Mathematics - 2022-12-10

leslie townes

12:50 AM

bhpbhpt

Ted Shifrin

2:06 AM

@leslietownes Feeling your usual grumpy self?

leslie townes

getting back there, yes.

Ted Shifrin

Joy to the world!

user4539917

2:33 AM

Jeremiah was a bullfrog

one potato two potato

4:18 AM

peep

Waited for 2hours to do this

shintuku

5:40 AM

@nickbros123 what's nice about Spivak's book is that there are a ton of resources for it! you can also check out lectures on youtube for the exercises

user 726941

10:20 AM

18

Q: Should teachers encourage good students to help weaker ones?

On the one hand, weaker students benefit greatly from studying with good students, and it's helpful for the good student to be able to explain a concept to weaker students as well. On the other hand, good students benefit from studying with other good students (example), and it's why universities...

teaching

nickbros123

11:38 AM

@shintuku yeah true, but the intention of Spivak would've been that we study his writing, and solve the problem with what he's given us. We are breaking this by looking up solutions and seeing lectures on yt

Btw turns out my teacher is using Bartle and sherbert as his main textbook for teaching. He didn't even tell us, a PhD student of his told me

one potato two potato

2:24 PM

-1

Q: Seeking a teacher specialized in the Riemann Hypothesis, to write to him privately

Please this is my dream I wish I had a teacher specialized in the Riemann Hypothesis that I could write to him privately to answer some questions. Believe me, I will not disturb him, but I have some questions and I need to write to him through his account on the communication sites. Thank you......

sequences-and-series monty-hall

Someone edited the question after the question is closed

Never seen such happening. This kind of question is usually just closed immediately and nobody cares.

Shaun

4:40 PM

Hi. I'm getting a lot of downvotes for no apparent reason lately. See above and the following answer of mine . . .

1

A: Let $H = \{g \in S_6 : g(a)\equiv a\pmod3 \text{ for } a\in\{1,2,3,4,5,6\}\}.$ Is this a subgroup?

Use the finite subgroup test. You have $e\in H$, so $H\neq \varnothing$. By definition, $H\subseteq S_6$. For closure, suppose $g,h\in H$. Then for all $a\in \{1,2,3,4,5,6\}$, we have $\color{blue}{g(a)\equiv a}\pmod{3}$ and $\color{red}{h(a)\equiv a}\pmod{3}$. Now $$\begin{align} (gh)(a)&=g(\c...

Curio

5:19 PM

Consider this function:
f(x,y) = (siny)/x for x!=0
f(x,y) = 0 for x = 0
Is f(x,y) derivable in (0,0)?

smth

5:35 PM

find smallest n such that there exists monomorphism between $\mathbb{Z}_{pq}$ and $\mathbb{S}_n$ where p and q are prime. I have got that n must be 10, am I right?

Ted Shifrin

5:48 PM

@Curio is it continuous?

Lukas Heger

@smth you can go smaller

the exact value of $n$ depends on if $p=q$ is allowed or not

Curio

@TedShifrin It is not in (0,0)

Ted Shifrin

So, doesn’t that answer your question.?

smth

@LukasHeger if p is not equal to q

Lukas Heger

the minimal choice is $p=2,q=3$. Now the question is for what $n$ is there a monomorphism from $\Bbb Z/6\Bbb Z$ to $S_n$?

smth

6:00 PM

@LukasHeger my way of thinking was following:

Curio

Well, no :(
Wait, I don't mean "differentiable" with "derivable".
"derivable" = the partial derivates exist

smth

first I tried the "minimal" case, if f is actually isomorphism, that is possible if n=3, because 3! = 6, but I realized that with that I will have a problem with something commutation of permutations, like if its homomorphism f(1+2) should be equal to f(2+1) so f(1) and f(2) should be disjoint permutations am I right?

Lukas Heger

hmm I'm not quite following your reasoning

smth

for example, let f(1) = (1,2) and f(2) = (1,3)

Lukas Heger

This is a helpful observation: if $G$ is any group, than a monomorphism $\Bbb Z/k\Bbb Z$ exists iff $G$ has an element of order $k$

smth

6:05 PM

f(1+2) is not f(2+1)

Lukas Heger

right

in fact, the whole homomorphism is defined by f(1)

this leads to the observation I mentioned above

If you know f(1), then you know that f(2)=f(1+1)=f(1)f(1), f(3)=f(1+1+1)=f(1)f(1)f(1) and so on

smth

I am not sure what is your hint

okay i get that

Lukas Heger

okay so f is a monomorphism iff f(1) has order exactly k

so the question reduces to what is the minimal n so that S_n has an element of order pq. You can use the disjoint cycle decomopsition of elements in S_n to determine the possible orders

Ted Shifrin

@Curio well, why can’t you answer that? Just compute.

