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6:30 AM
Is $C_n(A\cap B) = C_n(A) \cap C_n(B)$, where $C_n$ is the singular chain group?
 
how would you try to prove that?
 
@MikeMiller Is that for me? Well, an element of $C_n(A\cap B)$ is a formal sum of singular simplices whose image lie in $A \cap B$, whereas an element of $C_n(A) \cap C_n(B)$ can be written as a formal sum of singular simplices in $A$ and also as a formal sum of singular simplices in $B$.
So, the inclusion $C_n(A\cap B) \subset C_n(A) \cap C_n(B)$ is obvious.
I am not so sure about the other side.
 
6:56 AM
Morning all!
 
Maybe, we have to look at $C_n(A \cup B)$. If a chain has two decomposition: as formal sum of singular simplices in $A$ and as formal sum of singular simplices in $B$. Then, by freeness of $C_n(A \cup B)$, both the decompositions would have to have the same coefficients?
 
@Ted I very much hope so!
 
@feynhat That's what I wanted.
 
7:28 AM
morning !
 
Heya @Flowian
 
7:43 AM
May I present a rider with imaginary numbers:

Compute $\prod_{k=1}^{50} i^k$
I tried to solve it minutes ago, but when looked at answer from textbook, it was wrong.
 
Should just be $-i$, right?
 
correct
:)
I got -1
How did you do it ?
 
Well, $\prod_{k=1}^{50}i^k=i^{\sum_{k=1}^{50}k}$ because of the properties of exponents (I'm on phone right now. Cannot confirm how the latex looks)
 
My solution:

$\prod_{k=1}^50 i^k = i^{\sum_{k=1}^50 k}=i^\frac{50\cdot 51}{2}=i^5^5\cdot \sqrt{i}(\sqrt{i})^50=i\cdot i^25$
 
From there you substitute in the triangular number formula ($\sum_{k=1}^{n}=\frac {n(n+1)}{2}$ and since $n=50$ we get $i^{1275}
 
7:54 AM
ah yes, did the same thing
 
ahh nah ignore
 
Yeah, $i^n=i^{n\text{mod}4}$
 
how to maximize 2 variable function like this $ \frac{xy}{xa+yb} $ x,y are variables.
 
Haven't heard of order in group theory before
 
And 1275 mod 4 is 3.
 
7:56 AM
@flowian this is the smallest positive integer such that $i^k = 1$
 
oh, so 4
 
Right
so the exponent $n$ of $i^n$ only matters up to multiples of $4$
 
At the risk of confusion, order can have two meanings, also. The order of a group is the number of elements in that group. The order of an element is what Meg said.
 
if $n = 4k + r$ then $i^{n} = i^{4k + r} = i^{4k}i^r = 1 \cdot i^r$
I'm slightly saddened that the short version of my name has become Meg in this room hahaha
 
You can always change it :P
 
8:00 AM
For me it's because the symbols of the rest of your name are not on my phone keyboard.
I could call you "num," maybe.
 
yeah I know don't worry lol, it's just funny
nah continue with Meg
it makes my gender ambiguous
lol
 
Observation: numMeg sounds like nutmeg.
 
lol
i gotta stop saying lol
 
Lol
 
@Alessandro I got rejected from Heidelberg again!
 
8:06 AM
I did parial defferentiation , set them equal to zero.
 
@ÍgjøgnumMeg Wait what?
 
which gives $ x^2 b= y^2 a $
 
@Alessandro yeah, the admissions people last time I got rejected (which was my fault, I got my app in late) told me it was a sure thing if I just sent in the application again for the Wintersemester, so I did that and I just got a rejection letter through the door saying I was missing official seals on my transcript and my degree certificate
so they outright rejected me on a technicality, which means I also lose my scholarship!
(I'm ranting in this room because it makes it easier to parse the ridiculousness of the whole thing)
 
That's bs... So what are you doing?
 
Well Mathein is going to speak to his advisor (who is head of the board of examiners) and I have emailed the admissions team and the dean and the head of mathematics
to see if they can't overturn the decision
Otherwise I think I might top myself
 
8:12 AM
Anyone?
 
Urgh that sucks... good luck
 
Thanks :(
 
 
1 hour later…
9:34 AM
@ÍgjøgnumMeg come on pal, don't think that way. It's just a silly plastic seal; I'm sure they'll reconsider.
 
