Mathematics - 2012-03-10 (page 2 of 3)

Matt N.

11:09

This book contains too many occurrences of "G*d" and "praise be to G*d"".

Asaf Karagila

Are you reading the bible?

Matt N.

Almost: Arabian Nights & Days.

t.b.

@MattN that's a bizarre thing to complain about...

Matt N.

@tb Possibly. But I'm allergic.

t.b.

Then stay away from books that are related to medieval Arabia :)

Matt N.

11:12

I know, I know... : )

But I have to at least read 1001 nights.

Jonas Teuwen

I have received Duistermaat's Distribution Theory book!! Thanks Springer.

anon

Wouldn't it be Allah or am I off-based here? Was it just translated to "God"?

Matt N.

I assume it was translated to the G-word.

Jonas Teuwen

Hi Luke, I am your father.

Matt N.

: )

t.b.

11:18

Are you guys trying to annoy Jasper?

anon

$\color{Silver}{\text{(removed ... but not really)}}$

Damn those mods are ninjas.

Matt N.

: D

t.b.

Looks a bit like an accidental peek over at The Bridge when you don't close the flagging window fast enough...

Matt N.

?

Jonas Teuwen

@MattN Sobolev spaces become even cooler when using distributions!

Matt N.

11:26

Thanks, I'll add this comment to the long list of comments I'll read and understand later : ) (I don't know what Sobolev spaces are)

Jonas Teuwen

You don't?

Matt N.

I don't.

Poor me. Now teddy thinks I'm bizarre.

Asaf Karagila

I don't like the extensive use of the word "enumerable" in this question.

@tb I need your Trusted help.

t.b.

11:42

@AsafKaragila can't help...

Asaf Karagila

You were the first?

t.b.

Looks like it

Asaf Karagila

:-P

@anon @robjohn Maybe you guys start answering more and get to 20k already?? :-)

anon

lazy

robjohn

@AsafKaragila I am trying... I have been working on a hard problem.

@AsafKaragila but as with all the problems I work a long time on, they bear little fruit.

Asaf Karagila

11:45

@anon @robjohn Well, to both of you: LHF are easy, simple and quick.

robjohn

@AsafKaragila yeah, but where's the fun in that? :-p

Asaf Karagila

Well, you could delete answers. That's a lot of fun!

anon

I already go for lhf, with an (effort-to-type):(forecasted-upvote) ratio calculation. I just haven't been on SE that much the past weekish.

More specifically, I've been paying attention to other tabs instead of MSE.

Asaf Karagila

Well then, quit your real life already.

anon

Done.

Asaf Karagila

11:48

@anon Pornography?

anon

No, that wouldn't take that much time!

gag/chan/blogs/newsarticles/pdfs...

Oh, and Zelda and stupid family stuff.

Asaf Karagila

Well, cut the gag/chan out and you've freed about 99.999999% of your time already.

anon

You know what. Maybe sleeping during the day is making me miss lhf opportunities. Being 6am, I will attempt sleep now.

Asaf Karagila

And miss more? :-)

anon

I went to zz 9am yesterday. I figure it would be better to push it back instead of forwards.

Asaf Karagila

12:00

Ah. Then yes. Go now!

robjohn

12:26

I've made the argument in this answer more rigorous (and simpler). I wonder if the downvoter will revisit their downvote.

ymar

Hi. I have a question about where to ask my question and if I should ask it at all. May I ask it? (The first one.)

robjohn

Hmm... It seems that another downvote has been cast. I wonder why?

t.b.

@ymar sure, shoot!

Kannappan Sampath

@robjohn I upvoted it long ago once. I feel sad that an answer that requires thought has been downvoted.

t.b.

@robjohn I think it was cast before your edit

Kannappan Sampath

12:29

Spam, Sure it is.

Or probably a NARQ

ymar

I asked myself a question about something I have really no idea about. The question is this. Let $f$ be a function from the class of all semigroups with zero to the class of cardinal numbers which is defined by $f(S)$ is the "number" of all isomorphism classes of rings for which $S$ is the multiplicative structure. What Is there an upper bound to the values of $f$?

