Ah great, forgot the irrationals of the form $r+n,n\in\Bbb{N}$

Hmm, if we also remove elements of the form $r+q,q\in \Bbb{Q}$, will we end up removing all irrationals or there is still things left behind?

Is $r$ an arbitrary irrational? $0\in\Bbb Q$ so yes

Right, Hmm...

How about $\Bbb{R/Q} - \{r\in \Bbb{R/Q}, q\in \Bbb{Q}-\{0\} :r+q\}$ ?

That should only retain those irrationals that are not differ by a rational

every irrational is at a rational distance from another irrational

Ok I see. In that case I have no idea how the following is possible unless in those models the rationals are already uncountable to begin with

(Quoted from Wikipedia)

There can be such subsets of $\Bbb R$, but not the irrationals

The reals are made up of rationals and irrationals, so somehow such subset cannot have all the rationals, or all the irrationals. I have no idea if we can ever get a definable example of such subset

In such a model the irrationals will have countable subsets, but also a subset $X\subseteq\Bbb R\setminus\Bbb Q$ that has no countable subsets

Is it possible to get more descriptive about what X is like, or that is the best we can do?

no idea

Also since every irrational is a rational distance away from another irrational, it means for any subset of irrational, there has to exists at least countably many irrationals $r$ that can be expressed as $s+q$ where s is another irrational and q is a rational. And the existence of such countable representation will allow us to have a countable subset. Unless...

1. There are uncountably many such $r+q$ for some fixed $r$ or..,

2. There exists uncountably many unique pairs of (r,s) such that s=r+q

Since we need uncountably many, but there are only countably many q it means q will be used more than once (perhaps at least countably often)

@MatheiBoulomenos also ich probiere selbst es ableiten ohne Rucksicht auf Kummertheorie

@Secret it isn't possible.

@MatheiBoulomenos but who am I to develop the entirety of Kummer theory myself

@Secret unless there are finitely many irrationals of the form r+q where r is a fixed irrational and q is a rational

@Mr.Xcoder hi

@LeakyNun Hi

@Blue just because limit is $0$ on every straight-line path towards the centre doesn't mean that limit exists or is $0$ at all

Is there another easy proof that two consecutive integers are coprime other than considering a $d$ such that $d\mid n,\space d\mid n+1$ and then $\implies d\mid 1\implies d=1$?

@Blue "after substituting $(x,y) = (r\cos\theta,r\sin\theta)$"

that's a straight line path towards the centre

@Mr.Xcoder depends on how much you know

@Blue alright

@_@

Nvm then :-)

@Mr.Xcoder come on

@LeakyNun I mean, I will stick with that proof...

@Mr.Xcoder lol, use Euclidean algorithm

@Blue I'll shut up before you realize that I don't know what I'm talking about :)

I just learned about functions and stuff and I have a rather basic question: $f:\Bbb{R}\rightarrow\Bbb{R},\space f(x)=5-mx$. $m\in\Bbb{R}^{*}$ such that the point $M(\frac{1}{m};m)\in G_f$. What is the value of $m$... I don't really know how to approach this :-/

so f(1/m) = m

and you're done

@LeakyNun Let me write down what I have tried

My try: If $M(\frac{1}{m};m)\in G_f\implies f(\frac{1}{m})=m\implies m=5-m\frac{1}{m}\implies m=5-1=4$... Is this correct?

yes

Thanks :-)

welcome

This time I have like no idea: What are the elements of the set $A$ such that $A=\{x\in\Bbb{Z}^{*}|\frac{5x+19}{2x+1}\in\Bbb{Z}\}$

i tried a lot of things but didn't end up well, not even worth posting my attempt here

what do you mean by $\Bbb Z ^{*}$?

@MatheiBoulomenos The non-0 integers.

don't use that notation

why?

it's inconsistent with the conventions in ring theory

Idk ring theory, so idrc much... Will keep in mind though :-)

yeah, but everyone who knows will think you mean $\Bbb Z^{*} = \{1,-1\}$

Clarification: $\Bbb{Z}^{*}$ means the non-0 integers.

(can't edit)

@MatheiBoulomenos oh come on

I really have no idea how to approach the above

@Mr.Xcoder transform the question statement into equivalent statements

but those that you can more easily manage

@Mr.Xcoder no!

Wait...

@LeakyNun I didn't mean to write that

Better?

no!

Reverse order

?

no!

umm!

you've written the wrong statement three times

the same wrong statement

use words if you can't use symbols

there's an optimal word:symbol ratio, and it isn't 0:100

@MatheiBoulomenos talk about ratios... I think I can develop $\Bbb RP^1$ from linear transform of ratios lol

cool

Did you see my comment on the 420 solution?

kann ich einen Hinweis haben? Ich probiere aber ich kann nichts tun

@LeakyNun A is the set containing x non-0 integers, such that $\frac{5x+19}{2x+1}$ is also integer.

okay

@Mr.Xcoder that's better; now transform the latter statement

this time, you have to use centralizers

we talked about centralizers before

@Mr.Xcoder no, not better

don't use symbols if you don't know how to use them

What am I using incorrectly?

for every subgroup $H \leq G$, there's a homomorphism $N_G(H) \to \operatorname{Aut}(H)$ whose kernel is the centralizer

@Mr.Xcoder the $\implies$ of course

don't use symbols for now

they are distracting you

@MatheiBoulomenos alright wait

@Leaky Then you said I should transform the fraction into a simpler form?

