@Jeff I'm sorry, I have to go afk for a while

np

"I have to say, your entire "connection to Sha" reads to me like someone saying that they think they have a proof of the Riemann hypothesis, mentioning along the way that they aren't actually very sure about what "the Riemann zeta function" is, and saying that their proof just hinges on whether one can or cannot say that the product of two real numbers is always a real number,"

"or if it sometimes can be a complex number with nonzero imaginary part. In short, not something that can be taken seriously."

How many moderators do they need?

Hi , I need some help in relations question

(a,b) R (c,d) => a+d = b+c , for all (a,b), (c,d) belongs to N X N // I'm unable to interpret the relation

I've solved some a R b type questions only

@MarianoSuárezAlvarez Would it be possible to clear the clutter created by the user Gigili in Ben's proposal? The user makes rubbish claims there.

help me pleaseeeeeeeeeeeeeeeeeee

@x4d33746153706c306974 Can you visualise what the relation looks like?

Take a piece of paper and draw the co-ordinate plane. Pick a lattice point.

@KannappanSampath yes , It's kind of association btw set of objects

Find all those points that are related to it by $R$.

@x4d33746153706c306974 No, I am not asking you to visualize a general relation, I am asking you to visualise this one.

@JonasTeuwen can you help me with some integration-measure theoretic questions?

idk

@KannappanSampath aah , sec

@KannappanSampath, TheChaz removed his comments

@KannappanSampath so the relation is line joining two consecutive points ?

Ideally, everyone cleans up after him/herself :)

someone

@x4d33746153706c306974 Consecutive? With respect to what order?

No, a better description, please?

@KannappanSampath ok I'm unable to visualize the question

Dear @x4d33746153706c306974, have you considered a less uuidish name? :P

but why did you bring the place and co-ordinate system in the scene?

@robjohn can you helpme? if no busy

@MarianoSuárezAlvarez no, I love it :D

Well, a point (x,y)R(a,b) if and only if $x-a=y-b$ or $x-y=a-b$.

So, all points related to $(a,b)$ lie on the straight line $x-y=a-b$ in the co-ordinate plane.

@leo Ask it here or on main. I'm not joining another room.

@x4d33746153706c306974, to each his lot :)

@JonasTeuwen ok, but why?

......

@KannappanSampath A relation R on natural number is N X N , right ? so , elements of set R have to have syntax a R b or (a,b) , so all in all I'm not getting the syntax ?

@MarianoSuárezAlvarez What does that supposed to mean ? 0_o

remember the problem: $f$, uniformly continuous, $\int f\lt \infty$ implies $\lim_{|x|\to \infty}f(x)=0$

@leo my mother , that is like NP-complete problem?

@leo Because I don't feel like spending time that is only useful for you as I don't know you.

@leo Okay, how far did you get?

@x4d33746153706c306974 What do you mean by syntax? I don't follow. See--as I showed, all points related to $(a,b)$ lie on the straight line $x-y=a-b$ and of course, I mean, by all points, those that have both integral co-ordinates.

@KannappanSampath by syntax I mean way to write the right thing, like (a,b) belongs to N X N , but [a,b][a,c] belongs to N X N is wrong syntax

Yes, so....?

@x4d33746153706c306974, it simply means that you can choose whatever name you like, as you know :) I personally limit myself mostly to interactions with people with names I can pronounce and relate to, because I like to interact with actual people, even online.

@JonasTeuwen I want to know why exactly that don't holds if we assume $f$ continuous. I just finished a construction of a continuous function such that $\int f\lt\infty$ but $\limsup_{x\to\infty} =\infty$. But I'm attempting a proof with the hypothesis of $f$ uniformly continuous, and it seems the same arguments works in the case of $f$ continuous

I am old, though.

It looks like hex code for ASCII.

@JonasTeuwen I just did an other room because I have seen some users here that do so. I think that it can be useful for users other than me. However, no problem

@MarianoSuárezAlvarez It has nothing to do with oldness I guess, you're just 38 :), Anyway my nick is coded in hex form , and on SO it shows non-hex name :)

@leo What is your argument?

how does SO pick an encoding? :)

@MarianoSuárezAlvarez anyways I didn't know "To each his lot" expands to such a big sentence you said :D

One can say: it doesn't hold for $f$ continuous because there are counterexamples, but I'm not conviced

@x4d33746153706c306974, the most usual phrasing is "to each his own"

Heya

@KannappanSampath Well I still don't get your solution :(

meaning "each one chooses his own burden" or something, if you are in a biblibcal mood :)

You have not told us anything substantial! @leo

@KannappanSampath I'm writing, I'm on it, just a second

Since $f$ is integrable over $R$, we have $$\sum_{n=0}^\infty \int_n^{n+1}f=\int_0^\infty f\leq \int f\lt\infty$$

@x4d33746153706c306974 I cannot say anything further. For one thing, you can pick a point in $\mathbb{N} \times \mathbb{N}$ and find all those points related to it by $R$ and look at how that looks on the co-ordinate plane.

that says $$\lim_{n\to\infty} \int_n^{n+1}f=0$$

@KannappanSampath so (a,b) R (c,d) , R is set which contains element like {((a,b),(c,d)), ((e,f),(g,h)),... }

@KannappanSampath Can you list me some example elements of set R pertaining to question ?

