Semiclassic: You could learn to be a good cook as an alternative :P

lol

Think how useful and enjoyable that would be, and you'd acquire friends out of nowhere!

I've come to realize I continue to think / work while hanging out with other people, but rarely to the point it's an obstruction to chillin'

yeah, uh

I sometimes think most of my friends like me because of the food and booze I feed them.

that probably helps me with productivity lol

but dunno if it should be broadly advised

Mike, from what I could see on my visits, you are reasonably sociable!

:) That's partly my point - I often have reasonable success in multitasking (and said multitasking is not a conscious choice)

Oh, I met Deven and we discussed you briefly. He was one of the people running our training session.

Really cool thing that came up as an exploration: Characterize all numbers that can be written as the sum of at least two consecutive positive integers. [I'd never seen this before.] Hint: They're called trapezoidal numbers. @Balarka @Leaky @Akiva @anyone else

hmm

I think that there was a passage about those in Heath's history of Greek mathematics

It actually turns out way more interesting than I'd thought it'd be.

The hint completely answers it!

the first thing that comes to mind is n(n-1)/2 - m(m-1)/2

No, it doesn't, @MikeM. I figured out easily why the name. But only last night did I give a proof characterizing the necessary and sufficient condition for an integer to be trapezoidal.

@TedShifrin Everything but powers of two?

Right, @Leaky.

Have you seen that before, DogAteMy?

I have seen this before

Aha ... I never had.

I'm confuzed. Is the condition that $n+(n+1)=m+(m+1)$ for $n\neq m$?

(n-m)(n+m+1)/2

Anyhow, I worked out a proof and tried to send it to all of the people at our meeting, but they haven't forwarded it from the AoPS office.

It's an exercise in Weil's little book on number theory

I can't see that being right, though, because that's equivalent to $n=m$.

Oh, very cool, Balarka. My ignoramusness strikes again.

@Semiclassic. Leaky gave an algebraic formula. But characterize all the answers?

I had thought about the concept before, I don't remember if I ever asked or saw that specific question though

No, I mean, I'm not understanding what's being asked.

@Semiclassical characterise all numbers of the form m+(m+1)+...+n

I don't actually have a proof yet that was just a guess

"at least two consecutive positive integers"

I'm missing some details

@Semiclassical Like 13=6+7

Ah. So characterize what numbers can be written as arithmetic progressions with unit step

and 14=2+3+4+5

n-m > 1

and 15=4+5+6

DogAteMy: Once again your intuition is way better than mine. I'll let you give me a proof.

I can see visually why that's trapezoidal, yeah.

well

It's not super hard to see form the algebraic formula why every number not a power of 2 can be written like that

Note that "everything other than powers of two" is the same as "multiples of odd numbers", which is probably the best way to thing about it

Odd numbers other than 1

@Akiva yup

@TedShifrin Negative numbers are disallowed, yes?

@Balarka: Since I wrote a 1/4-page proof and typed it up, it couldn't be super hard :P

Right, just positive integers, DogAteMy.

Leaky: Don't give it away.

:)

@TedShifrin then how am I supposed to prove it?

I'm just saying not to tell everyone ... lots of people are thinking about it.

alright

Odd numbers are trivial, right?

I guess what I'd want to do is, given a positive integer which isn't a power of 2, show how to construct a trapezoid for it.

Yes

Yes, although it took me a little while to see that, Steamy.

@Semiclassic: And show you cannot possibly do so for a power of 2.

Right. If and only if.

Hmmm... I think I have a nice way to do the evens

at least, the possible ones

@TedShifrin well it's easy once Akiva gave it away that it's every number save the powers of 2

Oh I remember that exercise

I think I've proved it.

I sorta brute forced it I think

what comes to mind is forming a triangle, so to speak, and seeing if you can ever add the right-sized block to make that a trapezoid

Right, this is pretty much a one-line proof I think.

@LeakyNun I'm getting tripped up on how you can't have negative summands, though

@AkivaWeinberger what do you mean?

Like a naive algorithm might introduce negative summands

actually, though, the example I gave there is too easy since I can also do 4+5=9

so yeah, don't want to do odds

@AkivaWeinberger no it won't. just break 2k into 2^a b, and assign the smaller of (2^a,b) to (n-m) and the bigger to (n+m+1)

n+m+1 > n-m

I don't think that always works, Leaky. I had to do cases.

@Daminark we're the elementary number theory gods, thanks to Weil :P

(nah not really)

fantastic book though

@TedShifrin for which number does this not work?

Yeah Weil is a fun time, number theory is my favorite thing

Ohhh, wait. We have different notations.

Wait, are we posting solutions then?

We weren't supposed to.

But Leaky can't restrain himself.

I don't understand what you're trying to say, Leaky, but seeing "2k" there gives me an idea for a proof

@Leaky so is this naive algorithm something like, if $n = kp$ then write $(k - \lfloor \frac{p}{2}\rfloor) + (k - \lfloor \frac{p}{2} \rfloor + 1) + \ldots + k + \ldots (k + \lfloor \frac{p}{2}\rfloor)$?

