Mathematics - 2012-03-22 (page 3 of 4)

leo

15:01

hi there

t.b.

hi leo

leo

given any nonmeasurable set $V$ it is true that if $E\subset V$ is measurable then $m(E)=0$?

t.b.

Of course not.

leo

I'm able to prove this for Vitali sets

t.b.

For those it holds.

Else you can take a measurable set $E$ of positive measure disjoint from $V$ and consider $E \cup V$.

leo

15:05

yes

t.b.

This set isn't measurable but it contains a measurable subset of positive measure.

leo

you are right

My situation is this: $I$ an interval, $V$ a Vitali like subset of $I$, I want to prove that if $E\subset I\setminus V$ measurable, then $m(E)=0$

t.b.

How is $V$ constructed?

leo

by considering the equivalence relation

$x\sim y$ iff $x-y$ is retional

I mean rational

t.b.

(you can edit your posts by hitting the up arrow on your keyboard)

So it is a Vitali set?

leo

15:10

yep

anon

(\sim = $\sim$)

t.b.

and did you prove that it has full outer measure?

leo

I have proved that the outer measure of $V$ is positive

Jonas Teuwen

Nice.

t.b.

But you need to do some additional work in order to guarantee that it has full outer measure.

leo

15:12

$I\setminus V$ has positive outer measure as whell

I don't know what you mean by "it has full outer measure."

t.b.

Outer measure $1$.

leo

ah ok

So, my $V$ has outer measure $1$

t.b.

Does it?

without doing some work you can choose your representative set for $\sim$ in $[0,1/2]$ and of course then the outer measure will be at most $1/2$

Since you didn't tell me what exactly $V$ is, I can't help you, yet :)

leo

oh sorry

@tb pick an element of each one of the equivalence classes determined by $\sim$, put all them in a bag, $V$ is that bag :-)

t.b.

But how do you guarantee that $V$ has outer measure $1$ this way? As I said, nothing keeps you from picking only those in the $[0,1/2]$ half of the bag.

Matt N.

15:19

@tb I don't see how I can conclude $z^n$ from this. Are you sure this works?

leo

@tb at first glance, nothing

Jonas Teuwen

You want to construct a Vitali set with outer measure 1?

t.b.

@leo see this thread

leo

@JonasTeuwen yep

t.b.

@MattN What are the continuous homomorphisms $\mathbb{R} \to \mathbb{C}^\times$? Yes I am sure it works

Jonas Teuwen

15:21

Right, I've asked such a thing before as @tb has just linked to.

Robert Israel posted a solution on sci.math a long long time ago.

A long long time ago. I can still remember like the ... whatever I forgot the text.

leo

@JonasTeuwen thanks

t.b.

That was correct (if you restrict the target to $S^1$). Now you need to guarantee that this homomorphism descends to $S^1$ (since it came from there after all) by factoring over its kernel, so the kernel must contain $2 \pi \mathbb{Z}$. Now think about what closed subgroups of $\mathbb{R}$ contain $2 \pi \mathbb{Z}$.

Matt N.

No it was wrong, I think. They are $\varphi : \mathbb{R} \to S^1$ , $x \mapsto e^{ixy}$ for $y \in \mathbb R$.

Knu

Hi, how would you type 5n twice? (5 and 10 but not 15)

would 5n x 2 be correct?

some kind of repeat argument if you want

t.b.

Okay, I think this should be enough for the moment, I should do some serious work.

Matt N.

15:29

@tb Yes. Thank you and see you later!

leo

what means JDH by " It follows easily that V is not measurable and has inner measure 0, since otherwise the translates modulo 1 would have infinite measure inside [0,1]"

follows easily that $V$ is'nt measurable

that's true

got it

:-)

Knu

I don't want to use a limit which is less precise. And googling nth gives me nothing relevant.

leo

no problem

Knu

How would you express that repeat argument? Would a multiplication be correct or it would mean something else?

I guess no one cares.

Paul Slevin

15:45

@Knu what do you mean type 5n twice?

Matt N.

@JonasTeuwen Are you there?

What does "a homomorphism descends to $S^1$" mean? Is this a fancy way of saying it maps to $S^1$?