AMDG

Hello all. I'm not really sure what to query, so I'd like to ask: are there any series representations of $\ln(x)$ in which $x$ is the only term in the numerator of any series term?

smth

6:11 PM

if p=2, q=3, then S_n has element of order 6 if its S_6?

@LukasHeger

Lukas Heger

it's true that S_6 has an element of order 6, but is it the smallest S_n that does?

think about disjoint cycles

smth

S_5 sorry

Lukas Heger

no

S_3 has an element of order 2 and an element of order 3, but no element of order 6

evry element in S_3 is a 2-cycle, a 3-cycle or the identity

S_5 is correct

smth

S_5

Okay, just a moment please

Ted Shifrin

@AMDG If by $x$ you mean $x$ and not $x-1$, for example, then no. $\ln(x)$ isn't even defined at $x=0$ (or to the left of $x=0$), so has no analytic representation centered there.

@Lukas: Very sneaky using the fact that $6$ is not actually prime.

AMDG

6:19 PM

Hm, well then I suppose x - 1 can do. I forgot Wolfram Functions existed, so I'm browsing that right now to see what's available as well.

Lukas Heger

@Ted indeed

Ted Shifrin

Now do you mean no higher powers? @AMDG

Certainly $\ln(x)$ has an infinite Taylor expansion centered at $x=1$ and it converges on $(0,2]$.

smth

Okay so, if I know what f(1) is, then I know all of f(k); Z_pq has an element of order pq, so I would need to have an element of order pq in S_n also, for pq = 6, smallest such n is 5, so that is the answer. Is my conclusion alright @LukasHeger?

Lukas Heger

@smth yes, that looks good

AMDG

@TedShifrin Yes, I suppose I should have been more precise: whether there exists any series representation such that the highest order in the numerator of any series term is $\pm1$

Ted Shifrin

6:21 PM

Only for linear polynomials, then.

$-1$ gives you $1/x$ or $1/(x-a)$, and that's it.

smth

@LukasHeger Thank you so much!

Lukas Heger

@AMDG the problem is that $\ln(x)$ has no analytic continuation to a punctured neighborhood of $x=0$, so there cannot be a Laurent expansion around that point

Ted Shifrin

@Lukas The question is much more rudimentary.

Lukas Heger

Oh I confused numerator and denominator

smth

Can I ask you all one more question from Algebra (sorry for bothering you)?

AMDG

6:24 PM

@TedShifrin Yeah, and that is more or less permissible as the series terms in that case can easily be made first order if we end up with $-1$ order.

Curio

Actually, the original problem was:
f(x,y) = (siny)*x^a for x!=0
f(x,y) = 0 for x = 0
Check for which a f is continuos, differentiable in (0,0). For which a do the partial derivates exist?
It's continuous for a>=0. Then the textbook says: <<We don't consider a<0 for the "partial derivatives" condition in (0,0) because the function is not continuous>>. This statement is driving me mad.
In fact I got that the partial derivates of f in (0,0) are both equal to 0.

Lukas Heger

@smth just ask don't ask to ask

Ted Shifrin

Oh, wait. @AMDG. Your way of speaking totally confused me. You want a series expansion where all the coefficients are just $\pm 1$. Nothing to do with exponents.

smth

N is a normal subgroup of G, and N is isomorphic to $\mathbb{Z}$, $G/N$ isomorphic to $\mathbb{Z}_n$ and n is an odd number geq than 3. how to prove that G is abelian?

Lukas Heger

oh wait I missed the N is isomorphic to Z part

AMDG

6:27 PM

@TedShifrin No no. I'm terribly sorry for any confusion. The exponents of the principal input of the series being $x$ in $\ln(x)$ should be at most a magnitude of 1.

Ted Shifrin

@Curio Yes, the partial derivatives are $0$ when $a\ge 0$. However, when $a=0$ you have a domain problem with $x<0$. Continuity isn't the point.

AMDG

To save everyone time, I'm trying to find a way to break up an arbitrary sum in the domain of the natural logarithm in a way that is "nice", viz. $\ln(x + y) = f(x,y)$ for real arguments.

