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7:15 AM
1
Q: 2016 Moderator Election Chat October 4 16:00 UTC

David ZOur next regularly scheduled chat session, October 4 at 16:00 UTC/12:00 EDT, will be devoted to the current election. We'll be holding the session in our dedicated election chat room, not in the main site chat room as we usually do. The tentative plan is to split the session into two parts: A ...

discussion election
 
user116211
7:38 AM
Hmm, quite a quiet review queue... somewhat rare show ._.
 
Secret
Secret
8:10 AM
@JohnDuffield Ok I see. It's not really point like, but whatever it looks like as determined by scattering and other experiments suggest it is so small that all the interactions we have seen so far are basically indistinguishable from a point like particle.
 
Secret
Secret
8:31 AM
6
A: How does a magnetic monopole break time reversal symmetry?

Andre HolznerThe English Wikipedia article on magnetic monopoles has the following equation for the 'extended' Lorentz-Force of a magnetic field on a electrically and magnetically charged particle: $$ \vec{F}=q_{\mathrm e}\left(\vec{E}+\vec{v}\times\vec{B}\right) + q_{\mathrm m}\left(\vec{B}-\vec{v}\times\fr...

Well (if my understanding is correct), we cannot really tell apart an object that travels back in time or forward in time if the symmetry holds. Consider a Mikowski space with two particles as shown
Now imagine we take hyperspatial slice for each unit of proper time along the trajectory of the two particles to respectively build a time laspe movie. Note what happens:
 
user116211
8:49 AM
 
Secret
Secret
And then for the red particle:
(More correct version require to actually plot what each particle sees for each of their rest frame, however here we only plot basically what a particle that is stationary with the black frame shown here sees, with the intervals parametrised by the proper time of each particle, (which is not very correct. Will fix that in the future)) As you can see, the scenario is symmetric. We cannot tell whether the red particle or the blue particle is coming from the future, or just travelling ordinarily
forward in time and collide with the other particle
48
Q: Is anti-matter matter going backwards in time?

GerardOr: can it be proved that anti-matter definitely is nót matter going backwards in time? From wikipedia: There is considerable speculation as to why the observable universe is apparently almost entirely matter [...] the apparent asymmetry of matter and antimatter in the visible universe is on...

quantum-field-theory time antimatter
So in a (rough?) sense, antimatter could really be matter that goes back in time, and it just happens because of CPT symmetry, it has the same result as antimatter going forward in time. If that is true, then perhaps the existence of antimatter is already showing that CTC exists on a daily basis
I need to read more how people actually rule out that antimatter is matter travelling back in time...
@Tobia antimatter has been used for many things, and nobody has ever seen any evidence of it carrying information backwards in time. If anyone had an idea of how antimatter could be capable of carrying information back in time, in a way that would not have been noticed, someone would certainly do an experiment to test it, but that's never happened as far as I know. — David Z ♦ Jun 22 '15 at 8:53
> Cold neutral antihydrogen experiments[edit]
Since 2010 the production of cold antihydrogen has become possible at the ATHENA, ATRAP and ALPHA experiments at CERN. Antihydrogen, which is electrically neutral, should make it possible to directly measure the gravitational attraction of antimatter particles to the matter Earth. In 2013, experiments on antihydrogen atoms released from the ALPHA trap set direct, i.e. freefall, coarse limits on antimatter gravity.[8] These limits were coarse, with a relative precision of ± 100%, thus far from a clear statement even for the sign of gravity acting…
bleh, crossing fingers for their results. It will be cool either way for the outcome of it, because antimatter are already interesting to begin with
 
 
2 hours later…
user116211
11:03 AM
You may like it @secret:
 
user116211
30
Q: In classical logic, why is $(p\Rightarrow q)$ True if $p$ is False and $q$ is True?

user701510Provided we have this truth table where "$p\implies q$" means "if $p$ then $q$": $$\begin{array}{|c|c|c|} \hline p&q&p\implies q\\ \hline T&T&T\\ T&F&F\\ F&T&T\\ F&F&T\\\hline \end{array}$$ My understanding is that "$p\implies q$" means "when there is $p$, there is q". The second row in the tru...

logic propositional-calculus
 
Secret
Secret
noted. Btw what is the name of that Kreyszig topology book. Google once again only give me diff geom and functional analysis books?
 
user116211
11:20 AM
@Secret Functional analysis book; I have not meant a topology book; but he briefly explains the motivation of the criteria of a topological space and vacuous truth. You can check that. As I said, don't read Bourbaki for the first time; you can postpone it after you have an idea of what you are reading. Munkres is a very good introductory book.
 
