Mathematics - 2018-11-29 (page 1 of 3)

user330477

00:10

Where is the complex square root function analytic?

Semiclassical

welp. according to matlab, my 6D polytope has 14 vertices and 152 facets.

so that's...fun

Balarka Sen

rippity rippity rip rip

Semiclassical

oh, how I wish that were true

for better or worse, what matlab gives me is a 152-by-6 array

Ted Shifrin

oh oh ... is Balarka back to non-sleeping?

Semiclassical

each row tells me which 6 of the 14 vertices I specified belong to a given facet

Balarka Sen

00:18

@TedShifrin Passionately so, yes

Semiclassical

(six points in 6D space define a facet)

Ted Shifrin

See ... it's been so long, I forgot that it was unsleeping.

Leaky Nun

@Ted hi

Ted Shifrin

hi Leaky

Leaky Nun

@Ted will perfectoid spaces be textbook material 100 years later, like Lebesgue proposed his measure 100 years before?

what is textbook material?

Semiclassical

00:23

Depends on what textbook you're talking about, presumably

Ted Shifrin

Things of "general" enough interest that people teach them in many courses.

Semiclassical

(and depends on whether we'll still have textbooks at that point, lol)

Ted Shifrin

We barely have them now.

Semiclassical

Exactly.

Leaky Nun

@Ted and they would have to be extensively verified before integrating into curricula?

Semiclassical

00:25

Plus, what counts as textbook material is always shifting

Ted Shifrin

This is too vague a discussion for me.

Semiclassical

Some dimension counting: Each of my facets is some 5D hyperplane in 6D space. I intersect that with a hyperplane to get a 4D set; repeating that with two other planes, I get down to 2D. So the intersection of each facet with the three slicing planes should be a 2D edge.

(I feel like I'm tripping over my words there)

Ted Shifrin

I think that it's too awkward to have to refer to a 3D affine subspace as being obtained by repeated intersection with hyperplanes. We do know about dimension.

Leaky Nun

man, proofs in measure theory are very lengthy

is it because of not enough lemma extraction?

or must they be long?

Semiclassical

Not sure I follow. Do you mean: It's sufficient to describe my final 3D polytope as a 3D affine subspace of my 6D polytope?

Leaky Nun

00:31

It doesn't help that they divided it into 9 steps

@isquared-KeepitReal square it again to keep it positive!

Ted Shifrin

I just meant what you just did. If you have two (affine) subspaces $U$ and $V$ in "general position" (meeting transversely is the fancy term) in $\Bbb R^n$, then the dimension of their intersection is $\dim U+\dim V - n$.

Semiclassical

Ahh, okay.

Ted Shifrin

If that number is negative, then they don't intersect, of course :P

Semiclassical

Does that generalize to multiple affine subspaces? e.g. if I had affine subspaces U,V,W in general position in $\mathbb{R}^n$ then the dimension of their intersection is $\dim U+\dim V+\dim W-n$?

No, I guess it shouldn't be. Intersecting U and V gives the dimension you stated, and then I intersect that with W to get $(\dim U+\dim V-n)+\dim W-n$

Ted Shifrin

No.

Semiclassical

00:35

so that'd be $-2n$ rather than $-n$

Ted Shifrin

It gets trickier. That might be right. I haven't thought about it. Let's try 3 planes in $\Bbb R^3$.

Semiclassical

each subspace is dimension 2 and n=3, so my (revised) formula would give 2+2+2-2(3)=0 which is right

Ted Shifrin

But is it really?

They might not meet at all.

The line of intersection of two of 'em might be parallel to the third.

Semiclassical

hrm

yeah

so the issues re: general position get more subtle

Leaky Nun

What's a good self-contained measure theory book that would cover Haar measure, or at least Riesz representation theorem?

Ted Shifrin

00:39

But what I said isn't generic. You expect it to be $0$, yes.

Semiclassical

Yeah.

