Mathematics - 2017-01-11 (page 1 of 4)

jeff

00:11

Hi

Adeek

00:59

@TedShifrin I found both of my classes that is rings and modules and algebraic top awesome.

Hopefully this semester I will do really well. I will work super hard.

TheGreatDuck

@Adeek I edited this post and I think I fixed it. Do you personally see anything confusing with the post? I even put links in all the terms people didn't know (even though if they don't they should just google them). math.stackexchange.com/questions/2088483/…

I'm just asking your opinion. :-)

Adeek

I will check it out.

nah

I don't see anything confusing @TheGreatDuck

TheGreatDuck

I mean I suppose the axioms themselves may be confusing

but to be fair I think that's the nature of differential algebra seeing as how the last 3 of each set were copy/pasted from other answers regarding that subject.

(and I think that is the official set if I understand right)

Adeek

yeah

Akiva Weinberger

01:26

Apparently if $n$ points in the plane are not in all in a line, then they determine at least $n$ lines.

Semiclassical

@OneRaynyDay I get the same answer as you. I've run into similar things in the past, but I'm sorry to say that I don't remember the resolution.

The simplest solution would be for A to be symmetric, but I dunno if you're that lucky :/

TheGreatDuck

@AkivaWeinberger I'd need to think about that for a bit before I immediately trust that. Sure you don't mean that any 3 points are not in the same line?

nvmd

i see why

kinda trivial in fact

XD

OneRaynyDay

@Semiclassical Hmm yeah that is strange...

Okay, well on the test it gave me the choice $(A^T + A)x$ or $2Ax$ without any assumptions, and the teacher said it was $2Ax$.

$Triggered$

TheGreatDuck

01:42

don't be triggered

just ask the teacher

tell them that when working on your own you keep getting the first as a solution

chances are you just made a minor error and cant find it

happens to us all at times. triple read through your work line by line. if you still dont see it, ask the professor. there might be a context you didn't understand going in.

Akiva Weinberger

@TheGreatDuck How?

I mean any arrangement of the points, as long as they're not all in one big line

TheGreatDuck

@AkivaWeinberger if you arrange n-1 of the points in one line

and have one point off to the side

there are n-1 lines through the one point off to the side

obviously

Akiva Weinberger

Right, that arrangement gives us $n$ lines. But how to we know that no arrangement gives less?

TheGreatDuck

and then the other points are collinear

so that gives us n

@AkivaWeinberger read the rest of my earlier post. I said "nvmd. I realized it's a trivial thing to say is true."

Akiva Weinberger

I don't understand

It's not trivial. What is trivial is to find an arrangement with exactly $n$ lines, but that's not the theorem @TheGreatDuck

OneRaynyDay

01:48

@TheGreatDuck Yeah you're right, but I'm not supposed to discuss the quiz with my professor

it was a metric to see how qualified I am to take the class and he will soon discuss what the cutoff is

OneRaynyDay

02:03

Just walked through some series of steps and arrived at:

the gradient of the log of |A| is equal to A's inverse. I'm honestly pretty shocked. Anyone got any intuition on why this is true?

TheGreatDuck

no idea

i've never done whatever you're doing

@AkivaWeinberger it's trivial for me to assume that it cannot be lesser than n as the "least possible combination" I would assume is the combination where all of the points are collinear except one.

OneRaynyDay

here's a gentle walkthrough: docdro.id/58bKqQJ

TheGreatDuck

@OneRaynyDay not right now. I was merely saying I cannot help you other than to advise you to look carefully at the steps.

OneRaynyDay

Sure - I wasn't necessarily looking for help there, just wanted to share

Akiva Weinberger

Actually, maybe you can prove it by induction on $n$.

Ah, OK. So maybe it is trivial…

TheGreatDuck

02:11

to be fair, I've never done a proof by induction. I presume I will be learning that later this semester

also, I wasn't speaking rigorously. Just that one might be inclined to assume it rather than stumble upon it by accident while doing geometry work.

it seemed self-evident. :-)

Akiva Weinberger

So apparently there is a simple proof for the Euclidean plane… but the theorem actually also holds true for any projective plane

And it apparently takes the Sylvester–Gallai theorem as a lemma. Link to Wikipedia's proof.

