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06:01
$S$ is proper if $\forall y\in V, \exists x \in S, P \in V : F_P(x)=y$
$P$ exists. $P$ is class by default. $P$ is set if given $Q$ set or class, $F_Q(P)$ is defined
$\forall P \in V$, Boolean$(P) \in \{0,1\}$
Modify 1:
Axiom of regularity
Typo:
--axiom of limitation of size---
Replaced by:
$R := \{x,y | xRy \}$
$\exists P \in R$ s.t. Dom(P)=Dom(R)
(x,y) not x,y
$R := \{(x,y) | xRy \}$
Ok the above is all messy. Need to rewrite them again in more concise manner...
NB: I am copying stuff from this: https://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Bernays%E2%80%93G%C3%B6del_set_theory
06:17
what are you trying to establish?
I am trying to find a set theory that contains as many weird infinite sets as possible (not necessary well ordered)
> A statement in the language of ZFC is provable in NBG if and only if it is provable in ZFC
so you can't find them here
hmm...
yeah, I think I need something that includes ZF as a subobject, but otherwise go way beyond it
and AC gives a set theory that is not weird enough
and I also like to have a notion of proper class
So, something like:
ZF+proper classes+as many weirdness of infinite sets as possible
isn't Banach Tarski weird?
Sup @BalarkaSen
06:23
It and non measurable sets are weird, but compared to the non AC wilderness, it is too few
how's it going
It's goin
School starts in a couple days
The idea is that since AC is less weird compared to the wilderness, if I get used to thinking in the wilderness, than AC might be comparably easier since it is less weird
@EricSilva So I hear
@Secret ZF-C is quite interesting
amorphous sets, dedekind-finite sets, etc.
06:26
Is there a proper class generalisation to ZF-C?
Excited for math but annoyed because I have to work
#mixedfeelings
I need to learn more math
@Secret no idea
@LeakyNun Hey, still stuck on 9.4?
06:34
@TobiasKildetoft yes
@LeakyNun So you found the order of the automorphism group of the Sylow $p$-subgroup?
@TobiasKildetoft well, p lol
oh shit, thanks done
@TobiasKildetoft does the non-abelian group of order 21 (there's only one) have 7 Sylow 3-subgroups?
@LeakyNun Yes
06:41
:o
so many
gap> StructureDescription(SmallGroup(21,1));
"C7 : C3"
so a^7=b^3=1, ab=bba?
ok, google found absolutely nothing for a proper class non conservative extension of $ZF\neg C$. Guess ... have to construct it from scratch
the 3-subgroups are <b>, <ab>, <a^2b>, etc.?
@LeakyNun Hmm, I think that is the wrong presentation
you don't want the element of order $7$ to normalize the one of order $3$
hmm
so ab=ba^n
you want $bab^{-1} = a^n$ for any $n$ coprime to $7$
(and different from $1$)
06:44
and they're all isomorphic
let's say 2
(Construction will took place in Mathworks to prevent clogging current conversation streams)
ba=aab
yes, they are all isomorphic, as they are semidirect products and the Automorphism group of $C_7$ has a unique subgroup of order $3$
Great, it is deleted by the system
nvm then, reopen a new one
@Secret I'll help you keep your new one from being deleted
(the rule is 2 people 15 messages)
06:48
ok
07:29
I wonder if anyone has studied "fieldoids", like groupoids, but with an additive category where all non-zero morphisms are invertible.
Anyone familiar with BlueJ?
Is that a bird rapper?
@TobiasKildetoft depends on what you mean
I managed to prove the square of an odd number is odd without using commutativity of multiplication... one less step for me :P
I did use the associativity of multiplication tho, so I'll need to see if this requires that
 
1 hour later…
user308168
08:57
@Secret Can I know why you have created the Mathworks room? I mean I want to know what things you do in that room.