I'm just being overdramatic lol
 
::phew::
:-)
 
partially
 
:-/
Murray Gell-Mann said that on YouTube about getting into MIT
RIP
 
morning everyone
 
9:45 AM
morgen
 
Grüezi mitanond
 
10:03 AM
\o @ÉricoMeloSilva
 
im not here
 
lol
 
 
1 hour later…
11:17 AM
compulsory volunteer programs are an excellent initiative. Removes the hassle associated with these so called "human rights" conditions they try to impose on small business owners that need obedient slaves
 
Are you being forced to do volunteering @Adam?
 
@AlexClark no I'm not champ
 
Ah okay pal
 
@Secret lol poor Murray Gell-Mann perhaps he should have cultivated a belief regarding how easy the seizure of power is despite what he had previously been led to believe
That's the beauty of classical capitalism, what you believe to be called empathy is simply a synonym for misplaced guilt
 
Have you read any Nietzsche Adam?
 
11:29 AM
@Alex Clark yes I have one of his books I very much appreciate some of his ideas and find others to be complete bigotry so it's difficult to present an opinion of him that isn't contradictory
 
Sure
I haven't got to the bigotry yet I guess
 
Or haven't understood why it is, same thing really. That's why reading the same thing again at a later date isn't a pointless exercise
 
Yeah, I'm working through thus spoke zarathustra slowly atm
 
Ah ok. I only have Beyond Good and Evil
 
I think that one is meant to expand on zarathustra, so I'll read it next I think
 
11:32 AM
It's an incredible book he was a brilliant man despite his pride
The nature of his foolishness can been seen in contemplating his cause of death, if he had a better understanding of how irrelevant gender is to intellect he might not have had to pay for sex his entire life
 
12:01 PM
pfffff I need to keep track of time. Me and my old mate Satan have another nice guy convention to attend.
 
 
2 hours later…
2:05 PM
@Balarka I have some confusion concerning what precisely is the relationship between (small) $\infty$-groupoids and topological spaces
I would like to say that an $\infty$-groupoid is the same as a space, up to weak homotopy equivalence, but it seems to me that there's extra information in the groupoid, such as the cardinality of the space
Also is there a construction a construction that starts with an $\infty$-groupoid and spits out an honest to god space with that groupoid as fundamental groupoid?
I guess what I'm really looking for is an adjoint to the functor sending every space to its fundamental groupoid
 
@Alessandro geometric realization
$\infty$-groupoids are quasi-categories, so they're simplicial sets, so you can take the geometric realization
 
Oh damn is this the nerve of a category/simplicial set business?
 
I got sniped, yes it is
I got sniped again lol
 
regarding your concern about cardinality, there's a notion for homotopy equivalence between simplical sets as well
 
2:13 PM
What I was trying to say is that $S^1$ and the pseudocircle are weak homotopy equivalent, but have nonequivalent fundamental groupoids
So I guess we need to relax the equivalence of groupoids, which is the notion of homotopy equivalence you're talking about
 
so $\infty$-groups modulo homotopy equivalence correspond to spaces modulo weak homotopy equivalence
yeah exactly
 
I see, makes sense
 
yeah can do the whole homotopy business on the simplicial set side, you have fibrations, cofibrations, weak equivalences etc.
coverings of groupoids exist as well
I don't know much about $(\infty,1)$-categories or model categories, so take what I say with a grain of salt
 
I see
I just want to understand type theory tbh
 
I've got the impression that simplicial stuff is really en vogue right now, due to HTT and stuff
 
2:23 PM
Oh Jesus not this stuff
 
a friend of mine writes his undergard thesis on simplicial artinian rings and Galois representations
 
@RyanUnger lol
@Ryan I posted a still (still unanswered) analysis question on MSE a couple of days ago if you want some real math (it's not really your flavour of analysis I'm afraid)
 
I’ve never heard of s-infinite
 
s-infinite means that the measure can be written as a sum of countably many finite measures
A $\sigma$-finite measure is also s-finite: Suppose $\mu$ is $\sigma$-finite, cover the space with countably many $E_n$ and consider $\mu_n(A)=\mu(A\cap E_n)$, those are all finite measures and $\mu=\sum\mu_n$
 
@Mathein Would ya mind verifying smth for me? It's from the ANT sheets :)
 