Jonas Teuwen

16:13

@MattN Yes!

And now I'm a goner! Good night ding ding ding ding.

I'll go.

Back soon.

leo

if $m_i$ is the inner measure defined as here it is true that $m_i$ is at least finitely subaditive

Knu

16:32

@paul I was trying to express 5n twice (with a repeat argument) without using a range nor the result (5 and 10).

But after some research it seems the notation is not intuitive enough.

Unless you come up with something I didn't read.

Eric Gregor

it's a fact that the lie algebra of an abelian lie group is also abelian. can you show this using pushforwards? specifically using the pushforward of the inversion map (which is a group homomorphism)?

Knu

@paul {5*n | n in [2]} that's what I found. Do you have anything better?

Jonas Teuwen

@MattN I'm back. I'm melting.

Kannappan Sampath

I sincerely hope @MattN is doing extraordinarily well. I wonder why Matt drops in and out, so many times... Are you kinda busy now? @MattN

Knu

The stared imgur is for real? It's not photoshoped?

Eric Gregor

16:42

definitely photoshopped

it's making fun of sensitivity to islam

Paul Slevin

but what do you mean "express 5n twice". I don't understand what you mean? Have you got the explicit original question?

Eric Gregor

does anyone know some manifolds/lie algebras? i am just looking for a bit of direction in what i think should be an easy problem

Paul Slevin

@Knu $\{5n \mid n=1,2 \}$ is just the same as writing $ \{5, 10 \}$

Asaf Karagila

I am fairly certain that someone knows some manifold and Lie algebra material. I am not sure if someone in the channel knows, though.

anon

@Eric: What makes you think it's photoshopped? That means the actual words on the Fox news screencapture were altered, not that words were added onto the side.

Eric Gregor

16:59

it looks photoshopped, for one. two, it is an obvious parody of both fox news and the response to the incident

anon

hm. okay.

Eric Gregor

anon

wat :O

Eric Gregor

it is easy to troll partisans, apparently

Ilya

@EricGregor and what does it prove?

Eric Gregor

17:07

@Ilya, what does what prove?

Matt N.

@KannappanSampath I'm doing fine thanks. And yes, I'm busy! : )

Eric Gregor

@Ilya @anon demorgen.be/dm/nl/990/Buitenland/article/detail/1412767/2012/03/…

Matt N.

@JonasTeuwen : )

Ilya

@EricGregor you told that it is a photoshop and then put a screenshot. For which reason?

Eric Gregor

@Ilya, to show how easy it is to photoshop

Ilya

17:12

@EricGregor I think no one here has doubts

Eric Gregor

anon seemed to think the original was real

Ilya

I agree with anon, that's why I've asked you what should your picture prove. If it is easy to photoshop it doesn't say anything about the original

Eric Gregor

did you see the last link i posted?

you guys are gullible

with all due respect

anon

@Ilya It's clearly a photoshop to troll, as Eric has amply shown.

Ilya

@Eric: I saw it, and?

Eric Gregor

17:15

@Ilya, again: demorgen.be/dm/nl/990/Buitenland/article/detail/1412767/2012/03/…

Kannappan Sampath

@anon I need some TeXnical help on the main site. Can this be beautified? Please help me.

Ilya

@EricGregor do you think I will learn Dutch in 5 minutes?

Eric Gregor

there is no Fox News "watermark" on the right of the original, for one. second, it assumes a kind of monstrous stupidity of Fox News that only exists in a liberal caricature

google.translate.com

Ilya

thanks for the translation. So you rely upon DeMorgen

anon

@Kanna: I don't see what the issue is with your answer.

Kannappan Sampath

17:18

@anon Does it not look ugly? The fractions...

Eric Gregor

@Ilya, and common sense

anon

Looks fine to me, Kanna.

Kannappan Sampath

OK.

Eric Gregor

@Ilya, i just looked at your profile, it says you live in the Netherlands! how do you not know Dutch?

Jonas Teuwen

Because many people in The Netherlands don't speak Dutch.

Ilya

17:25

@EricGregor I appreciate your attention to myself, but I didn't learn it

Eric Gregor

@Jonas, really?? i didn't know that

i though Dutch was the national language

Jonas Teuwen

Yeah, people actually speak Spanish.

