The h Bar - 2016-06-08 (page 3 of 4)

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4:00 PM

@ChrisWhiteI just clicked your starred link and it crashed all of my SE tabs

did I just download a virus

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Does the patient to doctor ratio exist if there is no doctor? @0celo7

Acuriousmind said that in semirings there's an example such that 0n=/=0 does not imply distributive law is violated, thus disproving my conjecture

...Actually, looking at the article on semirings, if there are no identity elements, then how do we determine which element is the annihilating element? (since in semirings, the annihilating element is also the additive element 0)

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ No, this is the Unger Theorem.

@Secret You don't. There is no guarantee whatsoever for an "annihilating element" to exist.

@Secret Can you please write the full proof, in full sentences, with TeX?

I don't understand what you wrote above.

4:02 PM

It is usually added to the axioms of the semi-ring that zero annihilates other elements by multiplication because all examples of semi-rings you will encounter will be built that way.

But you can't prove it from the other axioms.

@ACuriousMind Is $0^2=0$ in a semiring?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

I'm talking about your mneumonic @0celo7

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ It's not mine

It was Dr. Briehl's

maybe no h in that name

Can't remember

@0celo7 yes since 0 anihilates any elements in the semiring and 0 is an element in the semiring

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Who is he?

4:03 PM

But he was a smug ass and took great pride in crushing my dreams of a STEM career

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ middle school math teach

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

I see.

@Secret Not by definition, no

ACM said you have to add that as an additional thing

@0celo7 Well, semirings are usually defined with $0\cdot n = 0$. The point is that in the absence of additive inverses $0^2 = 0\cdot 0 = 0\cdot(0+0) = 0^2 + 0^2$ does not imply $0^2 = 0$.

In any case, this is a boring question

Oh crap look at the time

Toodles ~~~

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

cya

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

4:15 PM

@ChrisWhite o/

is $A \times B$ the cartesian product of groups $A,B$?

Is anyone else experiencing that items in the review queue reload again after you have supposedly acted on them?

@Obliv What makes you ask that question? What are you reading that uses that notation without telling you what it means? What else are you thinking it could be?

lol.

It's telling me to prove $A \times B \simeq B \times A$ but isn't $A \times B$ the same exact thing as $B \times A$? What's the difference?

@ACuriousMind is the boundary operator a derivation of the algebra of topological spaces with product \times

4:19 PM

by same exact thing, I mean have the same elements

@Obliv The difference is that $A\times B$ contains ordered pairs $(a,b)$ and $B\times A$ contains ordered pairs $(b,a)$ where $a\in A,b\in B$.

@Obliv They have not. The ordered pair $(a,b)$ is not equal to the ordered pair $(b,a)$.

Note that the $(x,y)$ notation must be defined - the brackets $(-,-)$ are not equivalent to set brackets $\{-,-\}$.

@0celo7 It's a derivation alright. I'll refrain from trying to specify on what exactly.

(because, for instance, there is no "boundary operator" for abstract topological spaces)

user image

ok, that's indeed a counterexample

since we only have basically b = b + b thus it does not cause any contradiction

where b=0^2

That's not a counterexample. It's not an example at all. You've just written $0^2 = 0(0+0) = 0^2+0^2$ over the Wiki article on semi-rings.

How is that supposed to tell anyone anything?

user54412

@ACuriousMind Wait. "rng" is used for rings without identity now? I'm not sure if that's awesome or terrible.

@ChrisWhite Some people also use rig for the semi-rings because they don't contain negative elements.

The logical conclusion then is that a ring without identity and negatives is a rg

And...I think it's a funny idea but one shouldn't actually use it :D

4:27 PM

is $A \times B$ a group?

@ ACuriousMind In order to disprove the conjecture, we only need to find a counterexample where [given "0a=/=0" it does not imply "distributive law is violated"] (or the reverse of this statement) without any contradiction

The 0^2 example you gave satisfy "0a=/=0" but distributive law still holds and there are no contradictions like e.g. 0=1, a=0 etc.

