Mathematics - 2016-07-27 (page 2 of 2)

8:02 PM

@BalarkaSen please tell me how this depends on whether pi and e are algebraically independent or not .... "wwhich is probably an open conjecture - but I don't care"........

Balarka Sen

Z[pi, e] would not be a polynomial ring of pi and e were algebraically dependent.

I was just writing down an example of a polynomial subring of R.

CRAZY GAY SHERIFF

8:38 PM

tht is true

but algebraic dependence is totally irrelevant for the problem

Balarka Sen

I didn't say it was ...

CRAZY GAY SHERIFF

the subring in the problem is usually not a polynomial subring

also the version of the nullstellensatz you mentioned is of no use, since the field is not algebraically closed

Balarka Sen

To be honest, I don't know what "finitely generated ring" means. A finitely generated Z-algebra?

CRAZY GAY SHERIFF

yes exactly

i told him

Balarka Sen

@CRAZYGAYSHERIFF Huy asked about polynomial rings over fields. Please look back to the message I referred to.

CRAZY GAY SHERIFF

8:40 PM

he needs the statement that every finite type Z algebra which is a field

Balarka Sen

@CRAZYGAYSHERIFF A f.g. Z-algebra is of the form Z[x_1, ..., x_n]

CRAZY GAY SHERIFF

is a finite ring

true, but he doesn't understand what he needs to know

Balarka Sen

That may be. But you should talk to him about that, not me.

CRAZY GAY SHERIFF

i did talk to him, but apparently he didn't understand it

isaac9A

Thought u had to go to work @BalarkaSen

:)

Balarka Sen

8:45 PM

Then simply either (1) explain it better (2) don't bother.

isaac9A

yeah @BalarkaSen explain it better to me :)

Balarka Sen

@isaac9A I generally say it politely to imply "I don't want to help anymore". I thought that should have been quite clear.

isaac9A

yeah you should have just said that

would have made you seem like less of a douche

Balarka Sen

You need to work on your social skills, @isaac, seriously.

isaac9A

said the person who lied to me

the polite thing to do is say i dont feel like helping you anymore sorry

CRAZY GAY SHERIFF

8:48 PM

(2) is better haha

some people it is impossible to help

he has this attitude that since he has no clue about commutative algebra, someone just comes up and explains it in "elementary" (i.e., somethin he can understand) terms. but the result is actually non trivial, hence beyond his grasp

Huy

@CRAZYGAYSHERIFF: I asked for an elementary proof since I don't have any commutative algebra background, and then an elementary proof would be the best thing to have. it is not difficult to answer "I don't know an elementary proof" or "I don't think there is any elementary proof".

that's not an attitude, it's a waste of time if I need basically a single result of commutative algebra to study a whole textbook.

Balarka Sen

@isaac9A People have the right to say whatever they want as long as they don't offend the people they are talking to. You can't force people into helping them. Besides, I did have work, and guiding you out through group theory was distracting me.

isaac9A

@BalarkaSen people don't like being lied to

it IS offensive

Huy

@CRAZYGAYSHERIFF: and most importantly, if you truly didn't bother, you wouldn't still be talking about it, so clearly it is bothering you ;)

Balarka Sen

@isaac9A I didn't. But I am abandoning this conversation.

Semiclassical

8:54 PM

hi chat

Owatch

Is it possible to simply to this final expression without using some extra information, because I can't see how: $r_{11} = a_{11}\frac{a_{11}}{\sqrt{a_{21}^{2} + a_{11}^{2}}} + a_{21}\frac{a_{21}}{\sqrt{a_{21}^{2} + a_{11}^{2}}} = \sqrt{a_{21}^{2} + a_{11}^{2}}$

Akiva Weinberger

A triangle with side lengths in $\sqrt{\Bbb Q}$ will have area in $\sqrt{\Bbb Q}$

Balarka Sen

@Semiclassical Go away, too much drama.

Akiva Weinberger

where $\sqrt{\Bbb Q}$ means the set of things with rational squares

Semiclassical

@Owatch multiply the second and third parts of that equation by $\sqrt{a_{21}^2+a_{11}^2}$

Huy

8:55 PM

@CRAZYGAYSHERIFF: I think it's quite telling when someone without good knowledge of commutative algebra (Balarka) ends up being more helpful than someone with a good background (you)

Balarka Sen

Sure, Heron's formula, no?

isaac9A

http://i.stack.imgur.com/EnTxx.png
Can someone please explain to me why the first presentation is for a group of order 4 and the second one is infinite?