(is joke)

Ilya

@Jonas: is DeMorgen reliable? ;)

Jonas Teuwen

@Ilya The newspaper?

Ilya

@JonasTeuwen yes. The Spanish joke is thin

Jonas Teuwen

17:26

@Ilya It is something like NRC in NL.

@EricGregor No but if people notice you don't speak Dutch very well they will speak English.

Eric Gregor

wow, i didn't know that

that seems suicidal for a country. i imagine holland won't stay dutch very long

btw, to give this conversation a chance of becoming mathematical, do either of you guys know basic manifold/lie algebra theory? i have what i think should be a rather straightfoward question about lie algebras of lie groups that has been bugging me.

anon

You might as well just ask your question. The only thing I know about lie algebras/groups are their definitions, though.

Kannappan Sampath

Broke 5k! : )

Eric Gregor

the lie algebra of an abelian lie group is also abelian. this can be seen using basic representation theory. can you show this using pushforwards? specifically using the pushforward of the inversion map (which is a group homomorphism)?

anon

What do you get at 5k?

Kannappan Sampath

17:32

@anon Tag wiki edit can be approved! That's all (officially.) Unofficially, I am satisfied! : )

Ilya

@EricGregor what do you mean?

Kannappan Sampath

I can approve edits!

anon

Reviewing edits is a thankless task. Other than the badge you get for 100 actions on them.

Kannappan Sampath

Right. I'm happy with my recent answer you know!

Eric Gregor

@Ilya, Lee suggests that you can use a) the fact that inversion is a group homomorphism, and b) that the pushforward of inversion is $i_*X=-X$, to prove the claim

i don't see how to connect this to connect these ideas to the lie algebra itself. we want to show that the lie bracket is trivial, i suppose

anon

17:37

@Eric what if you do the inversion on a product and then take the differential? (I'm out of my depth here.) | Also if you check Ilya's reply link you'll see he was replying to your "suicidal" comment.

Eric Gregor

@anon what do you mean precisely by "taking the differential"? the pushforward of the inversion on the product?

anon

yes, that's what I mean.

Eric Gregor

@anon how would that connect to the Lie algebra?

@Ilya, regarding my comment, i suppose if you don't find it obvious that cultures die when languages die and people become insufficiently insular, it would take quite a bit of discussion. i am happy to have the discussion but i don't want to make this a political forum.

anon

I'm sort of shooting in the dark. | Also, you referred to "country" rather than "culture" previously.

Eric Gregor

i said "i imagine holland won't stay dutch very long". they may remain The Netherlands, but the Dutch culture and people won't survive, that's what i'm claiming. of course you could just rename the new people and language Dutch, but that's cheating

Kannappan Sampath

17:43

"This is a strange dialogue" writes, The Chaz, a M.SEr....(for those who did not know.)

Eric Gregor

@anon, i'm sure your idea is the right one, i just can't find that explicit connection. i just need to go back to the definitions, i'm kind of new to this stuff

Ilya

@EricGregor where are you from if I may ask

Kannappan Sampath

Hi @MarianoSuárezAlvarez

Mariano Suárez-Alvarez

Hello :)

Eric Gregor

@Ilya the American Northwest

Ilya

17:45

@EricGregor but you know Dutch?

anon

if $\varphi$ is the inversion map then $d (\varphi\circ\exp) X=-X$, is that what your note (b) is saying?

Eric Gregor

no. but i know how to use Google translate :P

Ilya

icic

Eric Gregor

@anon as long as your $d$ is the pushforward: $d: T_e G\to T_e G$

Ilya

Bill D. would like this guy

anon

17:48

oh man, overload on the \mathrm

Ilya

@anon nope, \text

he is just wasting ink

Kannappan Sampath

A closing related issue:

This was posed first.

This later.

I contend that the latter question can be merged with the previous one on all grounds.

What do we have to say?

anon

You linked to the same question twice.

Eric Gregor

i have to go to lunch. if anyone sees my question and feels like helping me out, please feel free to send an @ to me. hopefully i'll figure it out over lunch.

bye

Ilya

@KannappanSampath closely?