Asaf Karagila

Awesome. It seems that tomorrow is the first day of Armageddon, the end of the world. I am going to get that real-analysis specialist badge.

ymar

I'm just curious. I don't think I will understand the proof if someone gives it to me, but I'd like to know the answer and I think it's an interesting question. But I'm not sure if I should ask it. And if I should, then whether I should ask it here or on MO.

robjohn

@tb yeah, I just noticed that. I didn't see it until I refreshed after the edit.

Asaf Karagila

@ymar I'm not sure how you define that function. It seems to me that there is no upper bound?

ymar

12:35

@AsafKaragila What's unclear in the definition?

robjohn

@KannappanSampath Well, in defense of their downvotes, my first argument was a bit lacking in rigor. The newer answer is much better, but still follows the same idea.

Asaf Karagila

@ymar It would seem to me that you're asking if there is a bound on how many rings $R$ can have $R^\times\cong S$?

ymar

Yes.

But up to isomorphism.

Asaf Karagila

Of course. Otherwise it's not well-defined to begin with.

Jeff

Hey folks.

I'm just stopping in quick. I can't find a way to 'favorite', or save, a question. How do you do that?

ymar

12:39

@AsafKaragila Right. So why is there no upper bound? Is it simple?

Asaf Karagila

@ymar Let's see.

Kannappan Sampath

@Jeff The star beside a question allows you to make that question your favourite. Click on that to toggle...

t.b.

@Jeff there's a star right underneath the voting arrows on a question. Click it. You can then find your starred messages in the "favorites" tab in your profile.

Asaf Karagila

Let $S$ be a semi-group, and $R$ be a ring such that $R^\times\cong S$. Let $\kappa$ be a cardinal number, and define $R_\kappa$ as the direct sum of $R$ with $\kappa$ defined as the rng in which every element is nilpotent of order two...

t.b.

(otherwise there's always the bookmark feature of your browser)

Jeff

12:41

@Kan, @tb wow, and I couldn't find it.

is there a universal favourites where I can see my starred from several sites?

t.b.

@Jeff here is yours

Asaf Karagila

@ymar Also I forgot to say that all the elements are orthogonal in the sense that any multiplication is $0$. So no element from $\kappa$ intrudes $S$. I'm not sure why $S+\alpha$ for $\alpha\in\kappa$ wouldn't be in the multiplicative semigroup.

Jeff

@tb Ha, I already had some starred. Can you tell me briefly how to navigate there (I already have too many saved bookmarks)? Can I get there from my math.stackexchange profile?

t.b.

@Jeff to get there, go to your user profile on any site, click on "network profile" (upper right corner) and switch to the favorites tab there.

Asaf Karagila

I don't know. It's a question on the border between MO and MSE. Either site would be a good option to post that in. I can only recommend that you add alongside the definition the explanation I gave to the function (that is, what I asked to verify I understood the definition of the function).

Jeff

12:46

@tb perfect. muchos gracias (ditto you @Kan)

ymar

@AsafKaragila I'm not sure I understand how $R_{\kappa}$ is defined. Is it a direct sum of $\kappa$ copies of $R$?

robjohn

Oh darn, if it weren't for the downvote (which will hopefully go away) I would have a palindromic reputation.

Asaf Karagila

@ymar No, it's the direct sum of $R$ with $\kappa$ when $\kappa$ is chosen to have any abelian group structure and the multiplication of any two elements is zero. (Note that this would be a RNG no a RING)

Kannappan Sampath

@Jeff @tb took all the pain of typing in a lot of helpful directions for you. In any case, you're most welcome. : )

Jeff

@rob where's your downvote? I'll go counter it with an upvote! :D

t.b.

12:48

@Jeff since you seem to like proofs of the Pythagorean theorem: here's my favorite (Gerry Myerson calls it the "one line proof").

You see three houses. Since the roofs of the two small houses add up to the roof of the third house, the "bodies" of the houses must add up to the one of the third one by similarity.

robjohn

@Jeff This answer. However, it's the -2 from the downvote that kills the palindrome :-) Thanks anyway

Jeff

i replied too soon (before reading rest of your comment)

Asaf Karagila

@robjohn I don't see no downvote. Possibly was retracted after the edit...

robjohn

@AsafKaragila It was just removed. I guess whoever it was, is now satisfied :-)

t.b.