TheGreatDuck

by projective you mean...?

i know of the spherical and hyperbolic planes

are those projective?

nvmd

i see it would be true

a parallel line need not exist or be unique to show that to be true.

Akiva Weinberger

@TheGreatDuck No, it's actually more general

Look up finite projective planes

TheGreatDuck

yeah but are spherical and hyperbolic examples of projective?

Akiva Weinberger

Not hyperbolic, and only spherical if you identify opposite points (so that any two lines intersect in one point) @TheGreatDuck

TheGreatDuck

02:28

oooh

like the projective plane

p^2

TheGreatDuck

03:00

hi

@AkivaWeinberger I know it is an old question, but is this question unclear to you either?

1

Q: Iterating through the integerial points of $f(x)$

The floor function $g(x)=\left\lfloor f(x) \right\rfloor$ jumps at points where $f(x)$ is an integer. What I want is a function that gives the $n$-th jump point from $g(0)$. So, for let's say $g(x)=\left\lfloor x \right\rfloor$, those points would be $1,2,3,4...$ in that order. For negative value...

TheGreatDuck

03:28

Hello @arctictern . How is it going?

Akiva Weinberger

04:03

@TheGreatDuck Kinda

Ramanujan

Any one here solving millennium prize problems?

I am not going to steal your work :D just asking

anonymous

04:28

@Ramanujan Better to ask on Math Overflow :)

SAWblade

05:26

@Ramanujan yes i've got some really promising work on the poincare conjecture

Ramanujan

05:43

@SAWblade it has been solved

SAWblade

that is the joke

Ramanujan

Lol

SAWblade

hm

stalling in my research

though that's nothing new

perhaps i should try to read a paper

SAWblade

06:19

i wanna learn techniques for solving recursive functions

anyone have any suggestions?

Socrates

@SAWblade you mean, stating the explicit function? with solving

SAWblade

yes

in general i just wanna know more about recursive functions anf the math that's been done on them

they fascinate me

Socrates

sometimes it is not even possible, firstly

SAWblade

but i don't know any good books

well then i wanna know how to tell if it's impossible

Socrates

I wanna know that too :/

SAWblade

06:21

heh

like idk if $a_{n} = a_{n-1}^{a_{n-2}}$ is solvable

it's probalby not

Socrates

does it have a limit?

also, starting values are important, i guess

SAWblade

true

i just made that up tbh

like as an example

i know nothing about it

Socrates

certainly it converges to 1, for both starting values=1

@SAWblade I would search stating the explicit function, maybe you got more luck than me

SAWblade

no i don't care about it

i just wanna know more about recursive functions in general xD

Socrates

the explicit is a huge chunk I suppose

pilko

06:31

what is the mathjax for argth ? I had to \operatorname it, so it rendered properly.

Simplex

06:55

So this is probably trivial, but can you have a 2 dimensional surface with more than one tangent plane at a given point? I think I've constructed one but I'm not sophisticated enough mathematically to prove it rigorously. Google didn't seem to turn anything up

Balarka Sen

@Simplex This depends on what you mean by a surface.

If the surface does not "self-intersect", whatever that means, and is smooth, every point has a unique tangent plane passing through it.

Simplex

What if I take a line and rotate it 90° about its midpoint, then "finish" the surface by tracing out a Gaussian curve with the radius? Hopefully that made sense.

Balarka Sen

I don't know what you mean by "tracing out a Gaussian curve with the radius"

pilko

Note that sometimes there is no tangent plane at all, for example on a saddle point the gradient is 0.

Balarka Sen

On the saddle point there is a tangent plane.

Simplex

07:05

Taking one of the endpoints of the line I'm rotating and instead of completing the disk, moving it up and down (the other side goes down and up) so that there is no discontinuity as you rotate the line

I'm pretty sure what I'm describing is a saddle surface

Alessandro Codenotti

If you do this in a way such that the resulting surface is smooth every point will have a unique tangent plane

Balarka Sen

@Simplex If so you can just check that it has a unique tangent space at every point.

So I have no idea what you have in mind.

Hi @Alessandro

Simplex

@AlessandroCodenotti that's what my intuition tells me, I just can't seem to find the flaw in my example.