Secret
Secret
ok
 
Secret
Secret
11:56 AM
whenever I faced such questions, I just do an experiment. The nice thing about mathematical experiment is that the ycan usually be done with pen , paper and a computer, thus relatively accessible
Experimental mathematics
Experimental mathematics is an approach to mathematics in which numerical computation is used to investigate mathematical objects and identify properties and patterns. It has been defined as "that branch of mathematics that concerns itself ultimately with the codification and transmission of insights within the mathematical community through the use of experimental (in either the Galilean, Baconian, Aristotelian or Kantian sense) exploration of conjectures and more informal beliefs and a careful analysis of the data acquired in this pursuit." As expressed by Paul Halmos: "Mathematics is not...
One of my interest in this field is to investigate more about the cause of a given mathematical coincidence such as how it arises from the underlying expression
 
user3183724
user3183724
12:25 PM
In the lineshape of a two level system, does pure dephasing shift the center frequency? I can imagine that it broadens the lineshape and makes it more shallow, but I don't know why it would affect the frequency
 
John Duffield
John Duffield
12:46 PM
@HDE226868 : of course I read anna's answer. And it does not supply any evidence that the electron is a point particle. Can you supply any such evidence? No you can't.
 
ACuriousMind
ACuriousMind
@Danu ???
 
Danu
Danu
@ACuriousMind Que?
 
ACuriousMind
ACuriousMind
:P
 
Secret
Secret
Uh I think I have a logical question here:
 
John Duffield
John Duffield
@vzn : the Smoot interview was good stuff. And who can complain about relativistic baseball?
 
Secret
Secret
12:56 PM
We knew that there is no x in reals that satisfy the LHS thus the LHS is vacuously true, while for the RHS it can be true or false depending on whether my x=23
 
John Duffield
John Duffield
@KyleKanos : you watch your mouth, boy.
 
ACuriousMind
ACuriousMind
@Secret The l.h.s., assuming you mean $x^2 < 0$, is not "vacuously true". It's false for all real numbers.
 
0celo7
0celo7
@Danu I have a ping from you. What did you say?
 
Secret
Secret
??? But then what is munkres saying a few lines earlier?
 
Danu
Danu
@0celo7 I asked you to confirm that $SO(3,1)$ is not compact
 
ACuriousMind
ACuriousMind
1:00 PM
It's the statement $\forall x\in\mathbb{R} : x^2 < 0 \implies x=23$ that is vacuously true, not the $x^2 < 0$ alone.
 
Secret
Secret
Do he mean the whole statement?
 
0celo7
0celo7
@Danu Rapidity is a real number
 
Secret
Secret
Ah i see
 
0celo7
0celo7
It can take all real values
So yes, it is nonconpact.
 
Danu
Danu
Exactly---so I deleted the message :P
 
0celo7
0celo7
1:01 PM
Ah, ok.
There's probably a way better way to see that.
 
Danu
Danu
I think it's pretty nice and simple...
 
0celo7
0celo7
@ACuriousMind What is this?
 
John Duffield
John Duffield
@Secret : it isn't small. IMHO you should think like this: the electron's field is what it is.
 
ACuriousMind
ACuriousMind
@0celo7 It appears to be three question marks.
 
0celo7
0celo7
@ACuriousMind But, why?
 
ACuriousMind
ACuriousMind
1:14 PM
@0celo7 They commonly denote a state of confusion or surprise.
 
0celo7
0celo7
@Danu Thanks for the notes, but I want to do it myself using the crappy old books.
I'll probably write up my thoughts.
@ACuriousMind Why were you confused or suprised?
 
GPhys
GPhys
@ACuriousMind good morning
 
ACuriousMind
ACuriousMind
moin moin
 
0celo7
0celo7
Germab alert.
 
ACuriousMind
ACuriousMind
@0celo7 Because Danu asked a question whose answer I expected him to know :P
 
0celo7
0celo7
1:16 PM
Really? What's the answer?
 
GPhys
GPhys
I was thinking some more about symmetries, and I'm not sure it's true that $V\mapsto V$ (for transformations of $q_i$'s alone, not including $t$, for a $V$ dependent only on $q_i$) is even enough
@ACuriousMind Since I think the kinetic energy will not always differ by a total time derivative in general
 
Secret
Secret
Interesting, it appears the contrapositive is still doing fine even if the statement "if $x^2 < 0$ then $x=23$" is extended to include all multiples of the imaginary unit i.e. the set $\mathbb{R} \cup \mathbb{R}i$. One can then see that the contrapositive maps the row $T\hspace{1mm} F$ of the $P \rightarrow Q$ truth table to itself
 
ACuriousMind
ACuriousMind
@GPhys Hm, you are right. If the symmetry is not a symmetry of $q^2$, then $\dot{q}^2$ won't be invariant, either.
@Secret I have no idea what you are trying to say.
 
