Mathematics - 2017-05-21 (page 2 of 7)

Excalibur42

03:00

Well if anyone does, would love a brief explanation as to what the hell the tower of extensions $F$ $\subset$ $L=F[\sqrt{c}]$ $\subset$ $E=L[\sqrt{a+b\sqrt{c}}]$ $\subset$ $M=E[\sqrt{a-b\sqrt{c}}]$ looks like

Like, everything is degree two to the previous one, but $E/F$ is degree $4$ and so is $M/F$...

(and assuming $E/F$ is not Galois, but $M/E$ is)

TheGreatDuck

@Dair i dont know. It was just a random though. :p

Dair

what is that? lol.

TheGreatDuck

@Excalibur42 is that what you "mean by a tower of quadratic field extensions"?

Excalibur42

haha I wish @TheGreatDuck

Dair

@TheGreatDuck you have the weirdest taste in computer backgrounds...

TheGreatDuck

03:13

?

that's not my computer's background

Akiva Weinberger

@Daminark Hi! Did you look at problem 3? ("Spreading infection")

TheGreatDuck

I googled "field"

and then appended a "tower" of "quadrilaterals".

Dair

looks like some weird variant of the Windows XP background...

Akiva Weinberger

At the end, it essentially says "This problem can be solved be a single nine-letter word"

TheGreatDuck

meh

Dair

03:15

i tried.

TheGreatDuck

@AkivaWeinberger cleansing?

Akiva Weinberger

I've solved it, there indeed is a nine-letter word that works

@TheGreatDuck Do you know the puzzle set I'm talking about? It's this

Excalibur42

nah this is what a quadratic field looks like

$user image$

@TheGreatDuck

TheGreatDuck

ooo aaaah

Akiva Weinberger

I've done 1,2,3, 5, and 6 (not 4 yet)

TheGreatDuck

03:16

is that a field produced by appending a solution to a second order polynomial to the set of rationals?

Dair

is everyone here at U Chicago or something?

Akiva Weinberger

Not I

Excalibur42

Not always, that one in particular is the Eisenstein integers en.wikipedia.org/wiki/Eisenstein_integer which is the third cyclotomic field, and actually is adjoining a root of a second order polynomial to Q

Akiva Weinberger

I got the sheet from Daminark, who is

Dair

ok. dodsy also mentioned the worksheet.

Akiva Weinberger

03:17

@Excalibur42 What is this picture

Excalibur42

(third root of unity actually, but has minimal polynomial x^2+x+1)

reddit.com/r/math/comments/2xs4t7/… comments in that thread

visualizing arithmetic of quadratic fields

This is $Q[i]$ math.colorado.edu/~kstange/arrangements/gaussianmod2.png

TheGreatDuck

@AkivaWeinberger spreading

in other words, 1 is sufficient

the squares keep infinitely spreading

Akiva Weinberger

@TheGreatDuck Read the problem again.

TheGreatDuck

oooh

Akiva Weinberger

A square needs to be next to at least two infected squares to get infected.

TheGreatDuck

03:20

the diagonal is sufficient

anything less is insufficient

Akiva Weinberger

Right.

And that's eight.

TheGreatDuck

mmmhm

Akiva Weinberger

"Prove that 7 are not enough."

TheGreatDuck

at most it can form a 7 by 7 square in the end

not sure where a 9 letter word comes into play

Akiva Weinberger

Or perhaps two disconnected squares

Daminark

03:21

@AkivaWeinberger I haven't yet

Akiva Weinberger

or maybe seven squares that aren't near each other

TheGreatDuck

at most

as in, that's the most i see it claiming

i dont get it

Akiva Weinberger

Also, the diagonal isn't the only way to do it with eight

@TheGreatDuck You haven't given a proof yet

TheGreatDuck

@AkivaWeinberger i know but it is the heart of this

@AkivaWeinberger im not going to go write a whole proof by contradiction and use combinatorics and automata

Dodsy

dodsy is a loser

who is not currently in school

TheGreatDuck

03:23

i could just brute force it

Akiva Weinberger

Yeah, but brute force is ugly.

Dair

@TheGreatDuck You could but then you would have to write a program...

Dodsy

I mentioed the worksheet because somebody brought up one of the problems

Akiva Weinberger

That might have been me

Dair

Why write a program when you can write 9 letters?

TheGreatDuck

03:24

@Dair cause I don't know a relevant 9 letter word

Dair

@TheGreatDuck Dankmemes

Akiva Weinberger

In any case, think about it, TheGreatDuck, it might take a while to come up with the answer

TheGreatDuck

and the program would be fun to right

Dair

you could write a bot that posts every nine-letter word and Akiva could then tell you which word it is.

Akiva Weinberger

The next two problems I need to do, 4 and 7, aren't brute-force-able, unfortunately

Dodsy

03:25

I've heard of this problem before

about infecting squares

Dair

Induction

TheGreatDuck

@AkivaWeinberger meh. Right now I am writing a minesweeper solving program.

