Mathematics - 2018-12-18

Migos

12:19 AM

hello

Secret

1:55 AM

3 hours ago, by Ultradark

What is a space in terms of mathematics

2 hours ago, by Ultradark

Who is the auther of "Geometry: A Geometric Approach"

2 hours ago, by Ultradark

Are there parts of mathematics that are not useful

It's inaccessibly annoying to see unanswered and questions that are deliberately ignored from answering in this chat, that it will be better to never have those questions asked in the first place

I hate RHV, but I hate seeing dangling questions a lot more

If I see anymore of these, I am putting the source out of view

> If you knew you are being ignored, STOP ASKING, you are only annoying us with loads and loads of question that reminds the existence of being ignored

Ultradark

2:26 AM

Hey...I asked those questions

I asked questions three

:)

what didst thou just delte

Secret

DON'T Subscribe to Pewdiepie

►

the Vox link does not work

Ultradark

why not?

Secret

white supremacy and nazism flirting again, said vox

But then, I have no problem watching this world burn if it can ensure I can never ever see people ignoring people again

Migos

@Secret Does that include my question? I couldn't possibly tell beforehand if it was gonna get ignored (deliberately or not).

Secret

@Migos I don't know, the community here tend to be more readily answering abstract algebra questions

I recall in the past that sometimes questions of your form get answered the next day when some users stumbled upon it

Ultradark

2:36 AM

This chat does seem to skew hard away from analysis

and graph theory

and biholomorphic 2-forms. (jk)

Secret

I don't think there are many graph theory experts here. I have seen many graph theory questions got unanswered, at times users saying they knew nothing about it

For analysis it depends, usually when Daminark is on, there is some spike in analysis

Ultradark

I've asked a bunch of questions on here

Migos

@Secret I only asked because I certainly wouldn't want to annoy anyone.

Secret

@Migos I think you are fine, you seemed to be recently on

Migos

@Secret Thanks. Yeah, pretty much still trying to figure out the 'norms', if that makes sense. :]

Ultradark

2:40 AM

I have asked 61 questions

Secret

Nov 6 at 0:00, by Ultradark

Hi chat

Nov 6 at 0:30, by Ultradark

Hey Ultradark!

Nov 6 at 0:42, by Ted Shifrin

Please don't spam the room like this.

Ultradark

Oh know

The past is coming back to bite me

Secret

Tbh, unlike most users, I can ignore huge reams of messages because I am the most prolific reams of message poster. However:

If the message contains a reminder of someone ignoring someone or the concept of hell ban, it is so annoying that should I have expert hacking skills, I will hack the chat out of existence

I am that kind of person that will start WWIII just to make absolutely certain no human beings can ignore each other anymore (and also the inverse concept, trolling)

My resentment on ignore and disengagement is not something a Berkeley cardinal can capture

F.Tasos

anybody now how to explicitely find the integral submanifolds of a distribution ?

Secret

Manifold people are not on atm

Adam

3:51 AM

what's a Berkley cardinal?

ok I must retreat to the adam box

Secret

4:07 AM

Berkeley cardinal

In set theory, Berkeley cardinals are certain large cardinals suggested by Hugh Woodin in a seminar in Berkeley in about 1992. A Berkeley cardinal is a cardinal κ in a model of ZF with the property that for every transitive set M that includes κ, there is a nontrivial elementary embedding of M into M with critical point below κ. Berkeley cardinals are a strictly stronger cardinal axiom than Reinhardt cardinals, implying that they are not compatible with the axiom of choice. A weakening of being a Berkeley cardinal is that for every binary relation R on Vκ, there is a nontrivial elementary embedding...