Secret
Secret
Elaboration: Suppose we consider the set of reals along with real multiples of the imaginary unit i.e. $\mathbb{R} \cup \mathbb{R}i$, then
The LHS is false for all reals and true for all imaginary numbers. Meanwhile the number in question can be 23 or some other number for the reals, but it must not be 23 if x is imaginary. Therefore:
Consider the case for reals, LHS is false for all real x, while the RHS can be false or true. Therefore using the truth table we get True for the case. Now
Consider the case for imaginary x, the LHS is always true, while the RHS is always false. Using the truth table , we get False
Now consider the contrapositive of the original statement: "if $x\neq 23$, then not $x^2 < 0$". For all real x, the LHS can be true or false while the RHS is always true, therefore truth table will give True for this case. Now for all imaginary x, the LHS is always true, while the RHS is always false, therefore truth table will say False for this case. So
 
vzn
vzn
1:38 PM
@JohnDuffield miraculous found something you liked o_O
@Secret huge fan/ practitioner/ advocate etc of "experimental math" myself, think we are in for amazing times in future wrt it, eg machine learning is closely connected, etc... see also vzn1.wordpress.com/category/experimental
 
Secret
Secret
Comparing the rows of the truth table for each case with the truth table of the contrapositive, we see the triplet of entries FT T, FF T becomes FT T, TT T respectively in the contrapositive case, and TF F remains unchanged as TF F in the contrapositive case
 
ACuriousMind
ACuriousMind
...are you trying to state the elementary observation that an implication and its contrapositive are equivalent?
 
Secret
Secret
That is going from $P\rightarrow Q$ to $-Q\rightarrow -P$ only the rows TT T and FF T of the truth table are swapped, but the rows FT T and TF F are not swapped (which reminds me of fixed points in functions...)
 
ACuriousMind
ACuriousMind
What do you mean they "aren't swapped"? The order of lines in a truth table is completely arbitrary.
 
Secret
Secret
(Elaboration came shortly...) Suppose we have the proposition P is true, then not P must be false, similarly for Q. Therefore while the order of lines of the truth table are arbitrary, the order of lines in the contrapositive case seemed to be correlated to the original one
So regardless of how the truth table is ordered, any statements in the $P\rightarrow Q$ truth table that corresponds to the row TTT must become FFT in the contrapositive case, and vise versa
Whereas the row TFF and FTT are unchanged by "performing" a contrapositive to the table
 
ACuriousMind
ACuriousMind
1:53 PM
And your point is?
That's just how negation works, why is that surprising or interesting?
 
Secret
Secret
Because the behaviour of the rows TFF and FTT under contrapositive reminds of fixed points in function, and I knew fixed points are interesting and useful
But perhaps not very interesting for this case...
(or unusual, surprising etc.) * continue to read book *
 
0celo7
0celo7
@ACuriousMind tbh I don't know how to prove that
@ACuriousMind Boo, you proved A-A on $\Bbb R^n$ in your notes!
 
Secret
Secret
2:19 PM
Random stuff found on internet: Seemed to save a lot of trouble drawing in perspective
 
0celo7
0celo7
that's cheating
 
Secret
Secret
Well, does not matter cause I am not running for an architect exam, XD
Well, thinking deeper, the reason why all of this works is because our classical world is governed by principle of stationary action, which also applies to light as the (forgot name ) principle
and also the way our binocular vision evolved
 
user116211
@Secret Fermat's Principle of Least Time?
 
Secret
Secret
yes, that's the name
Someone obviously put too many boxes at home and forgot to pack them up
Ordered pair
In mathematics, an ordered pair (a, b) is a pair of objects. The order in which the objects appear in the pair is significant: the ordered pair (a, b) is different from the ordered pair (b, a) unless a = b. (In contrast, the unordered pair {a, b} equals the unordered pair {b, a}.) Ordered pairs are also called 2-tuples, or sequences (sometimes, lists in a computer science context) of length 2; ordered pairs of scalars are also called 2-dimensional vectors. The entries of an ordered pair can be other ordered pairs, enabling the recursive definition of ordered n-tuples (ordered lists of n objects...
I cannot keep track of so many nested objects! arrgghh
 
0celo7
0celo7
oh my god Secret
get Munkres
whatever you are doing is not working
topology is not this hard
 
Secret
Secret
2:34 PM
(I read any books and journal article from cover to cover, that's why all these apparently non topology stuff ramblings)
 
user116211
@Secret There is a very good discussion on set theory in Munkres; didn't you read that?
 