Ted Shifrin

So you have to be careful about "general position."

user330477

Where is the complex square root function analytic?

Semiclassical

In my case, my three slicing planes are just x1=1, x2=1, x3=1

Ted Shifrin

@user330477: What is the complex square root function? Is it even a function?

Semiclassical

00:40

so it's probably just easier to say that i'm intersecting with the affine subspace (1,1,1,x4,x5,x6)

which is definitely 3D

Ted Shifrin

Right.

user330477

@TedShifrin It is actually the principal branch of the complex square root function.

Semiclassical

Does that $\dim U+\dim V-n$ at least provide an upper bound on the dimension of the intersection?

Ted Shifrin

What does that mean, @user330477? If you know what you just said, you have the answer to your question.

No, @Semiclassic. In non-generic situations you can have too big an intersection. Think of two identical planes in $\Bbb R^3$ or two identical lines in $\Bbb R^2$.

user330477

@TedShifrin It is defined on $\mathbb{C} \setminus {(-\infty, 0]}$. But stiil how do I show that it is analytic on its domain?

Semiclassical

00:42

ah, drat

Leaky Nun

you differentiate it...

Ted Shifrin

Inverse function theorem will do it, since $f(z)=z^2$ is analytic.

Or you compute the derivative from the definition. @user330477

Balarka Sen

@TedShifrin Rapid-fire: If you were to define general position of two stratified subsets of a manifold, what would be your most immediate definition?

(Mine would be transverse stratum-wise, but I am sure that has issues)

Semiclassical

@TedShifrin I start to see why 'transversality' is such a big deal in various contexts

Leaky Nun

and one must be careful with branching

Ted Shifrin

00:43

Ugh, @Balarka.

Leaky Nun

one must always choose a small enough neighbourhood

Ted Shifrin

@Leaky: There is no problem once he restricted domain.

Leaky Nun

I see

user330477

@TedShifrin I see what you mean now. Is it true that the principal branch of the complex sqare root function is analytic on its domain.?

Ted Shifrin

That starts to sound like a multitransversality issue, @Balarka. See Guillemin & Golubitsky.

Balarka Sen

00:44

@TedShifrin That is not an acceptable answer!

Hrm, I shall whip out that book

Ted Shifrin

I just gave two ways you could prove that, @user330477.

user330477

@TedShifrin Ok, thank you. Now, I am quite convinced.

@LeakyNun Thank you to you as well.

Leaky Nun

it's not very helpful that Radon measure sometimes needs to be inner regular on all Borel sets, and sometimes just on all open sets

are they equivalent?

Balarka Sen

@LeakyNun I have an exercise for you

Leaky Nun

what, define general position?

Semiclassical

00:47

@TedShifrin one point I'm confused on. A facet of a polytope isn't an affine subspace, isn't it? It's a subset of one, but not one unto itself

Balarka Sen

Write down a complete proof of the following fact: $X$ be a topological space with the Borel measure and $f : X \to \Bbb R$ a measurable function. Then for every $\epsilon > 0$ there is a continuous function $g : X \to \Bbb R$ such that $S = \{x \in X : |f(x) - g(x)| \geq \epsilon\}$ has measure $\mu(S) < \epsilon$.

Leaky Nun

what do you mean by the Borel measure?

Ted Shifrin

Right, Semiclassic. So, you may have smaller intersections (like empty) even if the whole subspaces intersect.

Semiclassical

Hrm.

That makes sense.

Balarka Sen

with *a

Basically this says every measurable function can be approximated in measure by continuous functions.