Akiva Weinberger

6:51 AM

https://www.quantamagazine.org/mathematicians-seal-back-door-to-breaking-rsa-en‌cryption-20181217/

Neat

Secret

@AkivaWeinberger ah no wonder why polynomials are so hard to factorise when it gets so large. Also besides encryption, I suspect this also have implication on number theory due to polynomials play a huge role in finding factors, sum of squares etc. of numbers

So pretty much most of the large polynomials are irreducible

Alessandro Codenotti

7:38 AM

@MatheinBoulomenos I understand why we want integral and separated thinking about the classical definition of variety, but I have no intuition for the finite type, why do we require it?

Leaky Nun

@AlessandroCodenotti I think you always work with P^n classically

the "n" is finite

Tobias Kildetoft

@AlessandroCodenotti finite type means (classically) that we only consider polynomials with finitely many variables.

Leaky Nun

hey that's what I said :P

Alessandro Codenotti

I see, makes sense, thanks

Tobias Kildetoft

@LeakyNun You wanted the variety to be a projective space for some reason.

Some of us prefer affine varieties

Leaky Nun

7:53 AM

that's an open set of P^n

Alessandro Codenotti

So integral separated schemes of finite type over $k$ are equivalent (as a category) to varieties over $k$ (variety=separated prevariety). If I drop the "separated" I get prevarieties over $k$. What kind of bad things happen by dropping integral or finite type?

Tobias Kildetoft

dropping finite type essentially gives you "infinite dimensional" varieties (which I think are probably not called that)

Also, "finite type" can be weakened to just "finitely generated" which is weaker when not working over a field

which gives some other "weird" things

(actually it only matters when working over a non-noetherian ring)

Alessandro Codenotti

So for example $\mathrm{Spec}k[x_1,x_2,x_3,\cdots]$ is integral and separated over $k$ but not f.t.

Tobias Kildetoft

I am not actually certain it is integral and separated since I don't recall ever seeing that example in detail

(it might be obvious, but I am still waking up)

Alessandro Codenotti

Separated should be free since every morphism of affine schemes is separated

And if $A$ is a domain $\mathrm{Spec} A$ should be integral since all of its localizations are also domains so the sheaf is a sheaf of integral domains on a basis

Tobias Kildetoft

8:29 AM

The way to enable developer settings on Android really feels like it is "secret", even though it is perfectly well documented. It just has the feel of an easter egg.

Leaky Nun

8:42 AM

do you guys like kakuro?

Tobias Kildetoft

@LeakyNun those are the number puzzles where you need to put in different numbers that add up to a specified number?

Leaky Nun

right

Tobias Kildetoft

yeah, those are neat usually

though mostly the ones I have tried have been very easy, since they are still not as widespread as sudoku, so people find them hard

Leaky Nun

@TobiasKildetoft kakuroconquest.com

they're available at every difficulty

Tobias Kildetoft

cool

I usually only do such puzzles when I happen to come across them in a newspaper or similar and need to pass the time

@LeakyNun Another good solo game is Ganz Schön Clever brettspielwelt.de/Magazin/…

Leaky Nun

8:52 AM

@TobiasKildetoft oh es ist nicht kostenlos?

Tobias Kildetoft

Sure, it is free

(as the solo variant on computer that is)

The physical game obviously isn't free

The physical game is good too though, and good for playing with people who might otherwise not play board games but might want to play Yatzee

Leaky Nun

@TobiasKildetoft kann ich es spielen aus meine computer?

Tobias Kildetoft

You should be able to. Doesn't it work?

It works on both my computer and my phone just fine (using Chrome on both)

Leaky Nun

oh ok

ich dachte dass das play button eine video ist :P

Tobias Kildetoft

ahh, I see. It does look like it.

Leaky Nun

9:04 AM

also... wann lerntest du deutsch?