@robjohn I've just finished reading it. Very nice work!

Jeff

12:51

@robjohn i glanced at that question. if i upvote you get a different amount of rep than removing a downvote.

t.b.

@robjohn It took me a moment to convince myself of the reduction steps :)

robjohn

@Jeff yes, a downvote takes 2 rep from you an upvote gives 10 rep to you

Jeff

@tb the red and purple 'houses' are upside down. right? the 'houses' are squares with a triangle roof. yes?

t.b.

@Jeff exactly

Jeff

@robjohn i won't do it then.

robjohn

12:52

@tb You mean the assumption that $f''(x)\le0$?

@Jeff don 't worry about the palindrome :-)

Jeff

@tb red's 'right' roof and blue's 'left' roof equal green's 'ceiling'

Kannappan Sampath

Call for votes.

ymar

@AsafKaragila Ah, OK. But I don't understand how this could prove that there is no bound. You take $R\oplus\kappa$ and $\operatorname{card}(R\oplus\kappa)=\operatorname{card}(R)\times \kappa.$ The multiplicative semigroup of a ring (or rng) has the same underlying set as the ring, so in particular has the same cardinality...

t.b.

@Jeff The important point is of course that the houses are similar (I forgot to mention that explicitly). Roof = the entire ceiling (triangle). Body = square.

Asaf Karagila

@ymar Ah. I did not know that. :-)

I actually think that a model theorist could solve that question in minutes.

Jeff

12:55

@tb ok. lemme see how we show that the roof's are similar: blue and green roof share an angle (at the right)...

@tb and both have a right-angle... done

Skullpatrol

@tb Are the people who live in these houses similar also?

Asaf Karagila

Something like the Morley theorem or the Vaught conjecture about number of models in a generalized settings - if has been proved for the relevant theories...

t.b.

@robjohn Both reductions took me a small moment, $f \geq 0$ and $f'' \leq 0$. The explanations are fine, I'm just slow :)

Jeff

@tb red and green roofs also share angle and both have a right triangle... done with that

@Skullpatrol they're from different planets :D

robjohn

@AsafKaragila I see Morley and think of the geometric theorem about trisected angles.

Jeff

12:57

all three 'roofs' are similar. red roof's area plus blue roof's area = green roof's area, too.

Asaf Karagila

@robjohn Possibly a different Morley... :-)

robjohn

Morley's trisector theorem

In plane geometry, Morley's trisector theorem states that in any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle, called the Morley triangle. The theorem was discovered in 1899 by Anglo-American mathematician Frank Morley. It has various generalizations; in particular, if all of the trisectors are intersected, one obtains four other equilateral triangles. Proofs There are many proofs of Morley's theorem, some of which are very technical. Several early proofs were based on delicate trigonometric calculations. The first published...

Skullpatrol

@Jeff Where does the similarity end then?

ymar

@AsafKaragila I think I must be misunderstanding something. I think it's clear that the multiplicative semigroup of a ring has the same underlying set as the ring. What I mean by the multiplicative semigroup of a ring $(R,+,\times)$ is $(R,\times).$ Does it mean something else?

t.b.

@Jeff That's something to remember: if you draw the height of a right-angled triangle at the right angle, you cut your triangle into two smaller similar ones.

Jeff

12:58

@tb ok. i think i'm done (yes?). if the triangles are similar (and all squares are, by def., similar) an if the area of red+blue=green, then the area of the red square+blue square = green square

t.b.

That's it, yes!

Jeff

@Skullpatrol the aliens are treated by veterinarians :D

Asaf Karagila

@ymar Ohhhhhhhhhhhhhhhhhh... I was thinking on the multiplicative group.

t.b.

@AsafKaragila that's why ymar asked about "semigroups" :)

ymar

@AsafKaragila :-D That was a moment of total confusion for me!

Jeff

12:59

@tb this one lost me. if i draw the height of a right-triangle at the right-angle, then isn't the height one of the legs?

Asaf Karagila

I am tired, it seems that I am really tired.

I think I'll take a nap.