Secret
Secret
That's where I am (all theo screenie above are indication where I was at
 
user116211
@Secret So, what's the problem?
 
Secret
Secret
I just had a habit of rambling when reading a new topic, not necessary genuinely frustrated
 
user116211
ohh ;/
 
Secret
Secret
2:38 PM
because if I am frustrated I will ask you guys questions
0celo7 and slereah sometimes does ramblings too when they gone through their text
 
0celo7
0celo7
I do not ramble.
 
ACuriousMind
ACuriousMind
Jul 17 at 1:45, by 0celo7
@Mikhail I've just spent an hour rambling
 
user116211
Might be another 0celo7 ;P
 
0celo7
0celo7
@ACuriousMind Not me. Try again
The vitriol in this chat is unsettling
 
user116211
And as I said...
 
Secret
Secret
2:45 PM
I will just treat ordered pairs as a primitive object for now rather than a bunch of nested boxes
It's an irony when you realise that I like recursive things yet I don't like seeing a nested expression
 
ACuriousMind
ACuriousMind
@Secret You're not supposed to treat them "like nested boxes" in the first place! Just like you're not supposed to treat natural numbers like a bunch of nestes sets having the empty set as innermost element - the primitive set-theoretic definition is rarely useful once you've shown it behaves like the thing you wanted to define.
 
Secret
Secret
For example $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\pi+\cdots}}}}}}$ is fine, but $\{\{\{\},\{\}\},\{\{,\}\}\}$ is not
@ACuriousMind I see, noted
 
John Duffield
John Duffield
3:41 PM
@vzn : I like physics. I love physics. And cosmology and everything in between. But what I don't love is popscience pseudoscience and quacks peddling woo. There's a lot of it about.
 
ACuriousMind
ACuriousMind
4:00 PM
@HDE226868 General advice: Having the last word in a conversation is not the same as "winning" it. Many people with whom rational argument is impossible however see it as that, and will take the lack of response as the other side being unable to refute their "arguments". Nothing you can say will ever not lead to a response from them, so the only winning move is not to play their game.
 
HDE 226868
HDE 226868
But I'll follow that advice next time, preferably by not starting such a conversation in the first place.
 
Danu
Danu
@ACuriousMind Reminds me of that copypasta
About an AI running a shooter game for a really long time, with the final strategy being standing still
 
ACuriousMind
ACuriousMind
I've seen this pattern a lot this last week, and while I appreciate both you and others trying to engage with such people in my "defense", it's ultimately pointless. Have faith that other people can see the merits of arguments on their own without prolonged back-and-forth.
@Danu Heh. Funny thing: Some developer actually let loose the best AI they had in the test version of a shooter game. Players complained the computer "must be cheating" because they were unable to beat it, and for the final version they dumbed it down considerably because it turns out most players don't play the game to lose it to a computer.
 
HDE 226868
HDE 226868
@Danu Clever. What was the game?
 
Danu
Danu
@ACuriousMind That's hilarious :P
I have a hard time believing good shooter AI though
I mean... if it's a like a campaign thing where it's 1v100 sure
but 5v5, humans vs computers
easy win
 
ACuriousMind
ACuriousMind
4:07 PM
@Danu Not 1v100, but yes, I think the enemy outnumbered the players, not sure though, I'll see if I can find the article
 
Danu
Danu
@HDE226868 I'm trying to remember... It was one of the old ones---the copypasta was definitely bullshit
 
HDE 226868
HDE 226868
@Danu If it's old, I likely wouldn't remember it.
 
ACuriousMind
ACuriousMind
Flat Earther sighted!
2
 
user116211
@ACuriousMind Troll Alert!
 
user116211
Now they are linking facebook pages....
 
HDE 226868
HDE 226868
4:14 PM
Does that part count as spam?
 
ACuriousMind
ACuriousMind
Probably, yes.
 
Danu
Danu
Also
The Flat Earth Revolution is a deliciously ironic name.
 
HDE 226868
HDE 226868
It could be a nice band name.
 
0celo7
0celo7
I don't know where I stand on the flat Earth debate.
 
user116211
@0celo7 O.o
 
user218912
4:18 PM
@0celo7 which naber?
 
user116211
Are you trolling, @0celo7?
 