Ted Shifrin

OK, @AMDG, this is how I originally interpreted your question. The answer remains: Only a linear polynomial (or $1/(x-a)$). Never happening for $\ln x$.

Lukas Heger

@smth I'd try to prove that N is in the center of G and work from there

note that G/N acts on N by conjugation (this requires some justification)

Ted Shifrin

You have $f(x,y)$ already. What do you mean? $\ln$ is never going to be linear.

@Curio Sorry, typo in my previous comment. When $a<0$ is what I meant. If the function is defined for $x\ne 0$, then you can talk about partial derivatives. Or perhaps since $\sin 0 = 0$, you want to define $f(x,y)=0$ for $y=0$ and not give a formula for it.

Curio

6:31 PM

Where is the domain problem for a<0? If a<0 and x=0, then the f(x,y)=0 because I defined f in that way ( f(x,y)=0 if x=0)

AMDG

@TedShifrin Alright, well suppose due to certain limitations of computation we have some really large number for which we'd like to compute the floored logarithm of but symbolically to save space, and which is represented directly as a base and an exponent, e.g. $3^{2^{64}}$. We wish to simply compute $\lfloor\log_2(3^{2^{64}})\rfloor$.

A bifurcative algorithm that permits computing the floored logarithm of any number is preferable.

smth

@LukasHeger I am not sure what to do with "G/N acts on N by conjugation", understand that I have to prove that Z(G) = G, then G is abelian

Lukas Heger

@smth I'm saying that you should prove that N is contained in Z(G) first

Ted Shifrin

@Curio No, I'm worrying about the fact that for certain $a<0$ the function $x^a$ is undefined at negative $x$. Anyhow, I don't think you should waste any more time on this question.

AMDG

What we can compute directly for any real $x\in [0, 2^{64})$ is $\lfloor\log_2(x)\rfloor$, and that's it. We can't conveniently insert a coefficient between the floors.

Lukas Heger

6:36 PM

@smth so $N$ is normal in $G$, this means that we have a homomorphism $G \to \mathrm{Aut}(N), g \mapsto (n \mapsto gng^{-1})$. But $N$ is abelian, thus this homomorphism is trivial on $N$, so it factors over $G/N$

Curio

Oh, I just realized that I forgot something important...it's |x|^a instead of x^a

Lukas Heger

are you following so far?

smth

wait a second please

Ted Shifrin

@Curio Ah, in that case, there's no domain issue and they're wrong. As I said, move on.

Curio

Perfect, thanks. Sorry for the mistake.

AMDG

6:38 PM

(However, I did find a way to compute $\lfloor\log_2(y) - \log_2(x)\rfloor$ using a simple linear approximation of $\log_2(x)$ which is valid for reals.)

So the question at this point is what it would take to get something like prostaphaeresis without computing exponentials and logarithms directly, or even a table.

smth

@LukasHeger what you mean by it factors over G/n (sorry English is not my native language)

Koro

Strange macbook connected to wifi again

But didn't connect for last few months. I had to use cable to access internet.

probably the most untrustworthy device I have.

AMDG

See the funny thing about all this research that I've done on division is that I have learned that any Mersenne number can be expressed as the product of an odd number and another number, and even numbers, the same but with a power of two coefficient.

By this and a few other things that are quite obvious to me, it seems clear that there are in fact infinitely many Mersenne primes.

Now this is cool and all, but it isn't strictly related to what I want, so I digress.

What I'm trying to solve at the moment is $y (2a + 1) + 1 = 2^x$ for positive integer domain for $x$, $y$, and $a$.

Koro

Any suggestions for numbers theory reference please?

For a beginner.

AMDG

So is there at least a way to take the coefficient $y$ or $(2a+1)$ out of $\lfloor\log_2(y\cdot(2a+1) + 1)\rfloor$?

(We can treat $2a+1$ as a constant)

Hm, actually it's probably easier to work with $(2a+1)y = 2^x - 1$ since the floored log 2 of a Mersenne number is necessarily always one less than the power of two above it.

$\log_2(2a+1) + \log_2(y) = x - 1$

Koro

7:18 PM

What would go wrong if I defined $\frac {0}{\infty}=0$ in the extended complex plane?

leslie townes

koro, 'go wrong' is somewhat subjective. unless you have a very fixed idea of what it means to go right.

