Mathematics - 2012-03-21 (page 2 of 4)

Ilya

09:01

finally today I can tell students about stochastic signals

Asaf Karagila

But... why??

Ilya

why what?

these are not cool stochastic processes and methods are rather lame - but still at least some probability will appear

@Asaf: ?

Asaf Karagila

I had to attend some real life business.

Finally someone in this house is doing the dishes and I had to run around and gather all the dirty cups... :-P

robjohn

@Ilya are you complaining about my answer to the die rolling problem?

Ilya

@robjohn I didn't see that, no :) good night

robjohn

09:13

I had just worked out the matrix method for it. :-) I will add it to my answer.

@Ilya are you leaving?

Ilya

@AsafKaragila a simple inference is that you're not the one doing dishes, laze bastard :)

I mean good night as
$$
\text{Good morning}|_{\text{morning} = \text{night}}
$$

Asaf Karagila

@Ilya I'm cooking the meals!

robjohn

@Ilya like good morning, but for night?

Ilya

da

robjohn

@Ilya cool. I thought I was on another roll where everyone leaves when I show up :-)

Asaf Karagila

09:15

Hi @ymar

robjohn

I finally got my web server working, but then my local IP server crashed.

Ilya

@Rob: talking about probability - when I was doing the presentation on Monday in Paris, one professor told me that the word supermartingale is so cool that he wants now to include it at least in one of his papers. I promised him that if write the paper together, this word will appear there.

ymar

Hello!

Asaf Karagila

@robjohn I am going to leave soon.

robjohn

@AsafKaragila No! where are you going?

Ilya

09:16

@robjohn he is cooking

@ymar: hi

Asaf Karagila

@ymar On Monday I met with Haim and we discussed the answer. He was a bit annoyed that he couldn't have shown that there are $2^\kappa$ many non-isomorphic groups directly.

robjohn

@Ilya Ah, that is excusable :-)

Asaf Karagila

@Ilya Not today. Today we're going to a restaurant.

Ilya

@AsafKaragila so you lied

Asaf Karagila

@Ilya I made dinner last night.

ymar

09:17

@AsafKaragila That's a pity... I don't think I will understand his answer soon. But at least I know what the answer is! :)

Ilya

@AsafKaragila Matt will agree that then you should use "I cook the meals" rather than "I am cooking the meals"

robjohn

@ymar which answer is this?

ymar

5

A: How many non-isomorphic abelian groups of order $\kappa$ are there for $\kappa$ infinite?

It's known by Fisher, Eklof and Shelah (see theorem 2.1 in [1]) that there are abelian groups which are stable but not superstable. Another well known result of Shelah (see [2]) is that, roughly speaking (again, check the paper below for details), if $T$ is not superstable, then $T$ has $2^{\lamb...

Asaf Karagila

@Ilya Matt isn't here.

Ilya

@AsafKaragila btw, do you know, why? I hope nothing serious has happened

Asaf Karagila

09:22

@Ilya No idea. I went to bed around 4am last night and woke up only an hour or so ago.

Probably needed time without distractions for studying stuff which is not set theory.

Now I am going to trim my beard and take a shower. Then I am going to the university to work on my Whitehead thingie and then I am coming back home.

Ilya

@AsafKaragila Report accepted.

t.b.

@AsafKaragila still here?

Asaf Karagila

Yeah.

t.b.

What exactly is that theorem of Brunner you want to apply?

Ilya

@tb: good morning! haven't seen you for a while

t.b.

09:30

Morning Ilya. I saw you liked Paris :)

Ilya

@tb yet another time - but I think I only like to visit it, not to live there

t.b.

It's hard not to.

Ilya

@tb I wasn't pretending I am original in that :) To be honest, I was in Paris only Sunday's and Monday's night (after 9pm) and in the morning - the whole Monday I spent in INRIA.

Asaf Karagila

@tb In Solovay's model every Banach space is dreamy.

t.b.

So every Banach space is isomorphic to $l^1(D)$ with $P(D)$ Dedekind finite? I don't follow.