@Obliv That only makes sense to ask once you have defined a multiplication on it.

does this make sense: Let $\varphi:A \times B \to B \times A$ be defined as $\varphi(a,b) = (b,a)$ then $\varphi((a,b)\star(c,d)) = \varphi(a,b)\triangle\varphi(c,d)$ Since $|A\times B| = |B \times A|$, $\varphi$ is a bijective homomorphism. ($\star$,$\triangle$) are the group operations, since I don't know what they are?

@Secret I didn't give an example. I showed where the usual proof fails, i.e. which step doesn't work. I didn't actually show that there is any single semi-ring (wihtout the $0\cdot n = 0$ axiom) in which $0\cdot n = 0$ doesn't hold, and neither did I prove that there are no contradictions to the other axioms (that's a rather difficult proof for any statement).

@Obliv That's the right isomorphism, but your argument for why it's bijective doesn't work.

oh i have to include an argument for injectivity too

no wait

shouldn't that be sufficient to say it's bijective?

4:32 PM

the map $\phi(a,b) = (1_B,1_A)$ is clearly a homomorphism (and still $\lvert A\times B\rvert = \lvert B\times A\rvert$), but it is neither injective nor surjective.

You cannot conclude from two sets being the same cardinality that maps between them are bijective.

but I said $\varphi(a,b) = (b,a)$ so there is a unique element in its codomain for every element in the domain

and since they are the same size and injective, it must be bijective, no?

@Obliv 1. You didn't say that it's injective. 2. That's not true if one of the sets is infinite.

1. I thought it was to be inferred, mb. 2. Why not? They are of the same cardinality still

@Obliv $\mathbb{Z}\to\mathbb{Z}, n\mapsto 2n$ is injective but not surjective.

wait really? Surely that's surjective.. :O

damn infinity screwing with my head again. I see why it's not surjective nvm.

4:38 PM

@Obliv What do you think the preimage of, say, 1 is?

oh that is a good counterexample

that doesn't rely on $\mathbb{Z}$ being infinite, though.

What? For finite sets, injectivity does imply surjectivity.

(and surjectivity implies injectivity, too)

OH. yes, I see.

so how can I conclude $A \times B \simeq B \times A$ if $|A\times B|$ is infinite?

...by proving that that thing you defined there is injective and surjective?

okay I see. I have to show that a 2-sided inverse exists for $\varphi$

4:55 PM

@MAFIA36790 hi, dont really know if the "host" idea will play a role. slereah did great job of introducing himself in the meta post (which has high visibility/ upvotes). any mods if present can play some host-like role or DS may if there. suggested to slereah he help define/ describe what format he wants to follow at beginning/ intro. eg he could present, followed by questions, etc.; am intending to be there myself but maybe only to ask questions (esp if no one else is, hopefully unlikely).

@Slereah do you have a ref for that?

user image

Therefore, there exist semiring (S,+) and (S,*) (no annihilation) for which the conjecture is false

5:21 PM

user image

Correction: You can actually escape from the set {b} by +1

@Secret Why do you have an aversion to presenting a linear, coherent piece of text? It's really hard to guess what you're doing in these pictures, and if you want to effectively communicate with other people they should not be guessing to begin with.

@ACuriousMind that was randomly invoked in my algebra class and it melted my brain

@ACuriousMind #rekt

@ACuriousMind what do you mean there's no boundary operator for abstract topological spaces

It's just cl-int

AFAIK those are both defined for any topological set

@0celo7 Closure and interior only make sense for subsets of a topological space.

5:45 PM

So fellas

How do I solve the Helmholtz equation in bipolar coordinates

I can just rewrite the cartesian solution in those coordinates, but then I assume that basis won't be orthonormal anymore

I can try a Laplace transform, but I'm not sure how easy that would be

Or I can pretend it's analytic in $\lambda$ and see where this goes

Hi everybody

I'm not sure if I can say greetings here. more sounds like a class room here :D

Sure, why not

so I wanna share another quote, as a break for you fellas :D

and interested in hearing your opinions on it

may I????