That room is mainly to address Ted's and steamyroot's comment that I tend to flood the main chat
thus I just put some of my long [random] there
user308168
@Secret Do you know that you can also do math works in the Mathematics Workshop room? That room has been created to do specific math works there.
But mathematics workshop is a room dedicated to some kind of tutorial like format involving well established maths, if I recall your initial vision, which does not fit the purpose of Mathworks, because that room is more like a roughwork sheet room for me to test ideas (and is created simply because too many people are complaining that I flood the main chat)
user308168
09:13
@Secret No, it has not been dedicated to only tutorial things. That rooms has been created to do any math works from elementary to advanced level and also for testing new math ideas. You can do any specific math works there.
1
A: Prove if $n^2$ is even, then $n$ is even

Kenny LauDefinitions $x\text{ is even} := \exists y[y+y=x]$ (Denote as $E(x)$) $x\text{ is odd} := \exists y[y+y+1=x]$ (Denote as $O(x)$) Axioms $\forall x[S(x) \ne 0]$ $\forall x \forall y[S(x)=S(y) \implies x=y]$ $\forall x[x+0=x]$ $\forall x \forall y[x+S(y)=S(x+y)]$ $[\varphi(0) \land [\forall x...

truly my magnum opus
lunatic
@BalarkaSen :c
i thought logic induces rational thinking
you are the very counterexample to that idea
> This is surprisingly provable in intuitionistic logic.
@BalarkaSen :c
09:23
I gave an upvote. I want to die now.
thanks for your appreciation :)
@Jasper
user308168
09:41
Can we prove the existence of an indefinite metric on an arbitrary manifold?
Not every manifold admits a strictly Lorentzian metric though.
user308168
I can not understand your answer.
For example $S^2$ does not admit a Lorentzian metric of signature (1, 1).
user308168
Why?
Suppose it does
Then upto change of coordinates locally it looks like the billinear form $x^2 - y^2$ on $T_p S^2 = \Bbb R^2$
09:51
someone downvoted :c
The zero locus of the billinear form are the axes $x = \pm y$
So the Lorentzian metric decomposes $T_pS^2$ as direct sum of two 1-dimensional subspaces.
You can do this for all $p$, so you'd have decomposed $TS^2$ as direct sum of two 1-dimensional subbundles.
But that's impossible, by hairy ball theorem.
user308168
You know that this proof is very heavy for me?
If you don't know hairy ball theorem you can't prove this.
user308168
So only definite metrics can be extended to the whole of a manifold. Right?
Every manifold admits a Riemannian metric, yes.
user308168
09:58
Every manifold or every compact manifold?
> Let $G$ be a finite group and $H \le G$. Then, there is a maximal subgroup $K$ of $G$ such that $H \le K$.
Isn't this obvious? @BalarkaSen @TobiasKildetoft
@WDNWBM Every paracompact, Hausdorff, second countable manifold.
@LeakyNun Sure
@TobiasKildetoft so I can skip? :P
@LeakyNun That is up to you
10:01
@TobiasKildetoft I mean, all proper subgroups containing $H$ form a partially ordered set by inclusion, and apply Zorn's lemma
@LeakyNun Yeah, don't use Zorn, as it fails for infinite groups in general
@TobiasKildetoft right
user308168
Thanks.
11:28
Hi @AlessandroCodenotti
11:44
:O Astyx is here
Did uni start yet in france?
Yup
Except for me it wasn't Uni, but military training :p
@Astyx How long will you be doing that?