2:28 PM
@ÍgjøgnumMeg sure
 
Alright.. let $S$ be a multiplicative subset of $R$ with $1\in S$ and let $\tilde{S} = \lbrace s \in R : \exists s^\prime \in R\ \text{with}\ ss^\prime \in S\rbrace$. Then $\tilde{S}$ is multiplicative (which is obvious I think?). To show that $\tilde{S}$ is saturated, you can show that $R\setminus \tilde{S}$ is a union of prime ideals. Right?
(Also this $\tilde S$ looks remarkably like the inverse of a fractional ideal)
(but that's not relevant here js I noticed that lol)
 
hmm, yeah you could that
 
Do you have an easier way? Lol
 
I think I would just do it directly. If $st \in \tilde{S}$, then $\exists s' \in R$ with $sts' \in S$, but this shows that $s \in \tilde{S}$ by definition of $\tilde{S}$
 
Ah yeah
because $ts^\prime$ puts you in $S$
lmfao, well I can prove that $R \setminus \tilde S$ is a union of primes so whatever flips table
Cheers!
 
2:39 PM
singing: It's the never ending stoooo-wee ahhha ahha ahhhaaaa… the never ending stoooo-wee ahhha ahha ahhhaaaa…
 
@AlessandroCodenotti What's an example of something that's s-finite but not sigma-finite?
 
I only know a stupid example, from wikipedia
Consider the space $\{a\}$ with the only possible $\sigma$-algebra and let $\mu=\sum \mu_n$, where every $\mu_n$ is the counting measure
 
@AlessandroCodenotti aka every counterexample in measure theory
 
It is s-finite by definition, but not $\sigma$ finite since $\mu(\{a\})=\infty$
 
oh dear
 
2:45 PM
@RyanUnger Sometimes they're more satisfying than this one though
(I've also been doing some honest analysis in the weekend though, Schatten norms and Dixmier traces)
 
I only ever deal with Radon measures on (separable) manifolds
 
As the dual of some space of functions?
 
the Riesz theorem appears frequently yes
 
Are your manifolds also compact or do you like Stone-Cech compactifications?
Ah, no need for fancy compactifications, they're locally compact in any case
 
Stone Cech requires AoC so I don't use it
in any case I've never seen it used in geometric analysis
 
2:54 PM
The point is that if you take $C^b(X)=\{f\mid f\colon X\to\Bbb R,f\text{ bounded}\}$ then there is a nice description of $C^b(C)^\ast$ when $X$ is locally compact
 
Yes I know
 
If $X$ is not locally compact $C^b(X)$ is still a commutative $C^\ast$-algebra so it is isomorphic to $C(K)$ for a compact Hausdorff space $K$. Turns out that $K=\beta X$ is the correct space here
So $C^b(X)^\ast$ is the space of Radon measures on $\beta X$
I'm probably forgetting adjectives around measures
 
I like $\beta X$ because for discrete $X$, it's the spectrum of a product of fields
 
I prefer to think about it as a space of ultrafilters
But there was this question on an AG problem set last semester on whether an affine scheme can be totally disconnected, $\beta \Bbb N\simeq\mathrm{Spec}\prod\Bbb F_2$ was handy
 
2:58 PM
compact Hausdorff spaces are just algebras over the ultrafilter monad anyway
 
@RyanUnger LOL
 
by Whitney and Nash I'm only ever looking at subsets of $R^n$ haha
well this is not quite true, for Ricci flow you need to know some things about Alexandrov spaces
 
totally disconnected is equivalent to 0-dimensional and to Hausdorff and to T_1 iirc
for affine schemes
 
actually you might be able to embed nice Alexandrov spaces into R^n
 
3:14 PM
 
Are your Alexandrov spaces separable? In this case they can be embedded at least in $[0,1]^{\Bbb N}$
 
@AlessandroCodenotti I imagine they are. Alexandrov spaces show up when you take a limit of Riemannian manifolds without an injectivity radius estimate
so you get collapsing
 
 
2 hours later…
5:05 PM
Hi All. I'd let you know about my question and I would appreciate help
5
Q: Use Fatou Lemma to show that $f$ takes real values almost everywhere.

galleta Let $(f_n)$ be a sequence in $L^p$ such that $$ \| f_{n+1}-f_n\|_{p} <\frac 1{2^n} \ \forall n\in\mathbb{N} $$ and let $$ f: X \to [0,\infty], \ x \mapsto \sum_{n=1}^\infty| f_{n+1}(x)-f_n(x)|. $$ Use Fatou Lemma to show that $f$ takes real values almost everywhere. We must show the...

