Mathematics - 2012-01-29 (page 2 of 3)

3:00 PM

@BenjaminLim I know, but I shouldn't make assumptions, especially if the assumption is about someone who might enjoy a disagreement.

Benjamin Lim

@robjohn spot on.

@Srivatsan Have you seen the answer that Alex B. posted on a question of Eric Naslund on meta?

robjohn

@Srivatsan there might be a small group of people (or one person with different emails) joining together to vote each other?

Srivatsan

@BenjaminLim - which one?

Kannappan Sampath

What are we talking about, those upvotes, and .....?

Benjamin Lim

@Srivatsan here: meta.math.stackexchange.com/a/1722/5783

robjohn

3:03 PM

:-)

Benjamin Lim

I noticed the thread on BD. That has ruffled a few feathers....

robjohn

@BenjaminLim That is for sure!

Srivatsan

@BenjaminLim I am not losing sleep. After all, I didn't even post/comment in that thread. But my question still stands: has there been a recent spike in the number of such questions? Or no?

Benjamin Lim

I am not sure, but yeah I have noticed that people have posted some easy questions receiving a lot of upvotes. For example one question I answered on why the quotient of a ring by a maximal ideal is a field.

Btw guys Novak Djokovic has just won the 2012 Australian Open :D

@Srivatsan I don't get why typing calculating $99^{100} - 100^{99}$ and posting it as an answer gets 21 upvotes....

robjohn

@BenjaminLim That one in particular really makes me wonder.

Benjamin Lim

3:08 PM

@robjohn Me too an answer that requires very little thought put in gets 21 upvotes....

robjohn

At least JavaMan's answer is a real answer (other than claiming $\left(1+\frac{1}{99}\right)^{99}<3$)

Srivatsan

@BenjaminLim I am yet to upvote it, but I will upvote that answer. The reason why I upvote it is simple. The answer makes a fair point that there is nothing to be ashamed of in computing.

Benjamin Lim

@robjohn I agree with you.

Srivatsan

@BenjaminLim In fact, I believe that the answer has a lot of thought put in.

Benjamin Lim

@Srivatsan Why do you say that?

Srivatsan

3:11 PM

Ok, that is a contentious thing to say. But it is true that that answer was the most thought-provoking for me.

"Nice reminder that the world has changed since I first did mathematics."

robjohn

@robjohn actually he claimed 99 not 3, but same difference :-)

Jonas Teuwen

I have Glen! Glenkinchie.

Srivatsan

"I suppose it doesn't use logarithms..."
"technically correct is the best kind of correct."

I might be over-romanticising this, but this just displays the power of brute-force when it works.

Matt N.

@BenjaminLim Wot? Do we know who Bob is? It can't possibly be BD, he wouldn't be so stupid as to do such an obvious thing, or would he?

Benjamin Lim

@Srivatsan You should see the comments that a certain user called "Sniper Zaitsev" leaves. In one thread (which I believe is now deleted) he asks Arturo to "go make career as insulter".

@MattN What do you mean?

Srivatsan

3:14 PM

@MattN "It is not anonymous down voting, we know it is you." (see Eric's comment that begins thus)

robjohn

@Srivatsan I don't know; when I first saw that answer, I just shook my head. I do not get the same romance from it.

Matt N.

@BenjaminLim In that thread he calls the down voter Bob. BD's answers are usually "sophisticated".

Srivatsan

@robjohn Note that I quoted not the answer, but the comments. :=)

Matt N.

@Srivatsan Oh : D

I had only read the question and one of the answers.

Benjamin Lim

@MattN I notice you have account on the cooking.SE site?

Matt N.

3:16 PM

@BenjaminLim Puzzle solved. : )

Benjamin Lim

huh?

Matt N.

@BenjaminLim Yes but I don't use it.

Benjamin Lim

@MattN I absolutely love cooking.

robjohn

@Srivatsan Ah, so the last line refers only to the comments and not back to the answer. Okay. :-)

Matt N.

@BenjaminLim The puzzle of the "anonymous" down voter.

Benjamin Lim

3:18 PM

@MattN Ok

Matt N.