Tobias Kildetoft

From 7th grade

(had German in school for 5 years in total)

Secret

10:22 AM

[Random]

Let S,T be infinite sets with the following properties:

$S\cup T $ is finite

Sorry typo

Let S, T be finite sets with the following properties:

$S\cup T$ is infinite

Leaky Nun

10:46 AM

A : R-alg induces res : A-Alg -> R-Alg

Secret

Then there exists a bijective map $f\not\in \Bbb{N}$ such that $f: A \to S\cup T$ for some $f$

Sorry typo

$f$ such that $|f| = |S\cup T|$

and each of $|S| =n,|T|=m$ for some $m,n\in \Bbb{N}$

Now pick some $a\not\in S,T$, then $|S\cup T \cup \{a\}|$ is computed by;

Tobias Kildetoft

11:10 AM

@Secret What sort of set theory can have a finite union of finite sets be infinite?

Secret

That is something I and Holo were discussing in another room. We conclude a model for that will be very weird. What we knew however is the analoguous case where arbitrary union of finite set is finite can be modeled by the set of all absolutely convergent sequence since they are closed under pointwise multiplication and pointwise addition

Tobias Kildetoft

@Secret That does not answer the question. What sort of set theory are you working with underneath this model?

Secret

I would want that to be ZF (ZFC will not allow such sets) but I am not sure whether there is no contradiction

Tobias Kildetoft

ZF also do not allow such sets. If $S$ injects into the natural number $n$ and $T$ embeds into $m$ then $S\cup T$ embeds into $n+m$.

Secret

Hmm, so Holo might be correct that the whole notion of being finite might have to be altered (thus it will no longer be in ZF):

in :--O--:, Dec 10 at 0:53, by Holo

This is a bit harder to achieve without throwing away our definitions

Meanwhile for the "arbitrary union of finite set is finite", based on our discussion, we might be able to model it as follows:

Consider the set of all absolutely convergent series that sums to some finite natural

Then we can identify each limit of the series with a finite set

Now since the series are absolutely convergent, their point wise sum is also absolutely convergent

and likewise their point wise product

Tobias Kildetoft

11:25 AM

What do you mean by "model"? It is a false statement in ZF

Secret

Isn't a model just an algebraic structure that evaluates to true for a given set of propositions? Then it seems the ring of absolutely convergent series with their limits identified with finite sets and addition identified with set union can evaluate the statement "arbitrary union of finite set is finite" to true?

actually wait...

the concatenation of two series that converge to different limits will produce a divergent series

o nvm then

There is a trivial model though if the empty set is the only set that exists in the universe. Clearly, an arbitrary union of the empty set is empty and the empty set is finite

Tobias Kildetoft

How is that a model?

Secret

We can identify the empty set with each element in the ring of all convergent series that limits to zero. Then any concatenation, pointwise addition and pointwise multiplication will always give a series that sum to zero

and we can identify these binary operations with set union

Tobias Kildetoft

Isn't a model actually supposed to satisfy your axioms? Otherwise it seems fairly pointless.

Secret

Well concatenation, summing or multiplying any series in that ring will only produce a series that sum to zero and we identify this with the empty set. That seemed to satisfy the desired axiom that "arbitrary union of the empty set is empty", no?

Tobias Kildetoft

11:40 AM

But what about all the other axioms? Are you trying to build a new set theory with only that one axiom?

Secret

I want to build a set theory with that axiom, and try to retain as much ZF as possible. For that we are not sure how much of ZF axioms we can retain if we insert this weird axiom

Thus I am still experimenting in the other room

Tobias Kildetoft

I am fairly certain you just need to remove the axiom of infinity and everything else should work out

Alessandro Codenotti

$V_\omega$ models ZF(C)-infinity and there surely any union (of necessarily finite sets) is still finite, no?

Secret

What about unions of $\omega$ many finite sets, do we take the sup and thus we still end up with finite sets?