Ilya

09:34

But then with my supervisor we had a 2-hours walk on Monday's night, visited Seine and walk around the city center. That was so cool that even in a weekday, after 11pm there were still a lot of people there. So unusual for the Netherlands

Infinite compact discrete spaces?

t.b.

@AsafKaragila oh, silly me. You don't necessarily have a Hamel basis.... :/

Asaf Karagila

@tb I'm not following you... :-)

Ilya

@AsafKaragila Follow the White Rabbit

Asaf Karagila

@Ilya I am the White Rabbit.

Ilya

@AsafKaragila O RLY? have you already trimmed your beard?

t.b.

09:40

@Ilya I am not the white rabbit if anything then I'm the white teddy!

@AsafKaragila I messed up completely. So the real point is that you want to argue that $\ell^1$ is reflexive but not isomorpic to its dual?

Asaf Karagila

@Ilya No, not yet. My girlfriend is about to leave and she goes in and out of the bathroom... it would make showering and trimming one's beard quite less comfortable.

@tb No, I want to argue that $\ell^1$ is algebraically reflexive.

robjohn

@Ilya trimming is allowed. I haven't shaved mine for 35 years.

Ilya

@AsafKaragila I won't comment

@robjohn trimming is allowed for rabbits?

@robjohn or, you have a beard! nice :) me and teddy we're beardless

Asaf Karagila

@Ilya You shouldn't because you don't know how our apartment looks like and what is the dynamics of our relationship in that sense.

robjohn

@Ilya with greens and some vegetables :-)

t.b.

09:43

@AsafKaragila okay, but won't every $\ell^p$ do?

Ilya

@robjohn for Asaf that should sound quite evil :D

Asaf Karagila

@tb Because in Solovay's model topological duals are algebraic duals?

t.b.

If that is true then yes. DC is certainly enough to identify the topological dual of $\ell^p$ with $\ell^q$.

Asaf Karagila

Holy Scheiße.

So $\ell^2$ is isomorphic to its algebraic dual.

And that is what I wanted all along!

t.b.

Provided that it is true that Banach => dreamy you get a continuum of mutually non-isomorphic Banach spaces that are algebraically reflexive and I would bet that you can combine them using $\ell^p \oplus \ell^q$ to get a continuum of non-isomorphic examples that are isomorphic to their algebraic duals.

Asaf Karagila

09:51

@tb Yes, that is quite the idea!

Now, why did you write Provided in such way?

t.b.

Because that's the proviso for me (I don't understand that statement -- that is I just believe you this point). You'd probably need the open mapping theorem from DC which I believe should be true because it only relies on the Baire category theorem.

Asaf Karagila

Why would I need the OMT?

t.b.

One of the ways I know how to tell those spaces apart would be to derive a contradiction from Pitt's theorem and the OMT.

There's certainly some checking to do that I didn't use more than DC anywhere, but I'm quite confident that something should be doable.

Asaf Karagila

Well you would only need that for asserting there is a fora of interesting examples.

t.b.

Yes.

Asaf Karagila

09:58

My arguments for $\ell^2$ being self-dual and $\ell^\infty$ having a "smaller" dual stand.

t.b.

Yes. It should work for all $\ell^p$'s.

Asaf Karagila

Well, in the case of $p\neq 2$ we have the algebraically reflexive property but not necessarily the duality property. Right?

I have never seen this before, although all the needed axioms were out there already.

t.b.

Duality property? Isomorphic to the dual?

Asaf Karagila

Yes, that's what I meant.

I'm just excited because I've been trying to find this sort of space for almost a year now.

t.b.

I'm pretty certain that you can exclude the duality property for $p \neq 2$, but that would need some checking, as I said.

Asaf Karagila

10:02

@tb Of course. :-)

Well, I am going to take a shower and all that and head out to the university.

t.b.

Okay, have a nice day!

I'll probably be here tonight.

Asaf Karagila

I will probably be away.

t.b.