@BenjaminLim I think it's too much work : )

Benjamin Lim

@MattN I live in halls so I have to cook my own dinner everyday. I really love it especially the thought that I can cook whatever I want....

Srivatsan

@robjohn (a) Well, I think of it like this: when we are getting more and more technologically advanced, we are pushing the boundaries of what can be brute-forced. So just because my ancestors needed some fancy algorithm/estimate/inequality for doing X, I need not. In fact, after some time, these fancy techniques -however charming in their own right- might appear obscure to me.

robjohn

anonymous downvoters should be shot until we find the one responsible.

Jonas Teuwen

"Let's kill the Belgian just to be sure!"

robjohn

3:21 PM

@Srivatsan indeed, but this works only for certain kinds of problems. If you need to show that there are infinitely many primes, brute force might fail you.

Benjamin Lim

Good nights guys, I'm off. Really late now here in Australia.

Matt N.

@BenjaminLim Good night!

Benjamin Lim

bye :D

robjohn

@BenjaminLim Good night, I never knew you were down under :-)

Srivatsan

@robjohn So in some sense, you are doing the fancy stuff a disfavor by applying it to a "simple" problem. I take this to be the underlying message of the answer.

Benjamin Lim

3:22 PM

@robjohn You're here down under too? On my profile it says my location :D

Matt N.

(Dearie me, I didn't know "Bob" was that twisted.)

robjohn

@BenjaminLim Really? no I am in West Hill, California facepalm

Benjamin Lim

Ah peace. Anyway may you all be happy. Bye!

Srivatsan

@robjohn Sure, but that is not the question posed. :=) If you want to weed out such brute-force answers, then ask the correct question!

Kannappan Sampath

@Dylan Hi!

robjohn

3:25 PM

@Srivatsan Well, that question can be brute forced, so there is no way to say that it can't and that only pure methods work

Srivatsan

@robjohn Well: "Show that if $3 \leq a \leq b$, then $a^b > b^a$." sounds like a reasonable generalisation for which your method works and the brute-force doesn't.

Jacopo Notarstefano

/me is an occasional anonymous downvoter

Srivatsan

Or even: "$a^{a+1} > (a+1)^a$ for $a \geq 3$."

Matt N.

@JacopoNotarstefano On answers that got accepted instead of yours?

Jacopo Notarstefano

Ah, no, never done that

I downvoted two "puzzle-y" questions that went "craft the most numbers from these numbers using +,-,*,/"

Srivatsan

3:28 PM

@MattN I know what I will do next time to thank the OP for accepting my answer: downvote the question.

robjohn

@Srivatsan okay. That's true. I'm sorry, I am typing and getting ready to go to the grocery store. I should just stop typing and go :-)

Matt N.

@Srivatsan For not accepting your answer you mean?

@JacopoNotarstefano Not even in kindergarten?

Srivatsan

@robjohn No problem. I think I overdid it anyway. (Actually, in my head, I am arguing both for and against that answer. :=))

@MattN No, for accepting my answer. It's anonymous anyway, right? (just kidding)

Matt N.

@Srivatsan Right. : )

Jacopo Notarstefano

@MattN Wait. What?

Matt N.

3:31 PM

@JacopoNotarstefano Downvoting answers that got accepted instead of yours.

Srivatsan

@MattN Oh my. If only kindergarten kids could "downvote" stuff anonymously.

Jacopo Notarstefano

Can't say. I barely remember my kindergarten days.

Henning Makholm

@Srivatsan These days there's probably an app for that.

Matt N.

@Srivatsan I guess they do it non-anonymously, like in this case mentioned on meta.

Srivatsan

@MattN I have a feeling nobody will pass any exams in that case.

Matt N.

3:37 PM

Never mind : )

Srivatsan

:3174761 I thought the kids can upvote and downvote each other. I didn't think that the teacher was grading them.

Matt N.

@Srivatsan Oh I see! You just invented a new school system. I see doom.

@RajeshD Click on "delete" in the drop down menu.

Rajesh D

Hahha !!

Asaf Karagila

4:41 PM

@Martin, @Srivatsan: There's a new algebra question which I'm not really sure how to retag.

