« first day (2520 days earlier)      last day (2507 days later) » 

10:00 PM
hm i don't think so
 
(where $[A,[A,B]] = [B,[A,B]] = 0$)
 
Wait, group morphism? You mean binary product? Like the operation ("multiplication") for groups?
 
I have no clue
A classmate of mine had an exercise where the examiner asked him to qualify this operation
 
Ah ok. Is it associative?
There is clearly an identity and an inverse.
 
The examiner said it was a group morphism (and probably meant that)
That doesn't make any sense to me though
You think it's the group operation ?
 
10:03 PM
yeah a group morphism is a map $G \to H$. i don't understand from where to where it is
I think it's a group operation yeah
 
Doesn't look very associative to me
 
$(A, (B, C)) = (A, B + C + [B, C]) = A + B + C + [B, C] + [A, B + C + [B, C]]$
 
Oh actually it might well be
 
I think yeah by Jacobi identity or something
 
Because $[A,B]$ commutes with every element of the vector space we're working with
 
10:06 PM
$((A, B), C) = (A + B + [A, B], C) = A + B + [A, B] + C + [A + B + [A, B], C]$
 
So yeah, I guess you're right
So it's not a group morphism
But a group operation
 
Zee
@BalarkaSen Dostoevsky is my favorite author
 
Actually, I don't see it. Sounds like if it was associative then you'd end up with associativity of $[-, -]$, which does not hold.
Am I being dopey?
 
In the space we're working with, $[A,B]$ commutes with everyone
So the last term alway vanishes ($[C,[A,B]]$ for instance)
 
10:08 PM
So you end up with $A+B+C+[A,B] + [B, C] + [A,C]$
 
@Zee Glad to see you not calling Dostoyevsky a pleb for a change.
 
Lol
 
Zee
@BalarkaSen no he's one of the good guys
Along with John Nash and gromov
Montaigne too
 
Interesting list you are putting him in there.
Gromov is my favorite but I would call him as one of the weird guys.
 
Zee
Well these people are honest
I like honest people
Chopin is honest too
 
10:13 PM
I can see where you are coming from.
I sort of? agree.
 
Zee
Ya, I can't verbalize it too well but am glad you are getting a sense of what am saying
 
I like the authentic too. I just think one of many problems with that is that it's hard to tell the difference between authenticity and authenticity-for-show.
Or, for that matter, whether there even is a difference.
It comes down to a "gut" thing that's very very hard to define.
 
Yah that's exactly what I am worried about. When a man acts, I doubt he acts out of pure authenticity or pure sham.
There's something in the middle, almost a compromise.
Otherwise he would not act. (As an example, writing)
 
Much country music strikes me as authentic-for-show. Modest Mouse strikes me as authentic. To give one example.
 
Montaigne one of the good guys ?
"One the finest throne upon the earth, one is ever only sitting on his ass"
 
Zee
10:19 PM
I like to think vulnerablity is a sufficient condition for authenticity, you can fake that as well but then you would be exposing yourself to danger and that takes courage and courage is a nessecery condition for authenticity
 
@Astyx True statement bro
 
How many storey to a house without storey ?
 
On, not one
Proof I need to go and sleep
Seeya chat
 
G'night @Astyx
 
Bye. I have to run soon too
 
10:23 PM
How much bag is there in an empty bag?
Is same think, How many storey to a house without storey ?
 
I think every work of writing has some act of "dishonesty" (I don't identify that with originality) in it, which has it's roots on the very will of writing down one's ideas. One writes to communicate - that is showing yourself off to the audience to some degree.
Well, not only to communicate, but to document one's communications, rather!
I'm just ranting.
 
Zee
Why can't it be fully honest?
Not arguing just curious
 
Now there are not a good reponse you can choose one or zero
But I prefere one
 
I think Balarka's saying that even merely the intent to share an idea artistically will necessarily color it differently than sharing it in mere conversation.
 
The empty bag is in the bag, and the empty set was not empty
 
10:28 PM
@Zee Well, that's what I explained. A 100% honest man would not preserve his ideas, in my opinion, just because he wouldn't want to show himself off as having even slightly different ideas than the "mass", or whatever he perceives as a slightly different idea. An artist's primary intent, I think, is to realize his or her ideas as having some form of originality (from that motif they document their ideas: painting/writing/etc) a pedantic man like me would think of this as the root of dishonesty.
 
$$\forall A, A\in A$$
 
@Dattier I disagree. The set of natural numbers, for example, doesn't contain the set of natural numbers as an element.
 