0celo7
0celo7
@IceLord 1.
@MAFIA36790 No.
I have no reason to believe either side.
 
user218912
@0celo7 why?
 
user116211
okay ._.
 
user218912
@0celo7 seriously?
 
user218912
4:18 PM
all the evidence is there.
 
0celo7
0celo7
I have not seen any evidence with my own two eyes
 
user218912
did you ever fly in a plane?
 
0celo7
0celo7
Yes
I did not see any curvature
 
user218912
wow.
 
user218912
you weren't sitting by the window seat then.
 
0celo7
0celo7
4:19 PM
And even if there is curvature, that does not make the Earth spherical.
@IceLord I was, and I looked.
What convinced you all?
 
user116211
36
Q: What is the simplest way to prove the Earth is round?

tQuarellaAssume you've come in contact with a tribe of people cut off from the rest of the world, or you've gone back in time several thousand years, or (more likely) you've got a numbskull cousin. How would you prove that the Earth is, in fact, round?

education earth faq
 
user116211
Don't know why I even searched this T__T
 
Danu
Danu
@0celo7 You momma! :P
 
user116211
22
Q: Backyard experiments to falsify the Flat Earth theory

Mario CarneiroI recently became aware that the flat Earth theory still exists in the 21st century, and has colored the views of a friend of mine. Roughly speaking, the tenets are: The Earth is a flat disk, with the south pole blown up into a circular "ice wall" where one would expect Antarctica to be.        

everyday-life earth planets home-experiment geophysics
 
0celo7
0celo7
@Danu That doesn't even begin to make sense.
Are you saying she's so fat, a flat Earth would be a contradiction?
 
Danu
Danu
4:23 PM
She curved it
 
0celo7
0celo7
> The shadow of the Earth on the Moon during an eclipse and the way masts of ships are visible when they are out of sight are the classical reasons.
I have no proof of this.
> Shadows differ from place to place
This can be explained in the flat Earth theory.
Besides, why should sunlight work in that intuitive way to begin with?
Maybe I should use $\wp$ for the bundle projection
 
Danu
Danu
GOD no
$\pi$ and $p$ are standard.
 
0celo7
0celo7
@Danu don't tell me what's standard
 
Danu
Danu
Weierstrass $\wp$ is Weierstrass $\wp$ and nothing else
 
0celo7
0celo7
I'm making new notation
a bundle is $(\mathcal B,\mathscr M, F,\mathsf G, \wp)$
 
Danu
Danu
4:32 PM
disgusting
$\pi:E\to M$ or bust
 
0celo7
0celo7
that's for vector bundles, Danu
when making associated bundles there's a lot to keep track of, actually
 
Danu
Danu
Okay, $F\to E\overset{\pi}{\to}M$ or bust.
 
0celo7
0celo7
$(B,M,F,G,\pi)$ is reasonable.
 
Danu
Danu
That's completely general for any fiber bundle
 
0celo7
0celo7
Then the associated bundle is $E(B,M,F,G,\pi;\lambda)$
@Danu doesn't specify the structure group
 
Danu
Danu
4:34 PM
$F$
fiber
isomorphic to group
 
0celo7
0celo7
The fiber need not be the group...
 
Danu
Danu
ya know
 
0celo7
0celo7
What?
 
Danu
Danu
In a principal bundle
each fiber is isomorphic to the group
 
0celo7
0celo7
@Danu I'm not talking about principal bundles.
 
Danu
Danu
4:36 PM
Even wikipedia defines in the way I do
 
0celo7
0celo7
Defines what?
 
Danu
Danu
Fiber bundles
Your structure group is extra data, AFAIK
 
0celo7
0celo7
Oh, sure.
But I define all of mine with the structure group
I have no need for bundles without structure groups
I think my notation is quite nice.
 
Sir Cumference
Sir Cumference
4:53 PM
All right, this question may sound stupid, but it's driving me insane
Correct me if I'm wrong: operations are ways to manipulate a number, like addition, subtraction, taking roots, exponentiating, etc.
And functions take an input, apply operations and give an output, right?
 
ACuriousMind
ACuriousMind
@SirCumference Ehhh, on a formal level, operations are just special functions.
Assuming you talk about the mathematical notion of function
In programming languages one might have other distinctions
 
Sir Cumference
Sir Cumference
@ACuriousMind Well, if I understand it correctly, functions can be described through operations, but not the other way around?
 
ACuriousMind
ACuriousMind
How would you describe $f(x) = \sin(x)$ through "operations"?
 
0celo7
0celo7
taylor series, @ACuriousMind
take $x$, subtract $x^3/6$, etc.
@ACuriousMind That one is harder.
 