Alessandro Codenotti

Hi @Balarka

Balarka Sen

Did you learn anything new, @Alessandro?

Socrates

07:13

appereantly "axiom of choice" is a nice title for music tracks

Alessandro Codenotti

Not really @Balarka I did a few exercises from G&P yesterday but that's it

There's a movie called Zorn's lemma, I don't think it can be equivalent to a song

pilko

@simplex if it is not smooth, a cone summit have multiple tangent planes.

Balarka Sen

@Alessandro Did you do the stacks of records one?

Alessandro Codenotti

07:28

Not yet

Ramanujan

I didn't noticed x \to 4 + and - :P sorry

pilko

anyway I would have answered 1 and 3.

Ramanujan

And what about 2nd option?

Limit x\to 4+ F(x) = Limit x\to 4- F(x) so isn't it continuous?

pilko

you have f(4) which is different from both 4+ and 4- limits.

noticed the blue point in negative y-plane ?

Ramanujan

Yeah,i noticed

But isn't it continuous?

pilko

07:35

It could be prolonged to a continuous function, but f(4) is defined explicitely to another value. So in this case it isn't continuous.

Kasmir Khaan

hi chat ,how to show that a change of variable is bijective?

eg. u =xy , v = y/x

am doing double integrals change of variable ( jacobian )

pilko

@kasmir by expressing x,y from u,v

Kasmir Khaan

thanks but what is the idea behind that?

does that show that for each (x,y) there exist only one (u,v) ?

pilko

This is because you change also the bounds of the integral. So the interval has to be transported to another interval (in n-dimensions from an hyperrectangle to another hyperrectangle). If the change is not bijective the value of the integral may change.

Secret

but how to reason that if y=0, then with any nonzero x will map to the same u and v, so the function is not necessary bijective because they are not injective?

so basically, y/x and xy have a kernel for the points (x,0)

pilko

07:51

Sorry, this is for the change to be continuous. The bijectivity is only required from primitives because you have to find back the original form in x,y.

Daminark

Hey everybody

pilko

@Secret there are also issue with negative numbers since xy=(-x)(-y) and so is the fraction.

Secret

(I have not return to analysis yet thus pardon my rusty experience) I am guessing double integrals of the form
$$\int_{-a}^a\int_{-b}^bf(x,y)dxdx$$ will potentially ran into trouble when we try to use the u=xy and v=y/x substitution (supposing that f(x,y) is a neither even odd function thus preventing simplification of the integral via symmetry arguments)

Mary Star

At a game machine the probability of winning is 25%. A player plays 20 rounds.
1. Which is the probability that he wins exactly five times?
2. Which is the probability that in ten rounds he wins five times?
3. Which is the probability that in ten rounds he wins ten times?
4. Which is the probability that in twenty tries he wins zero times?

Do we use at each case the binomial distribution?

OneRaynyDay

does anyone have some way of explaining how lagrange multipliers work in 1-d? i can visualize for 2d and above, but not sure about 1d.

@MaryStar 20 choose 5 / 2^20

10 choose 5 / 2^10

1/2^10

1/2^20

Mary Star

08:04

How did you find these? What formula did you use? @OneRaynyDay

OneRaynyDay

# of possibilities / total possibilities

it's more logic than anything considering each game is independent

Oh sorry - my bad, didn't see the 25%

20 choose 5 * 0.75^15 0.25^5

10 choose 5 * 0.75^5 0.25^5

0.25^10

Mary Star

and the last is 0.75^{20}, right?
So, we use the binomial distribution, right? @OneRaynyDay

OneRaynyDay

yes - chat wouldnt let me type further

and if you think about it

the answers i gave before are for 50%'s

i dont tend to memorize - it makes logical sense

anyways gnight!

Mary Star

Ok, thank you very much!! :-) @OneRaynyDay

Secret

demonstrations.wolfram.com/LagrangeMultipliersInOneDimension

Lagrange multiplier in 1D: turns out the problem reduces to one of finding parallel tangents

Mary Star

08:27

I am looking at the following:

The average tallest men live in Netherlands and Montenegro mit $1.83$m.
The average shortest men live in Indonesia mit $1.58$m.
The standard deviation of the height in Netherlands/Montenegro is $9.7$cm and in Indonesia it is $7.8$cm.