Still another 6 month
Damn Ao Oni is hard
But from now on the training will be more specific to my branch of the army (gendarmerie)
11:48
Hi @Astyx
@Astyx Wow, that's still a long time. I fortunately only had to do 4 months (and in the mean time they have completely removed the compulsory military training, so now it is all voluntary)
Why do you have to suffer a military training
@BalarkaSen Because military training is not pleasant?
hah I meant what's the rationale behind having military training as compulsory
@BalarkaSen I suppose the same reason, with the rationale that you want enough people to do it
11:58
Heh
Thank for the Belgian army being so broke they couldn't afford drafting :P
I give up
This game is too hard
1
Q: Summation involving double factorials

user42819$${ \sum_{r=1}^{\infty}\frac{(2r)!!}{(2r+3)!!}}$$ What approach do we need to solve such type of summations? My attempt: \begin{align} \sum_{r=1}^{\infty}\frac{(2r)!!}{(2r+3)!!} &= \sum_{r=1}^{\infty}\frac{(2r)!!\cdot (2r+2)!!} {(2r+3)!!(2r+2)!!} \\ &= \sum_{r=1}^{\infty}\frac{2^r\cdot (r)!\cd...

Someone please help me with this question
12:25
if you put $2^{63}$ in the Googel-Calc, you will get $9.223372e+18$. What does $e+$ mean?
Exponent
Basically, $e+18$ means $\cdot 10^{18}$
@TastyRomeo so, Google thinks that the rest of $9223372036854775808 - 9.223372e+18 = 36854775808$ is just junk? What is the name of a such representation?
Erm...
This is just floating-point arithmetic?
Computers and infinite precision don't mix well in general.
@TastyRomeo understood. Is that a scientific notation or just a representation using the floating point number?
It's pretty standard notation
Like, basically any graphic calculator I know uses it
12:38
@TastyRomeo wow!
Hello
I have seen the mantissa-story only once, and it was in the Java course. Never in math.@TastyRomeo
hi @Daminark
How's it going?
12:41
Yo
Hi everyone
Sup Norman
@TastyRomeo may I also ask you about these two? $4e-45$ and $4028235e38$? I can lead out that $+e$ stands for multiplication from our Google example. So, maybe $e-45$, too. What's about just $e38$?
I spoke with the responsible people and it looks like I might be allowed to take a grad course this year
I imagine e38 is the same as e+38
4e-45 is just negative power
@AlessandroCodenotti Oooh, grats!
12:44
@Daminark are you already in Germany? :)
Wait Germany?
Am I German now? That's pretty sick
(It would actually be cool to be German but I'm going to Chicago, actually)
@AlessandroCodenotti which one?
Chicago? Is that near Frankfurt?
@Daminark Logic (it's not an advanced logic course or anything weird, it's just that we don't have an undergrad one here)
@Daminark not necessary. I thought you got some place in Hamburg, sorry if I am mistaken.
Come on Alessandro look at my name, I'm from Denmark!
@Kirill nah it's fine, don't worry about it :P
@Daminark oh, of course, my bad
12:46
@TastyRomeo Norman the Normie
@BalarkaSen you're so normie I think you're gonna functional analysis
But yeah I'm actually deeply conflicted on logic still. Honestly I might hold off bio to 4th year to do both
Both = grad analysis and logic
Like, grad analysis seems to be the "smart" or "practical" choice
my reply to you is on discord
what does grad analysis mean in terms of topics covered?
analysis + logic = analogic
2
But as I said, the first half of the class will be relearning stuff like Radon-Nikodym and all and I just prefer logic.
So I know we'll be using Bass Real Analysis for Graduate Students
12:50
analysis + logic = normie, okay? It's the most bland blend of vegetarian soup a normie can drink
You're already taking model theory so having logic as well would seem like a sensible choice
Logic is model theory
Well it's a 2-quarter class
First quarter is model theory, second is computability
normie is too rude. I like analogic.
12:52
Then there's a grad computability as well
everything below higher topoi is normie
analysis+logic+algebra+geometry=
NaN
Algebraic Geometry
Well I dunno how much logic comes up
elementary topoi, @Daminark
12:53
proves Hodge via ultrafilters
In mathematical logic, algebraic logic is the reasoning obtained by manipulating equations with free variables. What is now usually called classical algebraic logic focuses on the identification and algebraic description of models appropriate for the study of various logics (in the form of classes of algebras that constitute the algebraic semantics for these deductive systems) and connected problems like representation and duality. Well known results like the representation theorem for Boolean algebras and Stone duality fall under the umbrella of classical algebraic logic (Czelakowski 2003). Works...