@2physics You may talk about whatever you like in this room (as long as it's not rude or offensive), we don't really have any sort of topic restrictions here

@ACuriousMind alright, sure. thanks

@ACuriousMind I just don't wanna be talking to myself here :D that's why I asked before

lol

True ignorance is not the absence of knowledge, but the refusal to acquire it. (Karl Popper)

5:56 PM

@2physics ...other regulars of this chat room have no problem at all talking to themselves :P

rude

@ACuriousMind well I've been talking to myself out of here, so rather not to do that here anymore :D lol

(I am just very bad at presentation. Ok let me try again...)
We define a semiring with no annihilation elements $(S,+)$ and $(S,*)$ where $S=\{0,1,b\}$ with the usual additive identity 0 and multiplicative identity 1 and $0^2=b$

We can then compute some of the entries of the + Cayley table by the following:
\begin{align}
& 0+1=1=1+0 \text{(Additive identity and commutative monoid)}\\
& 0+0=0 \text{(Additive identity)}\\
& 0+b=b=b+0 \text{(Additive identity and commutative monoid)}\\
\end{align}

@ 90celo7/Acuriousmind Therefore the above semiring disproves the conjecture

user image

Why does this equation have a star of david

@Slereah lol...I guess it's supposed to be a superposition of $\nabla$ and $\Delta$?

6:07 PM

probably

I really believe in that quote, I think, first result of that is that no matter how much knowledge you have, as long as you stop learning, you are surrounded by true ignorance. and the second result is, no matter how much you know, as long as you are trying to learn more, you are a true wise man.

Here's another quote from a wise man, though

user image

btw, why does it shows codes for me?? it shows it like this: $ \ Delta ...

you need mathjax

17

A: Any chance of MathJax in chat?

As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including: ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room. An altern...

@2physics There's nothing to "believe" there. The quote defines "true ignorance", just like you defined "true wisdom" there. Saying one "believes" in a definition seems rather pointless.

6:10 PM

@Slereah can't see the quote actually: s-media-cache-ak0.pinimg.com refused to connect.

Acuriousmind, I am not sure if the \begin{align} version above is more clear to you, but this is a lot more linear than the picture stuff

@ACuriousMind by "believe" I mean "I think it's true" or "It's a rational statment"

@ACuriousMind in my opinion actually

If Karl Popper is so smart why is he dead

@2physics I know, but it's a definition. It defines what "true" ignorance is. Unless you had some prior notion of "true ignorance" and now can say "hey, yes, this matches with my prior definition", there is no assertion of truth here.

@ACuriousMind like now that I don't believe in what you say that :" there is nothing to believe there". I think there's sth to believe or not

@ACuriousMind well maybe it's sth more than a mere definition. It can be the result of some revelation or inspiration. who knows

6:18 PM

@2physics What? For it to be a "revelation" all concepts used in it must have had prior meaning. But I don't know of any agreed-upon meaning of "true ignorance" in common language, so any phrase that says "True ignorance is X" is a tautology defining "true ignorance" for the person speaking. Only after one such phrase was uttered becomes "True ignorance is Y" a statement that can be false.

@Slereah maybe he's not smarter than nature?

But @ACuriousMind, in actual language, words are not defined by formula but by prototype

however from the cayley table derived above, b act almost like 0. I am having trouble, however, to find a way to show that $b=0$ if there exist one (since my workings have accounted for all the possible elements in the cayley table)

If one let $b=0$, the results are still obeyed, but the above results showed that even if $b\neq 0$, the structure is still closed

@Slereah Maybe, do I look like a linguist!

Well you argue about word meaning :p

6:22 PM

@ACuriousMind "For it to be a "revelation" all concepts used in it must have had prior meaning." nope, I don't agree with that because revelation means : the revealing of something previously unknown.

@Slereah (test for linear communication) Does the above workings make sense to you?

Dunno, didn't read!

@ACuriousMind "But I don't know of any agreed-upon meaning of "true ignorance" in common language" that's also not general. when you don't know sth, or haven't heard it, that doesn't mean no body else hasn't discussed it before. and also many people may think that the true ignorance is absence of knowledge because these two words are antonyms. And maybe that's why he says true ignorance is Not "absence of this it's it's sth else"

@2physics I feel you're not getting my point. The sentence "Fladbgetrd is refusing to learn" cannot be a revelation because "Fladbgetrd" did not have meaning prior to me writing that statement - that statement is either a definition or non-sensical. Likewise, "True ignorance" is not a concept that exists outside of that sentence: While "ignorance" is a word with a well-defined meaning ("not knowing something"), "true ignorance" is not.