Semiclassical

00:50

yeah, I get it. Suppose I have a filled square and I draw a line through its center (without being parallel to the sides of the square). the intersection will be some line segment within the square

Balarka Sen

I thought about it a couple days ago and I think I have a fairly strong sketch, but putting all the details togather is really annoying

Semiclassical

If I now slide the line away from the origin, though, I'll see that segment shrink until it collapses to a point and then disappears

(that's not quite the right way to put it, since i chose to put my line through the origin initially. but the picture as the line gets slid out of the square is right)

Balarka Sen

@Ted Mostly I think that having every pair of strata from the two respective subsets intersect transversely is an insanely strong condition to have

user330477

How do I see that $\frac{\cos(z)-1}{z^2}$ is not analytic at $z=\infty$? Don't tell me that replace $z$ by $\frac{1}{w}$ and observe that this has a singularity at $w=0$. In this case, it is not even defined at $w=0$, so this seems circular to me.

Semiclassical

what I should probably do at this point, just to have something concrete: Pick one of my 152 facets, write down the equation of the 5D facet (convex combinations of the vertices), then intersect it with my 3D affine subspace and see what 2D subset I get

so I guess I get to have fun with that :P

Ted Shifrin

00:59

The question is whether you have a removable singularity at $w=0$, @user330477.

@Balarka: I sorta used to know this years ago, but now I totally don't remember anything.

Balarka Sen

Haha it's ok, your insights are still very valuable

user330477

@TedShifrin When I write the power series, this does not tell me anything whether the limit exists or not.

Ted Shifrin

We actually used various Whitney conditions seriously in a paper or two.

Balarka Sen

I need to get my hands into G&G but I want to finish reading this section on L^2 cohomology from Kirwan

Ted Shifrin

Sure it does, @user330477. If there are terms with negative exponents, you don't have a removable singularity.

Leaky Nun

01:03

@BalarkaSen what's a good book on measure theory?

Ted Shifrin

You also have to decide if you mean analytic as a map to the Riemann sphere (so the value of $\infty$ is allowed) or not ...

Balarka Sen

I haven't read a proper book on measure theory

Ted Shifrin

Royden is a classic, @Leaky. Modern people like Folland and Stein/Stakarchi.

Balarka Sen

I learnt whatever I learnt from Durrett, which is more like a probability book. I looked at Stein-Shakarchi's real analysis text, which has a good deal of measure theory.

I spent some time glancing over chapter 2 in Federer

That's all

Ted Shifrin

LOL, NO to Federer.

Leaky Nun

01:05

@TedShifrin are there proofs that span 3 pages?

user330477

@TedShifrin The power series is $-\frac{1}{2}+\frac{1}{24z^2}-\frac{1}{720z^4}+\cdots$.

Semiclassical

oh, joy

i just tested the first facet matlab gave me

and the intersection of it with my 3D affine subspace is a point

Balarka Sen

told ya

Ted Shifrin

So not analytic.

user330477

@TedShifrin "If there are terms with negative exponents, then we don't have a removable singularity." Why is this true?

Leaky Nun

01:07

@user330477 complex analysis

Semiclassical

somehow I think I'm going to be seeing a lot of that

Balarka Sen

I warned you man I warned you before anyone did

Semiclassical

oh, wait

Ted Shifrin

Work that out at $z=0$ for yourself (forget about thinking about infinity). Understand how to see pole, removable singularity, and essential singularity from the Laurent expansion.

user330477

@LeakyNun This was not what I was looking for.

Semiclassical

01:08

hmm, maybe I'm being too restrictive.

Leaky Nun

you essentially want the result that Laurent series expansions are unique

user330477

@TedShifrin This is just a question about power series. We have not discussed Laurent series, pole, etc. so far.

Ted Shifrin

Oh, that's unfortunate.

Leaky Nun

and various other results that I can't be bothered to list

user330477

@TedShifrin How do I this question, without invoking what these results about Laurent series?

Ted Shifrin

01:09

I don't know. Ask your professor what he expects.

OK, I'm leaving for tonight. Bye, all.

Rithaniel

Hi Ted, bye Ted.