 
5:35 PM
Hello guys! I have a very basic question
Suppose we know that $p\to q$ is true. We have $\neg p$. What can we conclude?
This question is flying around my mind all the time and I do not know how to answer it
I have examples and things like that, but idk what to conclude if we have $\neg p$ (if we had $p$ then by Modus Ponens we have $q$)
 
@manooooh You can't conclude anything from the implication
 
Hi @Tobias
 
@MatheinBoulomenos Hi
 
Hi @Tobias and sleepless @Mathein :)
 
Hi @Ted
 
5:39 PM
Hi @TedShifrin
 
@TobiasKildetoft the problem was that in a exam, the statement says "Find something". As always, one has to check if the hyphotesis are true. But the hyphotesis were not true, so the statement fails. So I answer "It is not possible to find it because the hypothesis does not hold". Did I answered correctly?
Watch that the teacher ask supposing that we could answer the question.... :(
 
@manooooh I am not sure I follow
 
@manooooh: Sometimes (not always) it helps to try a "real-life" sentence (either mathematical or not). If it is raining, then the ground is wet. OK, so if it isn't raining right now, can I conclude anything about the ground? ...
I didn't follow that, either.
 
@TobiasKildetoft shortly, the teachers gave a statement of the form $p\to q$ (it was true) and $p$ was not true, so..?
@TedShifrin well, no
 
Well, if it is also true that $q\to p$, then you could conclude something.
 
5:41 PM
@manooooh I am not sure what you would have been meant to conclude in any case
 
So it depends on the particular $p$ and $q$.
 
@TedShifrin no, $q\to p$ is not ture
 
As in my rain/ground example, then, you have no conclusions to draw.
 
Do you know Jacobi method?
@TedShifrin the problem is that the statement asks for an answer, and that blocked my mind!!
 
I think there's something more specific going on in your teacher's case.
 
5:43 PM
Does anyone know of research or OEIS entries dealing with integer partitions of both negative and positive numbers? And including zeros? The negative and positive numbers are bounded.
 
I think it was something like "Find $x-y$ if $x=1$ and $y=2$ in the set of positive integer [natural] numbers"
We can't give a positive answer ("Find")
 
@MatsGranvik Hmm, given a bound on the negative ones, does that not reduce to another case with non-negative numbers, where you then apply appropriate new upper bounds?
 
@Tobias I came up with a proof for the number of real irreps of a finite groups using Galois descent
 
@manooooh: So either the teacher made a silly mistake or the answer is "no solution."
 
@MatheinBoulomenos nice
 
5:47 PM
@TedShifrin the teacher did not make a silly mistake because he said "Guys, remember the class about Logic, if we know that $p\to q$ holds, what can you have whether is $p$ true or not?" So I think he puts that exercise to despite some students :P
So I think is your second option, "No solution"
 
Interesting, what are $p$ and $q$ here?
 
@TedShifrin well, I do not remember exactly (it was part of an exam), but something like:
 
@Tobias so the idea is this, if we have a finite Galois extensions $L/K$, then one can form the twisted group algebra $A=L^*[\mathrm{Gal}(L/K)]$ where one defines $\sigma \lambda = \sigma(\lambda)$, then one can show that $L^*[\mathrm{Gal}(L/K)]$ is Morita-equivalent to $K$. In fact $L$ is an $A$-module and the map $A \to \mathrm{End}_K(L)$ is injective by Dirichlet's theorem on independence of characters and it is surjective by dimension, so $A$ is a matrix ring over $K$
 
I'm just saying that the problem you just posed doesn't easily fit that format you're trying to use.
 
@MatheinBoulomenos Where $k$ is what?
 
5:51 PM
@TedShifrin do you mean the problem "Find $x-y$"? Well I just gave an example but did not know if it fits in our situation, let me remember about the original statement
 
oops
$k=K$
 
so basically from the construction of $A$-modules over $A$ are $L$-vector spaces equipped with a $\mathrm{Gal}(L/K)$-equivariant action of $\mathrm{Gal}(L/K)$ in the sense that $\sigma(\lambda v)=\sigma(\lambda) \sigma(v)$
the functors from the Morita equivalence are given by extending scalars along $K \to L$ and keeping track of the Galois action on $L \otimes_K V$ and for the other direction, taking invariants under a semilinear Galois action $W \mapsto W^G$
 
@TedShifrin I think it was something like: "Let $AX=B$ be a system with $A=\begin{pmatrix}3&5\\5&20\end{pmatrix}$ and $B=\begin{pmatrix}19\\79\end{pmatrix}$. Using Jacobi Method's, find the approximate solution know that $X^{(0)}=\begin{pmatrix}3\\4\end{pmatrix}$ is the initial condition. Find the absolute error of each iteration using the $\rho$-norm"
 