Sir Cumference
Sir Cumference
@0celo7 Exactly what I was thinking
I can write $f(x) = 3x$, and that'd be a function described by an operation
 
0celo7
0celo7
4:59 PM
I'm not sure what you're trying to get out of this conversation, though.
 
Sir Cumference
Sir Cumference
@0celo7 There's a greater question I have, but first I need to make sure I've got this right
 
ACuriousMind
ACuriousMind
@SirCumference What is your definition of "operation"
 
0celo7
0celo7
@ACuriousMind My question too.
 
Sir Cumference
Sir Cumference
A way to create an output number from an input number, but it can't be written in $f(x) =$ form.
 
0celo7
0celo7
What?
 
Sir Cumference
Sir Cumference
5:00 PM
Like multiplication
 
0celo7
0celo7
Lol.
 
ACuriousMind
ACuriousMind
Addition is just a function $\mathbb{R}\times\mathbb{R}\to\mathbb{R}, (x,y)\mapsto x+y$. We call it an "operation" because a function of that form is intrinsically required for the notion of the real numbers as a group/field/ring
 
0celo7
0celo7
What is it with people in this chat being confused by basic mathematics?
Like, really basic
 
ACuriousMind
ACuriousMind
@SirCumference That's not actually a meaningful definition, although I know that the way math is taught in school might make it seem as if it is
But to get to the root of the problem, one must recognize that saying a function is "something that takes an input and gives an output" is not a formal definition either.
 
Sir Cumference
Sir Cumference
@0celo7 Almost as basic as knowing the difference between a cardinal and ordinal. :/
 
0celo7
0celo7
5:03 PM
A "function" $f$ from the set $X$ to the set $Y$ is a relation $f\subset Y \times X$ with domain $X$ such that $(x\in X,y\in Y,y'\in Y,yfx \land y'fx)\implies y =y'$
 
Sir Cumference
Sir Cumference
@ACuriousMind All right then, my new question is: is differentiation an operation?
 
0celo7
0celo7
Yes.
 
ACuriousMind
ACuriousMind
@SirCumference I just said that "operation" is not a meaningful definition, so I don't know what that question means.
 
0celo7
0celo7
The derivative is a map from differentiable functions to functions.
 
ACuriousMind
ACuriousMind
@0celo7 Please try less to show off your knowledge and try more to actually help the confused person.
 
0celo7
0celo7
5:05 PM
I'm not showing off my knowledge.
 
Sir Cumference
Sir Cumference
@0celo7 Kind of coming off that way, tbh
 
0celo7
0celo7
Fine, I'll leave.
@ACuriousMind Why would I bother showing off my knowledge about trivial things?
Give me a break.
 
Sir Cumference
Sir Cumference
@HDE226868 Howdy
 
HDE 226868
HDE 226868
@SirCumference Hi.
 
ACuriousMind
ACuriousMind
@0celo7 It's evident that SirCumference lacks the proper definitions to appreciate what "the derivative is a map from differentiable functions to functions" actually means. I can't know why you said it, but it's definitely not helping.
 
0celo7
0celo7
5:07 PM
I just explained what a function is.
 
HDE 226868
HDE 226868
@ACuriousMind They're attacking en masse.
 
0celo7
0celo7
4 mins ago, by 0celo7
A "function" $f$ from the set $X$ to the set $Y$ is a relation $f\subset Y \times X$ with domain $X$ such that $(x\in X,y\in Y,y'\in Y,yfx \land y'fx)\implies y =y'$
 
ACuriousMind
ACuriousMind
@HDE226868 I know, I just protected the question
 
HDE 226868
HDE 226868
@ACuriousMind Ah, now I see. Thanks.
 
Sir Cumference
Sir Cumference
@ACuriousMind Probably obvious, but yeah, still stuck retaking calculus
Haven't learned much over the last two years.
 
ACuriousMind
ACuriousMind
5:09 PM
Well, maybe we should try this another way: Why is it relevant to you whether differentiation is an "operation"?
 
Sir Cumference
Sir Cumference
I don't have any real need to know. But it just bugged my mind.
 
ACuriousMind
ACuriousMind
Yeah, so the answer is that this question doesn't have an answer because mathematics does not actually neatly distinguish functions into "operations" and "not-operations".
 
Sir Cumference
Sir Cumference
Simple and concise. Thanks.
 
0celo7
0celo7
@ACuriousMind So if you can't know why, then why did you make the assumption? I don't know what I've done to piss you off, but please let me know so I can correct it.
 