Prashin Jeevaganth

For a theorem like" any tree with n vertices(n>0) has n-1 edges", am I right to conclude that "has" doesn't have an implication on the exact number? In general, "has n items" can never restrict a quantity to "exactly n items"

user330477

@TedShifrin A friend of mine told me when he went to his office that he just said the exact same thing that you said about negative exponents, but without any explanation.

Rithaniel

I have encountered a problem which is apparently expecting an infinite sum as the value of a residue

Leaky Nun

@Rithaniel that's not unheard of

Semiclassical

01:14

What's the problem?

Rithaniel

Yeah, the exact problem is that I need to show $\int_{\partial B_3 (0)} e^{z+1/z}dz=2\pi i\sum_{n=0}^{\infty}\frac{1}{n!(n+1)!}$

Leaky Nun

oh no I don't like that function

Semiclassical

Oh hey, Bessel function stuff

Rithaniel

So, just as a general question: what makes something like this arise?

Semiclassical

It may be better to look at that as $e^z e^{1/z}$

Rithaniel

01:16

Hmmmm, the product of two infinite sums?

Leaky Nun

@Rithaniel composing functions

Semiclassical

Yeah. But you're not interested in the entire infinite product

Rithaniel

Yeah, just the multiples where I get a -1 in the exponent, which there will be infinitely many?

Leaky Nun

right

Rithaniel

Neat

Leaky Nun

01:19

Let $\Bbb R_d$ be the real numbers with the discrete topology

let $X = \Bbb R \times \Bbb R_d$

this is a locally compact hausdorff topological group so it gets a Haar measure

let $f : X \to \Bbb R : (x,s) \mapsto \operatorname{sgn}(x)$

Semiclassical

@Rithaniel I forget: what does the subscript 3 mean here?

Leaky Nun

$f$ is obviously measurable

Semiclassical

circle of radius 3?

Leaky Nun

@Semiclassical yes

Semiclassical

mmkay

Rithaniel

01:21

That whole bit is the boundary of ball radius 3 centered at origin,yeah.

It threw me for a loop when I first saw it, so I guess I should be more careful tossing it around, myself.

Leaky Nun

but then for any continuous function $g : \Bbb R \times \Bbb R_d \to \Bbb R$, for any $s \in \Bbb R_d$, $\mu(\{x ~:~ |f(x,s) - g(x,s)| \ge 0.5\})$ must be positive, so $\mu(\{ (x,s) ~:~ |f(x,s) - g(x,s)| \ge 0.5 \}) = \infty$?

@BalarkaSen did I make a stupid mistake?

Semiclassical

The reason I was mentioning Bessel functions, btw: If you plug that sum into WA, what you get is $\sum_{n=0}^\infty \frac{1}{n!(n+1)!}=I_1(2)$ where $I_1(x)$ is the 1st-order modified Bessel function of the first kind

And if you were to run that story again with a factor of $z^m$ in the integral, you'd again get some $I_k(2)$

Prashin Jeevaganth

Does anyone know how to prove this?

Two subgraphs of a graph are said to be edge-disjoint if
no edge is common to both subgraphs.
Show that, for all integer n ≥ 4, Kn has at least two edge-disjoint spanning trees.

Semiclassical

(I forget what the relation between m and k is. It's either $k=1+m$ or $k=1-m$.)

Leaky Nun

@PrashinJeevaganth play with a few examples

Balarka Sen

01:27

@LeakyNun I'll look at this counterexample tomorrow morning. It's possible you want $X$ to be nice

Leaky Nun

@BalarkaSen locally compact T2 abelian group is very nice

Balarka Sen

It's not first countable.

Leaky Nun

oh well

Balarka Sen

Try to prove it for $X = \Bbb R$, Lebesgue measure first

Leaky Nun

I believe you mean second countable?

Balarka Sen

01:29

I'll check your example tomorrow and see what exact hypothesis should suffice

Leaky Nun

ok thanks

Balarka Sen

@LeakyNun $\Bbb R \times \Bbb R_d$ contains $\Bbb R_d$, an uncountable discrete subset, which is not first countable, so neither is $\Bbb R \times \Bbb R_d$.