Okay, then see you another time :)

Asaf Karagila

We're celebrating a year together today, so I'm probably gonna be preoccupied.

Ilya

@AsafKaragila congratulations! I think I've known you for more that a year and you always had this girlfriend

Asaf Karagila

10:05

Thanks... :-)

t.b.

Okay, I should go, too. See y'all!

Ilya

oh no

@tb: I was waiting for you to finish with Asaf to ask you to take a look on this question of mine

maybe later now - if you have time, of course

t.b.

@Ilya I just got your ping. What exactly do you want to know from me?

Ilya

@tb: 1. what do you think about the method I've described there
2. how to resolve the issue there
3. maybe you can advise some references on more neat methods

but that might take some time - even the question is not too short. The answer I received is quite ok, but I also wonder about your opinion. Just in the case you have time - I can also wait, that's not urgent at all

t.b.

Then I suggest we discuss this at another point in time because I really should be gone in a few minutes. Two very useful things in this context are Dynkin's $\pi$-$\lambda$-theorem and the monotone class theorem. That is what immediately comes to mind given your question. But probably you look for something else.

Ilya

10:20

@tb: I thought of them already - and I was looking for some exercises on that topic, just to understand if it would help or not. OK, let's discuss it another time, good luck with things you need to do

t.b.

I think I learned that stuff from Durrett's book, available here

I hope this helps for the moment. Sorry for not having more time right now.

See you soon!

Benjamin Lim

10:54

$\ubsetneq$

$\subsetneqq$

Kannappan Sampath

Hi @Ben,

Benjamin Lim

Hey Kannappan I'm kinda busy typing an assignment now. I'll try to talk to you later.

Kannappan Sampath

Sure!

Zhen Lin

The fact that we even have a symbol $\subsetneq$ says that something has gone horribly wrong with $\subset$...

Ilya

11:11

@ZhenLin yes, it annoys me - I'm so used for $\subset$ as just a weak inclusion

Benjamin Lim

@ZhenLin I am trying to see why the degree of $\Bbb{Q}(\sqrt{3} + \sqrt[2]{3})$ is exactly 6, I know it is at most 6.

Zhen Lin

It's contained in an extension of degree 6, so you just need to show it's not 2 or 3. I think you can do this by brute force.

anon

What's the difference between $\sqrt{3}$ and $\sqrt[2]{3}$?

Zhen Lin

He means $\sqrt[3]{3}$, I'm sure. :p

anon

oh okay

Benjamin Lim

11:23

Sorry, typo I meant why $\Bbb{Q}(\sqrt{3} +\sqrt[3]{2})$

square root of 3 plus cube root of 2

N3buchadnezzar

Heya

Benjamin Lim

I know it is not 2

because the quadratic formula cannot produce cube roots

Skullpatrol

Hi @ZhenLin could I get get your opinion on the mathematical content of the book "The Road to Reality" by Roger Penrose please?

ymar

@BenjaminLim Take a look at this question of mine.

Skullpatrol

Hi @N3buchadnezzar What's up?

Benjamin Lim

11:26

@ymar Thanks,

N3buchadnezzar

@Skullpatrol Integrals as usual. You?

ymar

@BenjaminLim Don't miss André Nicolas' comment!

Benjamin Lim

@ymar Not so easy to see why $\Bbb{Q}(\sqrt{3})$ is a subfield of what I wrote above

besides our questions are not exactly the same

Zhen Lin

@Benjamin: Then show it's not 3. Show that no non-trivial linear combination of $1, \alpha, \alpha^2, \alpha^3$ can be zero, where $\alpha = \sqrt{3} + \sqrt[3]{2}$.

Skullpatrol

@N3buchadnezzar Not much, just trying to get Zhen's opinion on the mathematical content of the book "The Road to Reality" by Roger Penrose.

anon

11:29

What about the math content?

Benjamin Lim

@ZhenLin A lot of algebra crunching, I'll do that

ymar

@BenjaminLim They are very similiar. Try taking the same path and it should work.