Kannappan Sampath

17:53

@anon @anon Edited.

@Ilya Pun?

Ilya

@KannappanSampath if only I knew what does Pun mean

Kannappan Sampath

@Mariano Your opinion is useful too.

anon

It should probably be closed/merged.

Kannappan Sampath

But, how do we merge?

anon

only mods can do that.

Kannappan Sampath

17:59

So, flag it for a mod, then?

anon

nah. just vote to close or something.

I don't think I've even seen a merge happen before.

Kannappan Sampath

OK. I'll vote to close the second one as a dupe of the first. I'll leave a comment there.

Mariano Suárez-Alvarez

so... the unit sphere tangent bundle to S^2 is a projective space. anyone can tell me what the homeomorphism is?

t.b.

@MarianoSuárezAlvarez identify RP^3 with SO(3) via quaternions and the cover SU(2) -> SO(3). the first column of a matrix in SO(3) gives you the base point, the second one gives you the direction in the unit tangent bundle.

Mariano Suárez-Alvarez

18:15

Ahh

thanks

stupidly, I started with the Stiefel manifold :)

t.b.

Euler angles would probably offer another way to put it, I think.

Mariano Suárez-Alvarez

SO(3) \cong SS^2 was the one map missing

i like to see SO(3) \cong RP^3 as follows: a positive rotation gives you an axis and an angle

that pair is a point in the closed ball of radius \pi in R^3

(in "polar coordinates")

but one has to identify rotations of angle pi and -pi

so one gets RP^3 by identifying antipodal points on the boundary

t.b.

That's nice!

Mariano Suárez-Alvarez

one avoids quaternions that way :)

(which is of course bad!)

t.b.

heh :)

Skullpatrol

18:40

@tb Something is wrong with the line of gravatars for people in this room.

@MarianoSuárezAlvarez I closed my browser and shut my computer down for 4 hours and when I came back my gravatar was still in the line.

N3buchadnezzar

Heya !

Kannappan Sampath

Hi!

N3buchadnezzar

Does anyone have the link to different ways to evaluate the gaussian integral?

I rember seeing it on the site, but can not recall where

ymar

Hello!

anon

18:55

30

Q: $\int_{-\infty}^{+\infty} e^{-x^2} dx$ with complex analysis

Inspired by this recently closed question, I'm curious whether there's a way to do the Gaussian integral using techniques in complex analysis such as contour integrals. I am aware of the calculation using polar coordinates and have seen other derivations. But I don't think I've ever seen it done...

@N3bu this one?

Skullpatrol

Free at last. I was stuck in the gravatar line for over 4 hours.

anon

14

Q: Proving $\int_{0}^{+\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$

How to prove $$\int_{0}^{+\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$$

calculus integral improper-integrals

or this one

N3buchadnezzar

@anon There was another one where one recalled the integral was sillimar to a cylinder or something. A method I had never seen before.

It is perhaps the second one, I will have a look.

=)

anon

second answer on the second one

N3buchadnezzar

thanks

=)

Skullpatrol

19:08

Free, free, free at last... I was stuck in the gravatar line for over 4 hours.

robjohn

19:29

@Skullpatrol can't see that. It is blocked.

Skullpatrol

try this one

robjohn

@Skullpatrol same thing

anon

@robjohn: youtube.com/v/PMndcQI4l9k&feature=related

robjohn

@anon has that question not been answered before?

anon

dicks

robjohn

19:34

@anon yeah

anon

well, it gave me a thumbnail

Skullpatrol

Who is EMI?

robjohn

It might be a campaign to get EMI to do something.

it is a British music company

Skullpatrol

@robjohn Thanks.

Kannappan Sampath

Hi @robjohn

robjohn

19:37

@KannappanSampath hey there. What's up?

Skullpatrol

@robjohn Can you youtube "Born Free" yourself?

hhh

Suppose cylinder and Dirichet's conditions. Are $u(x,y)=\frac{Q}{4} \left(R^2-x^2-y^2\right)$ in $\mathbb R^2$ and $u(x,y,z)=\frac{Q}{6} \left(R^2-x^2-y^2-z^2\right)$ in $\mathbb R^3$ the only solutions?