Semiclassical

(The upshot, to my way of thinking, is that a perfectly good definition of modified Bessel functions of the first kind is as Laurent coefficients to the function $e^{t(z+1/z)}$)

Prashin Jeevaganth

@LeakyNun I'm quite lost to even find examples, if I consider Kn to be a linear tree of 4 vertices, where are the 2 disjoint spanning trees? There's only 2 ways to making a spanning tree, which is left to right or right to left, but aren't they using the same edges?

Semiclassical

Isn't Kn usually taken to be the complete graph on n vertices?

Prashin Jeevaganth

01:31

Is that a standard notation?

Semiclassical

i think so, c.f. en.wikipedia.org/wiki/Complete_graph

(though the origin of that convention seems pretty obscure)

Balarka Sen

@LeakyNun Oh, first is the local condition, not the second.

You're right, I did mean second :P

Leaky Nun

@BalarkaSen I think you should sleep lol

Semiclassical

Under that assumption, the smallest example is the complete graph on 4 vertices

Balarka Sen

They should just call it globally countable and locally countable

Semiclassical

01:33

in which case I think you indeed have two edge-disjoint spanning trees

Leaky Nun

@BalarkaSen I agree

Balarka Sen

Petition to change topology terminology right now

just cuz my whim

Semiclassical

@BalarkaSen well, you've already got a change of the first kind. now you just need a change of the second kind.

Leaky Nun

so maybe prove it for compact $X$

and then extend to $\sigma$-compact using the geometric sequence trick

Rithaniel

$e^{1/z}$ has an essential singularity at 0, right?

Leaky Nun

01:36

@Rithaniel yes

Prashin Jeevaganth

@Semiclassical for K4, which is a square with a cross in the middle, is the (disjoint spanning tree the tree with 3 consecutive sides of the square) and (the tree with the side left out and the diagonals?

Balarka Sen

@Leaky Good plan

I'll think about this carefully tomorrow, I'm too bummed right now

Rithaniel

Alright, and $\frac{sinx}{x}$ has a removable singularity?

Leaky Nun

@Rithaniel yes

Balarka Sen

We should study measure theory properly togather

Leaky Nun

01:36

good idea @BalarkaSen

Semiclassical

@PrashinJeevaganth seems right, yeah

Rithaniel

(Alright. Just testing instincts, danke schon)

Leaky Nun

kein problem

Semiclassical

alternatively: 1) top side, bottom side, one of the middle diagonals, 2) left side, right side, and the remaining diagonal

Prashin Jeevaganth

@Semiclassical right I missed that out

Semiclassical

01:38

i imagine both of those are isomorphic up to a rearrangement of the indices

mostly I like that choice because it's alternating on the outer edges

Prashin Jeevaganth

I think there's some algorithm out there that says the number of minimum spanning tree for a complete graph is n^(n-2)?

Semiclassical

could not tell you

Leaky Nun

@MatheinBoulomenos what on earth

Semiclassical

Anyways. Next one up is K5

i.e. commons.wikimedia.org/wiki/File:4-simplex_graph.svg#/media/…

Leaky Nun

Semiclassical

01:41

there we go. I can never remember how to get wiki image links to work

@PrashinJeevaganth I think the disjoint trees you gave for K4 may be the example to imitate here

draw one tree along the outer edges, excluding one edge

Prashin Jeevaganth

Hmm, what's the best way to prove this? I would like to assume that there's definitely at least 2 disjoint cases in K4, then from K4 to K5 we have to add 4 edges to the complete graph, and from the 4 edges we can choose any 2 and it will still be disjoint

Semiclassical

hmm, but is the other graph you'd get actually a spanning tree

Prashin Jeevaganth

erm

what do you mean?