Koro

In extended reals it is $0$ then why is it not defined in extended complex plane?

leslie townes

well, you can just define it that way? isn't the question more, what do you want to hold, would it or would it not be compatible with that?

even choosing to work in the extended reals at all is a choice. nobody says "by the laws of the universe, the extended reals." you do that because you want to.

Koro

I see. Thanks. I'll see for what purpose the ratio is left undefined as I learn more complex analysis.

"The neighborhood of a young child consists of the people very close on the left
and right. As we get older we think in terms of two-dimensional neighborhoods
(the people around the corner) or even three-dimensional neighborhoods (the
people in the world). In this chapter we do likewise."

leslie townes

it seems like if you'd want to define it to be anything, 0 would be a good choice. i don't know what might go wrong.

Koro

7:30 PM

hmm, the extract makes sense.

leslie townes

the usual way of formalizing something like this, after adopting a definition, is picking your favorite thing you are worried about not being true anymore (some law of arithmetic) and then performing the case analysis to check whether it is still true as you have extended it. it's maybe 'thematically' unenlightening, but it answers the question.

if the law holds then hooray, your 'extended' system still has that law. and if not, you can say 'well, the fact that this law will fail to hold is one reason why you might not want to extend these operations in this way.'

the extended reals/complexes tend not to have multiplicative inverses for infinity, so you can expect to have some goofiness around products and things being equal to 1. other than that i don't know what might go wrong.

fwiw en.wikipedia.org/wiki/Riemann_sphere says it is "customary" to define 0/infty = 0 in the extended complexes. as with all wikipedia, who knows what that means, but it's a data point

Koro

Leslie, I noted that Ahlfors also defines $z/\infty =0$ for any $z\ne \infty$. But the other book that I have defines $z/\infty =0$ for $z\in C-\{0\}$.

leslie townes

7:46 PM

ah OK. stuff like this is generally a sign that it probably doesn't matter which way you do it, or if you do it at all.

or we can submit a claim to IUPAM and have this formally adjudicated once for all time.

Koro

yeah

Ted Shifrin

Seems more like an oversight — or something an incompetent editor did. Personally, I never define extended arithmetic.

How are Munchkin and sick pet?

leslie townes

in a masterpiece of comic timing, munchkin's mother tested positive for covid yesterday.

Koro

Leslie and Ted: I want to ask one thing.

Lately, I've been getting lot of news notifications like 'someone was healthy and was going home and then collapsed due to heart attack.'

The concerning thing is that the age of the victims is less than 30. It is strange because this disease is not common in this age group.

Reading more into such news, I observe that most of them were tested covid positive once.

Ted Shifrin

@leslietownes She succumbed. Not surprising.

Koro

7:57 PM

I was also tested positive once. So this concerns me a lot :(.

leslie townes

those kinds of deaths occurred pre-covid, but did not get as much media attention. i don't think there is enough data to associate it with covid. a lot of the 'heart attack and just died' stuff is also anecdotal, the result of reporting without medical examination.

Ted Shifrin

Coincidences happen, and sometimes ostensibly healthy people have underlying issues.

Koro

and what's more is there are experts comments on such news. The comments are like this - these days walking has reduced in this age group hence the cases.

Ted Shifrin

Pretty much everyone I know has tested positive and/or been sick.

Young people should have physicals but typically never do.

Koro

by walking I mean people are not moving enough like mostly sitting in chair etc.

leslie townes

7:59 PM

it will probably be a long time, like decades, before people can meaningfully parse out any potential long term effects of having covid. it's pretty difficult to document that somebody hasn't had it.

koro: maybe just read less news? the 'experts' they get to comment on such things tend not to be busy doing anything except talking to the media.

AMDG

Yeah, because there's definitely nothing about spike proteins being manufactured and put out into the bloodstream that would have anything to do with any of the recent deaths of perfectly healthy people, and if you question it, then, well, you're a conspiracy theorist who doesn't listen to reason or something.

Hashtag "Democracy" and "Freedom of Speech"

Koro

@leslietownes yeah. I think I might have looked up something related to some disease and my search data may have been sold to some third party and hence I'm getting 'related' news.

leslie townes

koro: i'm sure the helpful professors at youtube university can provide further background.

Koro

haha

or some experts on tiktok.

Ohh but tiktok is banned here.

Ted Shifrin

Or all the experts on FOX “News”

Koro

8:05 PM

it's good that tiktok is banned here.