Martin Sleziak

@AsafKaragila Yes, I've noticed that question too.

Srivatsan

Thanks, I will look at it in a little while. (In case you guys don't tag it already)

Asaf Karagila

It had [elementary-set-theory] which was wrong, I'm not sure about the third tag.

Martin Sleziak

I'm not sure what the OP is asking.

Asaf Karagila

I'm not sure about that either.

Srivatsan

4:44 PM

(Asaf: Ah, of course, I didn't intend the Heine-Borel theorem comment to be addressed to you. I forgot to ping the OP in my first attempt.)

Asaf Karagila

:-P

Jonas Teuwen

Hmm, learn Mathematica. If I have a $3$-vector $i_p$ and I have prod1 = Cross[Times, ip, ip] (the tensor product) and then I integrate prod1, how do I apply PolynomialReduce to the whole of the integrated prod1?

Asaf Karagila

@Martin: Your answer to Porton's question implies that you are nominating him for Abel prize?

Martin Sleziak

@AsafKaragila It does not at all.

Henning Makholm

@Martin, isn't your e_0 the wrong way around?

Srivatsan

4:49 PM

@MartinSleziak Would you reconsider if I nominate Porton for a nomination for the Abel prize?

Henning Makholm

(And yes, an Abel-nomination self-nominee really ought to be able to do this kind of things himself. Have voted to close as too localized).

Martin Sleziak

@HenningMakholm Thanks, I've corrected it.

Actually, I thought that giving the answer might be faster way to deal with that question than closing. (I did not vote to close.)

Asaf Karagila

I voted to close.

Oh look, one more and we're done with it.

Srivatsan

done.

Henning Makholm

Unlike Didier, I don't sense any hidden agenda from Porton. I just see the Dunning-Kruger effect dialed up to 11.

2

robjohn

5:02 PM

@HenningMakholm I should pay more attention; I haven't noticed Didier's hidden agenda. Perhaps because it is hidde. :-)

Henning Makholm

@robjohn He claims to sense one here

N3buchadnezzar

What is going on here with all the plusses ?

1

Q: Which of the numbers $99^{100}$ and $100^{99}$ is the larger one?

Which of the numbers $99^{100}$ & $100^{99}$ is the larger? Solve without using logarithms.

robjohn

@HenningMakholm Didier senses one. Are you saying that it takes one to know one? :-)

@N3buchadnezzar I was asking that earlier today.

N3buchadnezzar

@robjohn Perhaps it was linked to at some popular website, perhaps a blog. Who knows.

robjohn

@N3buchadnezzar That's a possibility.

Srivatsan

5:14 PM

@rob: Can you read my comment here and check if it is rude / condescending?

N3buchadnezzar

I discovered something amusing today.

Henning Makholm

@N3buchadnezzar It would be nice if the software tracked the most common off-site Referer for questions with high number of views. (Even the most common among the last 40 page loads, for example, would be useful).

N3buchadnezzar

I am active on another site where I answer math related questions. And someone asked how to get good at probability. I gave him a few pointers some links, and told him to do huge amounts of questions. He then said thanks, and told me he had already done math for several hours.

I thought that was qute ^^

Srivatsan

@N3buchadnezzar Oh wow. That's several hours more than what he usually does. =)

N3buchadnezzar

=)

Oh and I need to learn Topology guys, Brb 10 minutes

robjohn

5:20 PM

@Srivatsan I might just said: "It would be much easier to read your questions if you used LaTeX. It is not that hard."

Henning Makholm

@Srivatsan I don't think it's rude. (But it's not something I'd have bothered to write).

robjohn

@Srivatsan However, I don't see anything rude in you comment

Srivatsan

@robjohn Ok. =)

That would've been much easier to say also. :/

Thanks, Henning and Rob.

robjohn

I was trying to get a link to a comment by Pete L. Clark on this question by going to his third comment page. The comment starting "The question is not about anomalous voting patterns" has a link to the actual comment. The previous comment (next on the page) has only a link to the original question.