Look what I found
@BalarkaSen except engineering math. Then you're just a square, maaaan
how about algebraic logic geometry
algebraic probability theory?
But yeah grad analysis is first quarter real + basic functional, second quarter functional + PDE, and this year in particular third is gonna be analytic number theory
12:55
i prefer mathematical mathematics
Algebraic statistics is the use of algebra to advance statistics. Algebra has been useful for experimental design, parameter estimation, and hypothesis testing. Traditionally, algebraic statistics has been associated with the design of experiments and multivariate analysis (especially time series). In recent years, the term "algebraic statistics" has been sometimes restricted, sometimes being used to label the use of algebraic geometry and commutative algebra in statistics. == The tradition of algebraic statistics == In the past, statisticians have used algebra to advance research in statistics...
in math? @BalarkaSen
Categorical statistics
on that note, let us stop mixing terminologies and branches of mathematics up
@BalarkaSen pointless topology
I actually wish I understood free probability theory
The hairy ball theorem of algebraic topology (sometimes called the hedgehog theorem in Europe) states that there is no nonvanishing continuous tangent vector field on even-dimensional n-spheres. For the ordinary sphere, or 2‑sphere, if f is a continuous function that assigns a vector in R3 to every point p on a sphere such that f(p) is always tangent to the sphere at p, then there is at least one p such that f(p) = 0. In other words, whenever one attempts to comb a hairy ball flat, there will always be at least one tuft of hair at one point on the ball. The theorem was first stated by Henri Poincaré...
12:58
I still can't get over TQFT though, and it's actually super dank. At some point @Semi I'm gonna need to ask you why this bordism category stuff actually has anything to do with quantum mechanics
I'm the wrong person for that tbh. Look for stuff by John Baez though
lol Hairy-ball theorem vs. No-hair theorem
Yup
Different kind of hair and different kind of ball, tbh, but still
@Daminark I did offer that we should learn from Lurie's TQFT paper
True, and once I get through some more stuff I do wanna check that out
13:02
For instance see the second page of this: math.ucr.edu/home/baez/rosetta.pdf
(There's probably more in there but the sentences there are pertinent)
Yeah, search for 'bordism' in there and you'll get some nice stuff
cubordism = no result found
Next: Found the most cubic mathematics discipline
cobordism.
13:27
[Random]
Some arbitrary axioms to construct a choice free set theory
Original: (ZF-C)
1. Check whether two sets are equal
2. Existence of unordered pairs
3. Performing unions
4. Existence of $\omega$
5. Produce sets by using maps or relations
6. Produce power sets
New: (???)
What number systems exist that anyone can think of where $0.9999.... \neq 1$
1. Check whether two sets are equal
2. Existence of unordered sets of arbitrary number of elements
3. Performing unions
4. Existence of $\omega$
5. Produce sets by using maps or relations
6. Produce power sets
7. Infintary logic
2. is a beefed up version of the Axiom of pairing, so that given an arbitrary number of sets S, one can always form a new set containing all elements S
(though I had a suspicion it is equivalent to the axiom of choice...)
but again, this will be analysed later after we construct the ordinals in ZF-C
The usual construction of the ordinals is done in ZF
ok, so we don't need choice to construct them
Actually, I am currently trying to figure how to construct $\omega_2^{CK}$, in particular, since this ordinal must contains non recursive but countable ordinals, how to ensure all non recursive ordinals < $\omega_2^{CK}$ were constructed since we cannot produce them from a recursive map applied to the naturals?