Which might be the same as yours but possibly not

@TedShifrin Akiva asked me directly... how could I have explained otherwise?

Can always open a private room :P

(This is extremely unelegant so I'll feel free to say)

I'm not the tsar here. You guys do whatever you want.

I have an elegant solution

At least, for showing that all non-powers of 2 are possible

let's spoil the solutions here

OK I'll go there

Wrong

looool

loooooooooooool

epic fail

you posted a leafy vid

@Daminark Soooo secretive

I sent it to a friend last and for some reason it didn't copy the v secret # theory room

@WillHunting Right?

solution spoiled

hey, guys. Quick question: if I take a finite subset out of $\mathbb{Q}$, is the result still dense in $\mathbb{R}$?; I think it is, because a finite amount of elements does not affect the convergence of a sequence.

Oh for frick's sake my naive algorithm actually includes negative numbers

Of course, @Miguelgondu.

@Daminark So does mine.

But you can always get rid of them

For the Weil exercise it wasn't a problem

Negatives are verboten!

@TedShifrin Thanks, Ted!

$$\sum_{k = -\left\lfloor \frac{n}{4}\right\rfloor}^{\left\lfloor \frac{n}{4}\right\rfloor} (2+k)$$

since we're spoiling stuff anyway, apparently

and then just cancel out the negative ones left and right of the term that's zero.

I have no earthly idea what you guys are doing ... but that's fine.

Oh that's slick

@TedShifrin They are doing heavenly things.

@TedShifrin Ok, I misunderstood what "characterize" means. I'll leave this one to you.

Weil is good for the soul

It's out of my hands, @MikeM.

Well, that's the sum for $n = 2r$ with $r$ odd...

But Deven had a good laugh at my expense, because we (and, in particular, I) had talked about the importance of reading carefully, and my head forgot the word "consecutive" ...

@Daminark Jurgen Neukirch's ANT is the best, really.

I say that in a manner that sounds trolly but I'm actually being kinda serious, that stuff was some of the most fun I've ever had in math

@Daminark idubbz was the one who #rekt leafy, it seems

@TedShifrin Akiva posted an elegant solution in the room

This still isn't the kind of math stuff I find exhilarating for me.

I have ignored some users in chat by clicking the ignore button, but there is no button on the main site, so I will have to manually ignore them there.

@Will It's your favorite, but I'd be rather cautious about saying it's truly the best, like Ireland/Rosen and Niven/Zuckerman are supposed to fantastic

My ignore list might grow beyond Chris'ssis, Jasper.

@Daminark Come on, is there really an objective meaning of good in the first place? Don't pick on my words. =D

I actually like Hardy & Wright's book.

But I'm no number theorist.

@Ted I think most of it is real boring.

I mean that's fair, people get excited by very different things

I've wanted to learn some number theory for a long time but every time I try I just end up doing something else

You too started off in number theory @MikeM.

I personally like Hardy&Wright

Not contest number theory :)

Wait, is it H&W you think is boring? I'm confuzled.

Imagine you tell your friend 'This ice cream is the best!' Do you want him to say 'I would be cautious about saying it is the best!'? LOL

Harder number theory I didn't understand and eventually gave up on, of course

This problem was chosen as a model of how to get kids collaborating and formulating a correct mathematical argument ... I think that's all great.

@TedShifrin I think contest math problems they teach with are boring.

@MikeMiller Yes, they are boring to me too because I can't solve them.

@TedShifrin Steamy just pointed out something that should have been obvious. We don't have to worry about negative terms at all, since if they occur, we can cancel symmetrically around the term that's zero

Well, I plan to give a few of my own exercises, but I think most of the exercises are more thought-provoking than any high school or college precalculus text's.

I never knew where you guys got negative numbers in the first place. I certainly didn't think about the problem with any negative numbers.

@Akiva That's what I was thinking of when I said you can fix it by hand

@TedShifrin You know how $(x-1)+(x)+(x+1)=3x$

Or $(x-2)+(x-1)+x+(x+1)+(x+2)=5x$

If you read Great Expectations, you will see that Pip was brought up "by hand".

Oh ... I never did that.

The naive thing to do is to try to expand around 1/odd factor in 2floor((odd factor/2)) steps

But that's where negative numbers may come in

@TedShifrin Hm in any case I would definitely call it the correct way to do it. How did you do it?

You mean for odd numbers?

Hm?

Hm?

@Akiva lol at the time I was like "I'm pretty sure is exactly the solution they don't want"

Like we get all multiples of an odd number by doing that sort of thing, cancelling out negatives that way

@TedShifrin How did your solution go

Sort of the way Leaky's went. But I gave the odd numbers explicitly in the form $2k+3$, so take one row of $k+1$ dots and one row of $k+2$ dots. There's your trapezoid.