AMDG

Legit saw the other day that there is--I kid you not--a "research study" that is actually published in a medical journal claiming that the cause of recent heart attacks etc. is in fact due to those who are asking and want more beyond "due to unknown causes".

/shrug

Ted Shifrin

Well, I’ve had two major heart surgeries. Maybe I’ll drop dead.

AMDG

The Experts:tm: be like: "Caution: seeking the truth leads to adverse outcomes involving the heart."

Last I checked, science is the formal pursuit of truth, usually by way of the scientific method, rigorous and with theories developed and refined by existing and new evidence. :thinking:

Instead, we observe that the scientists apparently would like to conform reality to their model instead of conforming their model to reality.

Koro

the reality is $\mathbb R$ or $\mathbb R^n$.

rest all are strange fields.

AMDG

So true. Everything else is in $\Bbb{C}^n$

Koro

8:11 PM

even $\mathbb R$ is not real. I think only $\mathbb R^3$ is real so to speak.

AMDG

Failing to conform one's model to reality is the cause of all human suffering.

Ted Shifrin

No clue what you think “real” means.

AMDG

Objective reality

Or rather what is in it

shintuku

but how do we know if it is discrete or complete >:(

how can you affirm with certitude it is $\mathbb R$ and not $\mathbb Q$

such effrontery

Ted Shifrin

Discrete?

shintuku

8:26 PM

yes discrete, it could be discrete

what if matter moves through very small teleportations, huh, what about that

it's how we do it on computers, and everything happening on computers looks pretty continuous but let me tell you that there is a world of discontinuity between 0 and 1

smh people confidently affirming reality is $\mathbb R$

Ted Shifrin

The usual topology on $\Bbb Q$ is not discrete. This is getting ridiculous.

shintuku

i have noticed that fact and was about to edit my statement, but the logic of the statement did, in fact, not imply that I was saying $\mathbb Q$ is discrete

Ted Shifrin

And spaces are not “discontinuous.” Just stop.

Lukas Heger

@smth there's a projection homomorphism $\pi:G \to G/N, g\mapsto gN$ by saying that the homomorphism $\varphi:G \to \mathrm{Aut}(N)$ factors over $G/N$, I mean that there's a homomorphism $$\overline{\varphi):G/N \to \mathrm{Aut}(N)$$ such that $\varphi=\overline{\varphi} \circ \pi$

I'm applying the homomorphism theorem

why is the LaTeX not working? MathJax is dumb

leslie townes

your \overline is outta control

Lukas Heger

8:37 PM

oops

there's a projection homomorphism $\pi:G \to G/N, g\mapsto gN$ by saying that the homomorphism $\varphi:G \to \mathrm{Aut}(N)$ factors over $G/N$, I mean that there's a homomorphism $$\overline{\varphi}:G/N \to \mathrm{Aut}(N)$$ such that $\varphi=\overline{\varphi} \circ \pi$

Koro

you missed } after varphi.

robjohn

@TedShifrin Get real, man.

Lukas Heger

getting real means applying the functor $\Bbb R \otimes_{\Bbb Z}$, right?

and if you want to be rational, apply $\Bbb Q \otimes_{\Bbb Z}$

robjohn

@leslietownes $\overline{\overline{\overline{\text{what?}}}}$

Ted Shifrin

I am too complex for that, @robjohn. You done with your tests?

@Lukas I wanna be irrational!

robjohn

8:42 PM

@TedShifrin I hope so, but I haven't talked to the doctor that ordered the last test.

@TedShifrin That is why we have transcendental meditation.

Lukas Heger

@TedShifrin well, your eyerolls certainly are (conjecturally) irrational most of the time

Xander Henderson

@robjohn $$\dfrac{\overline{\Xi}}{\overline{\Xi}}$$

Lukas Heger

@Xander great notation

let $\iota \in C^1(\Bbb R)$, consider $\dot{\iota}$ vs $i$

Koro

$\iota, a+\iota b, a+ib$

Ted Shifrin

@LukasHeger So I do $\otimes_{\Bbb Z}\text{eyeroll}$?

@robjohn I hope the news will be good.

Lukas Heger

8:46 PM

yes, assuming your eyerolls form and abelian group

robjohn

@TedShifrin do you get an eyeroll when you go to a sushi bar?

Ted Shifrin

No, eyeball nigiri.

robjohn

rice in the eyes is a bit distracting

the humor is getting a bit vitreous here

Lukas Heger

In German barley grain is a kind of sickness

in the eye

Ted Shifrin

Better than vi(s)c(i)ous.