Srivatsan

@robjohn Try this: meta.math.stackexchange.com/questions/1720/…

robjohn

5:30 PM

@Srivatsan I get that one. It is the next comment on the page where he mentions being unacceptably mean that I wanted to get a link to.

Srivatsan

@robjohn No, I couldn't get it either.

robjohn

@Srivatsan That has always worked for me before.

Srivatsan

What has worked for you? Linking to an arbitrary comment?

robjohn

@Srivatsan looking up the comments on the users comment page. I've never seen it link back only to the original question.

Daniil

Good evening (morning, day)!

Came back from William Blake's exhibition.

Srivatsan

5:37 PM

@robjohn Really? My memory is different. I remember having trouble usually with linking to specific comments. Nowadays I give up and just link to the answer.

robjohn

@Srivatsan By digging in the page source, I found this

Srivatsan

@robjohn You are a real Hacker.

robjohn

I had to paste in the appropriate numbers into the link for another comment :-)

Skullpatrol

Does anybody know, or care, what is the difference (subtle or not) between a numerical expression and a numeral?

Srivatsan

4+5 is a numerical expression, but not a numeral. A numeral is just a number.

robjohn

5:42 PM

The string "17" is a numeral

Srivatsan

@robjohn How is a string a numeral?

Skullpatrol

Thank you both.

robjohn

@Srivatsan A number is more of an abstraction. It is not its numeral. The numeral changes from base to base

Srivatsan

So: what is the meaning of 17?

robjohn

A numeral is the written representation of a number

Srivatsan

5:44 PM

Re depends on base, I compute with Roman numerals.. =)

@robjohn Ah, I see.

robjohn

"XVII"

Srivatsan

Hm, that took some time for me to process.

Skullpatrol

Also, there is no roman numeral for the number zero.

robjohn

" "

Skullpatrol

0

robjohn

5:46 PM

They used blanks to signify "nothing"

Srivatsan

Sorry, Skullpatrol, my digression about Roman numeral is just that - a digression. It might not be the most productive to focus on that example.

Skullpatrol

I realize that but it is a well justified use of the word numeral Srivatsan.

Srivatsan

I think so too. It's fine, I guess.

robjohn

@Skullpatrol According to Wikipedia, and I remember hearing it before, the word "nulla" was also used to mean zero.

Srivatsan

@robjohn An ingenious Roman could have represented 0 as "IIIIIV" by counting backwards. =)

Skullpatrol

5:50 PM

@robjohn Getting back to the distinction between a numerical expression and a numeral, the use of zero in the two is also very different.

robjohn

@Srivatsan Or "VVX"

Srivatsan

Yes, that sounds cooler. (But I was implicitly shooting for -1 as well. ;))

robjohn

Kinda lends a different meaning to "LLC" :-)

Srivatsan

"limited liability company"?

robjohn

@Srivatsan follow the link :-)

Srivatsan

5:53 PM

@robjohn Oh, I see you edited the message. I had googled for it earlier.

robjohn

@Srivatsan Yes, I added the link later so I put a smiley so that I wasn't being accusatory :-)

Srivatsan

@robjohn Right, I sensed that.

robjohn

I didn't want it to sound like I was saying "hey stupid!" :-)

Srivatsan

I don't think I would mind that very much. =)

robjohn

perhaps, but it wasn't warranted here. :-)

Srivatsan

5:56 PM

Hm, right.

I think Skullpatrol's question is left out. Maybe we should get back.

Skullpatrol

:-(

robjohn

Doug Spoonwood confirmed that he had been the downvote on my $99^{100}$ answer.

Skullpatrol

Getting back to the distinction between a numerical expression and a numeral, is there one?

robjohn

So a numeral is a written representation for a number.

I would say that a numerical expression is an expression that evaluates to a number

I don't know whether I can tell if an expression is the written thing or the concept

The way that I can between a numeral and a number

"3+2" is a numerical expression, but is it the wriiten thing or the concept of "3+2" that is the numerical expression

I think I lean toward the written thing.

Expression means how something is communicated, and here it is the written statement

Skullpatrol

@robjohn You lost me.

robjohn

6:04 PM

Do you see what I said about numeral vs number?

does it make sense?