@ACuriousMind and also, you don't look like a linguist but without language (of any kind) you don't look like a physicist too (necessarily).

6:34 PM

@ACuriousMind I am strongly suspecting that $b=0$ based on what is found in the above workings (not that pics working, the polished working typedseted with \begin{align}) but I cannot seemed to came up a working that can show that $b=0$

(As major supporting evidence that "true ignorance" doesn't have a common meaning I invite you to just search for "true ignorance" - most of the hits are this exact quote, which should not be the case for any compound that has a meaning outside the quote. And many of the other hits are other philosophers defining their own notion of "true ignorance".)

@ACuriousMind 1-

@2physics ?

@ACuriousMind 1- "revelation" according to its meaning and according to what people who has experienced it can also include any element which you don't know about before 2- "true ignorance" is not necessarily a quite different thing from ignorance, you know what true is, and you also know what ignorance is, and true is used as an adjective here. maybe he has tried to say "what is believed as ignorance is not truly the absence of knowledge.."

so maybe now I'm who looks like a linguist actually :D @Slereah@ACuriousMind

@Secret I think that this algebraic structure is really just too weak to show that.

It can have elements that "look like" they should be equal but aren't.

6:46 PM

Hmm...I see, looks like an interesting thing to study and expand on ( I have been hunting for algebraic structure where $0a\neq 0$ and not lead to contradiction for ages)

@2physics 1. That doesn't address what I said in any way. 2. "Ignorance" is defined as "the state of not knowing something". If you now say "Ignorance is not truly the state of not knowing something", that's not a "revelation", it's just non-sensical ("X is Y" and "X is not Y" cannot both be true in the logic we usually use).

and also "algebraic structure" is a definition so, maybe it doesn't make sense if Secret believes in what you mentioned about it or not except it's a revelation lol

you're really just playing word games - as philosophers are so wont to do.

@2physics I feel you're not appreciating the need for logical argumentation :P

@ACuriousMind of course I do appreciate it! expressing opposite ideas doesn't mean there is no appreciation for critiques :D

plus,why do you use your logic to express your ideas and critics but use your feelings (as you said I feel) to judge about my appreciation about your attedance in arguments??:D

@2physics Oh, I really only referred to your comment about the algebraic structure with that

@Secret But...why?

6:58 PM

@ACuriousMind about 1- I think it truly did address. Revelation is not sth scientific, It can cause sth out of nothing! a totall new creature :D also it can result in revealing new relationships between old and existing definitions . 2- about this one please wait I'd like some tea

I like to study highly pathological things, and work out conditions that need to be satisfied by them
If they cannot exist, I also like to work out exactly how and what prevent them from existing
Basically, I am an explorer and maths to me is like an alien space

In particular, it is well known that division by zero will lead to the trivial ring unless some axioms of the reals are violated. Studying the papers of Wheel theory, Meadows and the hidden 0 (john Cender), and my past attempts on constructing and destroying potential division by zero structures all seemed to have the common property of being not distirbutive.

This caused me to wonder whether the distributive law is the key to the nonexistence of division by zero structures (which we have moved one step towards that answer by showing an explicit example where a semiring without annihilatio…

@ACuriousMind Tell that to Whitman

@ACuriousMind about 2: think you and most people think X is Y for some reason. and X is not a mere letter! X can include a vast range of meanings and concepts. so we can say most people believe in that statement. they believe X is really Y! that then somebody comes around and says no! you people are not right, X is apparently Y! but in a deeper level , (based on his logic which his arguments support it or based on his revelation) X has another totally different meaning! and that's Z!

that's all

@0celo7 See super long wall of text for the sermiring with non anihilation elements that I discussed with acuriousmind earlier

@ACuriousMind and also I don't play word games. words are not as precise as digits.

7:13 PM

@2physics I'm not sure I'm following you, but I suspect you have a different notion of what it means for a statement to be "true" than I do, and in particular what a definition is.