13:58
@Secret amusingly, QBism is a thing
There's also cubism in fine arts
QBism basically reformulate born rule with bayesian reasoning, but I don't claim I really understood it
There's a rule called the urgleichung which specify how a state assignment is being updated after a series of measurements
same
i sorta like the idea that most of what seems weird about quantum mechanics is just reflective of information / subjective probability
but I haven't read it enough to have a handle on it
I, however, more lean on psi-ontic interpretations and thinking that measurement is a class of non unitary interactions that is still yet to be figure out
my own sense of where things are weird in QM is in the realm of Bell's theorem i.e. nonlocal correlations
14:14
Indeed, and they are the most interesting thing in quantum mechanics
and since the weirdness there is vis a vis the spirit of relativity theory, I kinda think that the right setting for that is QFT. so I sorta think that the weirdness of QM is really a question about the weirdness of quantum fields.
for instance, if you entangle a pair of electrons, then you're taking advantage of the fact that they satisfy the Pauli principle (and implicitly the fact that these electrons are quantized excitations of a quantum field)
(specifically, a fermionic field)
Is it valid to say that in Hyperreals and surreals, $1 - 0.9999.... = \epsilon = \frac{1}{\omega}$ ?
it seems strange to me that we talk about Bell-type experiments w/r/t relativity theory while at the same time we don't have in mind that the interaction between the entangled particles is due to them being part of the same quantum field
(at least, I think that's true. Is there a Bell-type experiment involving two distinguishable particles?)
@LeakyNun thanks i actually got it from that hint
15:03
15
A: Closed form solution to $\int_0^1\arctan^2(x)\,\sqrt{x}\,dx$

Cleo$$\frac{\pi^2}{24}-\frac{2\pi}3+\frac1{36\sqrt{2}}\left[5\pi^2+12\left(4+\ln\left(\frac{1+\sqrt2}2\right)\right)\left(\pi-2\ln\left(1+\sqrt2\right)\right)-48\operatorname{Li}_2\left(\sqrt2-1\right)\right]$$

@Semiclassical are you done with the question above?
@Secret did you try it?
Let me find one more integral
This one is super simple
20
A: Integral $\int_0^1\frac{\log(x)\log(1+x)}{\sqrt{1-x}}\,dx$

Cleo$$\int_0^1\frac{\ln(1+x)\ln x}{\sqrt{1-x}}dx=16-8\ln2+4\ln^2\left(1+\sqrt2\right)\\+\sqrt2\left[2\ln^22+8\left(\ln2-1\right)\ln\left(1+\sqrt2\right)-\frac{7\pi^2}3+16\operatorname{Li}_2\!\left(\frac1{\sqrt{2}}\right)\right].$$

it would be interesting to see since Cleo's answers are blatantly computer-aided
That's even less context than the average Dr. Sonnhard Graubner answer
Who dat?
sonnhard is pretty intense
@Daminark resident trolololol
Strangely enough, every now and then he writes a perfect answer, that includes the context/strategy and intermediate steps
15:11
rofl
iseewhatyoumean.jpnsky
@TastyRomeo Link one, I don't believe you :P
@GeorgeCoote I imagine that you'll find some by looking for his highest voted answers
It's kind of sad, everyone knows him as a troll, so a lot of his answers get automatically flagged
Most of them rightfully so, but the occasional good ones too :(
I only just investigating harmonic numbers and some riemannian zetas, I don't think I know how to deal with polylogs yet
thus I will pass for those two integrals for now
Troll isn't really the right word
15:16
Well, yeah, he tries to be helpful in his own way I guess
He does have 2807 answers :P
Even if only one out of twenty answers is actually helpful, he's helped more people than me
Does he give as short of an answer as possible or something?
Usually he just plugs stuff in maple or mathematica and posts the answer
let's just say that sometimes his answers become brilliant pieces of dadaist poetries
2
Wait, what?
to be fair he's not like cleo
Guys, say we have a random variable $X$ with density function $F$. And let $F^{-1}$ be the quantile function. It is true then that $P(X\leq F^{-1}(y))=y$?