Skullpatrol

@anon Is it pedagogically sound?

anon

Well, it gives you a fair feel for what math says and why, but it can't be used as a textbook even though it has "exercises." I only read/skimmed in a few hundred pages though, iirc.

Benjamin Lim

@ymar Thanks.

Skullpatrol

11:35

@anon That is what I wanted Zhen's opinion on, can it be used as a textbook for an independent learner.

ymar

@BenjaminLim There is a theorem of Abel which says something to the effect that you'd have to be very unlucky for that not to work.

Matt N.

11:50

@tb ...with $n \in \mathbb Z$, you meant. And for a finite cyclic group $G$ I don't get a nice duality, do I? bbl

Benjamin Lim

@ZhenLin Hey you told me to show that no non-trivial linear combination of $1,\alpha...\alpha^3$ can every equal zero. So when I write out this linear combination, is it valid to say that the coefficients of $\sqrt{3}$ must be zero, the coefficients of $\sqrt[3]{2}$ must be zero and so on? Can we compare coefficients like that?

robjohn

@ZhenLin you mean rational linear combination?

Zhen Lin

of course.

N3buchadnezzar

@Skullpatrol I would say no, but I think you can learn quite a bit from it

Zhen Lin

@BenjaminLim: Yes, provided you are willing to assume that $\sqrt{3}$ and $\sqrt[3]{2}$ are linearly independent. (They are, of course.)

N3buchadnezzar

11:59

@Skullpatrol The problem I see with it is that the content is very diverse, and only skimms large parths of the math. Eg, the study on manifolds.

Skullpatrol

@N3buchadnezzar Thank you for being polite and replying to my question.

hhh

Is there any general definition for the term "real function"? Is it $f: \mathbb R\mapsto \mathbb R$? $f: \mathbb R\mapsto \mathbb R^n$? $f: \mathbb R^m\mapsto \mathbb R$? $f: \mathbb R^m\mapsto \mathbb R^m$?

1

Q: Harmonic real function, $\triangle f =0$, but not in origin?

Problem in English (original problem 7 on page 813 here) Suppose $f(|\bar{x}|)=\sqrt{x_{1}^2+...+x_{n}^{2}}$. For what kind of real $f$ it holds that $f$ is harmonic everywhere but not in origin? If $f$ is harmonic, then $\triangle f=0$. Definitions The "real function" apparently here...

Benjamin Lim

Well in the expansion I have $1,\sqrt{3}, \sqrt[3]{2}, \sqrt{2}\sqrt[3]{2}, \sqrt[3]{2}^2, \sqrt[3]{2}^2\sqrt{3}$

Skullpatrol

@robjohn Hi, how did the blood tests go?

Benjamin Lim

@ZhenLin There are exactly six of them. Now are they a basis for $\Bbb{Q}(\sqrt{3},\sqrt[3]{2})$?

If I know that they are then linear independence is immediate.

Zhen Lin

12:03

They are, but again, this requires a little bit of work.

anon

@hhh What does $|\vec{x}|$ mean? If that's the same as $\sqrt{x_1^2+\cdots+x_n^2}$, then doesn't the supposition say that $f$ is the identity on $[0,\infty)$?

Benjamin Lim

@ZhenLin as in??

anon

hhh: Oh, nevermind, there's a typo in your problem. You forgot to put an f() on the outside, but you wrote the correct thing in the scan.

Zhen Lin

Well, showing that they're linearly independent!

Benjamin Lim

i know, but as in not in the usual way of crunching stuff

hhh

12:24

@anon yes in this case $f: \mathbb R^{n}\to\mathbb R$ but generally, "real function" is an open term or? Is this actually correct domain $\mathbb R^n$?

Zhen Lin

@hhh: Don't use $\mapsto$ for that, use $\to$.

hhh

@ZhenLin Thanks, fixed.

anon

\mapsto is for explicitly defining a map, e.g. $x\mapsto 2x$. anyway a real function is anything $A\to \mathbb{R}$ I think.