D.C. are:

$$\begin{cases}
-\nabla u = \rho & \text{when A is an inner point} \\
u=0 & \text{when on the border of } \partial A \\
\end{cases}$$

Kannappan Sampath

@robjohn Can you please look at a pdf and comment on the appearance and TeXnical suggestions if any?

hhh

Reading page 809 here, bottom part where it only states the solutions without mentioning uniqueness.

robjohn

►

@KannappanSampath sure

Skullpatrol

19:42

@robjohn Ahhh... the memories...

Kannappan Sampath

@robjohn Here is a link: mediafire.com/?9tn6ob75gbbv85i Thank you for agreeing to look at it. : )

hhh

I cannot understand what it means $u=0$ on the border $\partial A$ when $u(x,y)=\frac{Q}{4}\left(R^2-x^2-y^2\right)$.

Kannappan Sampath

Lot of incomplete work, but you ca judge the style probably! @robjohn

hhh

@robjohn amazon.com/Born-Run-Hidden-Superathletes-Greatest/dp/0307266303 <--- for some reason, this book came to my mind from the song (running is often associated with youth, freedom -- etc). Wish someone had done some song about that :P

Matt N.

@tb Then this one isn't sane either... : )

hhh

19:51

I am now not sure whether the Dirichet conditions are just premises...thinking.

robjohn

@hhh You're kidding, right?

hhh

@robjohn No, I am not. The songs there do not have such melody/feeling as the earlier one...many have tried to do a song about running (related topics) but failed miserably...or I have not found any good yet.

Jonas Teuwen

Tiling window managers are so cool.

robjohn

@hhh that you consider any good.

Jonas Teuwen

@robjohn My advisor told me how to do the homogeneous estimate :-)).

robjohn

19:54

@JonasTeuwen it's the grout that gets in my keyboard from them that annoys me ;-)

Jonas Teuwen

You just need to consider functions with mean $0$ as in the Poincaré inequality.

robjohn

@JonasTeuwen how did he say to do it?

hhh

Could someone point me to the Wikipedia article about the Dirichet conditions above?

Jonas Teuwen

@robjohn Just consider functions with mean $0$.

hhh

en.wikipedia.org/wiki/Dirichlet_boundary_condition <--- this one does not seem to be it or?

robjohn

19:55

I haven't looked into it

Kannappan Sampath

@robjohn Into my .pdf you mean?

hhh

I cannot understand what my book is trying to say about this

$$\begin{cases} -\nabla u = \rho & \text{when A is an inner point} \\ u=0 & \text{when on the border of } \partial A \\ \end{cases}$$

so some English would help...

robjohn

@hhh that seems to be what the DC means to me.

It is saying that $-\nabla u=\rho$ on the inside of of $A$ and and $u=0$ on the boundary of $A$.

hhh

@robjohn but what does it mean? They are just some assumptions but what about them? What happens if they are valid? Why are they called boundary value problem? I canot understand at all what is the goal here...

@robjohn but why? What is the purpose here? One defines such assumptions and then just states that the following $u(x,y,z)=\frac{Q}{6} \left(R^2-x^2-y^2-z^2\right)$ is the solution for cylinder -- smart but so what? Look how can I know that it is the only solution etc?

robjohn

@hhh do they say $\nabla$ or $\Delta$?

hhh

19:59

@robjohn $-\triangle u=Q$ is what is stated (p.809) and $\triangle=\nabla\cdot\nabla$

@robjohn This is not what the book is stating, it does not use nabla...

robjohn

@hhh what is the context here? what is the purpose of making this statement? There is no way to answer your question without some context.

@hhh that makes more sense.

hhh

"the connection between field and source $\nabla \bar{F}=\rho$ can be written as $-\triangle u=\rho$" (top part 809), I fail here miserably to understand what the book is trying to say...it is like saying that $\nabla=\triangle$ which is of course non-sense.

robjohn

@Kannappan the first definition seems a bit redundant. Is a linear operator on a vector space any less known than a linear transformation? Also, you write "$T:V\to V$ that maps $V$ to itself" which is also redundant.

Is it just me, or is the bookmark working better today?