So using the category equivalence of A-Mod and K-Mod, we obtain a category equivalence of the functor categories [G,A-Mod] and [G,K-Mod], i.e. we can think of representations of G on K-vector spaces as representations on L-vector spaces that commute with a given semilinear action of the Galois group
 
5:56 PM
The students know that the method of Jacobi converges if the matrix of coefficients $A$ is diagonally dominant, that is, $$\underset{i\neq j}{\sum_{j=1}^n}|a_{ij}|$$
Clearly the given statement asks for the converge of the aproximate solution, meaning that $p\to q$ is true
 
now take a finite group G and consider the regular representation $\Bbb R[G]$, every real irrep is a direct summand of that. Extending scalars to $\Bbb C$, we obtain $\Bbb C[G]$. Decompose $\Bbb C[G]$ into complex irreps, the Galois group action acts on those complex irreps
 
But they give us $\neg p$!!!! Since $A$ is not diagonally dominant, since $3<5$, or , permuting rows, $5\geq3$ holds but now $5<20$!!
 
as we obtain a order-preserving bijection from the subobjects of $\Bbb R[G]$ and the subobjects of $\Bbb C[G]$ as a $A[G]$-module, irreducible $A[G]$-modules corresond to irreducible $\Bbb R[G]$-modules
 
$$|a_{ii}|\geq\underset{i\neq j}{\sum_{j=1}^n}|a_{ij}|$$ sorry
 
Right, it's not diagonally dominant. But I'm asking if that condition is necessary, not sufficient, for convergence.
 
6:00 PM
as th Galois group has two elements, there are two possibilities: either it fixes an irreducible $\Bbb C[G]$-submodule, then it is an irreducible $A[G]$-submodule, or it takes a complex irrep $V$ to its dual, then $V \oplus V^*$ is an irreducible $A[G]$-submodule
 
You need to know that to say automatically that the iteration diverges.
 
@MatheinBoulomenos Nice way to do it
 
So it remains to compute the orbits of the action of the Galois group $\Bbb Z/2\Bbb Z$ by acting via dualizing on the complex irreps, or equivalently of the Galois group on the space of class functions via pointwise complex conjugation (as we have a basis by irreducible characters)
 
@TedShifrin well, I am reading my notes and we did not see the reciprocal
The teacher gave us the following theorem:
Theorem Let $AX=B$ be a linear system of equations. If $A$ is a diagonal dominant matrix $\Rightarrow$ the indirect methods have solution no matter the initial value $X^{(0)}$
I think the $\Leftarrow$ is not true (the teacher did not say anything about it)
 
Given an operator $T$ on a vector space $V$ and its minimal polynomial $m_T$, what exactly does it mean to say that $m_T$ is the largest invariant factor of $V$? What exactly is an invariant factor of a vector space?
 
6:05 PM
Yes, most likely it will converge if you pick a very clever initial condition (e.g., the answer) :P But I'm just saying that you have to do some computation here to show divergence.
 
@Tobias Let $G/\sim$ be the set of conjugacy classes, and let $\chi_1, \dots, \chi_c$ be the irreducible characters, then consider the map from $G/\sim \to \Bbb C^c$ via $[g] \mapsto (\chi_1(g), \dots, \chi_c(g))$, extend this linearly to an isomorphism $\Bbb C^{G/\sim} \to \Bbb C^c$ (as the $\chi_i$ are linearly independent).
Then because of $\chi_i(g^{-1})=\overline{\chi_i(g)}$, this is $\Bbb Z/2\Bbb Z$-equivariant, where the action on the LHS is by inversion and on the RHS is by complex conjugation
so after taking invariants, we see that both permutation actions have the same number of orbits
proving that number of real irreps = number of orbits of the action of Z/2Z on conjugacy classes by inversion
 
@TedShifrin uh oh, in the sheet that I have presented to the teacher I did not do any calculations but I made calculations on the draft sheet and the method actually diverged!!!
 
@MatheinBoulomenos Neat
 
Well, if I were the teacher, I would want to see the work. I don't know how things work in your university.
 
@TedShifrin me neither. I mean, I do not know how that teacher will evaluate, but what I remember saying is "Since the matrix is not diagonally dominant, then Jacobi's method diverges"
If the teachers ticks it as an OK, yay! But if not, it's also OK, I have learned a very important lesson :)
 
6:11 PM
See, that's wrong.
 
@manooooh To me it sounds like to have misunderstood what the stated theorem says
 
That's my whole point.
It could converge if you started at the right place.
 