The height of a giant of Indonesia is exactly 2 standard deviations over the average height of an Indonesian. He goes to Netherlands. Which is the part of the Netherlands that are taller than that giant?

I thought to do the following:

Krijn

Oh hey, I'm taller than the average Dutch guy :D

@MaryStar It's a normal distribution and you know the average and the standard deviation, just formulas from there on

Balarka Sen

@MikeMiller I went to sleep yesterday since I got rekd at the 3rd mission. The game's neat, next up is training air units.

Krijn

@BalarkaSen What do you do when you're stuck at a problem for too long

Balarka Sen

@Krijn I procrastinate on it :)

Keep stuff at the back of your head until you get a light bulb

(light bulb may not be a full solution, just a workable idea/approach)

Krijn

Yeah, but I don't have infinite time :(

Let's hope I get a light bulb before monday

Mary Star

08:42

@Krijn We have the following:

right?

Balarka Sen

There's also the tactics of burning yourself up over it: Sit down with the problem for a whole day and nothing else, twiddling with it even though you don't feel like you have a good enough approach. It doesn't work for me very well though

Mary Star

We have 1.733 at the graph but we want to compute $P(x\leq 173.6)$. @Krijn

Krijn

@MaryStar You are mixing up meters and centimeters

Mary Star

Yes, it should be $P(x\leq 1.736)$, right? @Krijn

Krijn

Yes

Either of them, just don't mix 'em

Mary Star

08:47

Ok. But at the graph we have 1.733 and we want to compute $P(x\leq 1.736)$. How could we do that? @Krijn

Krijn

@MaryStar Doing it by hand is tedious, when I did this in a course on probability they gave some formulas to compute this, but I don't know them by heart

Socrates

is there an email-program that can handle mathjax?

Krijn

09:01

@BalarkaSen This makes me feel so stupid, especially when the answer is not really all that hard

Mary Star

@Krijn Do you know where I could find these formulas?

Krijn

@MaryStar Probably in every source that discusses the normal distribution, wikipedia, or any basic high school curriculum

Krijn

09:41

@BalarkaSen Solved it :)

Balarka Sen

@Krijn Great!

Ramanujan

If we have a matrix which gives a vector then transpose of that matrix gives same vector?

Alessandro Codenotti

10:00

What do you mean "a matrix which gives a vector"?

Ramanujan

This ^ @AlessandroCodenotti

Alessandro Codenotti

That's a dot product

You can write the vectors on the rows instead of the columns if you want since the determinant of $A$ and $A^T$ are the same

Ramanujan

And it will not effect even direction of vector?

@AlessandroCodenotti it's a cross product

Alessandro Codenotti

Yeah, a cross product, I always mix the names up in English

Nope, try to calculate a couple of them

Ramanujan

Hmm,both gives same

Alessandro Codenotti

10:12

That's because $\det A=$\det A^T$

Tobias Kildetoft

11:06

It says pretty much everything anyone needs to know about vixra that number theory has 1388 paper, general mathematics has 2079 and all other topics combined have 1160

(topology only has 56)

Alessandro Codenotti

11:19

?

I didn't see the message

Socrates

11:36

@TobiasKildetoft what does it say then?

Tobias Kildetoft

@Socrates that people who post there stick to the topics they think they understand

Akiva Weinberger

ViXra is the one that's like arXiV but anyone can post anything they want as long as it vaguely looks like math, right?

Socrates

@TobiasKildetoft so it can be assumed that many of those papers are redundand? Or that "nothing new" comes from there?

Akiva Weinberger

Or crankery

Tobias Kildetoft

@Socrates Pretty much every single paper there is complete garbage

@AkivaWeinberger I don't even think it has to vaguely look like math

The newest paper there is a counter example to Fermat's last theorem for $n=3$ calculated using excel. But the number got too long to include in full, as they were 18 digits

Socrates

11:42

XD

this must be a joke

s.harp

@TobiasKildetoft vixra sounds like a certain dirty word in german .P

Tobias Kildetoft

@s.harp Interesting. My German is clearly not good enough then, as I did not know that

Socrates

there is even a film called like that

GPhys

12:10

hi @AkivaWeinberger

I actually just answered an NSA question, haha

Socrates

12:38

._O

Animesh Ashish

check continuity of x^2[X]