Lukas Heger

8:49 PM

so I'd rather have a grain of rice in the eye

than a barley grain

Ted Shifrin

The grains of humo(u)r.

Lukas Heger

aha, it's called stye in English

never heard that word before

or you can say hordeolum

leslie townes

i'll stick with stye

Ted Shifrin

Stye I know. Not to be confused with where pigs live,

Lukas Heger

we never learned the important words in English class!

like sheaf

or floccinaucinihilipilification

that seems like a word to convince Germans that English is a good language

Ted Shifrin

8:57 PM

Looks like sime Germans envied our coup attempt.

Sime = some

smth

Is it possible to write double transposition (a,b)(c,d) as product of 3-cycles of the form (1,2,k)?

leslie townes

and for some reason one of the ringleaders was dressed like a time lord from a lost 1970s series of doctor who

Ted Shifrin

@smth Have you tried?

smth

@TedShifrin I have tried but have not succeeded :(

Ted Shifrin

What about general $3$-cycles?

smth

9:12 PM

@TedShifrin (a,b)(c,d) = (a,b,c)(b,c,d)

Ted Shifrin

What about $(1xy)$?

smth

@TedShifrin I am not sure

I am stuck, hint?

oh maybe (a,b,c) = (1,a,b)(1,b,c)

user10478

9:33 PM

Hmm, is this an error in this 3Blue1Brown video (i.imgur.com/uoTlCN5.png)? These formula seem to be exactly the same as what Wikipedia calls the Positive and Negative Likelihood Ratios (in diagnostic testing), while the Wikipedia formula for Bayes Factor is slightly more involved. Likelihood Ratios and Bayes Factors aren't the same thing, are they?

Ted Shifrin

You need to ask at Stat, not here.

user10478

They're pretty dead IIRC

Ted Shifrin

The $0\in\Bbb N$ debate continues.

Thorgott

$\mathbb{N}\cup\{0\}$ is just awful

at least have the decency to be a $\mathbb{N}_0$ person or something

Ted Shifrin

I don't think I've every seen/used that.

I just think it's obnoxious to think that only a trivial minority of mathematicians don't consider $0$ a natural number.

3

Certainly on a site like this, such presumptuous behavior leads to misunderstanding.

Xander Henderson

9:41 PM

@TedShifrin I tend to use $\mathbb{N}_0$, though I don't find $\mathbb{N}\cup\{0\}$ to be "aweful". It is certainly not appropriate to edit the question in order to force a given convention.

@TedShifrin I'm trivial? :(

Ted Shifrin

Me too.

Xander Henderson

@TedShifrin You said it wrong. I think it is pronounced "hash tag me too".

Ted Shifrin

awful and aweful are quite different.

Oh, sorry for omitting the hashtag.

Xander Henderson

@TedShifrin Yes, I am not filled with awe by it.

I WROTE WHAT I MEANT, AND YOU WILL BELIEVE ME!

@TedShifrin For future reference, that is the kind of edit that I would encourage you to roll back. If a rollback war ensues, a moderator can step in. But sheesh... what an obnoxious thing to do.

Ted Shifrin

I thought Martin deserved a rebuke. But he's obviously ignoring it.

Xander Henderson

9:44 PM

Question: which is better style? $\mathbb{R}[x,y] / I$, or $\mathbb{R}[x,y]\,/\,I$?

leslie townes

"it's a standard, but some authors don't follow it for some reason" is very close to "it's a theorem, but there are counterxamples." chef's kiss.

Ted Shifrin

I think the latter looks horrible (not to be confused with aweful).

leslie townes

i agree with ted.

Ted Shifrin

Oh no, not again!

leslie townes

it's probably the coronaviruses impeding my better judgment.

Ted Shifrin

9:45 PM

Doubtless.

Xander Henderson

@TedShifrin Okay, just me, then. I think that the former looks all squished together and gross, but maybe that's why I avoid algebra and category theory. :D

But if leslie and Ted agree...

I must be wrong.

Ted Shifrin

Oh, you saved me from starting a war.

I think Munkres writes $\Bbb Z_+$ for $\Bbb N$ (probably to avoid these issues) and $\bar{\Bbb Z}_+$ to include $0$. I suppose I should go back and check.

He explicitly says to let $\bar{\Bbb R}_+$ denote the nonnegative reals. So I'm done.

leslie townes

that's not too bad, as long as it means what it ought to mean without the bar.