ACuriousMind
ACuriousMind
@0celo7 You began just dumping the (correct) definitions without the slightest regard shown for whether the person asking the question was prepared to understand them or even asking at that level of rigor
 
0celo7
0celo7
5:33 PM
@ACuriousMind The person in question claims to understand the difference between personals and cardinals and has made fun of (maybe not quite) me for not knowing what they are. I apologize for assuming he can read basic logical quantifiers.
 
user116211
Okay; that Flat Earth troll is using multiple accounts now.
 
0celo7
0celo7
@Danu I have three projections now. Can I use wp to simplify notation please?
 
user116211
-4
A: Why does sunset over a body of water cause a path of light stretching towards the horizon?

Summer SummerI agree to Nick Nicu.. This is because THE EARTH IS FLAT! nor curved, no.

 
Danu
Danu
@0celo7 $\pi$, $p$, $\pi_X$, etc...
 
0celo7
0celo7
p is a point on the bundle
Yeah I could write pi_P, pi_E, pi_G
But that's not very nice
 
user116211
5:40 PM
WTH! Someone upvoted both the Flat Earth posts O.O
 
Danu
Danu
Much nicer than $\wp$
> peterh
 
0celo7
0celo7
They probably upvoted themselves
@Danu why do you hate it?
 
user116211
C'mon; not again ._.
 
Danu
Danu
@MAFIA36790 That post needs to be locked
 
ACuriousMind
ACuriousMind
@0celo7 They can't , you need 15 rep to upvote
 
user116211
5:44 PM
@Danu yes.
 
ACuriousMind
ACuriousMind
@Danu I already protected it, calm down everyone :P
 
0celo7
0celo7
Mod abuse
 
user116211
@0celo7 How?
 
Danu
Danu
@ACuriousMind lol I wasn't in the least un-calm :P
 
user116211
Sometimes, I can never understand you @0celo; sigh T__T
 
Danu
Danu
5:46 PM
Sometimes never
 
0celo7
0celo7
I don't understand myself either
 
Secret
Secret
Venn diagrams for set theory can sometimes be misleading:
The $y=x^2$ example to illustrate item (g) cannot be represented by venn diagrams, because circles have positive radii
 
ACuriousMind
ACuriousMind
...what.
 
Danu
Danu
It's not called Venn diagrams, but simply "drawings of sets intersecting in $\Bbb R^2$" :P
 
Secret
Secret
Took me hours in trying to find an example to show that (g) is not an equality in general before I actually saw the question (g) and hence being informed of what type of "non equality" is
I do previosuly tried to find that example by attempting to write the LHS and RHS in set notation (as if I am writing a proof), but the stuff become too messy to find the example
It does seemed that similar to physics, one need to knew the answer somewhat to find the examples and counterexamples
It seems I am simply not clever enough to just start with $f(A_0 \cap A_1)$ and $f(A_0)$ and $f(A_1)$ and show that the operator that link the two statements is a $\subset$
 
ACuriousMind
ACuriousMind
6:02 PM
...the counterexample to (g) being an equality is trivial: Take any constant function, e.g. the function that maps $\{0,1\}$ to $\{0\}$. Then $f(\{0\}\cap\{1\}) = f(\emptyset) = \emptyset$, but $f(\{0\})\cap f(\{1\}) = \{0\}$.
 
Secret
Secret
...guess I need to think about the empty function (more generally, all things empty) more...
 
ACuriousMind
ACuriousMind
I didn't use the empty function, which would be the unique function $\emptyset\to\emptyset$.
 
0celo7
0celo7
Why can't you map a set into the empty set @ACuriousMind
 
ACuriousMind
ACuriousMind
Because by definition of a function each element of the source needs an element in the target, but there are no elements in the target for that case.
 
0celo7
0celo7
Oh I was kidding
Remember when I didn't understand that
 
ACuriousMind
ACuriousMind
6:09 PM
Faintly
 
Danu
Danu
oh god
that was super annoying
 
0celo7
0celo7
Since then I have learned some PhD set theory
 
Danu
Danu
hurr durr do not accept vacuous truth
 
0celo7
0celo7
@Danu You're super annoying
 
Danu
Danu
I was genuinely irritated by that :P
 
ACuriousMind
ACuriousMind
6:10 PM
@0celo7 How am I supposed to know when you genuinely forget things and when you are "kidding"? :P
 
0celo7
0celo7
Oh, I still don't accept that.
 
Danu
Danu
OK.
Super annoying.
 
Secret
Secret
vacuous truth = true by default because there are no counterexample
 
0celo7
0celo7
I accept it as a convention, so that we needn't have special cases everywhere.
 