Skullpatrol

A number is an abstract idea?

What is the expression of that idea, the numerical expression or the numeral?

robjohn

I think numerals are a subset of expressions

Skullpatrol

A simplified subset?

robjohn

numerical expressions are built from numerals and operators.

and grouping delimiters (parens etc)

and perhaps other things I have not considered.

"3+2" is a numerical expression which includes the numerals "3" and "2"

"3" is a very simple numerical expression composed of a single numeral.

Skullpatrol

hmm... thank you for your help.

@robjohn By the way, a number and a value both refer to the same thing right?

As "words"

robjohn

6:17 PM

@Skullpatrol I would say so.

However, these are not only mathematical, but also linguistic, concepts. One might need to be versed in both to render a more comprehensive answer.

I may not be knowledgeable enough in either to do justice.

Skullpatrol

@robjohn So when people say "1/0 is not a number because it has no value," they are saying there is no concept "the reciprocal of zero" ... right?

robjohn

That there is no number that when multiplied by $0$ equals $1$

There might be a concept of "the reciprocal of zero", but it is not a number.

since by the very definition of $0$, $0\cdot x=0$ for any number $x$.

Skullpatrol

Here we have implicitly used "number" to mean "real number" right?

robjohn

or more general. We could be talking about any ring.

$0\cdot x=0$ is true in any ring

MaX

Did anybody tried this problems before?

Is this even real?

robjohn

6:30 PM

@MaX That has to be a joke

MaX

I too think so, but the problems seems to be real at first sight.

Skullpatrol

@MaX Hey the wiki globe is now bouncing, how cool is that?

MaX

@Skullpatrol: It's uncyclopedia ;)

Skullpatrol

That explains it, thanks.

MaX

haha :) @Skullpatrol

Henning Makholm

6:42 PM

@MaX It fails already at the second sentence -- a triangle with the specified side lengths is not right.

MaX

@Henning: Do you mean this "You are given a triangle that has sides of 66cm, 73cm, and 94cm" ?

Henning Makholm

@MaX Yes, and the next sentence falsely claims that one of the angles is right.

MaX

@Henning:So, that's not a Pythagorean triple, hence the error. Thanks I understood.

Henning Makholm

Exactly.

I have two unexplained downvotes on this answer (note, recently edited slightly from the state that was downvoted). Anyone care to theorize what people don't like about it?

Skullpatrol

@robjohn In conclusion, would it be correct to say: both a numerical expression and a numeral name a particular number, with the numeral doing so in the most simplified form?

Kannappan Sampath

6:56 PM

I need some help guys!

anon

Sorry @Sri, I went zzz last night. I'll take a look when I have time, but I have to be off again right now.

robjohn

@Skullpatrol as long as the numerical expression is proper. In any language, there are improper expressions, too.

Kannappan Sampath

@robjohn I need some help now!

On Probability!

m. k.

@HenningMakholm: I dunno about random downvotes, but when you define $S^1$ I think you're missing $x^2 + y^2 = 1$?

robjohn

@KannappanSampath with what?

Kannappan Sampath

6:59 PM

@robjohn With random variables in Probability!

Henning Makholm

@mk Whoops. Thanks, fixed.

robjohn

@KannappanSampath again, with what? what is the problem?

m. k.

@Kannappan: Don't ask to ask :)

Kannappan Sampath

@robjohn Let me be explicit this time: I need to show that a pair of random variables don't have joint density.

@mk What does that mean? I just wanted to confirm if I can ask a question in probability!

robjohn

@KannappanSampath which random variables? and define joint density, please.

Kannappan Sampath

7:03 PM

$U \sim \operatorname{Uniform}[0,1]$ $V=16 U$

Henning Makholm

@KannappanSampath It means that asking whether you can ask a question is a waste of bits. Just ask the question and see whether you get yelled at or not. (You might, but inappropriate questions are usually not any more annoying than vacuous "can I ask a question" questions, so we all get out ahead anyway).