PS if $b=1$ we get a very boring structure because 1 will become an absorbing element in addition and multiplication (which means no escape once moving to 1)
This is caused by $b+1=b=b+b$ thus killing off the pathway $b+1$ to escape from the absorbing element

(I have no idea how to write the above paragraph in rigorous mathematical language)

if $b=0$ we get the trivial ring $\{0\}$. Therefore as long $b\neq 0 or 1$ the structure will be somewhat interesting since you can escape from the absorbing multiplicative element b by $b+1$

(Ok, so it turns out we still have annihilation in this structure, except that it is not 0)

@ACuriousMind also you know scientific true/false and moral (and other stuff) true/false and the logic behind them and their cause and effect system is quite different.

what the heck is the absorbing element

Absorbing element

In mathematics, an absorbing element is a special type of element of a set with respect to a binary operation on that set. The result of combining an absorbing element with any element of the set is the absorbing element itself. In semigroup theory, the absorbing element is called a zero element because there is no risk of confusion with other notions of zero. In this article the two notions are synonymous. An absorbing element may also be called an annihilating element. == Definition == Formally, let (S, ∘) be a set S with a closed binary operation ∘ on it (known as a magma). A zero element is...

any element a that multiplies any other element give a

0 is the best known example (which is why division by zero often lead to problems)

@ACuriousMind oh, right

derp

7:42 PM

@BernardMeurer quest???

I have to live for a week with a tube inside of my lung and out of my chest.

feels bad.

oh sorry

sounds really hard

yeah my lung collapsed.

why?!

no reason, it was a random collapse.

it's common in smokers but i've never smoked.

does it happen for everybody?

how old are you

7:45 PM

yeah but mostly tall and lanky males 18-35.

i'm 18 xD

@3075 They emailed me yesterday asking me to send them an 'outline of my course' and I was like "The heck's that mr" but politely of course

@BernardMeurer any day now.

and they sad it was an outlook on what my highschool level maths course covered

so I said what it covered but they asked for some sort of document to prove it, which I don't have and afaik I can't get it. And I mentioned that the CS people didn't ask for that either

@3075 maybe you need to do exercises..

so they seemed to be chill and just left that alone

so I'll know in a couple of days I guess

7:48 PM

@2physics nah it's due to irregularites on the lung you're born with.

a lot of people have irregularities but only some cause collapses.

@3075 oh I see. I hope you get better soon and start learning about string theory more and more with uncollapsed lungs ;) :)

@2physics thanks.

u welcome

@3075 But yeah, don't look too good

hey @bernard how are ya

@3075 dang maybe I should start exercising more lol. i'm a tall lanky 18 y/o too D:

7:56 PM

@Obliv Horrible depressed and dying

Jk I'm fine :)

why are we all 18?? :D

Good thing I'm short and lanky, my lung is safe :P

@bernard what kind of programs have you been writing lately?

@ACuriousMind I thought you were tall o.O

@ACuriousMind you're also 18 ?? :D

7:59 PM

@2physics Nah, I'm an old man

if he was 18 I'd feel even more inferior than I already do with @0celo7 being 18. @2physics

@Obliv you probably know more than me

I forget everything

@0celo7 ...why does everybody think I'm tall? oO

anyway, I suggest you guys be 18 too

@ACuriousMind tall people are smart

8:00 PM

just the fact that you know what you know at your age is enough to impress me :p @0celo7 regardless of if you forgot how to tie your shoes

@0celo7 or smart people are tall??:D

@Obliv Sad programs

@Obliv Actually I did forget that

I'm not convinced i do it correctly

@0celo7 ...just like short people are smug to you, I suppose? :P

@ACuriousMind YES

Also really thin people are smug

Like, where do the organs go?

Also really fat people are smug too

8:01 PM

I can imagine you wearing those sketchers with the velcro straps in grad school @0celo7 lol

@Obliv On a serious note not much, busy with some stuff

wat

@Obliv Nah, I wear Nike, Jordan, and Polo

What does the location of organs have to do with smugness?!