[X] is greatest intg func

Akiva Weinberger

12:53

What are the right- and left-hand limits at x=1? @AnimeshAshish

GPhys

@AkivaWeinberger I think you said hi to me some days ago and I never got back

I'm doing well though, but busy with graduate school

Ramanujan

13:14

What's so special in this picture?

yh05

Hello all, i have a question, why is it sufficient to prove a certain series is convergent/divergent by using series test such as integral test and comparison test for large n only.

robjohn

@yh05 The sum of an initial, finite number of terms is simply that, finite.

yh05

even if its like 1000 terms?

even a million terms?

SteamyRoot

Yes.

yh05

cool

Anyone knows a good reference text on mathematical problem solving?

for undergrad

Arrow

14:31

Differential geometry: the vertical bundle of a smooth fiber bundle is defined fiberwise as the kernel of the differential of the bundle map. Why is this a subbundle of the tangent bundle of the total space in general? Should there be some additional condition on the fiber bundle?

Jessy Cat

Really quick. If I had a heat equation PDE plus a constant, what kind of change of variables would I have to do to get rid of the constant?

@Semiclassical you'd knôw doubt know thus

thus

Balarka Sen

@Arrow The kernel of the differentials, fiberwise, are subspaces of the tangent spaces of the total space.

Arrow

But why are they globally a subbundle?

Balarka Sen

Literally by definition of a subbundle? It's a vector bundle with fibers being subspaces of the tangent bundle of the total space.

Arrow

I think the dimension function of the fiberwise kernel must be locally constant - sorry if this is obvious

LeGrandDODOM

14:35

wow the number of deleted answers on this

Balarka Sen

@Arrow It is constant. Kernel of $df$, where $f : M \to N$ is a smooth submersion, always has dimension $\dim M - \dim N$.

Vector bundles are smooth submersions, is the point.

Arrow

Ah, the last sentence is what I completely forgot about. Thank you!

Perhaps I can bug you with an additional question. I'm trying to understand the second paragraph of the comment section here. In particular, why does the cocycle condition on fiber isomorphisms imply the bundle is trivial?

Jessy Cat

0

Q: Heat equation plus a constant.

Suppose I have the following PDE: $\begin{align} u_{t}=au_{xx}+1, \quad 0<x<\infty \\ u(0,t)=0, \, u(x,0)=1, \quad x>0\end{align}$ In order to solve it, do I need to do a change of variables in order to eliminate the constant term? Or how would I approach solving it? Thank you.

pde heat-equation change-of-variable

Semiclassical

so, an inhomogeneous heat equation

physically, I think that's equivalent to saying that the entire system is in contact with a heat source

As for how to solve it, I think most of the standard approaches should be applicable. For instance, you can do a Laplace transform in space and/or time just fine.

You could also look for a particular solution to this inhomogeneous equation and then handle the general solution separately.

(scratch the statement about Laplace transforming in space. you don't know enough about the boundary condition for that to be much help. Laplace transforming in time should still be applicable but I'm not sure it's ideal.)

Mithlesh Upadhyay

14:58

The number of strings of length 4 that are generated by the regular expression (0^+1^+|2^+3^+)^*, where | is an alternation character and {+, *} are quantification characters, is:

I found that 0001, 0011, 0111, 2223, 2233, 2333, 0123, 2301,... but, answer is given total 10. Can you guess other two, please?

Balarka Sen

@Arrow Hmm, I don't have a quick proof off the top of my head. The point should be that if the vector bundle is non-trivial then there should be a sequence of points $p_1, p_2, \cdots, p_n$ lying on charts $U_1, U_2, \cdots, U_n$ with $U_i\cap U_{i+1} \neq \emptyset$ (with indices mod $n$)

such that the transition maps $\varphi_{i, i+1} : f^{-1}(p_i) \to f^{-1}(p_{i+1})$ give isomorphisms which compose $\varphi_{1, 2} \circ \varphi_{2, 3} \circ \cdots \circ \varphi_{n, 1}$ to an isomorphism $f^{-1}(p_1) \to f^{-1}(p_1)$ which is not the identity.

(that is, it has "nontrivial monodromy")

Semiclassical

@JessyCat I have to tip my hat at Pragabhava's comment to your question; that makes things very simple.