@TobiasKildetoft why do you say that?
@TedShifrin do you mean my answer is wrong?
 
He's saying what I said, @manooooh. You're using the converse of the theorem to say that.
Yes, your answer is definitely wrong.
 
@TedShifrin oh, we haven't make the converse. Ouch :(
What would you answer?
 
6:15 PM
I would do the estimates to show divergence.
 
Would you make some calculations and conclude that by Jacobi, the method diverges?
Oh, ok
 
This is why your general question (an hour ago) needs specific context. :)
 
@TedShifrin I don't understand. Didn't you (or Tobias) say that "If $p\to q$ holds and we have $\neg p$, then we can conclude anything we want"??
 
As I said, you need specific context.
We're not discussing mathematical logic here. We're discussing a concrete question.
 
@manooooh No, it means we can't conclude anything from the implication
 
6:18 PM
@TedShifrin but "If $p\to q$ holds and we have $\neg p$, then we can conclude anything we want" is true in general, isn't it?
 
Here's another instance where students fall in the trap.
We prove the theorem that if $\sum a_n$ converges, then $a_n\to 0$.
 
@TobiasKildetoft but you actually had a conclusion
 
From this we can of course conclude that if $a_n\not\to 0$, then the series must diverge.
 
@manooooh I never had any conclusion
 
@TobiasKildetoft well, Ted had a conclusion
 
6:19 PM
But what if $\sum a_n$ diverges. Can we conclude anything about the $a_n$?
 
@TedShifrin yup
 
I hope you're listening to me.
 
If we already know that $p$ is false, then the implication says nothing new, since it is implied by $p$ being false
 
@TedShifrin if works out over $\Bbb Q_p$ :P
 
@TedShifrin well, no
 
6:19 PM
smacks Mathein
You have to examine the particular series to conclude anything. Same with your question. You have to examine the iteration.
 
@TobiasKildetoft I don't understand these, and I think it is the most important thing to learn :/
@TedShifrin ok, learned
@TedShifrin how would you answer these question not knowing the expression of $a_n$?
 
You have to know.
 
@TedShifrin "We can't conclude anything"
Right?
 
Well, your English is bad, but right.
LOL, better.
 
6:25 PM
@TedShifrin thanks for an obvious observation. You level up my morale* (thanks Ted!) :) xD
 
You could have $a_n = 1$, you could have $a_n = (-1)^n$, you could have $a_n = 1/n$, you could have $a_n = 1/(n\log n)$, etc.
morale :P
I should quit teaching math and stick to English.
 
@TedShifrin yes, but we can't count an infinite set of possible expressions of $a_n$ :P
 
That's why you need to examine specific situations when you don't have a blanket theorem ...
 
@TedShifrin uhm, I would love to read a conjunction between "Math" and "English", not a negation
 
LOL
 
6:28 PM
@TedShifrin ok, many thanks!
@TedShifrin now I teached you: we can't count an infinite set LOLOL
 
We can enumerate a countably infinite set :P
teached $\to$ taught
 
Uh, I hate technicalities, dammit!
 
looks up "technicalities" :D
 
@TedShifrin is it referred to a legal term?
 
@TedShifrin you can enumerate every set with choice :P
 
6:38 PM
@Alessandro: Only if you're sophisticated and know cardinal arithmetic.
 
Dammit Google Translate!
 
LOL
 
@TedShifrin I am a poor student of Systems Engineering :(
My all knowledge of Logic is concentrated in one course of Discrete Mathematics, yay me!
 
LOL, as you know, I'm not so much a fan of overdoing formalities with logic.
 
@TedShifrin If thou sharest thy lore of the English tongue, I shall take thee up on thy word with rapture, but -alas!- if thou ceasest to quethe of mathematics and to share thy wisdom thereof, it shall be a direful day for this roomth and we shall lament thy decidement for evermore.
 
6:40 PM
@TedShifrin oh, wow. I want to learn a lot about Logic, and I think the formalities are necessary when learning Logic, but it is just a perception
 
ROFL @Mathein: Viel Spaß :P
 
My family is calling me to have lunch
Thanks guys!! Brb
 
Bye, @manooooh. I thought you were halfway around the world. It's lunchtime here.
 
enjoy thy meal @Ted
 
I'm not leaving yet :P
Nice try, though.
 
6:43 PM
oops
 
You doing better?
 
yeah, I really am
 
That's great. :)
 
I'm back to working on my undergrad thesis
and I'm productive on my blog
 
And back to mathing here, too, so we are glad to see that.
 

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