Kannappan Sampath

@HenningMakholm I will ask it straight away, next time!

m. k.

yes, exactly

robjohn

....@

Kannappan Sampath

Joint density is obtained from distribution as follows:

I am writing out a new version.

robjohn

7:07 PM

what is $v$ in $F(u,v)$?

the RHS is a function of $u$

Kannappan Sampath

$f_X(x)=\int_{-\infty}^{\infty}f(x,y)dy$

where $f_X$ is the density of $X$ and $f(x,y)$ is the joint density!

robjohn

and $f_Y(y)=\int_{-\infty}^{\infty}f(x,y)dx$?

Kannappan Sampath

Yes, true.

Another equality is: $$ F(a,b)=\int_{-\infty}^b\int_{-\infty}^a{f(x,y) dxdy}$$

robjohn

And what is the question?

Kannappan Sampath

@robjohn For $U \sim \operatorname{uniform}[0,1]$ and $V=16U$ show that joint density does not exist.

Asaf Karagila

7:25 PM

@Brian: I see that you've answered the residual set question. Could you check it and tell me if the definition for locally residual is correct? Often local definitions look a bit different.

Brian M. Scott

I’ve never seen anyone use the term before, so I’m taking the OP’s definition at face value.

Asaf Karagila

I see.

Kannappan Sampath

@rob Shall I tell you how far I've gone?

Skullpatrol

@mk Could not someone asking "May I ask you a question..." be interpreted as politeness on the part of the person asking to not interrupt someone in the middle of doing something else?

Brian M. Scott

@Skullpatrol It could. That’s certainly how I interpret it, barring unusual circumstances.

m. k.

7:30 PM

@Skullpatrol: Yes. It's not a big deal, but I think you get better results in a chatroom when you just ask your question (instead of waiting "yes, I'm here, please ask"). People are often idle, and someone might answer your question later.

I don't think it's rude to ask

Skullpatrol

So there is no real reason to get annoyed with vacuous "can I ask a question" questions.

Srivatsan

@anon Np

m. k.

@Skullpatrol: you're right, I wouldn't get annoyed about it. But I think things go more smoothly that way

Skullpatrol

7:48 PM

@BrianMScott In your opinion sir, do the unusual circumstances we have here in the chat room justify anyone getting annoyed with vacuous "can I ask a question" questions?

m. k.

did someone get annoyed?

Skullpatrol

46 mins ago, by Henning Makholm

@KannappanSampath It means that asking whether you can ask a question is a waste of bits. Just ask the question and see whether you get yelled at or not. (You might, but inappropriate questions are usually not any more annoying than vacuous "can I ask a question" questions, so we all get out ahead anyway).

Henning Makholm

That's not annoyance, it's an explanation of someone's (I forget whose) earlier remark.

Srivatsan

@Skullpatrol "not any more annoying" -- that is a relative statement. Probably Henning means that if some person is (e.g.) in a bad mood some time and gets angry at inappropriate question or question posed at an inappropriate time, then chances are that a "can I ask a question" question will also make them angry. [And this is my interpretation, so it is possible it is not what he meant.]

Brian M. Scott

@Skullpatrol Let’s put it this way: I don’t share the objection, but I do understand it. And I’m inclined to regard the whole thing as a silly tempest in a teapot.

Skullpatrol

7:54 PM

@BrianMScott Who is the "tempest in a teapot?"

Brian M. Scott

Not who, but what: the whole discussion.

Skullpatrol

Tempest in a teapot

Tempest in a teapot (American English), storm in a teacup (British English), is an idiom meaning a small event that has been exaggerated out of proportion. There are also lesser known or earlier variants, such as tempest in a teacup, storm in a cream bowl, tempest in a glass of water, storm in a wash-hand basin. and storm in a glass of water. Etymology Cicero, in the first century BC, in his De Legibus, used a similar phrase in Latin, possibly the precursor to the modern expressions, "Gratidius excitabat fluctus in simpulo, ut dicatur", translated: "Gratidius raised a tempest in a ladle...

Srivatsan

Can I ask a silly question?

I like that variant. =)

Brian M. Scott

I’m sure that you can. And we’ll even allow you to do so. :-)

Skullpatrol

1 hour ago, by Kannappan Sampath

@mk What does that mean? I just wanted to confirm if I can ask a question in probability!