Zee
Ok I see
So it's about the motivation
 
$\Bbb N \subset \Bbb N$, but $\Bbb N \notin \Bbb N$
 
Zee
So it may be the case that the top mathematcians to walk the earth were so honost they never shared their ideas...
 
10:30 PM
I talk in theorie set, all object is like a set, and the number too
@Fargle
 
Sure, but in the classical construction of "numbers as sets", no number contains itself as an element.
 
Ideally, such a thing might happen. But I am sure that such ideality does not exist within a man.
 
That is, $0 = \emptyset, 1 = \{0\}, 2 = \{0,1\}, 3 = \{0,1,2\}, \dots$
 
But, you can choose, a set theory with this principe
no?
 
@Dattier It couldn't be a model of ZFC.
 
10:32 PM
Yes, but another set theory
 
I mean, sure, maybe? I don't know enough about other axiom systems.
 
Zee
@Fargle I can't understand him but I think maybe he means a model where all the objects of sets are just sets, empty sets that is
 
@Zee Yes, but there's also the assertion that $\forall A(A \in A)$.
In such a system, what would a one-element set look like?
$A = \{A\}$?
 
Zee
It's like {{}}
 
The empty set couldn't exist, because it would have to have both no elements and one element.
It would have to contain nothing, but also itself as an element.
 
10:35 PM
@Fargle @Zee Here is a scene about the experience and paranoia of writing from one of my top 5 movies and favorite director, tangentially relevant to whatever we were talking about before.
 
When you impose $$A \in A$$, you can't use $$\{\}$$ the accolade
or there another sens
 
I guess my other question would be: what would a set mean in such a system?
If there can be no such thing as an empty set, what objects does this set theory even describe? Or, if you're just treating it as a formal system, could it be consistent?
 
There exist set theories where all sets are elements of themselves, I think
 
Now, I don't know
 
Also hi
 
10:37 PM
hi
 
I'm not trying to be negative or anything, this seriously isn't something I know very much about.
I just know that by the axiom of regularity such a model cannot be a model of ZFC, so it lies outside the scope of anything I've ever done.
 
I continue my reflection
 
Apparently a set $x$ that satisfies $x=\{x\}$ is called a "Quine atom." These are forbidden in ZFC, but it's possible to have more than one of these if the axiom of foundation is removed.
You'd think that there could only be one, since it seems like they're all $\{\{\{\dots\}\}\}$
 
for you : 1/How many storey to a house without storey ? 2/How much bag is there in an empty bag? 3/The question 1/ and 2/ have a same answer ?
 
but two sets are equal iff they have the same elements, so $x=y$ is true iff… $x=y$. Which gets us nowhere.
 
Zee
10:42 PM
@BalarkaSen damn that was a good scene
 
@Dattier: I would say "none" to both 1 and 2, and that they have the same answer.
 
Right? Andrei Tarkovsky is my jam.
 
Now, here two set are equal, if there are same elements and the bag is same
@Akiva @Fargle
 
Ok, I have to sleep now.
See ya'll later.
 
Zee
@BalarkaSen Good movie?
 
10:43 PM
Bye @Barlaka
 
Bye @Balarka.
 
Zee
Bye
@Fargle I don't know much formal logic but I seen it somewhere before
 
@Zee Stalker is said to be one of the greatest movies of all time, and it's on my top 5. I recommend it to everyone.
But yeah I'm gone now.
 
@Zee Seen what?
 
Zee
All the sets look like {{},{{{}}}},{}}
No objects
 
10:45 PM
there are many bag empty not only one
 
Zee
Except those sets
 
Yeah, that's possible. There's a construction of the natural numbers (that I alluded to above) where $0 = \emptyset$ and the successor of $n$ is $\{0,1,\dots,n-1\}$
 
there many set empty not only one
 
In ZFC, two sets are equal iff they have the same elements.
So there's only one empty set.
 
in ZFC
 
10:45 PM
@Dattier You could operate in such a system, but in the common axiom system we use (ZFC), the empty set is unique. No stories is the same set as no bags.
 