Kannappan Sampath

Test: $x \sin x$.

Works better today for me as well, hence by $\text{strong}^{\text{strong}}$ induction principle, we conclude that book mark works well for everybody.

Skullpatrol

@robjohn "render MathJax" becomes un rendered sometimes... are you having that problem to Rob?

robjohn

20:08

It doesn't require me to rerun the bookmark :-)

not now

Kannappan Sampath

For me as well.

robjohn

$\frac{1}{x^{1+\epsilon}+x^{1-\epsilon}}$

Whee!

Skullpatrol

@robjohn Maybe it's EMI again!!!

Kannappan Sampath

@robjohn So, how do I define it then.

The above thingy failed to render! : (

Induction doesn't tell you anything as $n \to \infty$.

robjohn

@KannappanSampath what do you need to define?

Kannappan Sampath

20:12

A linear operator on $V$ is a linear transformation from $V$ to itself, OK?

robjohn

$\frac{1}{x^{1+\epsilon}+x^{1-\epsilon}}$

Kannappan Sampath

Failed to render even now. : (

Skullpatrol

renders and then becomes unrendered

ussually just after I send a message

it rerenders

robjohn

It was working for me, and then I refreshed and re-ran the bookmark, now it is failing :-(

Skullpatrol

welcome to the club

robjohn

20:15

@Skullpatrol It has been doing that for a long time, but for the last bit it was working the way it used to.

Skullpatrol

@robjohn Indeed.

Periodic nonfuntionality?

robjohn

@KannappanSampath Just saying that would be sufficient, I think.

Kannappan Sampath

OK. I'll edit that, Further comments, please.

Skullpatrol

Look, Brian is stuck in the basement with Asaf.

I shouldn't have listened to "Born Free" twice because now I can't get it out of my head :-(

robjohn

@Kannappan: Lemma 2.1 seems to be missing some equivalent conditions.

Kannappan Sampath

20:24

@robjohn Yes in fact a lot of places, it is in complete, I am sorry. I think I chose to give you at the wrong time. : (

robjohn

@KannappanSampath perhaps I am looking at it too critically.

The general formatting looks fine

Skullpatrol

@MattN Hi

Matt N.

Hi.

Skullpatrol

What's up?

Kannappan Sampath

@robjohn Not at all. I'd like to know that, but given some are incomplete, I should have asked for it later as I observed. But how do complete proofs look. A few, I guess.

Asaf created a new Gallery room called " The Bin". I wonder why? (Likely, to transfer all my messages there?)

Matt N.

20:34

What's the "topological countable product" of an ordinal?

Kannappan Sampath

@Matt Glad to tell you, I just broke 5k! :-)

Matt N.

Congratulations!

Kannappan Sampath

The recent answer has caught the fancy of many people, : P

Skullpatrol

@N3buchadnezzar Hi, Rob found a non-sensored link to a classic

@tb Hi

t.b.

@MattN The ordinal is equipped with the order topology and you take the product topology on that countable product.

N3buchadnezzar

20:37

@tb Ahoy thar

t.b.

Hi Skullpatrol

Hey N3 runs away from bear hug

@KannappanSampath congrats

Skullpatrol

encrypted-tbn2.google.com/…

robjohn

@tb: you've shown up too early; I'm still here :-D

Matt N.

@tb Product topology of what? There is only one set, $\omega_1 + 1$.

Kannappan Sampath

Hi!

@tb Thank you! : )

t.b.

20:39

@MattN you take the countable product of $(\omega_1 + 1)$...

Matt N.

@tb the countable product?

How many times? $\omega$?

t.b.

Let $X_1 = \omega_1 + 1$, $X_2 = X_1$, ... Then take $\prod_{n=1}^\infty X_n$.

@MattN doesn't matter. only the cardinality matters

Brian M. Scott

@Skullpatrol Sorry about that: I went to lie down for a little and never made it back to the computer last night. No, I’m not familiar with Brown, Dolciani, et al.

Skullpatrol

@BrianMScott That's Ok.

ymar

@KannappanSampath In the proof of 3.6 you seem to be implicitly saying that $0^0=\operatorname{id}.$ You might want to state this convention explicitly.