@x4d33746153706c306974 Yes. Now, I am asking you to find a subset of $R$ that has a nice property that the first co-ordinate of the ordered pair is actually $(a,b)$.

@MarianoSuárezAlvarez: You mean we can't talk about what's being discussed there? I don't get it.

suppose that $f\not\to 0$ as $|x|\to\infty$

@Gigili, I don't understand the question :/

@leo Oh well, you should say what if the limit is non-zero and then cook up something to keep it positive for a long time.

@KannappanSampath so also means R = { (x,y) : x belongs to NXN and y belongs to NXN} , right?

@x4d33746153706c306974 Yes, sir.

@KannappanSampath cool , Thanks , now lemme reread your solution :)

@JonasTeuwen yes, that's was I did. In that way I achieve a contradiction with $\lim_{n\to\infty} \int_n^{n+1}f=0$. My point is that I think I can do the same with $f$ just continuous

@leo Make a post on main.

I mean continuity is enough to $$\lim_{\delta\to 0}\frac{1}{\delta}\int_a^{a+\delta} f=f(a)$$

@MarianoSuárezAlvarez, how can I delete a Chat Room?

no idea :D

did you create it?

yes

lemme ask

this

@JonasTeuwen is continuity enough to the above identity?

@MarianoSuárezAlvarez Okay.

@Gigili, I really do not know what you asked me about

@leo, I can delete it for you if you want

Is this tag necessary?

@leo, the procedure is, it seems, that you flag it for moderator attention

@MarianoSuárezAlvarez please

ok

good "whatever-time-of-day-it-is-for-you", folks

@MarianoSuárezAlvarez thanks

@leo: done!

Good "whatever-time-of-day-it-is-for-you", David.

it is mid-afternoon, here

@JonasTeuwen, I don't want bother you. Sorry if I did. It will not happen again

Good mid-afternoon then.

Anyone good with subsets?

today, i wish something exciting would happen. but not terrifying. that would be bad. just mildly exciting.

@JohnSmith Well, please please tell us subsets of what?

same question I had earlier, Kan

:(

@KannappanSampath i bet subsets of other sets.

I've asked so many people smarter than I am and for some reason nobody knows how to do it

@JohnSmith That is NOT a question about subsets, it is more about algorithms.

as i wasn't here earlier...i have no idea what you are talking about

It's about subsets

Well, okay, finding an algorithm that's still based in the logic of subsets

@JohnSmith Then, you have not understood the question! :P

I understand the question perfectly well

in a sense, ALL logic is based on the logic of subsets

Then your classification needs review.

well, at least the "classical" logics

@KannappanSampath what is his question? all this mystery intrigues me.

The way to GET the algorithm is derived from knowledge of how to work with subsets

@David consider a main set of arbitrary length N, say the set for N=10 is [1, 2, 3, 4, 6, 9, 13, 19, 28, 41] where a(n) = a(n-3) + a(n-1) with [1,2,3] being hard-set

Every reasonable 8th grader in my country knows what a subset is, can talk about union, intersection, subtracting sets from the other. I am not sure, there is more to subsets than this. You're interested in a combinatorial aspect of something.

i am trying to count the number of subsets where the largest element is less than the sum of the rest of the subset

e.g. [1,2,4] is not valid but [4,6,19,28] is

ok, but the only thing you have to go on is the recursive defintion

of the main sequence, yes

but i can calculate the answer with brute force

all set theory tells you is you have 2^N cases to check with brute force

yes

but for large N, that is not feasible

my brute output up until N=10: pastebin.com/raw.php?i=hsAtkYq1

so you need to think of a better way to think about which (sub)sets might qualify

i am having trouble deriving a recursive relationship I can calculate f(n) in

i've tried countless things

ok, let's look at this another way: will singleton subsets work?

no

3 or more

by definition of what i am looking for

ok, so start with the obvious: look at 3 consecutive terms

any three consecutives will work

now, what happens if you have 2 consecutive terms...and a "gap" (of 1,2, etc.)?

that will not work

why not?