Zhen Lin

A real-valued function is something with codomain $\mathbb{R}$. A function of a real variable is something with domain $\mathbb{R}$.

Kannappan Sampath

I had a downvoter. Random question. Very Very elementary but for all that is good, I don't see anywhere I can be a bit more explanatory. sigh But this time, I am going to wait for the next instance and bring it to the notice of mods.

hhh

12:36

@KannappanSampath I did not downvote you if you inferred that. I would care less about downvoters, they may have some meaning later but it is better to concentrate on one's own working and try to do the best. Do not let the gamification to steer your attention from math...

Kannappan Sampath

@hhh Hmm. Not even in my dream, did I guess that some name for who possibly would have downvoted. So, this strikes me as strange confession to make. But, anyway, thanks forr your advice, there. : )

Rajesh D

12:53

Hi guys

Kannappan Sampath

Hi!

Rajesh D

how az it been ?

Kannappan Sampath

Not bad.

Rajesh D

I can understand

t.b.

13:08

@MattN Yes of course $n \in \mathbb{Z}$. And yes, there is a nice duality, just not a terribly exciting one (but that is a matter of opinion).

Jonas Teuwen

How would LaTeX go using speech recognition?

"slash int"...?

Kannappan Sampath

@JonasTeuwen Just great. I would like it! : )

Height of Laziness.....

Jonas Teuwen

No, carpal crap.

Kannappan Sampath

What is Carpal?

Jonas Teuwen

New wrists would be kickass too.

@KannappanSampath This!

Kannappan Sampath

13:15

I see.

Daniil

Hi

Jeff

13:31

hi folks

Kannappan Sampath

13:49

Hi @DidierPiau

N3buchadnezzar

Could someone look at this and tell me what they think of it?

frm.utn.edu.ar/analisisdsys/material/funcion_gamma.pdf

Honestly I do not really like this document, i feel that it lacks rigour.

anon

Could you specify what aspects of it lack foundation? Also, why do people write ]a,b[ for intervals sometimes? Does that mean something special?

Zhen Lin

It means they don't have to use parentheses, which is helpful when parentheses are being used for so many other things...

anon

It's also helpful in giving me a brain aneurysm.

N3buchadnezzar

@anon I feel he is skipping out on the proofs.

Like why there is only one function satisfying $f(x+1) = x f(x)$ and $f(1)=1$

Jeff

14:05

@N3 I lose understanding of it at the derivatives on the second page. But I see the proof before it is one whole line.

BTW, if anyone wants some free repuation in real analysis, see: math.stackexchange.com/questions/122885/…

Nevermind. It seems there's two answers already.

N3buchadnezzar

I simply starred Zhen Lin`s comment.

anon

@N3bu: With log-convexity. Yes, it outsourced the proof - possibly because it's a rather involved proof the author didn't feel belonged, or the author didn't have the time/energy/motivation to go over it, etc. I would call it a lack of comprehensiveness rather than a lack of rigor.

Jeff: Was that a serious question?

Hm. Is there a connection between user mixedmath and user mixedmath?

Jeff

The question I posted? Yes.

@anon stupid question?

Dylan Moreland

Ooh, Asaf asking about the adeles.

Maybe he's coming around.

anon

Well, elementary.

N3buchadnezzar

14:13

@anon How can two users have the same name?

@Jeff It is a good question, but you phrased it as you were asking the question to give away reputation (eg, you already knew the answer before asking)

anon

@N3bu: The thing that differentiates users is their user ID #, not their handle. For example, there are at least four anons plus other variants.

Jeff

but isn't that circular proving? the proof that .9999....=1 is that there is no number between them (1-.99999....=0.0000......)

anon

@Jeff: What do you think 0.999... stands for?

Jeff

@N3buchadnezzar well, in here i was lighthearted about it. do you mean the question itself was worded poorly?

@anon i don't understand the question

anon

@Jeff: I'm asking what you think "0.9999..." means, or represents.

N3buchadnezzar

14:17

@Jeff No, just the way you asked about it in chat.

anon

Also, have you heard of limits?