@bernard busy with calculator memes? or irl stuff

8:02 PM

@ACuriousMind because they're like

"I don't need a big liver like you, loser"

@Obliv He's probably out hunting more calculators to torture :P

@0celo7 I'm sure that's what all those monks that fast think :)

"I only need acorn-sized kidneys"

@Obliv huh?

you know the really thin monks that try to reach enlightenment by starving themselves and meditating? @0celo7

@0celo7 lol, I'm sure that's exactly what's going through their minds

8:04 PM

@Obliv Mostly IRL stuff sadly

btw dear physicist please inform us how long(i mean high :D) a grown person should be in your area to addressed "tall" by average people of that same area :D

@bernard well I hope it clears up man :p I'll try to think of challenging programs for you to write until then 8D

@ACuriousMind doesn't matter if its conscious or not

@BernardMeurer got the baton

I now have a harry potter wand ;)

@2physics you should ask that on the main site. I'm not a physicist so i don't qualify to answer that ;)

@2physics 6ft I think

8:07 PM

@Obliv well let us correct it: Dear readers please inform us how long(i mean high :D) a grown person should be in your area to addressed "tall" by average people of that same area :D (please send your answers in centimeters)

@ACuriousMind Is it supposed to be obvious that the index of a vector field does not depend on the small sphere

@Obliv Soon it will :)

@Obliv that was a good idea, you think they won't block my question?

@0celo7 Remember my photoshop

@BernardMeurer yes that's why I said it

8:09 PM

@2physics "Tall" people are at least 1.85m

don't actually ask main lol @2physics I think it's pretty intuitive as long as they stand out (maybe >7.5 cm) than the average, they're tall.

@ACuriousMind I'm tall by your standards

@0celo7 I'm not the authority for deciding what is obvious, but if you know that it's the determinant of the Jacobian at the point, it's clear that it doesn't, no?

@Obliv it was a good answer. . but the problem is, where can I find the average

@0celo7 Thank you for this valuable piece of information.

List of average human height worldwide

Below are average adult human heights by country or geographical region. The original studies and sources should be consulted for details on methodology and the exact populations measured, surveyed, or considered. Note: Letters in grey indicate non-measured height. == Notes == == References... ==

8:11 PM

too quick holy crap

wikipedia has lists for everything

ok thanks wait

I wanna see if I'm tall or not

@ACuriousMind ...what

based on your standards

:D

One of my favourite lists is probably list of tautological place names

8:12 PM

@ACuriousMind Is tall one sd abv mean?

@0celo7 I'm also not the authority for defining tall

> "Tall" people are at least 1.85m

That was my personal guess/feeling of how tall a person has to be that I would call them "tall"

1.85m? thats 185 cm? isn't that only 5 foot 2

@ACuriousMind GP says if the sphere $S_\epsilon$ works and $S_{\epsilon'}$ also works then it will have the same index

8:15 PM

no wait thats 6 foot 2

where "works" means "includes only one zero"

because the vector field $v/|v|$ "extends to the annulus created by the two spheres"

Do you know what they mean?

Ladies and gentlemen I just realized that I'm tall. actually 6.6cm above male avg and 20.2cm more than females avg in my area. @ACuriousMind@0celo7@Obliv@BernardMeurer

I don't know what to do! be glad or ashamed of :D

not really tall you know

are you a standard deviation above the mean

but I'm sure I haven't experienced any lung collapse yet.

@0celo7 The degree of a map is a homotopy invariant.

8:19 PM

soooo maybe I'm not tall enough?

@ACuriousMind Yes, I know, so?

@0celo7 And the vector field on the annulus is a homotopy between the two vector fields on the spheres.

@0celo7 I love how you called ACM bajoran. reminds me of Lt. aldo rain in inglorious bastards. I think the correct spelling is bjorn, right?

@ACuriousMind Hmm...HMM...grumble...not sure I get it

@ACuriousMind I agree. soooo am I tall really??? :d

8:20 PM

@Obliv Bjørn

Björn, actually.

@ACuriousMind The Germans butchered it when they took it from the Norsemen

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

@ACuriousMind interesting redundancy

@0celo7 No butchering going on, we just use a different symbol for the sound.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

One of my favourites is "last and final call"

8:23 PM

@ACuriousMind Ok, I don't get it. In the theorem "degree of map is homotopy invariant" we don't homotopy the spaces the maps are defined on. i.e. if $f,g:M\to N$ and $f\sim g$, then their degrees are equal

but in this case the two vector fields are defined on different spaces altogether

$S_\epsilon$ and $S_{\epsilon'}$ are not the same spaces

@0celo7 Identify the two spheres by a homeomorphism of your choosing.