Tobias Kildetoft

@Secret You can't take that union in this model

Alessandro Codenotti

11:48 AM

The only unions you are guaranteed by the axioms are unary unions, with pairing this gives finite unions, without infinity that can be all

Secret

Right thus we never have $\omega$ in that unary union

Tobias Kildetoft

@AlessandroCodenotti I thought you were guaranteed unions over arbitrary sets (but all the sets here are finite hence only finite unions)

Alessandro Codenotti

Yeah that's what I mean with unary unions, for every $A$ you can take $\bigcup A$

But there is no binary union in the axioms $A\cup B$ needs to be defined as $\bigcup X$ where $X=\{A,B\}$ (and pairing is needed to show that $X$ exists)

I don't disagree with you, we're really saying the same thing, you do have unions over arbitrary sets, but they are all finite

My point was that $\bigcup_{n<\omega}A_n$ is really $\bigcup X$ where $X$ is a countable set containing each $A_n$, but no such thing exists in $V_\omega$

Secret

Yeah, that's why I am initially mistaken because I forgot that union $\bigcup_{n<\omega} A_n$ is a countable set hence is not an element in $V_{\omega}$

Anyway, the more interesting axiom I want to include is this paradoxical axiom "finite union of finite set is infinite". One version of phrasing this is the same as asking the question "Is there exists a model in ZF where $\aleph_0$ is singular". The discussion in the logic room seemed to suggest the question is still open. Tobias have pointed out ZF does not allow that, but what remains open is what axioms we have to discard to make $\aleph_0$ singular

I mean, we do have models of ZF where $\aleph_1$ is singular, so it seems weird we cannot extend that to $\aleph_0$

(clarification: Analogous to how singular cardinals are defined, $\aleph_0$ will be singular if its cofinality is finite)

Adam

12:52 PM

So I am a big fan of Hypergeometric terms and Series, I found the Wikipedia basically just vaguebooked a lot, yet also said very little, anyone else enjoy these things maybe wanna throw me a reference and or pay for the a book on the subject for me?

i guess id solicit to ted "smacking" me if it ends up in a peer reviewed publication on the subject mentioned above being sent my way, i mean its still a dirty that no shower will ever be able to wash away but it would feel better knowing it wasn't for money or crack, but a reliable academic resource

tatan

1:24 PM

Is |A^-1|=\frac{1}{|A|} ? How can I prove this? |.| stands for determinant and A is a 3x3 matrix

Tobias Kildetoft

@tatan What is $|AA^{-1}|$?

tatan

AA^-1 is I ,right?

So det of that = 1

Tobias Kildetoft

yes, by definition

tatan

Oh I get it, thanks!! ;-)

Leaky Nun

1:48 PM

@TobiasKildetoft I must object to "by definition" lol

Alessandro Codenotti

Why is it the case that there is no morphism from $\mathrm{Spec}\Bbb Z$ to $X$ if $X$ is a $0$-dimensional scheme? I see that this would give us that the identity on $\mathrm{Spec}\Bbb Z$ factors through a 0-dim scheme and this should be a contradiction, but I don't see why exactly

Leaky Nun

maybe because Z is one-dimensional lol

Alessandro Codenotti

Sure but I don't know how well if at all dimension is preserved by morphisms

Leaky Nun

any ring hom $f : A \to \Bbb Z$ would need to be surjective, and $\ker f$ is a prime ideal of $A$, hence maximal ideal, so $A/\ker f \cong \Bbb Z$ would be a field, contradiction...

so that clears the deal for affine schemes...

Alessandro Codenotti

Why is the kernel necessarily maximal? And there surely are nonsurjective ring homs if $A=\Bbb Z$

Leaky Nun

2:01 PM

assuming that $A$ is 0-dim, every prime ideal is maximal

and there is only the identity ring hom $\Bbb Z \to \Bbb Z$

you're thinking about group homs

Alessandro Codenotti

Oh, ok, I forgot you were talking about 0-dimensional $A$

Secret

[Random]

The following is most likely not ZF:

in :--O--:, 14 mins ago, by Secret

A set $T$ is subfinite if $T \subset A$ where $A$ is finite, $|T|=n$ for some $n\in \Bbb{N}$ and $|\mathcal{P}(T)|<|\mathcal{P}(n)|<|\mathcal{P}(A)|$