Kannappan Sampath

20:43

@ymar Hah! Nice, I did assume that. Thank you.

I'll add it in.

Matt N.

What an unproductive day. But at least I finally know why $z^n$. Or at least I think I do.

I still can't make sense of your hint though.

Skullpatrol

@BrianMScott Remember this?

Kannappan Sampath

@MattN No, One's mind can make an heaven out of hell and hell out of heaven! It was a productive but not as productive as you wanted it to be. : )

N3buchadnezzar

I spent yesterday trying to learn complex numbers, and learn about the Gamma function.

Matt N.

: )

N3buchadnezzar

20:46

Totally unrelated to anything to do with anything productive.

Matt N.

Heh. Disappeared again.

Brian M. Scott

You forget: I’m on a dial-up connection. It would take me half an hour to listen. But yes, I do remember the song, though I don’t think that I ever saw the movie.

N3buchadnezzar

Is there any theorem or proof that if a number can not be written as a perfect square, then the root of the number is irrational ?

Skullpatrol

@BrianMScott I didn't know you were on a dial-up connection.

Brian M. Scott

@Skullpatrol Ah, I thought that you were one of the folks around when it came up before. Never mind then.

Matt N.

20:48

What do you think of this:

robjohn

@BrianMScott Dial-Up!?

Brian M. Scott

@N3buchadnezzar Yes.

@robjohn ’Fraid so. Rather good, as dial-up goes, but still ...

Matt N.

Brian M. Scott

I’m going to have to switch this summer; too many sites are getting to be impossible.

Skullpatrol

@BrianMScott And it ties up your phone.

Brian M. Scott

20:50

@Skullpatrol That’s a feature, not a bug!

Skullpatrol

LOL

Brian M. Scott

But it actually doesn’t, because I’ve two lines. That dates back to when I was married, so that we could both be on-line at once.

Skullpatrol

Do you have a cell phone?

N3buchadnezzar

when you were married ? Sad storry =(

Brian M. Scott

As for the phone, I’m an unnatural creature: I don’t like ’em. And not only do I not have a cell phone, I’ve literally never used one.

Matt N.

20:52

I don't like them either. : )

anon

Amen.

N3buchadnezzar

63? Impressive

Skullpatrol

Wow, I thought I was the only one.

Brian M. Scott

@N3buchadnezzar It was quite a while ago now. But no, it’s not much fun to be told two weeks after your tenth anniversary that your wife has decided that she’d like to try being single.

N3buchadnezzar

Our professor last year just turned 83, or 87 I can not quite remember.

@BrianMScott Indeed.

Matt N.

20:53

@BrianMScott : /

Brian M. Scott

@N3buchadnezzar It was a good time for me to retire, both in terms of benefits and in terms of the way education is changing.

Matt N.

Well better than being dumped for someone else, no?

@TeddyBär
Thanks!

Skullpatrol

@BrianMScott Education is changing?

N3buchadnezzar

@BrianMScott Do you think education is changing for the better or worse?

@Skullpatrol Ofcourse it is changing, it always is.

Brian M. Scott

@MattN I’m not sure: being dumped for None of the Above is a bit disconcerting, too.

t.b.

20:55

@MattN why doesn't your $r$ depend on $x$ and $z$?

Matt N.

@BrianMScott Yes, I can see that. But nonetheless...

N3buchadnezzar

@BrianMScott Having to dump someone you still love is tough too.

Matt N.

@tb Yes.

t.b.

?

Matt N.

@tb Ok. It's wrong. Thanks for looking at it : )

Skullpatrol

20:57

@BrianMScott There are always plenty of more fish in the sea.

anon

Totally realistic, skull.

N3buchadnezzar

I hate when people say that

Brian M. Scott

@Skullpatrol At the college/university level it’s run increasingly as if it were a business with students as the product; this isn’t my notion of education. To me the good, small liberal arts college is as close to the ideal as an institution is likely to be able to get.

Kannappan Sampath

5151, @tb Shall we exchange our reps, numerologist? : )

Matt N.

@tb No actually. $z$ depends on $r$ and $x$. And $r$ is any old real.

@KannappanSampath : D

Transcript for

$Mathematics$ Mathematics