Dodsy

but I forget the ...

you fucker.

TheGreatDuck

wat

i dont get it

Akiva Weinberger

YOUFUCKER - 9 letters

TheGreatDuck

03:26

@Dair what does a proof by induction have to do with this?

Dodsy

Akiva wins

Dair

@TheGreatDuck Probably be like if it is one less than the diagonal you can't do it...

clearly doesn't hold if it is a 2 x 2...

TheGreatDuck

maybe...

Dair

assume it doesn't hold for k...

Dodsy

what 8 squares need to be selected to ensure that all of the squares become infected?

Akiva Weinberger

03:27

@TheGreatDuck (Induction has nine letters)

TheGreatDuck

oh i got it!

Dair

then something something...

Akiva Weinberger

@TheGreatDuck ?

That would be infinitely faster than me…

TheGreatDuck

Conjecture: if a matrix is n*n then the minimum infection needed to cover it is n squares

then use proof by induction

Akiva Weinberger

How?

TheGreatDuck

03:28

the base case is trivial: 1x1

the induction step is beyond me

but I bet it cannot be real hard

Dair

@Akiva How? With induction of course

Akiva Weinberger

So you don't got it.

"Eureka!"

TheGreatDuck

@AkivaWeinberger I never would be able to. I've never done automata proofs before. I just know of them to recognize them when I see them.

Dair

Proof: It is obvious to the most casual of observer.

Akiva Weinberger

(^Makes nine letters if you count punctuation)

Dair

03:30

recursion is also 9 letters. spooky.

TheGreatDuck

that might be more useful actually

automata are more relvant to computer science

Dair

@TheGreatDuck I highly doubt this requires much knowledge of automata.

TheGreatDuck

true, but they are occurring here.

Dair

so how do you plan on using recursion?

TheGreatDuck

i wasnt saying i was going to

just that it would be relevant to the whole parent -> child relationship automata have

:p

induction seems to be the solution

i just don't know the inductive step

Dair

03:35

i'm kind of surprised they post a problem set for an REU...

i would have thought it was more about projects than problem sets...

TheGreatDuck

a what?

Dair

REU: Research Experience for Undergraduates

The problemset is from his REU page.

TheGreatDuck

i thought you might've meant that

is this a current one going on?

Eric Stucky

Dair: it's pretty normal to have the first 1-2 weeks be problem sets

in a 10-week REU

TheGreatDuck

*9 week

Dair

03:37

no 2012.

@EricStucky Oh ok, didn't know that. Thanks.

TheGreatDuck

if this is a math REU than wow these look intense

Akiva Weinberger

I've been stuck on #7 (a variant of #6, which itself is a variant of #5. Which is a classic.) for a while now

Dair

Says to look for the AH-HA proof except for no 7 rip.

TheGreatDuck

the first 2 are based on modulo arithmetic

i.e. when you take away a piece the remainder must be divisible by 3. otherwise it is impossible to do

sorry

the first one is different

Akiva Weinberger

Right, but that only rules out $n\equiv0\pmod3$ (you're talking about the triomino one, right?).

Dair

03:42

people.cs.uchicago.edu/~laci/REU12/appr-prob193.pdf do they expect all the problems to be done in the first 2 weeks?

Akiva Weinberger

They want to show that $n\equiv-1\pmod3$ is impossible, and $n^2-1$ is a multiple of $3$ for those

(Ex: $8\times8-1=63$)

Prince M

Comin' in guns hot with a question: In Atiyah - Macdonald they say "An integral domain is said to be integrally closed (without qualification) if....."

Does anyone have an idea what is meant by "without qualification"?

Akiva Weinberger

Maybe "without more adjectives on it"

TheGreatDuck

@AkivaWeinberger I think you are doing it wrong. wait a sec

Prince M

I got the vibe that they mean it is not "with respect to some other ring"

TheGreatDuck

03:44

n = 3m - 1

Eric Stucky

Dair: If the ~100 students at the UChicago REU couldn't collectively solve those questions in 10 days, it would probably be the worst year they've ever had by a longshot.

TheGreatDuck

therefore $n^2 = 9m^2 - 6m + 1$

^^indivisible by 3

Prince M

Even though it technically is, the other ring being the field of fractions, but we dont talk about A being integrally closed with respect to some ring B

Dair

@EricStucky Oh so they can collaborate...

Eric Stucky

Well, obviously :P

oh, are you in high school?

Dair

03:45

no lol.

Akiva Weinberger

@TheGreatDuck You forgot to subtract 1

Eric Stucky

okay

Akiva Weinberger

The corner

Dair

but i've also never been assigned 193 to do in 2 weeks either lmao.

Eric Stucky

I tend to forget that HS students aren't encouraged to collaborate; which is why I asked.

Akiva Weinberger

03:46

You want to show that an $8\times8$ chessboard (for example) with one corner missing can't be covered in triominoes.

Dair

my school like has weird policies on it.