Rationale: Powerset axiom only said it will produce all subsets of a given set. It does not really restrict us as to how many subsets are there in a given set, thus leaving open a possible loophole to axiomise a set that has fewer subset than the finite set it bijects to

one possible consequence of this is that for these sets, many partitions will become empty

So I could have anormalous set with 5 elements {a,b,c,d,e} such that it's power set is only {{a,b},{b,c,d,e}}

The Star Wars Room is currently investigating just how many normal rules are screwed by the existence of these sets

In other news, we build new systems by first defining the objects we want to include, derive all consequences from them and then decide what axioms is needed at the last step

i.e. we work "backwards"

user525966

3:38 PM

I need some advice on taking notes for self study, what do you guys use and how do you store them

Abdullah UYU

3:58 PM

I want to share an exercise that I liked, of my real analysis course: Let $f$ be a continuous function on $[0,1]$. Show that there is a point $x$ such that $f(x)=x$. This is interesting because it's easy to convince yourself with the geometric meaning.

Leaky Nun

@AbdullahUYU $f(x) = x + 1000$

Abdullah UYU

On a square, there is no continuous curve from one edge to the opposite one that doesn't intersect the diagonal.

Leaky Nun

but then your real question is just a trivial application of the intermediate value theorem

Abdullah UYU

Yes it is, depending on your definition of trivial.

Abdullah UYU

4:14 PM

@LeakyNun What is that by the way?

Leaky Nun

a counter-example

Abdullah UYU

Is it though? It's not defined on $[0,1]$.

Leaky Nun

sure it is

it sends every $x \in [0,1]$ to $x+1000$

Abdullah UYU

We're considering the functions from $[0,1]$ to $[0,1]$.

Leaky Nun

well you didn't say that

Abdullah UYU

4:18 PM

On $[0,1]$ means that :)

Leaky Nun

nope, it means the domain is $[0,1]$

you'll find texts saying that if $X$ is a topological space then we define $C(X)$ to be the ring of continuous functions on $X$

and what they mean is functions $X \to \Bbb R$

Abdullah UYU

In French, it's said like that: "Soit $f$ une application sur $A$" means that both domain and the codomain are $A$.

Leaky Nun

well this is not french :P

Abdullah UYU

It's pratique :)

Is there a way to express it like in French? With just saying the set. I speculate that it's common on working with a sole set. There has to be a way to simply state it.

Eran

6:25 PM

Does there exist an infinite solvable group?

Tobias Kildetoft

@Eran Sure, take the integers

Eran

as much as I know $\mathbb{Z}$ is not solvable

Tobias Kildetoft

It is abelian and hence solvable

Leaky Nun

lol

rip

Eran

the definition of solvable is that the composition factors of any composition series of $\mathbb{Z}$ are abelian, but $\mathbb{Z}$ does not have any composition series

is it still solvable?

Tobias Kildetoft

6:34 PM

No, that is the definition of a solvable group which has a composition series.

Eran

okay.

Thank you sir.

I'm trying to refute the following claim:
If G is solvable and H has the same composition factors as G , then H is also solvable

Tobias Kildetoft

Do you allow infinite composition series?

Eran

If G is finite then the claim is clearly true, because a finite group is solvable iff it's composition factors are cyclic of prime order.

well, hmm idk what it is it, but let's say yes

Tobias Kildetoft

Hmm, actually it seems that allowing infinite composition series is not usually a thing

maths researcher

hello

anyone here?

Eran

6:42 PM

@TobiasKildetoft what do you suggest then?

Tobias Kildetoft

So since both groups have composition series, the only way they can be solvable is by being finite

maths researcher

guys is my proof for the following correct?