Jeff

@Skullpatrol i have no real idea what i'm saying.

robjohn

@AsafKaragila you're faking it

Jeff

@AsafKaragila you've been here long enough!

actually, most of you in here have been here long enough!

Asaf Karagila

@robjohn The Koenig need not fake things. He controls history!

t.b.

13:00

@Jeff No, I mean the perpendicular to the hypotenuse through the vertex of the right angle.

robjohn

@AsafKaragila rewrites it more like :-p

Asaf Karagila

@robjohn Tomato potato.

robjohn

@tb are we talking about similar triangles?

Asaf Karagila

Well, I'll see you guys later.

robjohn

@AsafKaragila indeed :-)

Jeff

13:01

@tb oh. the, uh <mental block!>.... altitude(?)

robjohn

@AsafKaragila l8r dood

t.b.

@Jeff I think so (sorry I never learned the English translations of the plane geometry terms properly).

Jeff

@tb that's ok, i forgot people on the web might not speak english as a first language (i forgot where you're from?).

@tb my favorite visual pyth proof is the very simple to understand and explain one with one big square and four similar triangles along each edge of the square.

t.b.

@Jeff Hint: Chocolate, cheese and banks.

@Jeff The one anon posted/explained?

Asaf Karagila

@tb: LOL! The AC question guy got confused by my AC/AD answer. :-D

Skullpatrol

13:05

@tb Could you do it in one line now please :-)

t.b.

@Skullpatrol I added one line to the Pythagoras picture.

Jeff

@tb switzerland? and i don't remember who posted it.

t.b.

@AsafKaragila was the choice not uniquely determined?

Jeff

@skull you know you don't seem to have a math profile

t.b.

@Jeff yes, Switzerland. Do you mean this proof below?

Skullpatrol

13:08

@tb Oh! that kind of line. I thought you meant one line of text.

t.b.

Jeff

haha... has anyone ever noticed the symbol math.se displays when you navigate to non-existent profile: math.stackexchange.com/users/5030

Asaf Karagila

@tb Apparently. This is why I hate "explain to me intuitively this concept which is about infinitely long processes which cannot be grasped as physical phenomenons". Alas, I am somehow drawn to answering those questions anyway. Oh well, time to write a long long answer.

Kannappan Sampath

This time I'll push harder for vote. This question needs to be closed. :-)

Jeff

@tb i think the other one was a little different. it had the inside square labeled with sides of length $c$ and the sides of the big outer sqare were length $a+b$... now i forget <darnit>

@kan what are you asking for? you want us to vote up or down? and why?

Skullpatrol

13:10

@Jeff That is the symbol for "Skullpatrol"

t.b.

@Skullpatrol One line: similar houses. QED

Kannappan Sampath

@Jeff I am asking for votes to close that question. You'd understand why it has to be closed if you read that.

t.b.

@Jeff oh, that one, then:

Jeff

@Skullpatrol i thought it was 'does not exist' symbol :D

@tb haha!

Skullpatrol

@tb With or without similar people?

ymar

13:11

Why I asked my question is that I know that MO is for researchers. I'm not a researcher and I haven't researched my question because I wouldn't know how. Is it still OK to ask it there?

Jeff

@KannappanSampath i did just read it. i don't understand the question. maybe write a comment to the OP saying how to improve it

Kannappan Sampath

@Jeff I tend to believe it is not a real question.

robjohn

@tb In that diagram, it's hard to see what's $a$ and what's $b$

Jeff

@tb yes. side of big square: $a+b$. area of four triangles, $2ab$, plus the area of the small square, $c^2$ = the area of big square: $(a+b)^2$. q.e.d.

t.b.

@ymar If you're unsure, it might be better to ask here first (it probably isn't urgent, is it?). After a few days or a week of no responses you can then go to MO and say: I asked this on SE and I know it's a borderline question but I didn't get good responses, so I decided to try my luck here.

Jeff

13:14

@KannappanSampath i would guess he's a student who misunderstood the question and so he paraphrased it wrong on there.

ymar

@tb No, of course it's not urgent. Thank you, I'll do that.

Jeff

@KannappanSampath and what the heck do $p$ and $t$ have to do with what needs to be proven?