Jeff

@anon yes

anon

You're aware 0.999... represents a limit, specifically the limit of partial sums 0.9, 0.99, 0.999, ..., no?

N3buchadnezzar

cough Geometric series cough

Jonas Teuwen

That always felt quite fishy to me.

It's not like the number is defined as 0.9 + 0.09 + 0.009 + ...

It is some Cauchy sequence, or whatever.

N3buchadnezzar

14:22

@JonasTeuwen I do not see the problem as defining numbers as sum of series, no?

Jonas Teuwen

Well... That series somewhat already assumes the existence of the real numbers right?

anon

Just like addition of rationals assumes the existence of the rationals.

N3buchadnezzar

@JonasTeuwen Does it? It always assumes we can find a number that is 10 times smaller than the previous number.

anon

Jonas is talking about series in general, N3bu.

N3buchadnezzar

I was thinking of defining numbers as series where each part is a digit

3.1415 where a_1 = 3 , a_2 = 0.1 ...

Jeff

14:28

ok. i'm just gonna go and delete the question before i embarrass myself more, i guess

too late. it won't let me delete it

Jonas Teuwen

Well, you define real numbers as equivalence classes where 0,999... and 1 are in the same one. Then you need to show both have a decimal expansion... etc.

Kannappan Sampath

14:56

I have not been having time to look into Topology for some time now. Hope I get more time in the days to come.

Asaf Karagila

Holy Cow! @Srivatsan!

anon

Long time no see.

Jonas Teuwen

Hiiiii.

Kannappan Sampath

Hi @Srivatsan

N3buchadnezzar

15:19

Is Srivatsan a girl ?

Jonas Teuwen

Here everybody is unisex.

N3buchadnezzar

Reason for asking is that I seem to recall watching some videos, on youtube. Where the person was a indian girl called Srivatsan, or something fairly close.

Jeff

@N3buchadnezzar sure it was youtube? ;)

N3buchadnezzar

@Jeff yes? It was about Complex analysis

anon

ysis

N3buchadnezzar

15:22

A link here

Srivatsan

hi all

Kannappan Sampath

@N3buchadnezzar Srivatsava is her Surname

Srivatsan

Sorry guys, my browser crashed, so I had to restart it

N3buchadnezzar

@KannappanSampath Its fairly close though

anon

just figured you were afk

Srivatsan

15:25

@KannappanSampath One of my profs once asked if some D. Narayanan she knew is related to me. =)

stoicfury

hey folks

Jeff

@Srivatsan your ex-prof had a serious case of thinking-all-foreigners-know-each-other (it's not usually fatal)

Kannappan Sampath

@Srivatsan I wonder if there existed unique Narayanan according to your prof.

@stoicfury Hi!

stoicfury

i was on the train today and some guy was wearing a shirt from the MIT coop, it said something like ∈o∅Eda= ∑(d)

Srivatsan

@Jeff fair assumption perhaps :) (BTW, Narayanan is my second name as well.)

stoicfury

15:28

what equation is that?

Kannappan Sampath

Makes no sense.

stoicfury

he said it was "maximals equations" or something

didn't quite hear him

N3buchadnezzar

this

Srivatsan

@stoicfury Maxwell's equations perhaps?

N3buchadnezzar

mstatic.mit.edu/nom150/items/maxwell.jpg ?

stoicfury

15:31

This shirt! physicsworld.com/blog/…

those are "Maxwell's equations"?

Kannappan Sampath

Maxwell!!!!

Yes, Srivatsan is very good in guessing. Intelligent guesser!

stoicfury

haha

Srivatsan

@KannappanSampath "equations" is a good hint. There aren't too many popular named equations.

N3buchadnezzar

Navier stokes equations?

Kannappan Sampath

:)

Jonas Teuwen

15:34

Hmm, so God does not use very low intense light? Maxwell's equations fail to hold there.

Asaf Karagila

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

I just found a first white hair in my beard!!!!!!