Akiva Weinberger

Yeah, though it's not immediately obvious from the formula

I mean, it comes from the fact that the thing under the square root is invariant under replacing $a$ with $-a$

So I guess kinda obvious

Balarka Sen

World is full of such weird people who thinks it's their right to be insulting and offensive towards people. We're turning into the physics chat.

isaac9A

?

Semiclassical

8:59 PM

@isaac9A if $x_1^2=1$ and $y_1^2=1$ then $(x_1 y_1)^2=x_1 y_1 x_1 y_1=1\implies x_1 y_1 = y_1 x_1$ (multiply the last equation by $y_1 x_1$ on both sides)

so $x_1$ commutes with $y_1$ . thus if i write down any sequence of $x_1$ and $y_1$ terms, I can rearrange the product so that it's of the form $x_1^m y_1^n$. but then $x_1^2=y_1^2=1$ lets me cancel most of these terms.

and then there are only four possibilities. hence, a group of order 4.

Pedro Tamaroff

@BalarkaSen What's going on?

isaac9A

@Semiclassical ahhhhhh thanks!!!

u rock

Owatch

@Semiclassical I don't see what you mean. If I multiply $\frac{a_{11}^{2}}{\sqrt{a_{21}^{2} + a_{11}^{2}}} + \frac{a_{21}^{2}}{\sqrt{a_{21}^{2} + a_{11}^{2}}}$ by $\sqrt{a_{21}^{2} + a_{11}^{2}}$, I don't get the final expression.

Balarka Sen

@PedroTamaroff There's a chatlog full of stuff. People offending people for no reason, people thinking it's their right ask other people to do their homework.

I can imagine why Mike (and Ted?) left.

isaac9A

whos doing whos homework?

Owatch

9:04 PM

I just get $a_{11}^{2} + a_{21}^{2}$.

isaac9A

This aint homework, this is my area of interest. I'm not even in school right now lel

Pedro Tamaroff

@BalarkaSen Could you flag the comments you find offensive?

Balarka Sen

I don't really want to, sorry.

Pedro Tamaroff

Then why complain?

Do you want something to be done about it?

Balarka Sen

I didn't complain...

Semiclassical

9:07 PM

@Owatch not sure what else to say---this is quite simple. Take a break then look at it again.

Pedro Tamaroff

sigh. =)

Goodbye then.

Balarka Sen

Byes.

Akiva Weinberger

Today is a day of random questions for me, but:

What is the Cartesian product of a triangle with itself?

Some 4d polytope…

Balarka Sen

Bleh.

isaac9A

@Semiclassical so the order 4 comes from the fact that any product of any number of x1s and y1s boils down to either x1, y1, x1y1 or 1?

Semiclassical

9:10 PM

Right.

Akiva Weinberger

Oh, it has a Wikipedia page: en.m.wikipedia.org/wiki/3-3_duoprism

isaac9A

awesome thank you so much!

Balarka Sen

It has "sides" which are prisms.

3 such sides.

Akiva Weinberger

The Wiki's picture of the net makes it look like there are five triangular-prism cells

Balarka Sen

Huh, weird. I can see three of them coming from triangles cross edges of the second triangle.

Oh, that's really what it is.

The two other come from looking the two smaller edges of the first triangle cross the second triangle. They share a triangular face, given by a vertex of the first triangle cross the second triangle.

Fair enough. Not particularly hard to see.

Owatch

9:16 PM

I don't see it.. multiplying by the denominator just means I cancel that with the denominator, and get left with the numerator. The numerator isn't a square root. I guess I'm really missing something essential then...

Balarka Sen

I mean, look at the 1-skeleton of each of the triangle and cross them. You get wikipedia's picture.

Akiva Weinberger

By messing with Heron, we get that the hyper volume of the duoprism obtained by multiplying a triangle with side lengths $a,b,c$ with itself…

is equal to a sixteenth of $2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4$.

Which is a bunch of 4d hypercuboids.

And it would be amazing if there were a way to show that these have the same volume geometrically, by cutting up these thingies and pasting them in some weird way.

Hypervolume.

Balarka Sen

How many hypercuboids, exactly?

Owatch

Unless I multiply both top and bottom.

Akiva Weinberger

Six?

From that formula.

It has six terms.

But I guess a bunch of copies of everything.

Balarka Sen

9:20 PM

Right, right, whoops.

Of course.

Akiva Weinberger

Since we have a bunch of coefficients and a "sixteenth" in front of everything.

Balarka Sen

Uh-huh.

The minus terms mean that you have to subtract the hypervolume though.