Kannappan Sampath

@tb I wanted to tell you I did section 12 and 13 and some exercises of Munkres. Topology is fun. :-D

robjohn

@KannappanSampath closed.

t.b.

@robjohn indeed. I just copied it from the first Google hit. (cut the knot)

Kannappan Sampath

13:16

@robjohn Thanks a lot. :-)

t.b.

@KannappanSampath It surely is :)

Kannappan Sampath

@Jeff These are what are classified as NARQ-Not a real question.

t.b.

@KannappanSampath It looked like a lot what I listed, but I don't think it's that much. A few days of work for someone like you :)

Jeff

@tb all those other proofs seem so complicated in comparison. they may be fun, but they should be kept out of a 9th grade geometry class (IMO).

@KannappanSampath oh. but i don't like voting down. i prefer to vote up for the Qs I like. so i'm gonna pass anyway.

Kannappan Sampath

Probably time to break 5k. I should swing into action and answer some questions.

Jeff

13:17

@KannappanSampath sorry

@KannappanSampath do you know expected value calculations? I'll post a question for ya

t.b.

@Jeff Agreed. The nice thing about the square with four triangles one is that it really doesn't need any sophistication. The downside is that it needs some computation, hence it isn't really geometric.

ymar

Thanks for your help, I'll be going now. Bye.

t.b.

Bye ymar, have a nice weekend

Kannappan Sampath

@Jeff Yes, I do. Are you given random variables and you want expectation....? Anyway, post it there are good answerers for probability and statistics.

@tb I think it would take about 3 weeks for me, given the present pace in which I do exercises. : )

Jeff

@tb by 'it needs some computation' you mean how you have to distribute the $(a+b)^2$ and then cancel its middle term with the other side? i think that's good for a 9th grade geo class (i am tutoring a 9th grade geo student is why i care). but i see your point about it not being an entirely geometric proof.

t.b.

13:21

@KannappanSampath well, 20 days are not much more than a few. Oh, you're young, I forgot.

Jeff

@KannappanSampath well, it's not a straightforward expected val quesiton. i will definitely post it some day as it is the question which motivated me to choose a math major and i have only figured out half the answer (it's really two questions).

@KannappanSampath i'll go start it now.

Kannappan Sampath

@tb Oh, I don't quite follow the last half of the comment, though.

t.b.

@Jeff Oh, yes, definitely, and by doing both variations of the proof, you remind them of the formulas for $(a \pm b)^2$... I think it would also be the one I'd present to ninth graders.

@KannappanSampath never mind. I was being silly.

Matt N.

@KannappanSampath He's saying that 20 days feel like one day for an old person but very long for a young person.

: )

While considering himself old.

t.b.

What Matt said (before the smiley) :)

Matt N.

13:26

: )

Jeff

@tb so we agree to agree. :D

Kannappan Sampath

@MattN I think I'll have to buy this argument, definitely. : )

Matt N.

@KannappanSampath You do. It will turn out to be true in a few years, like 10 or so. : )

I managed to spend another 2 days without doing any thinking/work : /

t.b.

What a silly answer

Kannappan Sampath

tmi \me thinks.

Matt N.

13:33

^I missed that.

Kannappan Sampath

@MattN Oh, well, that was really long.

Jeff

@kan still working on that question. do you know the game backgammon?

Matt N.

@tb Just to prove my point. : )

@KannappanSampath Ok, never mind : ) I was searching the transcript and hence not looking.

Kannappan Sampath

@MattN I'll share this some other time as well. : D

@Jeff No, :/

Matt N.

@KannappanSampath Ok : )

Jeff

13:37

@KannappanSampath i explain the necessary parts in the question.

@kan basically, each game is played for one point (let's say it's you vs me)...

Kannappan Sampath

@Jeff Sure.

@Jeff OK.

Jeff

@kan i can offer to 'double' the game and you can offer to accept (play for 2 points), or not (you lose one point).

robjohn

@tb: you think you're old...

Robert Sköld

how would i solve this to "z=...": x*1/z=y

Jeff

@kan the equity necessary to accept a cube is .25 (i'm trying to remember how you get that now)...

Kannappan Sampath

13:38

OK.

Jeff

@kan the big question is: what equity is necessary to offer to double

robjohn

:3771082 I don't know for sure, but I'd bet I'm older.