There are many set theories. The most common one, by far (and the only one I know anything about) is ZFC.
 
You could add types to your set theory, but that goes into a realm I know nothing about.
 
Yes, but I want found a theorie of set, with the model of bags
 
I would recommend studying axiomatic set theory and other axiom systems besides ZFC as you attempt to do this.
 
in this theory, a set are the same, if the element is the same and yhe bag is the same
 
10:47 PM
You may find that someone already has done this.
Or not, in which case that's exciting.
 
I have the reality like for physicien
And the bag is in the bag
so there are not empty bag
 
So your model requires that every bag is inside itself?
 
only a bag with one element
 
@Dattier I suggest you ask questions on the main site. Neither Fargle nor I are set theorists.
 
10:49 PM
@Fargle is just a convention
 
@Dattier The problem I have with it is: how do you distinguish between a blue bag containing a brown bag, and a brown bag containing a blue bag?
 
My english is very bad
 
Both of these "bags" would have the unique two elements "brown bag" and "blue bag".
 
can you help me for formular that in a good english ?
 
Admittedly, I can't because I myself can't understand it--not because of your English but because of the nature of it.
 
10:51 PM
@Akiva and @Fargle
can you help me for formular that in a good english ?
 
That doesn't mean it's a bad idea--I just can't wrap my head around it.
 
You must reason like a physicien
the model is the model of bag
 
Zee
lol how insulting
 
and I want give a theory of that
 
Zee
If he reasons like a physicist then you reason like a poet
 
10:53 PM
Why, are you see the empty bag (like the empty set in ZFC) ?
 
Physicists don't think that a bag is part of its contents.
 
@Dattier "Physicist"
 
The bag isn't part of the bag, it is the container.
 
@Akiva yes, sorry for my english
It's another point of vue @Fargle, but the model is unique (the bag in our world)
 
Is it a useful point of view if "nothing" can't even exist?
 
10:56 PM
Unrelated: Is $\frac12\begin{bmatrix}1&1&1&1\\1&1&-1&-1\\1&-1&1&-1\\1&-1&-1&1\end{bmatrix}$ the product of 45-degree rotations?
 
Yes, it is a point of view which leads to madness or malice @Fargle
 
What are you trying to do, @Akiva?
I don't think physics needs any help doing that.
 
Let's mature this questions, bye
 
11:12 PM
I want to check if the function f(x)= (x-2)(x+3) is injective. We have that $2\neq -3 \Rightarrow f(2)=0=f(-3)$. Therefore the function is not injective on ℝ. But what happens on the intervall [-0.5, ∞) ?
 
Taking a poll guys, please participate!
How do you eat corn on the cob? Do you go straight across or in circly spirals?
After you've decided, [please read this](http://bentilly.blogspot.ch/2010/08/analysis-vs-algebra-predicts-eating.html).
 
11:27 PM
Do we have to check the injectivity there with the monotonicity? We have that f'(x)=2x+1, which is equal to 0 at the boundary -0,5 and positive everwhere else. Is that function monotonically increasing or strictly monotonically increasing on [-0.5, ∞) ?
 
hi chat
 
11:47 PM
Must a cube in $\Bbb Z^3$ have sides parallel to the axes?
This is false in $\Bbb Z^2$
…Never mind
Must a cube in $\Bbb Z^3$ has at least one side parallel to an axis?
 
I want to say yes, on the grounds that I can do stuff like $3^2+4^2+12^2 = 5^2+12^2=13^2.$
(that might be unnecessary, but w/e)
 
I essentially want three mutually orthogonal vectors of the same length.
 
Right.
And you want them to have integer entries.
 
Cross products seem relevant
 
yeah.
 
11:54 PM
I guess we just want two mutually orthogonal vectors with equal integer lengths, then, which do not both lie orthogonal to the same axis.
Because then we can cross product them and divide by that shared length.
 
sounds right? but admittedly my brain is not sharp at the moment
 
$$(3) \, \, \, \, \partial_{hhzz}f(z) = \partial_{hh}(\frac{1}{h}(f(z+h)) - \partial_{zz}f(z))$$
^ Is this operation correct
 

« first day (2520 days earlier)      last day (2507 days later) »