Semiclassical

It really just amounts to x/sqrt(x)=sqrt(x)

Akiva Weinberger

So, sixteen duoprisms plus sixteen $a$-hypercubes, $b$-hypercubes, and $c$-hypercubes each…

should be the same hypervolume as 32 of each of those other thingies.

Owatch

@Semiclassical mfw I thought you wanted to multiply by $\frac{\sqrt{exp}}{1}$ and not the conjugate. :(

Balarka Sen

9:21 PM

Right. Yikes.

Owatch

I should have known better.

Can't just multiply over 1 and expect it to remain the same.

Balarka Sen

@AkivaWeinberger Good luck with that.

Semiclassical

Right.

Akiva Weinberger

And since the cut-and-paste theorem (whatever it's called) fails in dimensions three and higher, there's no reason to expect this to be possible anyway.

(The theorem that says that if two things have the same area, you can cut one into a bunch of thingies and rearrange them to get the other thingy.)

Balarka Sen

That Hilbert conjecture or something?

Akiva Weinberger

9:23 PM

Yeah.

Balarka Sen

Mr. Dehn has a counterexample, yes.

Good point raising that up.

isaac9A

@AkivaWeinberger "(The theorem that says that if two things have the same area, you can cut one into a bunch of thingies and rearrange them to get the other thingy.)" this doesn't work in 3 dimensions?

Balarka Sen

No.

isaac9A

neat

Akiva Weinberger

@isaac9A It's also interesting to show that it does work for two dimensions.

Polygons, anyway.

Semiclassical

9:27 PM

Looks to be Hilbert's third problem

Balarka Sen

Right, @SemiC.

Akiva Weinberger

(I think in three dimensions, the "angles" at the vertices muck it up somehow. I don't remember exactly. Proofs From the Book proves it.)

Balarka Sen

@AkivaWeinberger The idea is to triangulate, right?

Akiva Weinberger

I don't quite remember, but possibly. I can check later, they have a copy of the book here.

Semiclassical

Wikipedia gives it in terms of Dehn invariants

Balarka Sen

9:28 PM

For convex ones, anyway.

Semiclassical

I.e. it can only be done if the initial and final configurations have the same invariant

Balarka Sen

I mean, what do you expect, there's the Banach Tarski paradox.

Akiva Weinberger

@BalarkaSen Oh, I misunderstood that question

Yeah, for 2d, you can triangulate

Balarka Sen

Right.

Akiva Weinberger

I thought you were talking about the 3D version

Balarka Sen

9:31 PM

No, heh.

Akiva Weinberger

Turn polygon into triangles, turn triangles into right triangles, turn right triangles into rectangles, turn rectangles into one big square

Or something

and then two things of the same area can be turned into the same square and hence to each other

Balarka Sen

Ah-hah.

Akiva Weinberger

I think in 3D it stops working at the "turn tetrahedra to cuboids" stage

Balarka Sen

That makes sense. I was wondering, because triangulating shouldn't be a problem.

Akiva Weinberger

Rectangular cuboids

Boxes

Balarka Sen

9:33 PM

Yup.

Nice.

Akiva Weinberger

They should just be called boxes

isaac9A

@AkivaWeinberger really interesting

Akiva Weinberger

The "and hence to each other" takes a second to verify, also

isaac9A

I think in three dimensions, the "angles" at the vertices muck it up somehow. Hmmm this kinda makes sense, do you have the proof that it doesn't work in 3D? if you think of a 3d object as a bunch of 2d slices that you can integrate to get the 3d object, couldn't you triangulate the 2d slices even if there are an infinite amount?

Balarka Sen

No, anything like that couldn't possibly work.

Akiva Weinberger

9:39 PM

I didn't explicitly say it, but you can only use finitely many pieces

Balarka Sen

Integration doesn't slice higher dimensional object into lower dimensional things anyway, contrary to popular belief. One cuts up things into objects of the same dimension, but of smaller and smaller volumes.

Akiva Weinberger

Semi said Wikipedia has a proof

Hilbert's third problem

The third on Hilbert's list of mathematical problems, presented in 1900, was the first to be solved. The problem is related to the following question: given any two polyhedra of equal volume, is it always possible to cut the first into finitely many polyhedral pieces which can be reassembled to yield the second? Based on earlier writings by Gauss, Hilbert conjectured that this is not always possible. This was confirmed within the year by his student Max Dehn, who proved that the answer in general is "no" by producing a counterexample. The answer for the analogous question about polygons in ...

though it requires knowledge of tensor products and quotients of groups

or tensor products of vector spaces, which would be simpler if these weren't weird vector spaces like R over Q

Balarka Sen

Does the proof of that involve axiom of choice?