i have tried this approach though

i don't understand the proof that inverses are unique in groups: let x,y be inverses of b, then xby=x=y

i mean, if any 3 consecutive terms work, then two consecutive terms, and a "gap" (skipping numbers) ought to work even better

like for the proof that identity is unique, i can see it

i picture a group table, with the identity in it

@AbstractionOfMe it goes like this:

then i can see that adding another identity will break the first identity

let x,y be two (two-sided) inverses of b. so xb = bx = e = yb = by.

it doesn't work that way though

so x = xe = x(by) = (xb)y = ey = y

David: right there, i following the notation but i can't see it

[3,4,9] for instance has one gap (skipping 6) and does not work. This will not work for any such subset.

my trouble is in creating the generalized, recursive definition

oh, sorry, LESS than

@AbstractionOfMe where do you get confused?

Y U no use punctuation? T_T

All triplets (except [1,2,3]) work. All quads work. All group5's work. All group6's work, and so on.

@David: I feel like my understanding of that proof is based on treating a group like a formal system

But i want it to be more than a formal system

because i feel like if it treat it like a formal system i will have no intuition about how to prove new things

@AbstractionOfMe the trouble with thinking in terms of Cayley (multiplication) tables is that this is unwieldy for large groups (and totally impractical for infinite ones).

@David: what about an infinite cayley table?

who has paper that big?

who has paper big enough to enumerate the reals? but we still consider them

yes, but we don't calculate sums of real numbers by looking them up on a table, either

true...

we have nifty rules of arithmetic to help us

consider algebra of reals\

i can write some notation: x+y = y+x

the group axioms are really just the same rules you learned in early school, except for commutativity (we leave that out because some things don't commute. matrices, for example)

but it's MORE than just notation for me. i see two arbitrary real numbers as columns for varying height, and i can see that swapping the order of the two columns preserves the total height\

ok, well one way to think of what "group elements" really are, are some kind of function

and it usually isn't the case that doing f then g is the same as doing g then f

the one i'm having trouble with is that (xb)y = x(by) implies x=y

(if x,y are 2-sided inverses of b)

it's just a voice

but i want it to be a picture too?

do you know what an inverse function does?

@David I changed everything to X's and _'s denoting which items are marked: pastebin.com/raw.php?i=Yf2jPz70

@David: if you compose it with its inverse you get the identity

@David: can you explain what you mean by saying that group elements are a function?

you mean a binary operation?

ok, i will give a simple example

imagine a rectangle in the plane (or, if you prefer, a playing card, with both sides printed exactly the same, say the four of diamonds)

Holy cows. How I wish I could downvote comments.

now what can we do to the card that leaves us right back where we started?

well, we could rotate it 180 degrees (that's a function, right?)

right

we could flip it top-to-bottom

or we could flip it left-to-right

D_4?

or we could just do nothing

hmm okay no because its a rectangle

that gives us 4 operations (functions) we can write as {I,R,H,V}

we could write these as 2x2 matrices, it really doesn't matter.

When an equation has no real roots? $\Delta<0$?

so what do you get if you do RH?

@Gigili if you mean the discriminant, yes

@David: I

are there any other communities more centered around number theory/algorithms/etc

not to be rude but i feel like rarely does anyone here know how to help with these questions

@AbstractionOfMe um, no. let's label the sides of our rectangle: 1(top), 2(right), 3(bottom), 4(left)

RH means first do H, then do R

@JohnSmith Could you perhaps send me a link to your question?

H (a horizontal flip) leaves 2 and 4 where they are, and switches 1 and 3.

i mentioned it up above earlier in-chat

oh okay so it's V?

rotating (R) sends 1 to 3, 2 to 4, 3 to 1, and 4 to 2 (a "double-swap")

@JohnSmith Then, I suggest you ask it as a full question in Math.SE. In all probability, it will be answered. Incase it is not, you will get helpful comments atleast. Many users do not frequent these chatrooms.

tried already

but since we did H FIRST, 3 is where 1 used to be, and 1 is where 3 used to be.

@DavidWheeler Uhum, thank you.

@David: I have to go, is there a way we can continue later?

so doing R after H, puts 1 and 3 "back where they were", but 2 and 4 wind up switched.

so RH = V, the vertical flip

now you can check this on your own, but abstractly, we would say:

G = {e,a,b,ab} where a^2 = b^2 = e, and ab = ba

and that "abstract" info captures what is happening with the "physical" operations we are doing with our rectangle

that is, you can create the cayley table from just that, but it's a lot more compact.

how do you prove that corresponds to a unique cayley table?

using this: G = {e,a,b,ab} where a^2 = b^2 = e, and ab = ba, try to create a cayley table. see what happens.

ok here goes

so ab is a unique element right?

yes

you can use c = ab if you prefer, but it's not necessary

so i filled in a^2, b^2, and put two circles in ab and ba to mean they must be the same

i'm not immediately sure why i can't put arbitrary elements in the other spaces

@leo I was out; now I'm back :-)

well the first row and the first column should involve the identity, right?

so those are no-brainers

does that depend on the fact that G is defined as a group?

