

ce qui semble être le cas

Balarka Sen

(checking)

Mary Star

14:24

Hello!! We have that $\sigma$ is a reflection along a line and $\delta$ is a rotation. How can we check if $\sigma\circ\delta$ and $\delta\circ\sigma$ are reflections or glide reflections?

Leaky Nun

16:10

@MaryStar get an object, and do it yourself

Akiva Weinberger

@MaryStar I'm thinking, for the reflection first and then rotation

We can assume the reflection is along the $x$-axis for a moment

and then we can always write the rotation as a composition of a rotation about the origin and then a translation.

So then if you rewrite that as (reflect across x-axis + rotate about origin) + translation

the first two combine into like a reflection about some other line $\ell$

and if the translation is purely perpendicular to $\ell$, it's a reflection

and otherwise it's a glide reflection

and, looking back, there was absolutely no reason to require the first reflection to be along the $x$-axis.

But whatever, what's done is done

MiKiDe

Hi guys! I wonder if anyone had some knowledge in the percolation theory and could answer to some of my questions?

Akiva Weinberger

I have no idea what the word percolation means, sorry

Astyx

Isn't that thing about trees ?

BAYMAX

the proof is trivial site contains the three lines and nothing more?

Akiva Weinberger

16:22

But post the question just in case someone who knows sees it later

Astyx

And heaps

Akiva Weinberger

@Astyx Dictionary says it's passing slowly through lots of holes

Astyx

(which are specific heaps)

Akiva Weinberger

Like coffee

Kasmir Khaan

16:45

hi chat

Akiva Weinberger

hı hi hï

…hī

Waiting

@Hippalectryon o/

Hippalectryon

@Waiting \o How are you ?

Kasmir Khaan

@AkivaWeinberger I calculated a rook polynomial can you please check my work ?

Astyx

hî

I prefer bishop polynomials

Kasmir Khaan

16:52

the board is a 3x3 with the entry 3,3 deleted

Astyx

Oh it's not even a typo

Kasmir Khaan

Hi @Astyx

is there really bishop poly ?

Astyx

What's a rook polynomial ?

I have no idea

Kasmir Khaan

-.-

Akiva Weinberger

Yeah, what's a rook polynomial

Kasmir Khaan

16:53

its the number of ways to place n rooks such that they cant take each other

Waiting

@Hippalectryon I managed to do some outstanding discoveries lately, trying to investagate all the details related to them. How about you?

Astyx

Oh ...

Kasmir Khaan

:)

Semiclassical

To be more precise: It's the generating function for those numbers.

I presume?

Astyx

Then I'm pretty sure there is a bishop polynomial

Kasmir Khaan

16:54

Hi @Semiclassical

Yes you are correct

Semiclassical

hi

Akiva Weinberger

So I guess the constant term is $1$

Kasmir Khaan

yes allways =p

Akiva Weinberger

and the linear term is $8x$

Kasmir Khaan

yes :D

Astyx

16:54

What ?

Lucyfer Zedd

hey,I've a silly doubt, is a compact subset of an open set(in subspace topo) in the complex plane compact as a subset of the plane ?

Kasmir Khaan

those the easy cases =p

Astyx

Wait, why 1 ?

Akiva Weinberger

@LucyferZedd Yes, compactness is an intrinsic property

Semiclassical

zero rooks.

Kasmir Khaan

16:55

1 way to place 0 rooks on any board

Akiva Weinberger

It doesn't depend on the space it's in @LucyferZedd

Semiclassical

Though I'm not seeing why it'd be 8 ways if there's one rook.

Akiva Weinberger

We're deleting a corner apparently

Lucyfer Zedd

@Aki

Hippalectryon

@Waiting That's great :D nothing special on my end :(

Kasmir Khaan

16:56

1 rook is alone

Astyx

Oh and you're dealing with a $3\times 3$ board where one (bottom right) is missing

Kasmir Khaan

cant take anything =p

Lucyfer Zedd

@AkivaWeinberger How to show that it is closed in the complex plane ?

Semiclassical

Oh, 3-by-3 minus a corner.

Kasmir Khaan

Yes @Astyx

Yes!

Semiclassical

16:56

Yeah, okay.

Astyx

Scheisse

Akiva Weinberger

@LucyferZedd Do you know the "finite open cover" definition of compactness?

Kasmir Khaan

1+8x+14x^2+4x^3

Astyx

I just put molten chocolate all over my bed

Kasmir Khaan

was my answer but I need some conformation i done it right =p

Semiclassical

16:57

...

Lucyfer Zedd

@AkivaWeinberger and also, what do you mean by that it is intinsic?

Akiva Weinberger

I just counted 14 for two rooks

Kasmir Khaan

counted?

Lucyfer Zedd

@AkivaWeinberger yea, okay I will try using that

Akiva Weinberger

@LucyferZedd It doesn't depend on the surrounding space. (Closedness is not intrinsic)

(but compactness is)

@KasmirKhaan Just went through all the cases

Waiting

16:58

@Hippalectryon I go to bed for 15-30 min, very tired here. Some research takes away all the energy you have. I need to rest a bit.