Akiva Weinberger

Of what? That the Dehn invariant does what it says it does?

Balarka Sen

Of Hilbert's 3rd problem, I mean.

Akiva Weinberger

9:48 PM

Doubt it

Nah, it says nothing of the sort

Balarka Sen

I see R/Qpi, thus my question.

IIRC you need AC for choosing a basis of that.

Akiva Weinberger

Never mentions bases.

isaac9A

@AkivaWeinberger thanks!!

Akiva Weinberger

@BalarkaSen In fact, it looks at it has an abelian group, not a vector space, unless you want the converse

Balarka Sen

Interesting.

Akiva Weinberger

9:51 PM

(Converse being "Equal Dehn invariant $\to$ scissors-congruent")

Balarka Sen

@AkivaWeinberger Ah, OK.

(BTW, did you see my potentially correct description of the RP^3 fibration thing?)

Akiva Weinberger

Yeah

Balarka Sen

OK.

Balarka Sen

10:08 PM

It's late, I gotta head to bed.

But would I?

Akiva Weinberger

Nah, sunrise is only ninety minutes away, just go the whole thing

/s

Balarka Sen

I have school tomorrow though.

Akiva Weinberger

You should have thought of that before you stayed up to 3:45am.

Seriously, though, go to bed.

Bye

Balarka Sen

True. Have fun.

Meanwhile I do hope the chat becomes more civil and people who left would come back.

Owatch

This place always seems civil to me.

Balarka Sen

10:18 PM

It is more often the opposite, @Owatch.

With lots and lots of dramas.

Owatch

I dunno. I once made the mistake of going into the C++ room. There was no drama, they were just flat out rude.

Balarka Sen

Heh. We're (were?) better than other chats, yes.

Owatch

As in, literally cursing me out for asking in the chat. Here, everyone is much nicer.

It's We're btw.

Balarka Sen

No, I mean, "we are (were?) better"

Owatch

Ah.

Well, everything seems fine to me. I even see Ted around sometimes. He isn't gone forever is he?

Balarka Sen

10:24 PM

Nope, but he's not as frequent anymore, evidently he's annoyed by several people in here. Mike's gone too.

Owatch

Oh... I did not speak to Mike. But I did speak to Ted sometimes. It's very nice of him to help me out given he seems quite important.

Balarka Sen

He's helpful, yes.

Owatch

Hopefully he'll begin coming back more in the future. I probably shouldn't ask about the drama since I don't want to stir anything though.

Balarka Sen

Right. It's all in the chatlog anyway.

But best not messed with.

Owatch

I'm going to sleep now I suppose. Goodbye!

Balarka Sen

10:27 PM

Me too. Night.

isaac9A

The cardinality of the powerset of Integers is the same as the cardinality of the real numbers correct?

even though they are both uncountably infinite?

Balarka Sen

Yes. You can construct an explicit bijection.

isaac9A

cool, thanks! Made sense in my head but a lot of stuff that makes sense in my head turns out to be flat out wrong lol

CRAZY GAY SHERIFF

@Huy you still have no clue what the problem is about, and he misled you as well. he does not know either which version of the nullstellensatz the answerer is referring to

which i do, and i posted it. it is just you that is unable to understand it.

isaac9A

10:43 PM

be nice, let it go man

Balarka Sen

I agree with that, though. But I am not sure if I misled, because I was simultaneously asked about $k[x_1, \cdots, x_n]$ and subrings of $A$.

I stick to the "be nice" policy in any case, which you rarely follow.

isaac9A

I am the very model of a modern major general, i've information vegetable animal and mineral

CRAZY GAY SHERIFF

yeah he asked this cause he has no clue what the problem is about. he just looked up the definition "finitely generated ring", and he saw that they are defined via polynomial rings. as i said, just from the definitions he won't understand it anyway

i am just annoyed that i took some time to explain it to him, while he clearly proves that he is not worth it

i should have looked at his profile first

but i should have known it, i mean he always needs explanations from 3 different people before he understands anything

who is the third, though?

Balarka Sen

By "that", I meant having no idea about the version Qiaochu refers to. But I am outta this business.

Akiva Weinberger

@isaac9A I am the very model of a modern major general, the venerated Virginian veteran whose men are all…

(From Hamilton)

isaac9A

10:49 PM

@AkivaWeinberger what is that? I was thinking of the gilbert and sullivan song from Pirates of Penzance

I understand equations, both the simple and quadratical,
About binomial theorem I'm teeming with a lot o' news, (bothered for a rhyme)
With many cheerful facts about the square of the hypotenuse.
I'm very good at integral and differential calculus;
I know the scientific names of beings animalculous:
In short, in matters vegetable, animal, and mineral,
I am the very model of a modern Major-General.