Matt N.

I think you are.

robjohn

@MattN Thanks -_-

Matt N.

@robjohn : D That's not what I meant. But I know you worked for Apple some ages ago.

Kannappan Sampath

13:40

@Jeff I'm not sure if I'll be able to answer this question.

robjohn

@MattN And I have the fossils to prove it!

Kannappan Sampath

I think I am going AFK. I'll be back later.

Jeff

@KannappanSampath oh. well then we'll both learn. and i've been meaning to learn how to answer this Q for a while.

ok. i'm going back to entering it. i figured out the math for the first part.

Matt N.

@robjohn Wow. : ) Old macs, I assume?

robjohn

@MattN The Mac first came out (128K Mac) while I was in grad school.

Matt N.

13:42

I need to make some coffee. (Pretext to procrastinate the point in time where I actually might start working on the seminar thing.)

Skullpatrol

@tb I strongly agree with both of you about the "square with four triangles." May I also add that if you make a paper model of it, and cut the four triangles out and move them around even a fifth grader followed the argument when I presented it to him.

Matt N.

@robjohn I have a feeling I wasn't even born then. What year was that?

Jeff

@Skullpatrol that is great to know. cuz i always wondered about that. i'm gonna try that with my niece and nephew next time i see 'em (they 5yo)

<-- is happy :3771112 said that

Robert Sköld

i have this very simple algebraic formula: 1/z = x
how do i solve for z? z=x*1 seems incorrect...

t.b.

@MattN It seems like you're taking some things too literally :)

Jeff

13:46

@RobertSköld if you multiply both sides by $z$, the left side will cancel and leave you with a $1$. what would you have on the right side?

Matt N.

@tb : )

@robjohn Oh. I was born then!

Robert Sköld

@Jeff hmmm that would become 1=x*z so.. x/1=z?

Jeff

@RobertSköld yes!

Matt N.

@tb I take everything 100% literally and seriously. : )

Robert Sköld

@Jeff thanks :)

Asaf Karagila

13:49

Right. I finished the answer.

Jeff

@RobertSköld vg

Skullpatrol

No

Asaf Karagila

Long long answer. but considering the fact it took me 30 minutes... not bad!

Matt N.

: )

Asaf Karagila

Actually for comments.

This is too long for me to review on my own.

t.b.

13:51

@MattN I rest my case :)

Skullpatrol

@RobertSköld x/1=z means x=z

Matt N.

@AsafKaragila tl;dr

Asaf Karagila

@MattN evil eye.

Matt N.

: D

Robert Sköld

@Skullpatrol arghh

Matt N.

13:52

Must. Resist. Temptation. To write "tl;dr" as comment on Asaf's answer.

Asaf Karagila

@robjohn: You might enjoy that answer, actually. I did my best to explain the axiom of choice and the Russell's saying about socks without getting into set theory.

@tb Can you help me find the year in which Fraenkel proved the socks thing? I have several references all in French and German, though.

Skullpatrol

@RobertSköld Try again.

t.b.

@AsafKaragila give them to me, I'll take a look

Asaf Karagila

Right. I'll start digging.

Jeff

@RobertSköld yw

how do you make multline equation in asking a question?

Robert Sköld

13:54

@Skullpatrol @Jeff ok, got it now. it'll be z=1/w

@Jeff shift return

Jeff

@RobertSköld where did $w$ come from?

Asaf Karagila

@tb Oh wait! Maybe Herrlich has the answer!

Matt N.

@Jeff \begin{align} & blah \\ & blah \end{align} , maybe?

Robert Sköld

@Jeff haha, sorry, i meant x. w is the one i used in the code

Jeff

@RobertSköld i gave you a wrong answer (which @Skull pointed out). Sorry, I misread. Do you understand what was wrong?

Matt N.

13:55

: O

Jeff

@Matt two backslashes? i just tried that

Matt N.

I had no idea you didn't need to put $s for the plugin to compile the latex!

Robert Sköld

@Jeff yeah, doing the middlestep 1 = z*x made it make sense to me

Matt N.

@Jeff Yes, two backslashes.