@ACuriousMind and how does this not change the degree?

doesn't it actually have to be an orientation-preserving diff?

@0celo7 It should be a homeomorphism with degree 1.

Netherlands 183.8 cm (6 ft 1⁄2 in) Avg height. Really?!

@ACuriousMind which is an orientation preserving diff

8:27 PM

didn't know they're so much tall actually

nobody said anything about differentiability

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

@Danu o/

@ACuriousMind in Milnor everything is smooth

@2physics Right on cue, @Danu :D

how the heck are you defining this without smoothness

8:28 PM

lol

@0celo7 The degree of a map $S^n\to S^n$ induces a homomorphism on the top integral homology - the degree is the number by which this map multiplies every integer (such multiplications are the only homomorphisms $\mathbb{Z}\to\mathbb{Z}$)

you know that means nothing to me...

@0celo7 are you angry?

yes

@0celo7 you know that means nothing to me...

:D

8:31 PM

lol

you people are evil.

@0celo7 how the heck are you defining this without smoothness :D

@ACuriousMind how do I know that spheres of different radii are diffeomorphic, anyway

@2physics by blocking you

@0celo7 what

@ACuriousMind I need a diffeomorphism between $S_\epsilon$ and $S_{\epsilon'}$

8:33 PM

I firmly believe you're able to figure that out yourself :P

@ACuriousMind I'm sure you'll tell me I need to just slide points along rays through the origin onto the smaller/larger one

or something

@0celo7 was that a threat?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

yes

NOT!

:D

@0celo7 And once again you have asked me a question to which you already know the answer.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

lighten up @2physics

8:35 PM

@2physics Yup.

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ will you block me too?

At 175cm I'm a pretty short guy :(

@ACuriousMind not really

I don't know how to prove this is continuous, not to mention smooth

and with a smooth inverse

...shrinking all vectors by a constant factor is an invertible linear map, certainly you can prove that linear maps are smooth, no?

@Danu no worries, Acceleration just depends on Force and mass, according to Newtons laws. Height doesn't affect it.

8:38 PM

@ACuriousMind of course, I'm not a child

I just didn't know that it was "shrinking all vectors by a constant factor"

@2physics Errr?

I just don't understand why he want's to block me :( did I do sth wrong?? I'm just tall, that's all

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

@Danu Some insiration for you :P

He's just a bit "rough" around the edges @2physics ;-)

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ may I watch it too?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Sure!

8:43 PM

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Who?

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ I already did

:D

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:D

0celo @Danu

Hi @3075 how are you feeling?

bad but I'm hanging in there.

injectivity implies every domain element is mapped to the codomain 1:1 with no shared codomain elements right?

a few more days.

8:46 PM

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ He's just angry and I understand

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Hang in there @3075

:(

hang in where?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

He's sick

Surgery

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ really? Hope he'll get better soon, he's cool

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Yup

8:48 PM

@Obliv No-one says "codomain" (at least I have to think briefly every time when that word is used :P) Just say "no two elements have the same image (in the target)"

and hope his block button will break and can't use it anymore :D

what do you call it then, target?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:D

@Obliv yep

you know sometimes codamines work as the opposit of codeines

8:51 PM

@acuriousmind must it map the entire domain to this target set? I'm told to prove that if a homomorphism is also injective, then the image of this homomorphism is isomorphic to the domain of the map

but the isomorphism requires bijectivity which requires surjectivity and Idk how to prove it must also be a surjection. Do you want context?

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ you still think that was a threat by 0celo7??

@Obliv The image is the subset of the target to which at least one element of the domain maps. Every map is by definition surjective when this image is taken as its target.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

It wasn't @2physics

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ what then??

so if $\varphi: G \to H$, $\varphi(G)$ is injective, it is bijective by definition because surjectivity is implied?

8:54 PM

@Obliv no

er. I mean $\varphi(G)$ is bijective

You want to say that the map $\varphi : G\to\varphi(G)$ is bijective.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Don't worry about it @2physics

(where it is actually abuse of notation to use $\varphi$ for this map but everyone does that...)

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

He's just being himself @2physics

Btw, The quote in your profile @2physics is one of my favourites :-)

8:57 PM

@0celo7 Does ORNL celebrate the proposed name for element 117?

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