Tobias Kildetoft

Why are you trying to refute the claim?

maths researcher

Prove that the ball B(a,$\epsilon_1$) $\subset$ $\mathbb{R}^n$ is an open set.

Eran

Hmm, I thought it was true but then I thought what happens when G is infinite

maths researcher

6:45 PM

Let u ∈ B(a,$\epsilon_1$). Set epsilon_2=$\epsilon_1$-d(a,u). Then x∈ B(u,epsilon_2) implies that
d(x,a)≤d(x,u)+d(u,a) ≤epsilon_ 2 + d(a,u)≤ epsilon_1

Eran

So you're saying existence of composition factors implies finiteness?

Tobias Kildetoft

@Eran For solvable groups, yes

Not in general of course, since there are infinite simple groups

maths researcher

but why can't we say that d(a,u) is less than epsilon 2?

Eran

@TobiasKildetoft but those infinite simple groups do not have composition factors right?

Tobias Kildetoft

@mathsresearcher Because it might not be due to the way you defined it. You get $d(a,u) = \epsilon_1 - \epsilon_2$ which is not necessarily smaller than $\epsilon_2$.

@Eran Sure they do. They have exactly one composition factor, namely themselves.

maths researcher

6:55 PM

@TobiasKildetoft i meant to say e_1

Eran

@TobiasKildetoft Do you think the claim is true or false then? I'm confused >.<

maths researcher

is it because it is a strict inequality?

from the definition of the ball?

Tobias Kildetoft

@Eran The claim is true, since one group being solvable and having a composition series implies that it is finite, hence so is the other

@mathsresearcher Right, you might have equality. But when would that happen?

maths researcher

@TobiasKildetoft it would happen when the set is closed?

Tobias Kildetoft

@mathsresearcher No, I mean specifically with the way you have defined things. What choice of $u$ would give equality?

maths researcher

7:08 PM

@TobiasKildetoft i'm not sure

They shouldn

because the ball was defined with a strict inequality

GodotMisogi

I need some help finding the primitive element of ℚ(¹³√7, ⁵³√7) over ℚ. Usually the linear combination works out, but proving this one without an insane number of computations seems to be impossible

user193319

7:43 PM

I'm reading an MSE post and am confused about one point. In this post, they define the following equivalence relation on $\Bbb{R} \times \{0,1\}$: $(x,0) \sim (x,1)$ if and only if $x \neq 0$. What's confusing me is, how do I, for example, determine whether $(x,0)$ and $(y,1)$ are equivalent under this relation? I'm use to equivalence relations being defined for arbitrary points in the space, not specific points like $(x,0)$ and $(x,1)$.

Alessandro Codenotti

7:57 PM

This relation is defined for arbitrary points, but the only points with a nontrivial equivalence class are of the form $(x,0)$ or $(x,1)$, $x\neq 0$

Adarsh Kumar

Did anyone know what is the use of "community wiki" option ??

user193319

@AlessandroCodenotti So, if $x \neq y$, then $(x,0)$ and $(y,1)$ would not be considered equivalent? How about $(x,0)$ and $(y,0)$?

Alessandro Codenotti

They are not either

Adam

@GodotMisogi it really isn't my field of expertise excuse the pun but the wiki page says that the linear combination is the generator of the multiplicative group of non zero elements of the field, I must be missing something here definition wise because I thought that would make it the primitive element

Vrouvrou

Hello

Any help for my question:math.stackexchange.com/questions/3043545/…

user193319

8:25 PM

Consider the following equivalence relation on $\Bbb{R}$: $a \sim b$ if and only if $a - b \in \Bbb{Q}$. Is the projection map from $\Bbb{R}$ to $\Bbb{R}/\sim$ an open map?

Alessandro Codenotti

What do you think?

user193319

I don't think so. I think $\Bbb{R}/ \sim$ has the trivial (indiscrete) topology.