15:21
Windows 10 suggests "Minecraft: The Island" for me to play
often he posts the result in the comments then writes an answer, which is fair enough I suppose
this is kind of what I need in a proof
Hi Sha
$$\int \cdots\cdots^{\cdots^{\cdots}}dx$$
hey, wow
it's been a while @Astyx
15:22
Indeed
How are you doing ?
I'm alright, how about you?
did you get enrolled?
(Attempt to integrate an uncountable set failed miserably)
https://math.stackexchange.com/questions/799391/uncountable-set-from-riemann-integral
In some sense, yeah :p
uh, in a good sense?:)
I'm just back from 3 weeks living in a military camp
15:23
oh right
It was an interresting experience
sounds intense
that explains why I didn't see you here the past couple of weeks:P
What about you ?
not much really. I was kind of weak the whole summer, so I was keeping everything at a slow pace. but now I'm practising push ups and I do fitness stuff:P so I'm getting stronger
Weak ? Why ?
15:26
because she does not eat healthy
smoker too
I just had lowered energy for some reason
and drinks lots of wine
like at 6pm I would already be crushed. but now I'm feeling much better
lol wut @Kasmir XD
don't listen to him :P @Astyx
If you want energy, try some bacon & eggs :^)
She does not eat bacon
15:27
lol, no thanks, I don't need dead animals on my plate :P
kind of barbaric :P
Why does nobody understand :^) anymore ;_;
I never use that :P
that's why we need normal smileys here
like I already suggested
if you want energy use cocaine
Commando bars are good too
so people can understand each other
15:34
@Daminark Yo I have a gravatar suggestion for you
Gold
Anyhow, say we have $X_1,\dots,X_n$ continuous random variables that have distribution function $F$. Let $U_i=F(X_i)$ for all $i$. I have to show that $U_1,\dots,U_n$ form a sample from a uniform distribution on $[0,1]$. I was thinking of first considering $y\in[0,1]$. Then we want to show that $P(U_i\leq y)=y$. So I was thinking of something like this: $P(F(X_i)\leq y)=P(X_i\leq F^{-1}(y))=y$, but I’m not sure if this is correct.
$F^{-1}$ would be the quantile function, btw
Beautiful
Also @TastyRomeo I now want to make the same meme I made of @Sha about you regarding emojis
Or emoticons or whatever the fuck
I'm on my phone so I can't do it
loolzz
I'm not sure I'll ever get used to this name change:P
I betrayed Steamy on a bet, so credits should go to me
15:43
tasty romeo is a great name - I think it's a rerun?
you said you could find another anagram? @balarka
@Sha I said if he changes his name to this, I'll change mine to N J Wildberger
haha alright
I did actually change my avatar to the front cover of njw's rational trigonometry book for a while
Yeah Balarka's a dirty liar. Not gonna trust his papers
Also Romeo you gotta send me an audio recording of yourself saying "My name is Sam Tertooy" so I can change the ooy to Big Smoke's thing
15:49
looool
My nama Sam Tertoohhhhh
(In Jeff's voice)
Hi @Secret. Wanted to ask you is it okay if I study chemical kinetics before chemical thermodynamics? (need your guidance coz you are a chemist :))
@Abcd no you will be executed if you try
(Jk I dunno, but I imagine it's okay? Still ask the expert)
Lolol
@Daminark I am really confused to be honest. I asked two seniors and they said I won't understand aught in kinetics if I do that but my teacher said that I will. Now I need expert advice from Secret.
If anyone else here has an answer, please let me know?
Yeah you need to understand the equilibrium constant and friends
@BalarkaSen Yes, I have studied the chapters "Chemical equilibrium and Ionic equilibrium".
15:58
Not thermodynamics at it's fullest strength, but interacts a lot with it (K has direct affinities with the Gibbs' free energy)
@BalarkaSen Which one do I begin with- Kinetics or thermodynamics? I am done with equilibrium stuff.
I say go for kinetics.
Okay, thanks for your advice. I'll follow it.

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