Kasmir Khaan

@AkivaWeinberger hmm that would be hard when dealing with bigger board =p

Hippalectryon

Kasmir Khaan

well how many ways of placing 3 rooks by your method? =p

Hippalectryon

@Waiting Alright, rest well :-)

Akiva Weinberger

Oh, they all need to be in a diagonal with wraparound

so that'd be 6, but two contain the deleted corner so it's 4

Astyx

16:59

I also have 14

Kasmir Khaan

:D

so I did it right:)

thanks all ! :)

Astyx

@Hippalectryon Thank you for that, I'm releaved !

Lucyfer Zedd

@AkivaWeinberger, oh ,yeah, silly me ... hey can you help me with convergence of harmonic functions?

Astyx

And 4 for three rooks

0 above

Akiva Weinberger

@LucyferZedd Probably not

Maybe someone else here could, though

Alessandro Codenotti

17:03

@Akiva is $\aleph_\alpha=\beth_\alpha$ independent of ZFC for all $\alpha\neq 0$?

Akiva Weinberger

@AlessandroCodenotti Isn't that the generalized continuum hypothesis?

Oh, the conjunction of all of them would be

Alessandro Codenotti

GCH says that this holds for all $\alpha$, I'm wondering whether each instance is independent

Akiva Weinberger

I don't know if there's a specific $\alpha$ for which it's provable. I doubt it

Alessandro Codenotti

for $\alpha=1$ that's just CH

Akiva Weinberger

<-- Not a set theorist

I have no idea how forcing works

@KasmirKhaan My guess for how to compute it for larger boards

So you'd want an $n\times n$ board to depend on the $(n-1)\times(n-1)$ version

Secret

17:08

[Chemistry] The horror: Realised that 10 of the structures I calculated are not the lowest possibel energy, meaning that a total of 30 calculations need to be redone!

Akiva Weinberger

If it's the bottom-right corner that's dropped,

either none of the rooks are on the top or right edges,

or one of them is,

or two of them are.

Alessandro Codenotti

Speaking of set theory I don't remember exactly who was interested in the set theory study group, but I'll create a room for that on Thursday or Friday @Mike @Dami @Perturbative @Astyx @Balarka

Akiva Weinberger

For each of those cases, you could do it in terms of smaller boards

looking at what other squares are available

Astyx

Count me in (I might not be here on thursday and friday though, but then on should be fine)

Balarka Sen

@Alessandro I probably won't be available for 10 or so days, but definitely do it!

Akiva Weinberger

17:10

If there were a set called $\in$

we'd have ${\in}\notin{\in}$

Astyx

$\ni$

Alessandro Codenotti

@AkivaWeinberger being able to write $\in\in\in$ is a good enough argument to reject the axiom of foundation to me

Astyx

$\notin \ni$

$\niton$

Sad

Kasmir Khaan

@AkivaWeinberger Thanks but we have 2 formulas so far, one by recursion and other by product , I have to keep reading and do some examples =p

Akiva Weinberger

$\notni$

Alessandro Codenotti

17:11

or regularity maybe, I don't remember what's the usual name

Balarka Sen

$\in$_$\in$

Akiva Weinberger

$\in\!\ni$

$\in\!\equiv\!\ni$

Astyx

I was lied to

Akiva Weinberger

Oh what

Pitchforks! Get your pitchforks here! $-\!-\!\!\!\in$

Astyx

$\in\!\equiv\!\equiv\!\equiv\!\equiv\!\equiv\!\equiv\!\equiv\!\equiv\!\equiv\! \ni$

Balarka Sen

17:13

@Akiva $\pitchfork$

I got mine.

Akiva Weinberger

Wait that's a thing??

Astyx

$\psi$

Balarka Sen

Yeah it's used to denote "transverse intersection"

Akiva Weinberger

$\madeupbullshit$

Balarka Sen

$\newcommand{\madeupbullshit}{END OF WORLD}$

@Akiva edit that message

but don't change anything

Akiva Weinberger

17:16

I see through your tricks

Balarka Sen

$\madeupbullshit$

Alessandro Codenotti

the spacing in that "END OF THE WORLD" is super wonky on my screen

Balarka Sen

mine too

Semiclassical

.endofline

Balarka Sen

@Astyx check discord

Secret

17:30

$\newcommand{\madeupbullshit}{\text{BRACE THE INFINITE VOID}^{\text{BRACE THE INFINITE VOID}^{\text{BRACE THE INFINITE VOID}^{\text{BRACE THE INFINITE VOID}}}}}$

$$\Huge{\madeupbullshit}$$

$$\infty\infty\infty\infty\infty\infty\infty\infty\infty\infty\infty\infty\infty‌\infty\infty\infty\infty\infty\infty\infty\infty\infty\infty \infty\infty\infty\infty\infty\infty\infty\infty \infty \infty\infty$$

Leaky Nun

17:52

@Secret for the love of God

Transcript for

$Mathematics$ Mathematics