Balarka Sen

$\phi_{b, b'}$'s do not satisfy that. But that doesn't mean anything. I dunno.

Arrow

@BalarkaSen do you have any geometric intuition for this? It's as if the cocycle condition should somehow "flatten" the base space..

Mithlesh Upadhyay

Well, I got that other 0101 and 2323.

Balarka Sen

15:12

@Arrow Essentially you should try to build a set of transition functions on the bundle using those isomorphisms which are all identity.

Arrow

@BalarkaSen I don't have a picture in my head :\ I asked this here though.

Akiva Weinberger

15:30

In the free group on $\{a,b\}$, are the elements of the form $a^nb^n$ independent?

Meaning, is the subgroup generated by $\{a^nb^n:n\in\Bbb Z\}$ isomorphic to the free group on $\Bbb Z$?

Mike Miller

@Balarka Glad you like it. When you finish with field training you should go back and A-rank those.

Semiclassical

@AkivaWeinberger Is that actually a subgroup? If $a,b$ don't commute then I don't see how that set would be closed under multiplication.

$(ab)(ab)\neq a^2 b^2$, after all.

Akiva Weinberger

The subgroup generated by it @Semiclassical

Semiclassical

derp.

SteamyRoot

@Akiva Wait, are you trying to construct $F_\infty$ as subgroup of $F_2$?

Akiva Weinberger

15:39

Yeah

SteamyRoot

It's been a while since I did that, but I think elements of the form $b^n a b^{-n}$ work

Akiva Weinberger

I believe it's the kernel of the $b\mapsto a^{-1}$ map — that is, the set of all elements with an equal number of $a$s and $b$s. So, if I'm right, that subgroup is not finitely generated.

($ba$ is in it because it's $(a^{-1}b^{-1})^{-1}$, and I had $n\in\Bbb Z$.)

Semiclassical

Hm, hadn't heard of this before: Nielsen–Schreier theorem

the index formula they cite is particularly neat: if $G$ is a free group on $n$ generators, and $H$ is a subgroup of finite index $e$, then $H$ is free of rank $1+e(n-1)$.

SteamyRoot

Hrmf. Seems like I handed in that assignment on paper rather than type it out in Latex.

Semiclassical

I don't know much about group rank, though, so it doesn't actually tell me much :/

Mahmoud

15:57

Hi.

Mary Star

Hello!!
We have a 90%-confidence interval. I want to check if the following statements are correct.

1. If double the sample, the possibility that the value that we are looking for is out of the confidence interval is smaller.

2. The bigger the derivation error, the smaller the confidence interval.

Since the confidence interval is $\left (\overline{x}- Z_{a/2}\cdot s_x, \overline{x}+ Z_{a/2}\cdot s_x\right )$, where $s_x$ is the derivation error, I think that the second statement is wrong and it should be that the bigger the derivation error, the bigger the confidence interval.

Mike Miller

@AkivaWeinberger Every normal subgroup of infinite index is infinitely generated.

The easiest one is the one generated by commutators

SteamyRoot

Oooh, I did find that old assignment

"For $n \geq 2$, let $F_n$ denote the free group on $n$ generators, where $n = 2,3,4, \dots, \infty$. For every $n = 2,3,4, \dots, \infty$, show that $F_n$ contains $F_m$ as a subgroup for every $m = 2,3,4,\dots,\infty$."

Balarka Sen

It's easy to see that via covering space theory.

Semiclassical

Which actually reminds me that there's an obvious algebraic-topology meaning to it: The fundamental group of the double-punctured plane is $F_2$, whereas the homology group is $\Bbb Z^2$.

Balarka Sen

16:11

@MikeMiller Right. I have been working on the naval unit trainings.

SteamyRoot

I did end up proving it by taking the group generated by $\{a^nba^{-n} |mid n \in \mathbb{N} \}$

Semiclassical

cough

SteamyRoot

Since for those generators it's pretty trivial to see you get $F_\infty$.

Mike Miller

most choices of generators are independent

@BalarkaSen Gotcha. You get special messages for beating some of the missions in especially short numbers of turns :)

Semiclassical

I know that particular example well, since it's how I got interested in homotopy vs. homology in the context of complex analysis.