Asaf Karagila

8:07 PM

Brian, you got 36 seconds ahead of me!

Kannappan Sampath

I seem to have broken ice over the problem! Let me see if I can take it to completion!

Bye all!

m. k.

See ya

Henning Makholm

@AsafKaragila And I'm horribly late.

Asaf Karagila

@HenningMakholm Quite!

It's because you don't have a homework assignment due two weeks ago. So you actually work instead of procrastinating. :-)

Srivatsan

@Brian: I have a question about this axiom thing, but I don't know if it is silly or meaningful. There are these rules of inference in logic, no? [For instance, constructivists deny certain rules taken for granted by classical mathematicians.] What is the status of these things? Are these to be treated as axioms? Or something else?

Asaf Karagila

8:15 PM

Yes.

Rules of inference are axioms of logic.

Srivatsan

That makes sense.

I get warped in circularity whenever I think about these things.

Asaf Karagila

Oh yeah, it can certainly make you cross-eyed :-)

Henning Makholm

However, the customary usage in logic is to say that an "axiom" is a special kind of rule of inference, namely one with no premises.

Srivatsan

Can you give an example?

Asaf Karagila

$\forall x.x=x$?

Srivatsan

8:19 PM

Ah, right.

Henning Makholm

@AsafKaragila I don't think that is usually an axiom -- it is proved by inference from $x=x$, which sometimes is an axiom.

Of course in a greater context, all of the rules of inference occupy a similar status as, for example, the group axioms do in group theory.

Asaf Karagila

I guess you're right. I often forget that $\forall$ doesn't usually appear in the logical axioms.

Henning Makholm

@Srivatsan For example, in some presentations of classical logic, $\neg\neg P \to P$ is a logical axiom -- which means that you're allowed to conclude it from nothing for any $P$.

On the other hand, modus ponens is a rule of inference (from $A$ and $A\to B$ you're allowed to conclude $B$) which is not usually spoken of as an "axiom".

Srivatsan

I see.

Skullpatrol

I don't.

Srivatsan

8:23 PM

I guess $P \lor \lnot P$ is an axiom too. Is that right?

Asaf Karagila

That is just the definition of the truth :-)

$$2\mathcal B\lor\lnot2\mathcal B\ ??$$

Srivatsan

Is that a serious comment, Asaf? :=)

Asaf Karagila

Do you want it to be?

Henning Makholm

@Srivatsan Depends. I think there are few formal developments that will take both $P \lor \neg P$ and $\neg\neg P \to P$ as axioms at the same time, since they imply each other in the presence of an axiom set for intuitionistic propositional logic.

Asaf Karagila

How can you be intuitionistic if you allow $\lnot\lnot P\rightarrow P$?

The whole point of intuitionistic logic systems is that you cannot eliminate double negations.

Henning Makholm

8:26 PM

@AsafKaragila You can't. But it seems to be popular to construct logical axiom sets by taking an axiom set for intuitionistic logic and then adding a single axiom that makes it classical.

Srivatsan

@HenningMakholm Oh right, I did not pay attention to that. Thanks. // But: I am not particularly interested in developing a minimal set of axioms for (e.g.) classical maths. I was just asking if $P \lor \lnot P$ fits the definition of an axiom.

Henning Makholm

That way it is easy to explain to your students what intuitionistic logic is...

Srivatsan

Now I think I got that question clarified.

Il y a

пщщв умутштп

good evening to all of you who has an evening now. good morning to @Srivatsan

Henning Makholm

@Srivatsan The thing is, it only makes sense to speak about whether something is an axiom or not in the context of a particular formal development.

Brian M. Scott

8:28 PM

@Ilya Gesundheit!

Il y a

@Brian: I though, you're away. good afternoon

Brian M. Scott

I’m in and out, since I’m doing several things at once.

Srivatsan

@HenningMakholm Yes, I understand that. Let me put it this way (mimicking your previous comment): "If I took a bunch of standard rules defining intuitionistic logic, and added the rule $P \lor \lnot P$, would this final rule be called an axiom?"