Akiva Weinberger

That's $63$ squares, which is a multiple of $3$.

TheGreatDuck

@AkivaWeinberger the triominoe puzzle is equivalent to proving the conjecture: "if $a$ is not divisible by 3, then $a^2$ is not divisible by 3"

Dair

i guess it is more of our CS department that has an issue with it.

Akiva Weinberger

@TheGreatDuck YOU FORGOT TO SUBTRACT THE CORNER.

TheGreatDuck

03:47

@AkivaWeinberger oh sorry

Akiva Weinberger

It really isn't that simple.

TheGreatDuck

:p

Dair

or is it?

Akiva Weinberger

Divisibility isn't enough.

TheGreatDuck

dun dun DUUUUN

seriously though

have you solved the infection one?

i really want to know how that one is solved

Dair

03:48

induction

Akiva Weinberger

$n^2-1$ is a multiple of $3$ when $n\equiv-1\pmod3$

@TheGreatDuck I have.

TheGreatDuck

i meant the inductive step

Akiva Weinberger

Can I spoil something tiny

TheGreatDuck

sure

Akiva Weinberger

about the infection one

(Hey, "infection" has nine letters!)

TheGreatDuck

03:49

O.O

Akiva Weinberger

The word isn't "induction."

Dair

darn

TheGreatDuck

but would a proof by induction be successful?

Akiva Weinberger

I dunno. That's not how I did it.

TheGreatDuck

alrighty

how did you?

Akiva Weinberger

03:50

I really don't want to spoil it.

Think it over.

TheGreatDuck

(feel free to use an irrelevant chat to randomly ping me)

ok

was it a mathematical proof as in like number theory and stuff or was it more puzzle-like proof?

Akiva Weinberger

It's not full of computation, if that's what you mean

or algebra

It's puzzle-like, yeah.

TheGreatDuck

ok

would I be wrong in seeking out a maximal number of squares that can be claimed or is that irrelevant?

i.e. like how I said 7 can only take a 7*7 grid?

Semiclassical

Thinking about some smaller cases seems like a good idea just to get a feel for the problem.

TheGreatDuck

@Semiclassical not at all what I meant

Semiclassical

03:54

Didn't say it was.

Akiva Weinberger

^Good tip in general

TheGreatDuck

do you know what we are doing?

Semiclassical

Yes.

TheGreatDuck

ok

Semiclassical

Triminoes.

TheGreatDuck

03:54

no...

Semiclassical

or however it's spelt.

Akiva Weinberger

Not right now @Semiclassical

Problem 3 ("spreading infection")

Dair

we're spreading diseases now

TheGreatDuck

♥
(Spreading Infection)
Some of the 64 cells of a chessboard are initially infected. Subsequently the infection spreads according to the following rule: if two neighbors of a cell are infected then the cell gets infected. (Neighbors share and edge, so each cell has at most four neighbors.) No cell is ever cured. What is the minimum number of cells that need to be initially infected to guarantee that the infection spreads all over the chessboard? It is easy to see that 8 are sufficient in many ways. Prove that 7 are not enough. (This is an AH-HA problem. The main idea of a …

Semiclassical

oh

Woops.

Dair

03:55

can 2 guys infect a 3 x 3?

TheGreatDuck

@Dair no

Dair

ok

just trying smaller cases...

Semiclassical

Regardless, considering a smaller case seems like a smart idea.

Just to get a feel for how the problem works.

TheGreatDuck

well 1 as a base case for an inductive proof is trivial

obviously

0 cannot claim 1

Akiva Weinberger

For the record, I know how to do this and they don't @Semiclassical

Semiclassical

03:56

Understood.

TheGreatDuck

im trying to think of how if n-1 cannot claim n implies that n cannot claim n+1

that would demolish the problem

Dair

so you claim that you can only infect the hole n x n square if you have n guys...

Akiva Weinberger

It would "cure" the problem, so to speak @TheGreatDuck

Never mind, that makes no sense

Semiclassical

lol

TheGreatDuck

@Dair no I claim that n-1 squares cannot claim an n*n square grid.

Dair

03:58

@TheGreatDuck I'm working on a slighly different train of thought.

TheGreatDuck

i got it I think!

Dair

@TheGreatDuck I'm thinking as to what extent the n x n is claimed by n-1 squares...

TheGreatDuck

it might be a series problem

Akiva Weinberger

If any of you have ever played Minecraft, this is similar to how the water mechanics work.

TheGreatDuck

@AkivaWeinberger lol akiva let me show you something

Dair

03:59

i never understood their fluid mechanics lol

TheGreatDuck

@AkivaWeinberger indiedb.com/games/block-builder

Akiva Weinberger

It's really complicated but really interesting @Dair

TheGreatDuck

@Dair it's the same as the infecting tiles but with only diagonal spread

Akiva Weinberger

I found out a way to get a 1x1 column of air in the middle of the ocean (stretching from the ocean floor to the surface) @Dair

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$Mathematics$ Mathematics