Mike Miller

16:14

I'm thinking in particular of the one aerial map where you have your units in the top left, there's an ocean going from bottom left to top right, etc.

Balarka Sen

Yeah, I saw. I finished on in 4 or 5 turns IIRC

Semiclassical

So I'm a tad embaressed about forgetting ^2 :/

Advance Wars?

Mike Miller

oh yeah

I think there's another message if you do it in three turns.

Semiclassical

Nice.

Balarka Sen

@MikeMiller I think I know what you mean. It was the easiest one.

Semiclassical

16:15

the bonus finishes could be pretty crazy.

Mike Miller

one mission where Olaf has a ton more troops than you and you're supposed to capture the HQ can be won by rout ;)

Balarka Sen

Olaf had Mid Tanks but I could just transport a man for capturing the HQ

Semiclassical

I think there was one map you had to beat without losing any units.

Balarka Sen

too late, yep

Mike Miller

if you're anything like I was when I first played, I'm going to break your heart when I tell you that there aren't big tanks

Balarka Sen

16:18

lol

Akiva Weinberger

> [False theorem:] Every infinite group has an infinite proper subgroup.

Mike Miller

huh, seems like that should depend on choice

Akiva Weinberger

Nope.

Semiclassical

Counterexample?

SteamyRoot

Hmmm

Aren't the Prüfer groups the only counterexamples to that?

Akiva Weinberger

16:26

I already saw the answer, but I should have thought about it more first

Mike Miller

@SteamyRoot Abelian counterexamples

SteamyRoot

Oh, yeah, of course.

Akiva Weinberger

Yep, that was it.

Mike Miller

I'm also not 100% sure that's true

Semiclassical

Is there a 'nice' non-abelian counterexample?

Mike Miller

16:28

dunno!

Akiva Weinberger

@MikeMiller There are finitely many elements of a given order

SteamyRoot

I'm about 99% sure they're the only Abelian counterexamples. I must admit I've never seen a non-abelian counterexample; though.

Akiva Weinberger

so an infinite subgroup contains elements of arbitrarily high order

which means it's the entire group.

Mike Miller

No, I meant I'm not 100% sure Prufer are the only abelian counterexamples.

Semiclassical

That sounds like a nice question, then. "Is there an infinite non-abelian group whose proper subgroups are all finite?"

Akiva Weinberger

16:29

Ah, OK, sorry.

Mike Miller

I am plenty happy with the Prufer groups.

Akiva Weinberger

Prüfferfish

SteamyRoot

Oooh

Tarski Monster

Are those abelian?

Semiclassical

"Tarski monster group" sounds like something out of anime.

6

SteamyRoot

"Tarski groups give examples of infinite simple non-abelian p-groups," according to GroupProps

Semiclassical

16:33

If it did show up, I'd almost demand something like "Operation Monstrous Moonshine" to show up eventually

SteamyRoot

So I guess that works

@Semiclassical I'd fund it (if I could).

Semiclassical

lol

Akiva Weinberger

Are any ones known (other than the ridiculously large ones)?

SteamyRoot

According to MO, "The largest known prime for which existence of Tarski monster is not known is 997".

Mike Miller

sure, but the groups are going to be huge, no?

Akiva Weinberger

16:42

Right, but that could mean that for all smaller ones it is known that they do not exist @SteamyRoot

SteamyRoot

Well, the groups are infinite anyway.

Yes, for $p = 2,3$ it's known they don't exist. I think $p = 5$ is still open.

Akiva Weinberger

Wait, the largest for which the existence is not known is 997?

SteamyRoot

Yes.

Akiva Weinberger

So for 1003 (or whatever the next prime is), we do know that it either does or does not exist.

SteamyRoot

Yup

Mike Miller

16:45

are they countable?

Akiva Weinberger

*1009. 1003 is composite.

SteamyRoot

It was already known since $1979$ that for all primes $p > 10^{75}$ there exist Tarski Monsters

Akiva Weinberger

Right, 'cause it's 1020-17, and 1020=2*510...

SteamyRoot

@Mike Tarski Monsters are finitely generated

(but no examples are known that are finitely presented, though)

Mike Miller

oh cool

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$Mathematics$ Mathematics