And I guess the answer is yes.

Henning Makholm

@Ilya G'aften.

Il y a

@BrianMScott quantum Caesar?

@HenningMakholm how are you doing?

Srivatsan

8:31 PM

@Ilya Hi Ilya.

Henning Makholm

@Srivatsan Yes.

@Ilya Nice enough. Been managing to do a lot of interesting stuff at work after I stopped coming into the chat on weekdays ...

Brian M. Scott

@Ilya More like quantum Bozo the Clown, I’m afraid.

Il y a

I have a questions. There is a linear equation of the form $\mathscr A f = 1_B$ where $1_B$ is an indicator function of $B$ and $\mathscr A$ is a linear operator. The trick is that I am not sure whether the solution of this equation exists and is it unique.

Henning Makholm

@Ilya: In the vector space of pointwise real arithmetic on functions $C\to\mathbb R$, where $C$ is some superset of $B$?

Il y a

@HenningMakholm thanks, yes

Henning Makholm

8:35 PM

Does $C$ have structure that your functions must respect?

Il y a

On the other hand, if for a fixed $B$ I denote the class of solutions as $[B]$ then $0\in [\emptyset]$ and if $f_n\in [B_n]$ where $B_n$ are pairwise disjoint, $\cup_n B_n = B$ then $\sum_nf_n \in [B]$

Srivatsan

What does $[B]$ mean?

Il y a

@HenningMakholm in general, $C$ is a measurable space and $f$ are measurable functions (with possibly infinite values)

@Srivatsan this is the class of solutions, I wrote above $$ [B]:=\{f \text{ measurable }:\mathscr Af = 1_B\}$$

this construction gives a clue that solutions of this equation have a certain measure-kind-of structure

Srivatsan

@Ilya Wait: "possibly infinite values" and the vector space structure are not compatible with each other, are they?

Il y a

@Srivatsan well, that is something which also bothers me. Namely, I would like to solve this equation - and then it's me who has to provide the class of functions where the solution should be

e.g. if I will ask for the solution to be bounded and measurable (Banach space) then it may not necessary be there

Srivatsan

8:41 PM

But the class is not necessarily a linear subspace..?

Il y a

@Srivatsan linear subspace of what? the space of all functions $f:C\to\mathbb R\cup\{-\infty\}\cup\{\infty\}$ is not linear

Srivatsan

(I meant linear subspace; I edited the above comment.)

Il y a

on the other hand, I guess, that if I will look for the cone of functions $f:C\to[0,\infty]$ then the solution does always exist

Henning Makholm

Google Translate is rather confused. When I set it to "Russian to English" and paste in "пщщв умутштп", it translates it to "pschschv umutshtp", and suggests that the Russian original should have been "good evening"...

Srivatsan

@Ilya Exactly. Ya, ok, this is something we already clarified.

Il y a

8:43 PM

@HenningMakholm it seems that you're not very busy now:D in fact I was typing good evening without turning on English keyboard

Henning Makholm

Ah, that explains it. I did think those final consonant clusters were unusual, but who knows what might happen in Russian?

Il y a

The question is the following: let us consider the linear equation $\mathscr Af = 1_B$ for $B\subset C$. Suppose that for any measurable $B$ the solution exists in the class $f:C\to[0,\infty]$ and we denote all such solutions as $[B]$. We know that

1. for any $f\in[B]$ it holds that $f\geq 0$

2. there is $0\in [\emptyset]$

3. if $B_n$ are pairwise disjoint and $f_n\in [B_n]$ then $\sum_nf_n\in[\cup_nB_n]$

where the limit exists in the class $C\to[0,\infty]$

now, if $0$ is not an eigenvalue of $\mathscr A$ then the solution is always unique for each $B$ and hence it defines some measure on $C$

Henning Makholm

It looks like (1) gives new information whereas (2) and (3) are consequences of the definitions (at least for finite sums in 3), right?

I agree that "0 not an eigenvalue" gives you uniqueness, but how do you get from there to a measure?

Ah, I see, you just take the value of f at some fixed point?

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