CRAZY GAY SHERIFF

@Huy here is it:

-Jac(A)=Nilrad(A) since A is a Jacobson ring (finite type algebra over a field).

-Nilrad(A)=0 cause A is an integral domain

-A/m is a finite type Z-algebra, hence finite as a ring (by the question i linked to)

Balarka Sen

@CRAZYGAYSHERIFF It's a finite type algebra over Z tho.

Not a field.

CRAZY GAY SHERIFF

(A/m is a field)

true

i meant Z

Z is a Jacobson ring, since Z is a pid

Balarka Sen

Fair.

OK, now really it is too late.

CRAZY GAY SHERIFF

there is not more to say about it i think

what you mean by that balarka?

"it is too late"

isaac9A

10:52 PM

hes is getting sleepy

thats what he means by its getting too late

CRAZY GAY SHERIFF

i see

isaac9A

im in california so its only 4pm

U guys are in europe?

CRAZY GAY SHERIFF

yeah

i think balarka is from india

so it is like 6 am for him haha

isaac9A

ahdeeeeeemn

haha suns coming out

I'm jealous, India had such good food

everything was so fresh

Akiva Weinberger

@isaac9A Yeah, Hamilton references it in its first act

(Hamilton is another musical)

isaac9A

10:55 PM

its pretty new right @AkivaWeinberger I've heard of it but dont know much about it

CRAZY GAY SHERIFF

so you don't think the food at mac donalds/burger king is fresh? :D

isaac9A

hahaha :)

@CRAZYGAYSHERIFF

Akiva Weinberger

The writer made the reference because he thought he could do a better rhyme for "general"

(The sentence doesn't end at that line)

isaac9A

better than gilbert and sullivan? Impossible XD

Akiva Weinberger

(The character speaking is George Washington there.)

Only for that one rhyme :P

isaac9A

10:56 PM

Then again as a huge Gilbert and Sullivan nerd I am clearly biased

Now I have to see Hamilton @AkivaWeinberger

Anything that has a G/S reference is good enough for me :P

Akiva Weinberger

I am the model of a modern major general / the venerated Virginian veteran whose men are all / lining up to put me on a pedestal / writing letters to relatives embelleshing my elegance and eloquence, but:

(IIRC)

the elephant is in the room. / The truth is in your face when you hear the British canons go BOOM!

@isaac9A If you can afford tickets :P But the soundtrack is online for free, and I highly recommend it. It's amazing.

isaac9A

@AkivaWeinberger awesome, found the soundtrack online! I'm at a business meeting right now but I will definitely listen later :)

Akiva Weinberger

Also, there's commentary to the lyrics at genius.com or something if you want

isaac9A

Are there a countably infinite number of Bijections from the set of integers to itself? I would think it is uncountably infinite and even has the same cardinality as the set of real numbers.

?

arctic tern

11:19 PM

@isaac9A it's uncountable, and yes same size as R. try proving it. (hint: consider the 2-cycles (12), (34), (56), (78), ...)

Leaky Nun

Who the hell flagged this?

Rainbolt

Someone is flagging old chat messages that are not starred on the transcript

Seems like a waste of time

Leaky Nun

exactly

Krijn

So how does flagging work?

Leaky Nun

Is there a way to conjugate with just exp, ln, and subtraction (and constants)?

Krijn

11:23 PM

Because I flagged that, because that level is just too low for chat

isaac9A

@arctictern thanks I am going to try proving it but proofs don't come easily to me. At least not yet.

I am about to start college and don't know if I should be a math or physics major

arctic tern

@LeakyNun what do you mean by conjugate?

Leaky Nun

@arctictern eh, complex conjugate

arctic tern

well, considering composing those things yields something holomorphic on its domain, whereas complex conjugation is antiholomorphic, I would think not

Leaky Nun

@arctictern thanks

Samuel Yusim

11:31 PM

depthofthesun.wordpress.com/2016/07/27/… I wrote a new blog post, so I figured I'd plug it here

isaac9A

@SamuelYusim I think a lot of people have similar problems with motivation. Math at this level is hard and taxing on the brain. Sometimes one must read over and over again for days until one gets something. Do you have any active hobbies?

Samuel Yusim

11:59 PM

@isaac9A I watch a lot of tv shows, which mostly includes cartoons. I'm also trying to spend some time learning to play piano lately. I just started but it's fun

Transcript for

$Mathematics$ Mathematics