Secret
Secret
Danu's comment is a heated version of what I said yesterday: I fully understood it, but that does not mean it is not weird
There are things that are weird no matter how much you understood how it works and why, because they are just ... weird
 
0celo7
0celo7
6:12 PM
@ACuriousMind I will label my posts with [kidding] from now on.
 
ACuriousMind
ACuriousMind
Making them actually funny would also be a good indicator - pretending to not remember something isn't really amusing :P
 
Secret
Secret
Caption: I knew a couple of examples in matrices, but I am currently too lazy to find function examples
There are matrices that when multiply to another, does not do anything to that matrix in question
 
0celo7
0celo7
I think it's amusing as hell.
Maybe the most fun thing I've done all week.
 
ACuriousMind
ACuriousMind
@Secret ...and what has that to do with being an inverse?
 
Secret
Secret
o wait a sec, I might have mix that up with neutral elements
 
ACuriousMind
ACuriousMind
6:16 PM
@0celo7 What a terrible week you must've had
 
Secret
Secret
bleh XC
 
0celo7
0celo7
@ACuriousMind No, I think it was pretty good.
Proving things about associated bundles is a bitch though. I can't even draw pictures
No clue what's going on any more
 
Secret
Secret
Attempt on 5d: Let $g_1,g_2$ be left inverses, then $g_1 \circ f=\text{id}_A=g_2 \circ f$. Then unless $f$ has at least one right inverse $h$ such that $f \circ h =id_B$, $g_1\neq g_2$ in general
Once that happens, the left and right inverses are unique and we call that the inverse
 
ACuriousMind
ACuriousMind
You haven't actually shown that $g_1\neq g_2$ is possible, you just asserted it.
 
Secret
Secret
hmm.... I had to think about that one...
 
Slereah
Slereah
6:33 PM
Heyhey
I think I found the solution
A grate is probably the best item to juice those apples that I have
 
user116211
@Slereah You haven't still crushed those apples?
 
Slereah
Slereah
Well crushing them is out of the question
 
user116211
Oh Lord ;/
 
Slereah
Slereah
I don't have anything close to being able to crush that many apples
Grating works fine, tho
then filtering the paste
 
user116211
There must be TNT some devices that can make your work faster @Slereah; check Amazon...
 
Slereah
Slereah
6:39 PM
I don't want to buy farm equipment either
 
user116211
@Slereah Then you can make one...
 
user116211
 
Slereah
Slereah
Grate is fine
 
user116211
._.
 
Secret
Secret
Attempt on 5d again:
Consider $a \in A$
$g_1 \circ f = g_2 \circ f \Rightarrow (g_1-g_2) \circ f = 0$
$f(a)=b \in B$. Using 5a, f is injective as left inverse exists, thus $\forall a \in A$, there is only one $b \in B$. In particular, the kernal of f is $\{0\}$. Now $g_1 \circ f=g_2 \circ f=\text{id}_A$. It follows that $\forall a \in A, id_A(a)=a$, that is $\forall b \in B, g_1(b)=g_2(b)$. Therefore $g_1=g_2$
 
ACuriousMind
ACuriousMind
6:49 PM
@Secret That doesn't make any sense. "$\forall a\in A$, there is only one $b\in B$" is not what injective means - injective means that the $b$ is different for each $a$. That there's "only one" is for each is part of the definition of a function. also, what is 0 supposed to be, and what is a "kernel", you're only talking about functions of set here.
How you concluded at the end that for all $b\in B, g_1(b) = g_2(b)$ is beyond me entirely.
 
Balarka Sen
Balarka Sen
@0celo7 I reject your notation
 
Secret
Secret
Sorry, I am trying to write in notation "For each a in A, there's a unique b in B such that f(a)=b" but I screwed up somewhere. For the $g_1(b)=g_2(b)$, I get it from $g_1(f(a))=g_2(f(a))$ and then using $f(a)=b$. Since every a corresponds to a unique b, thus for all b the $g_1$ and $g_2$ agree with each other, then the two left inverses must be the same thing
 
ACuriousMind
ACuriousMind
No, that's wrong.
That there is a unique $b\in B$ for each $a\in A$ such that $b=f(a)$ does not mean that all $b\in B$ are $b=f(a)$ for some $a\in A$.
In formal notation, it's the difference between $\forall a\in A\exists ! b\in B: f(a) =b$ and $\forall b\in B\exists a\in A : f(a) = b$. The two statement are not equivalent.
 
Secret
Secret
ok, noted, guess I need to think harder on how to deal with the non surjective property of f (since f does not necessary have a right inverse, by 5a, it is not necessary surjective). That means I need to consider the preimage of some b being empty, hmm...
 
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