Semiclassical

I mean that i'm not sure the construction I was giving for the edge-disjoint trees in K5 is right

I think your inductive approach has merit, though

Prashin Jeevaganth

hmm

But I will have that problem of "not considering all cases" right?

Semiclassical

01:45

well. presumably, your proof will look like this

Prashin Jeevaganth

I learnt somewhere for geometric problems like this, I have to write it as consider K(n+1) then remove the edges first

instead of going bottom up starting from Kn and say K(n+1)

Semiclassical

hmm

you may be right. I haven't done any graph theory proofs in a while

Prashin Jeevaganth

It's quite weird because usually graph theory I do contradiction

construction or induction seems rare

I'm praying on good faith that induction works :)

Semiclassical

lol

another approach would presumably be to give some specific construction for the two trees on Kn

I dunno if that's feasible tho

i mean, if you work it out for something like K8 and there's some reasonably nice pattern to the trees, then you may be able to do that

Prashin Jeevaganth

@Semiclassical the suggested answer for this problem did that, but I don't understand it. Definitely can't produce that under exam stress

It came up with a general construction and divided into cases that there is always a walk

Semiclassical

01:51

hmm

Leaky Nun

02:16

ugh, intersection of unions, the favourite trick in measure theory

Rithaniel

02:30

Do you guys ever find yourself looking at a theorem or expression and find yourself seemingly unable to parse what it states? The mathematical equivalent of reading the same line of text over and over because you're just not absorbing it.

maths researcher

02:47

hello

I have a question

what would be an intuitive, non formal explanation of a manifold?

Prashin Jeevaganth

Could someone be kind enough to look at my proof to see if there's anything too vague with my proof? paste.ofcode.org/LP8x5YxnMsN9Fciyieywnm

Rithaniel

The best way that I've learned to understand what a manifold is, is that it's a type of space which mimics Euclidean space when you "zoom in" far enough.

maths researcher

Thank you Rithaniel, so what would a manifold be formally defined as in euclidean space R^n?

so R^n is a manifold

is there way to prove that a vector space is a manifold in R^n?

Rithaniel

Well, I'm just copying and pasting the formal definition from wikipedia, because I've not yet studied this stuff extensively, but: "A topological space $X$ is called locally Euclidean if there is a non-negative integer $n$ such that every point in $X$ has a neighbourhood which is homeomorphic to the Euclidean space $E^n$ (or, equivalently, to the real n-space $\mathbb{R}^n$, or to some connected open subset of either of the two). A topological manifold is a locally Euclidean Hausdorff space."
Also, yes, $\mathbb{R}^n$ is a manifold.

maths researcher

without topology, just analysis, using the standard metric abs(x-y)

?

Rithaniel

02:58

It's difficult to show that there exists a homeomorphism between two spaces without using topology, so I'm unsure.

maths researcher

Why is homeomorphism a sufficient condition rather than a bijection?

Rithaniel

A homeomorphism preserves openness, which is a major thing in how mathematical systems behave.

maths researcher

I mean, if I had a structure where I wanted to show a homeomorphism exist how would a homeomorphism differ from a bijection?

oh

i see

Rithaniel

Well, I didn't explain what "openness" is, but you get the idea.

maths researcher

I'm currently searching it up

well, since I havent found a good explanation, what would openness be?

Rithaniel

03:04

A homeomorphism must be continuous and have a continuous inverse. Where continuous means that if a set is open in the image, it's preimage is also open in the domain.

Well, take an example in $\mathbb{R}$ under the standard metric topology. $(0,1)$ is open, and $[0,1]$ is closed.

maths researcher

How would I prove a set is open?

and how would I show that (0,1) is homeomorphic to [0,1]?

i mean

homeomorphic to R

Rithaniel

Well, that's the thing about topology. When you're starting out defining a topology, you simply decide what's open, so long as you follow the rules of a topology.

Rules are: The whole space is open. The empty set is open. Arbitrary unions of open sets must still be open. Finite intersections of open sets must still be open.

maths researcher

Oh, so if I assume R^n is open then that would mean any set in R^n is open

then where would the concept of close sets come into place?