4

Srivatsan

@JonasTeuwen Ohh, why? Quantum effects or something?

Jonas Teuwen

Yeah.

@AsafKaragila Cool!!!

N3buchadnezzar

Only if it had been a green hair

Asaf Karagila

What? It's a point of pride that I haven't got a single white hair yet.

Alas, this point is gone and will never return.

Srivatsan

15:36

@AsafKaragila Oh my god, I must now re-picture you in my mind!

Asaf Karagila

@Srivatsan I also trimmed my beard today, which is how I found out the white hair.

Srivatsan

@AsafKaragila Ha. You trimmed your beard after a long time?

Asaf Karagila

@Srivatsan Something like a month or so.

Maybe two?

Something in between, that's for sure.

Srivatsan

Alright

Jonas Teuwen

I wish I had gray hair.

N3buchadnezzar

15:44

green

Jeff

17:11

@AsafKaragila Solution: Shave your beard

Skullpatrol

@Jeff Are you fully convinced that there is no number between 1 and 0.999...?

David K.

Can someone help me come up with a (hopefully simple) function $f$ of a real variable $x$ with the property that $\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=0$?

I've considered the normal distribution function, but that's slightly more complicated than I'd like...

Jeff

@Skullpatrol Yeah, (i guess). I liked the 2nd explanation (with geometric series) better. Do you have another explanation?

David K.

17:29

Crickets. I suppose $f(x)=\frac{1}{x}$ mostly works, but I really need $f$ to be continuous and bijective over all of $\mathbb{R}$

Skullpatrol

@Jeff As you know I prefer simple explanations, and the one I like is "Try and write down a number between them?" If you can't they must be the same number correct?

Jeff

@Skullpatrol Actually, I thought of that, but I found that circular reasoning. "If I can't write down a number between them then they're the same. If two numbers are the same then you can't find a number between them." I wanted to be convinced by a different argument that they are the same number (is possibly another way of stating my question).

Jonas Teuwen

Hmm, say given the natural numbers, then "1" and "2" are the same?

Skullpatrol

@Jeff Actually, that is not circular reasoning. It is called a statement and its converse, ie with terms the other way round, also called vice versa.

@JonasTeuwen Hi @JonasTeuwen

Jonas Teuwen

Hi.

Jeff

17:43

@Skullpatrol Well, both statements may be true. But using one to prove the other, then the other to prove the one still seems like circular reasoning.

anon

The whole thing is founded on the established convention that infinite digital expansions represent infinite sums, which are limits of partial sums. Limits, when they exist, are unique, and 1 is obviously a (and thus the) limit of 0.9, 0.99, 0.999, ...

Skullpatrol

Thanks for offering me a drink ;-) @JonasTeuwen

Jonas Teuwen

8-).

Skullpatrol

@Jeff Well, one is not used to prove the other. The question is simply first: "Try and write down a number between them?"

Jonas Teuwen

@anon Yes, that's what is confusing to me. It at least requires some work to get this...

anon

17:52

The uniqueness of limits is straightforward, though constructing R with Cauchy sequences and showing well-definedness and yadda yadda is decidedly not. I guess students of math are tasked with taking the reals and their properties for granted before they go into the fundamentals.

Daniil

ustream.tv/penguincam :)

anon

haha, still researching the RH, still using horrible grammer and punctuation

actually that's not bad, but that guy is persistent!

Jonas Teuwen

@anon That's right.

@anon Hmm, that's the Riemann Hypothesis?

anon

Yes.

Skullpatrol

@Jeff One reasonable attempt to find such a number between 1 and 0.999... would be to try and find the average "number" between them. So adding 1 and 0.999 gives 1.999... now divided by 2.

Jeff

17:59

@Skullpatrol I see your point. But my original question was that "I can't write down a number between these two numbers even though theory says I should be able to". So your argument, while correct, seemed like circular to me.

@Skullpatrol I thought of that one, too, before posting.

@Skullpatrol But it still seemed circular.