@robjohn Hi

yes, i'm not asking you to prove it is, i'm telling you ahead of time it is

so a*e = a, for example

and a*b = ab (its customary to omit the *)

so you should be able to fill 10 out of the 16 squares with no problem at all

are there any other communities?

@leo I think you're wrong unless you are taking limits.

okay so e means it's the identity

yes, that is standard notation

:-)

i count 11 squares filled out no problem?

(because with different definitions e might be the function f(x) = x, or 0, or 1, and we want a "neutral symbol")

ok, so you deduced that b*a = ab, good :)

@robjohn I'm about to ask something. I'll let you know when the post is ready

what could (ab)a be?

well ab = ba, so (ab)a = (ba)a = b(aa) = b(a^2)

what is a^2 again?

wow

e

so (ab)a = b(a^2) = be = ?

b

now you have 12

but how do i know theres not some other derivation that ends with a different element

because i told you it was a group (and the cayley tables for groups have unique entries)

i'm not asking you to PROVE its a group, that would be a different matter. i'm telling you it's a group, and you are just "calculating" its cayley table.

but how do i know there's not another group that satisfies the eqns

there might be...i didn't say this was the "only" group with this cayley table

how do i know theres not another cayley table that satisfies the eqns

let me put it this way: if you find an answer for an entry (by hook or by crook), you're good...all you have to do is complete the table

for example, what is a*(ab)?

(aa)b = eb = b

ok, so you must nearly be done, now

what is b*(ab)?

is it possible there's another cayley table that satisfies the eqns?

when you finish, i will answer that

b(ab) = b(ba) = (bb)a = ea = a

and (ab)a= ? and (ab)(ab) = ?

let me rewrite this

are you finished?

not yet

ok done

i got {{e,a,b,ab},{a,e,ab,b},{b,ab,e,a},{ab,b,a,e}}

now: the answer to your question is, this cayley table is unique....except for a possible "re-naming" of the elements.

what i want you to do is: copy this table, but use I for e,R for a, H for b, and V for ab.

@JonasTeuwen are you there?

@MattN Yes.

Aces : )

which would be {{I,R,H,V},{R,I,V,H},{H,V,I,R},{V,H,R,I}}

@David, one minute...

@JonasTeuwen Regarding leo's integral question above: does $f$ have to be uniformly continuous? It seems to me that if $f$ is continuous and $\int_{\mathbb R} f(x) dx < \infty$ then $\lim_{|x| \to \infty} f(x) = 0$.

i agree with leo, using the fundamental theorem of calculus

@David: done

@MattN I'm practically dead. Gief me proov.

@JonasTeuwen : D Ok, hang in there.

now, get a "real" rectangle, like a piece of blank printer paper, and verify your cayley table, by actually DOING the operations.

R = 180 degree rotation, H = horizontal flip, V = vertical flip, I = do nothing (or rotate 360 degrees)

@JonasTeuwen No, never mind, I'm too tired. I'll ping you tomorrow when I know more.

@MattN You too! Sketch it 8-).

@David: i believe that the eqns will not lie about the rectangle, so if i deduce from them that VV=I, i believe that's true about the rectangle

yes i verified it

@JonasTeuwen I hadn't actually produced a proof.

so you see our "abstract" description actually describes a "real thing"

hmm

wow

so, that means the cayley table is unique?

the "formal system" is just a SHORTER way to say it

you can see that it is...there's only so many ways to put the piece of paper back where it was, right?

Good night!

@David, you mean as a composition of 2 elements?

you can try RHRHVRRVH, if you like, the "end position" will always be the same as one of our four operations

@MattN Sleep well.

@David: right

that's what "closure" means

since we can simplify any pair to a single element, we know we can simplify anytyhing

it means that those 4 elements are always the result of any product (composition) no matter how long and ugly

what does this have to do with my earlier question?

i really will have to leave at 5 after :(

unless... my iphone..

well, if the cayley table for a group is unique, then if ab = e, that means in the row a*_

b is the ONLY column in which "e" occurs

why

i mean, yeah, the associatve argument

but...

for some reason i don't fully believe it

look at the table you just created, do you have 2 occurences of e in ANY row or column?

in fact, do you have 2 of the same of ANYTHING in any row or column?

it's sort of magic, isn't it?

i can see why particular examples of groups have unique inverses

but yeah, it's hard for me to see how it works for a "general" group

well, the "group axioms" were created to "handle all the examples at once"