Rithaniel

Now, the standard metric is called "standard" because it has all the rules you see in everyday mathematics. A lot of topology involves finding homeomorphisms between $\mathbb{R}^n$ in the standard metric topology and other topologies.

closed sets are sets whose complement is open.

maths researcher

how would I prove (0,1) is the complement of [0,1]?

Rithaniel

03:11

It's not.

maths researcher

what do you mean by complement then?

Rithaniel

$[0,1]$ is the complement of $(-\infty , 0)\cup (1,\infty )$

The complement of a set is everything not in the set. (I'm teaching stuff out of order, to be honest)

Similarly, $(0,1)$ has a complement of $(-\infty , 0]\cup [1, \infty)$, which is closed.

maths researcher

So what I understood is that if I have a topological space, I should assume that the whole space is open, and so any union of subsets of the space should also be open. If I want to show continuity in a set, I should first assume its open in the image then if its preimage is open then its continous. What would be the preimage of the function 1/x

on the interval (f(0),f(1))

?

i apologize for any inconvenience

Rithaniel

03:37

You're okay, I was just afk. The function needs to define what it is from and what it is to. I assume this function 1/x is from and to $\mathbb {R} $ minus the point 0. In which case the preimage is $\mathbb {R} $ minus the point 0.

maths researcher

I see, I have another question with regards to probability in which I haven't been able to solve

may I ask you?

Rithaniel

I would also assume you mean real numbers under the standard metric topology.

maths researcher

yes

Rithaniel

Well, feel free to ask. If I don't answer, someone else might.

maths researcher

Assuming I have a set {1,2,3,4,5,6,7,8} and a number of subsets satisfy two conditions, each subset has exactly 4 elements and each element of {1,2,3,4,5,6,7,8} belongs to exactly 3 subsets

how many subsets would i have?

I know its 6 but i'm not how to get that answer

Semiclassical

03:42

i'd call that combinatorics rather than probability but that's just semantics

Prashin Jeevaganth

Does anyone know in general what kind of transitive relations are out there? I have an upcoming test and I can't detect transitivity fast enough

maths researcher

sorry, you're right. It is combinatorics

Semiclassical

alas, that's the only comment I've got. I don't really know where to start with this

Prashin Jeevaganth

@Semiclassical you know anything about transitive relations?

like a quick way to detect some

Semiclassical

not beyond the definition, no

i mean, if you want to detect something being non-transitive, you look for an element which is related to two un-related elements

Rithaniel

03:47

You probably want to rewrite it as {1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,8,8,8} and, boom, you have 24 elements. Divide it up into sets containing 4 elements each and you have 6 sets.

Semiclassical

4 distinct elements, but yeah, that's a good strategy

Rithaniel

The number of ways you can divide these up is a little bit more difficult

Prashin Jeevaganth

At least they didn't ask the exact ordering of the groups.

Semiclassical

Yeah. It's a strategy to finding the answer; it's not the answer itself.

Rithaniel

Well, the question was "how many sets do you have" not "how many ways can you construct these sets"

Semiclassical

03:50

true

maths researcher

well, I know there are 2^8 subsets, and also there are 8C4 subsets with 4 elements, is there a way to proceed after this?

Buddhini Angelika

05:39

@TobiasKildetoft thanks for the reply. But I do not have a good knowledge about it. By congruent I think they mean that st^(-1) belongs to (Z_5 X Z_5). I am referring the proof in page 17 of this document. cs.uleth.ca/~morris/Research/low-order.pdf

There under case 1 second bullet point and the first line under case 2 is mentioning about such a congruence.

I would like to understand how we can think of such a congruence when the order of s and t are equal to 3

Anyway thanks a lot for helping :)

P.S. How can we say that st^(-1) belongs to (Z_5 X Z_5)?

Prashin Jeevaganth