Skullpatrol

@Jeff Let's leave circularity aside for a minute. What do you get when you divide 1.999... by 2 Jeff?

Jonas Teuwen

So, $0.999... + 0.999... = 1.999...$ now substract $0.999...$ from both sides 8-).

Eric Gregor

Is it obvious why $T$ has a unique smooth structure making it into a smooth rank-k vector bundle over the $G_k(V)$ here: math.stackexchange.com/questions/90069/…

Jeff

@Skullpatrol I get $1$. I understand the issue now. My point is that before I understood it, I thought that was circular reasoning.

Skullpatrol

@Jeff I don't get 1. I get 0.999... to be honest.

Jonas Teuwen

18:06

That's the bloody same.

Skullpatrol

@Jeff So, 0.999... is the average number between 1 and 0.999... correct?

Jeff

@Skullpatrol Whoops, I answered too quickly (also known as without thinking). You're right (of course).

Jonas Teuwen

@anon Hmm, in the Cauchy construction, how do we define the product of two Cauchy sequences?

I thought it would be the same as with completions of vector spaces, but now I was wondering "how about the product?".

Jeff

@JonasTeuwen I believe there is a theorem which states that the product of two Cauchy sequences is the product of their limits

anon

Componentwise, as in $(a_n)\cdot (b_n)=(a_nb_n)$?

Jonas Teuwen

18:08

That's all?

Oh man.

anon

Jeff, Jonas is talking about definitions.

Skullpatrol

:3907209 How can the average of two numbers be one of the numbers? Answer: they must be the same number, correct?

Jeff

@Skullpatrol yup

Jonas Teuwen

@anon Oh, right, it was in the case of Dedekind cuts that products were tricky.

Jeff

@Skullpatrol interesting. that comment did not highlight (in orange) the @Jeff

Skullpatrol

18:14

@Jeff Nothing circular about that.

Jeff

@Skullpatrol nope

Skullpatrol

Thank you for your attention ;-)

Jeff

@skull thank you for your explanation ;)

Skullpatrol

Np

Jeff

@Srivatsan LOL.

anon

18:28

Skullpatrol

@anon What is that?

anon

slowpoke

Jeff

I accidentally voted up a bad comment (first comment here: math.stackexchange.com/q/118397/22544). It won't let me undo it. Is there another way to retract my vote?

anon

@Jeff: Nope.

Skullpatrol

@anon Why are you referencing slowpoke?

anon

18:35

@skull see kym and the time delay in jeff's response to srivatsan

Jeff

@anon :)

Skullpatrol

0.999...

In mathematics, the repeating decimal 0.999... (sometimes written with more or fewer 9s before the final ellipsis, or as 0.9, \scriptstyle\mathbf{0}.\mathbf{\dot{9}}, 0.(9)) denotes a real number that can be shown to be the number one. In other words, the symbols 0.999... and 1 represent the same number. Proofs of this equality have been formulated with varying degrees of mathematical rigor, taking into account preferred development of the real numbers, background assumptions, historical context, and target audience. In fact, every nonzero, terminating decimal has an equal twin represen...

t.b.

call for votes: two dupes

anon

I voted the second before your call :)

Jeff

vote up? down? (sideways?)

anon

18:42

vote-to-close (you need, what is it, 10k?)

N3buchadnezzar

just 10k? Then it is no problem!

Asaf Karagila

@Jeff No beard makes me look like a child.

N3buchadnezzar

Arturo could triple-double-double-double vote then!

Jonas Teuwen

18:57

@AsafKaragila Because you're so tiny?

Jeff

@AsafKaragila that's an advantage. especially if you're still single

Asaf Karagila

First, no - it's not an advantage. It's not that I'm 30+ that I need to look younger. I'm still under 30.

Second, as it happens today we're a year together and we're going out to dinner soon.

Skullpatrol

@AsafKaragila ;-)

Asaf Karagila

@JonasTeuwen And no, it's not because of that (I'm not tiny at all. It is you who is a gigantum).

Transcript for

$Mathematics$ Mathematics