Otherwise, I broke things into two cases.

Probably not the most elegant, but it was a proof.

I guess it's hard to show the converse (e.g. 32 is impossible) with that method

The "(x-1)+x+(x+1)"-y method

Hi chat

Chi hat

$\hat\chi$

I had to use the formula $n(n+1)/2 + kn$ and factor it two alternative ways to see you must always have an odd factor.

hi Alessandro

I am going to bed, you all misbehave without me, and also misbehave with me, LOL.

Jasper: You're stealing my lines. I want royalties.

I have another question regarding Hausdorff measures

Oh, did you ever get that Cantor set thing to work?

I think it should work if at every stage you remove enough of the segment so that the remaining area after $n$ steps goes to zero faster than $2^{-n}$

@AlessandroCodenotti About 5 foot something

something like that sounds right, Alessandro.

You should give that to DogAteMy to work on :P

What are you doing, making a Cantor set with positive measure?

(You know what would be interesting, a Cantor set with negative measure)

No, that wouldn't be interesting.

No, making an uncountable set whose Hausdorff dimension is 0

@AkivaWeinberger That's easy, just find an appropriate signed measure :D

What's the Hausdorff dimension of the usual Cantor set?

something like $\log 2/\log 3$? Oops. Has to be less than $1$.

Ah makes sense

vice versa I think?

the inverse

can't be bigger than 1 :P

I fixed it. Stop your bitchin'. :P

So... first a sanity check, if $E\subseteq\Bbb R^n$ has Hausdorff dimension stricly smaller than $n$ then $E^c$ must have Hausdorff dimension $n$, right?

I believe so, @Alessandro :)

It must have empty interior, right?

Because $H^n(E)=H^n(E^c)=0$ and it would break the additivity

So you want to restrict to a cube or ball or something ...

@TedShifrin right, that works too

So if we call $H^d$ the $d$-dimensional Hausdorff outer measure (working in some $\Bbb R^n$ with $n>d$) and $M_d$ the $\sigma$-algebra of $H^d$-measurable sets what's the relation between $M_r$ and $M_s$ when $n>s>r$?

I've never heard of inner/outer Hausdorff measure.

I am not going to be of any help for this. I don't know any of it.

@TedShifrin How do you define the Hausdorff measure? In my course we defined the outer measure first and then got an actual measure when restricting to its $\sigma$-algebra of measurable sets, but I don't see a straightforward way to do it without going through the outer measure

Wait a minute, @Alessandro. If a set has positive $H^s$ measure, then its $H^r$ measure has to be infinite.

I am never a measure theory person.

yeah

I just learned stuff (which I've forgotten) about taking appropriate powers of measures of balls.

$M_r\subseteq M_s$ since all $H^r$-measurable sets have measure $0$ with respect $H^s$ and if we have an outer measure the sets with measure $0$ are measurable

Ah, actually I'm not sure. $M_r$ is full of sets that don't have Hausdorff dimension $r$

Less than or equal to ...

Right.... I get to supervise an exam in a computer room aka sauna tomorrow, so I'd better prepare by getting some good night's rest. Night all.

It's also full of $n$ dimensional sets to be closed under complements

Oh that seems garbage.

Those are annoying

And I'm not sure whether a countable union of $r$ dimensional sets needs to have dimension $r$, but I'd bet that's true

Seems right by considering $r+\epsilon$.

Hmmm, I'll think about this tomorrow, I should go to sleep now

bye!

Night!

and thanks for your help!

I wasn't any help.

Thanks for your time then? It feels wrong to just leave without thanking since you spent time thinking about my problem

LOL ... g'night.

@EricSilva: Now that you missed your lecture, you feeling better?

Unfortunately not :/

I haven't gone to bed yet, LOL.

Okay so, apparently conjugacy classes in $S_n$ correspond to partitions of $n$

I'd suspect it's that if you break up a permutation into disjoint cycles and add up the lengths, you get back $n$

So you want to show that the classifying permutations like so corresponds to conjugacy classes (e.g. all $k$-cycles are conjugate, etc)

If you can't solve a problem, maybe write it down, put the paper below your pillow, go to sleep, and you will dream of the solution.

I know someone who dreamt the proof of pigeonhole

But it's quite rare, and I forget dreams

Hi. Let $2^A$ denote the power set of a set $A$, and let $\Omega=[0,1]$. It is true that $(\Omega,2^{[0,1]})$ is a measurable space, but we typically consider, e.g., the Borel $\sigma$-algebra of $[0,1]$ rather than $2^{[0,1]}$ because no reasonable measure $\nu$ can be defined so that $(\Omega,2^{[0,1]},\nu)$ would be a measure space?

That is, I understand in what way $2^{[0,1]}$ is too big, but am I right that it is still a valid $\sigma$-algebra and the problem appears only when looking for a ''reasonable'' measure?

@Daminark I am going to eat and then sleep, good night, I will see you in my dreams, LOL.