Ted Shifrin

Yes, he says that in the preceding sentence.

leslie townes

operator theory people love it when 'positive' means 'nonnegative,' and i get it in our little room in the playhouse where we're doing operator theory, but it's no reason to be a dick about normal things like sets of numbers.

Ted Shifrin

9:56 PM

Actually, saying an operator is positive definite when it is only semidefinite confuses the hell out of me.

And to me, a positive definite operator is assumed to be self-adjoint (same with finite-dimensional matrices), so the occasional author/OP who speaks of positive definite NON-symmetric matrices bugs me even more.

Xander Henderson

@TedShifrin I believe that I have adopted the notation $\mathbb{Z}_{>0}$ and $\mathbb{Z}_{\ge 0}$ in a few places. It is not great, as the symbols are small. But it also so very rarely matters. :D

@leslietownes I side with the operator theory people. :D

$0$ is positive. :P

It is also negative, but neither strictly positive, nor strictly negative.

Ted Shifrin

shrieks and runs insane from the room

Xander Henderson

Heh.

In other news, my family was out for Thanksgiving. Because we are unlikely to all be together again for a few years (my sister is moving to New Zealand in the spring), my mother decided to hold some Channukah celebrations a month early. It was nice. And then she gave me two baseball cards.

Ted Shifrin

Ooh, but you get to go visit New Zealand!

Lukas Heger

inb4 Xander teaches linear algebra and talks about non-negative definite operator

Xander Henderson

10:08 PM

@TedShifrin Maybe. I would need to get a new passport (mine expired, like, a decade ago, I think). And I don't have a lot of extra money for travel.

@LukasHeger I avoid "non-negative". Better to just say "positive".

leslie townes

ooh, that reminds me my passport is also expired.

Lukas Heger

is it true that a lot of Americans don't have a passport?

I heard that

leslie townes

most do not

Xander Henderson

@LukasHeger There is generally very little reason for a typical American to have a passport.

Lukas Heger

and do you have some kind of id card?

Xander Henderson

10:11 PM

@LukasHeger Most Americans have a driver's license, issued by their local department of motor vehicles, which serves as a general purpose ID.

leslie townes

i have a state id, drivers license. common but by no means required to have one.

Xander Henderson

Most DMVs will also issue an ID card for those that don't drive.

Ted Shifrin

I still hope my back/neck will allow me to use my passport.

Xander Henderson

Still, no one is required to have such an ID by law.

Lukas Heger

oh okay

I think you're required to have an ID card here in Germany

I don't use it very much

Xander Henderson

10:12 PM

Yeah, not in the US.

leslie townes

the non-driver id card isn't [edit: necessarily] free, either. so if you don't drive or travel you might not get one.

Lukas Heger

but if you go to some kind of state agency (how do say that?) you need your id card

or if the police wants to identify you

leslie townes

yeah, you just don't do that.

Xander Henderson

@leslietownes In Arizona, state issued ID is $12, unless you are over 65, in which case it is free. :D

leslie townes

for what it's worth, there's also no general requirement that you show id to police upon request.

Lukas Heger

10:15 PM

in Germany if you don't want to show your id card, the police is allowed to take you to the police station to identify you there

Xander Henderson

@LukasHeger If you are driving, you are generally required to produce your driver's license (some states have databases which are accessible in real time, and will look you up, so you may not be required to carry the physical card with you---there is a lot of variation).

Otherwise, if a cop asks you for ID, you are welcome to say "Sorry, man, I don't carry ID with me."

leslie townes

lukas: what are you going to get at some kind of state agency? health care?

[this is american humor]

Xander Henderson

(assuming that you don't actually carry it on you; if you say you don't have it, and they search you, and you have it, you could possibly get in trouble for lying to the police, which is a crime in certain places and contexts)

Lukas Heger

@leslietownes could be that your applying for wellfare benefits (oh wait, you don't have those :P)

in Germany you're not required to care your ID with you, but most people do, and technically as I said the police has a right to identify you somehow

Xander Henderson

11:43 PM

@LukasHeger According to the state department, there are about 150 million valid US passports in circulation. The current US population is around 330 million. Making the very rough assumption that the number of non-US citizens living in the US is on the same order of magnitude as US citizens living outside of the US, about half of all Americans have valid passports.

Lukas Heger

11:54 PM

I see

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$Mathematics$ Mathematics