Jeff

@matt nevermind, it worked (i musta made a typo the first time)

@RobertSköld right. and if $1/z=x$, then $1/x=z$, too (which is what i thought you typed)

@kan, you still here?

@kan afk? well, i'm gonna post soon anyway (i bothered to type it up)

t.b.

14:01

@AsafKaragila 1919. The paragraph starting at the bottom of page 125 here

Skullpatrol

@RobertSköld Once you get to the equation 1=x*z, you have two numbers, x and z, whose product is 1. This is, by definition, a reciprocal relationship. Do you follow?

Asaf Karagila

@tb Herrlich's book specifies the quotation from 1907, and used "boots" instead of "socks".

Jeff

@RobertSköld ok, good to know i didn't screw you up with my wrong answer. i always tell students to stop me ASAP if they think i did something wrong! :D

Robert Sköld

@Skullpatrol yeah i follow, and replacing x with 5 for instance made it quite obvious :)

t.b.

@AsafKaragila Hm. The passage I gave you has boots versus socks.

But the exact phrase is not in there.

Jeff

14:04

@RobertSköld now you owe us an answer to a math question! :D

@RobertSköld (kidding)

Robert Sköld

@Jeff hehe, that's ok, the results didn't make sense so i came back :P

haha

Asaf Karagila

B. Russell. On some difficulties in the theory of transfinite numbers and
order types. Proc. London Math. Soc., 4:29–53, 1907.

Robert Sköld

19

(is my answer)

Asaf Karagila

From Herrlich's bibliography list.

Skullpatrol

@RobertSköld And replace z with 1/5 will give you the true statement 5*(1/5)=1.

Robert Sköld

14:08

@Skullpatrol i started from y/(y*z)=x which i refactored into 1/z=x and then i got stuck. long time no algebra for me :P

@Jeff hehe, you can come up with the question for it ;)

t.b.

@AsafKaragila yes. I still think that my passage is a better reference for the paraphrased quotation:

> In our case it can be done with the boots, but not with the socks, except by some very artificial device. The reason for the difference is this: Among boots we can distinguish right and left, and therefore we can make a selection of one out of each pair, namely, we can choose all the right boots or all the left boots;

> but with socks no such principle of selection suggests itself, and we cannot be sure, unless we assume the multiplicative axiom, that there is any class consisting of one sock out of each pair. Hence the problem.

Asaf Karagila

@tb Well, I'm only looking for a year. I wanted to know whether or not Russell made the comment in retrospect of Fraenkel's second model; or was Fraenkel driven to make Russell's idea formal.

t.b.

I see

Jeff

@KannappanSampath K, I posted my question (although I think you are AFK).

if anyone knows expected value calcs, have at it: math.stackexchange.com/questions/118566/…

now that i took so much time here. i'm going to go do my work. see y'all

t.b.

see you

Matt N.

14:22

@Jeff Bye.

@tb Are you not going to catch some sun today?

(Not that mind!)

robjohn

@AsafKaragila Excellent answer! (+1)

Asaf Karagila

14:37

@robjohn Thanks!

Matt N.

(^That should've been "Not that I mind!")

robjohn

@MattN I was just about to ask, "which mind?"

Asaf Karagila

Heh. My reputation is 2-5-three-5s.

On the other hand, in 30,000 points I'll have a nice clean 55,555. Which is five-five!

t.b.

@MattN Sorry was on the phone. Luckily I did decide to catch some sun this morning...

robjohn

@AsafKaragila that's noteworthy :-)

Asaf Karagila

14:51

@robjohn Which is why I made a note of it here. :-)

robjohn

@tb the sun is just about rising here.

It is probably kind of chilly outside.

Asaf Karagila

And you're "waiting for the sun"?

Oh hi.I didn't take my nap!!

Kannappan Sampath

@AsafKaragila I have a conceptual point about AC to clarify. Are you around?

robjohn

@AsafKaragila I am taking Lilly out for a walk in half an hour (we go an hour later on Saturday)

@KannappanSampath it can be transmitted long distances more easily than DC.

Kannappan Sampath

@robjohn Yes. True. : ) (The loss in stepping up and down is also lesser. )

Asaf Karagila

14:59

@KannappanSampath Can it wait an hour or so?

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