Alessandro Codenotti

It does

Semiclassical

I'm writing up a problem and I'm realizing (yet again) I don't know my definitions properly

What I thought I knew: A $d$-dimensional affine subspace of $\mathbb{R}^n$ is defined as $\Gamma=\{x\in \mathbb{R}^n: Ax=b\}$ for some choice of $A\in\mathbb{R}^{d\times n}$ and $b\in \mathbb{R}^n$

But that doesn't seem to be the wikipedia definition

(I'm assuming $A$ is full rank I guess)

Shaun

I've made a start on answering an old, unanswered MSE question. I could do with some help, please. My next post directly below is a link to a recent post of mine in the group theory chatroom about it.

in Group Theory, 21 mins ago, by Shaun

I'm trying to answer the following . . .

Semiclassical

8:51 PM

I'm forgetting set-builder notation. Suppose I want to write the set of $y\in A$ such that $y=f(x)$ for some $x\in B$ subject to some condition on $x$.

Most obviously, I could just define a subset $B'\subseteq B$ on which said condition is satisfied.

And then said set should be something like: $\{y\in A| \exists x\in B(y=f(x))\}$

But I can't remember if that's the right notation.

Alessandro Codenotti

I would write $\exists x\in B\text{ such that } y=f(x)$ which seems just clearer

Semiclassical

looking at wikipedia, $\{y\in A| (\exists x\in B)[y=f(x)]\}$ is probably more appropriate than what I had above

Alessandro Codenotti

The fully formal thing would be $\exists x(x\in B\land y=f(x))$

Semiclassical

Yeah

writing my condition for $x$ as $\Phi(x)$, I think the full notation would be $\{y\in A|\exists x[y=f(x) \wedge x\in B \wedge \Phi(x)]\}$

chandx

Is my reasoning right? I have $f_n(x) = \sqrt{x^2 + \frac{1}{n^2}}$ for $x \in \mathbb{R}$, so I conclude that it's pointwise convergent $f_n \to |x|$, and moreover it's uniformly convergent to $|x|$, because $\left | \sqrt{x^2 + \frac{1}{n^2}}- |x| \right | = \frac{1}{n^2(\sqrt{x^2 + \frac{1}{n^2}} + |x|)} \leq \frac{1}{n^2} \to 0$

and that's because $\sqrt{x^2 + \frac{1}{n^2}} \to |x|$ and it's decreasing for any $x \in \mathbb{R}$, therefore I can make denominator smaller by $\sqrt{x^2 + \frac{1}{n^2}} \leq |x|$, so I'd get in denominator $n^2 2|x|$ and if $|x| > \frac{1}{2}$ then $\frac{1}{n^22|x|} \leq \frac{1}{n^2}$, otherwise if $|x| \leq \frac{1}{2}$ then $\frac{1}{n^22|x|} \leq \frac{2|x|}{n^22|x|} = \frac{1}{n^2}$

Eran

9:10 PM

Let $p,q$ be prime, I need to prove that (1) any group of order $p^a q^b$ is solvable for $a,b\in \mathbb{N}$ if and only if (2) there are no simple groups of order $p^a q^b$ for $a,b \ge 1$. Already proved the first direction, struggling with proving (2)--->(1)

GodotMisogi

@Adam The field isn't finite. To be precise, the primitive element is defined as the single "generator" of a simple extension.

Shaun

The question I'm trying to answer has only been viewed on MSE 59 times in two years and nine months.

Thus I think it's safe to assume that perhaps it was just overlooked and/or forgotten about, so no detailed answer was provided. The old comments are helpful though. Maybe the OP got an answer elsewhere but didn't share it here.

user76284

11:23 PM

Are there any good bounds for $\mathrm{E}[\min_{i:X_i=1} a_i]$ where $X_i \sim \text{Bernoulli}(\theta_i)$?

and $\{i : X_i = 1\} \neq \varnothing$

Nobody recognizeable

11:41 PM

Transcript for

$Mathematics$ Mathematics