@Mr.Xcoder I didn't

I said transform the statement "$\dfrac{5x+19}{2x+1}$ is (also) integer" into a simpler form

transform statements, not fractions

simplify statements, not fractions

Ok

simplying fractions is grade 5 material

Haha true

$\dfrac{5x+19}{2x+1} = d\in\Bbb{Z}$

what does it mean?

That $5x+19=d(2x+1)$

writing it in symbols isn't simplifying it

@Mr.Xcoder use words

maths isn't a compilation of symbols

they use symbols to assist words, not to replace them

I am not sure what you want me to do rn :)

I'll do this for you

"$\dfrac{5x+19}{2x+1} \in \Bbb Z$" means that $5x+19$ is divisible by $2x+1$

Ok, that is true.

Oh crap... gtg Will be back shortly

the mathematical educational system is a failure and I've witnessed it many times

@Daminark Potato comet

it has produced people who can manage symbols and know a lot of facts

but understand nothing

also known as "learning from wikipedia"

Hi @BalarkaSen

@BalarkaSen lol

Hi @Mathei

@BalarkaSen do you think this argument is too categorical: "As the suspension functor is left adjoint to the loop space functor in the pointed homotopy category, we see that suspension commutes with coproducts, i.e. wedge products" ?

Hey, I like that argument.

I guess just because I don't think $[\Sigma X, Y] \cong [X, \Omega Y]$ is a categorical tool.

I use it in my daily livelihood.

I used that once to compute some cohomology ring of a wedge of some spheres

my tutor wrote it was overkill and I should just have argued geometrically

You probably need Yoneda somewhere though. You'll probably end up with $[\Sigma (X \vee Y), Z] \cong [\Sigma X \vee \Sigma Y, Z]$ for all $Z$

At which point you'll want to Yoneda

to get $\Sigma (X \vee Y) \cong \Sigma X \vee \Sigma Y$

@MatheiBoulomenos LOL, your tutor doesn't like abstract nonsense

I mean it's general theorem in category theory that left adjoints commute with coproducts

So the Galois group of $\sqrt{2+\sqrt2}$ is $\Bbb Z_4$ while that of $\sqrt{2+\sqrt3}$ is $\Bbb V_4$?

yes

@MatheiBoulomenos and it took you exactly 30 seconds to calculate that

@LeakyNun I'm back rn...

@Mr.Xcoder hi

what can you say if you know that $5x+19$ is divisible by $2x+1$?

well, the latter is obvious from $\sqrt{2+\sqrt{3}} = \frac{1}{2}(\sqrt{2}+\sqrt{6})$

@MatheiBoulomenos which also took you less than 30 seconds to calculate

thinks

@MatheiBoulomenos You could also use $H^*(X; G) \cong [X, K(G, -)]$ :)

does that give me the multiplication as well?

Yep

Smash product of $K(G, n)$'s give cup product on cohomology ring

It's a little tricky, but it works

oh nice

We didn't do smash products at all in our rush trough homotopy theory

Ugh, that sucks. Smash product is the holy grail of stable homotopy theory

ok so $5x+19$ is divisible by $2x+1$...

(Of course $\Sigma^n X$ is $X \wedge S^n$.)

Not that I claim to know any stable homotopy theory...

we actually did some stable homotopy theory, for 15 minutes or something like that

Can't remember much

@BalarkaSen what's the easiest explanation that $\pi_4(S^2)=\Bbb Z_2$?

@LeakyNun There is none :)

@BalarkaSen then just give me any explanation

@LeakyNun I don't know a proof, to be honest. But the generator of $\pi_4(S^2)$ is $S^4 \to S^3 \to S^2$ where the second map is the Hopf map and the first map is the suspension of the Hopf map.

@Mr.Xcoder easier problem: what can you say if $3x+7$ is divisible by $x$?

@Mr.Xcoder no, 2x+1 is odd doesn't mean 5x+19 is odd

7 is odd doesn't mean 14 is odd

Maybe it's easier to compute $\pi_4(S^3)$? I have no idea

ಠ_ಠ I am dumb

$$\frac{3x+7}{x}=3+\frac{7}{x}$$ :p

@BalarkaSen do you like visualize the Hopf map at all?

The Hopf map is very visual

I can say that if $3x+7$ is a multiple of $x$, then $3x+7 = nx$, where $n$ is an integer.

$\pi_4(S^3) \cong \pi_4(S^2)$ by the LES of Hopf fibration

@BalarkaSen how do you visualize $S^3$?

@MatheiBoulomenos Oh, that's a good point.

@LeakyNun $\Bbb R^3$ with a point at infinity

@Mr.Xcoder if p is a multiple of x and q is a multiple of x, what can you say about p+q?

And $\pi_4(S^3) \cong \Bbb Z/2$ by Pontryagin-Thom theorem

@LeakyNun $P+Q$ is a multiple of $X$.

It's just $\pi_1(SO(3))$ I think

@LeakyNun What Mathei said.

@LeakyNun Well, x divides $3x$, but $7$ must also be divisible by $x$, and as $7$ is prime, $x$ is either $1$ or $7$.

@Mr.Xcoder good

so can you do the same for $5x+19$ and $2x+1$?

I can say that if $5x+19$ is a multiple of $2x+1$ then $5x+19=(2x+1)n$, where $n$ is an integer (first part)

@MatheiBoulomenos I guess my point was that I don't know a proof that $\varinjlim \pi_{n+2}(S^n) \cong \Bbb Z/2$ directly.

@Mr.Xcoder is there anything that is a multiple of $2x+1$ that you know?

just a sec

@BalarkaSen is $\pi_3(S^2) = \Bbb Z$ easy?

@LeakyNun Depends on what kind of tools you want to use.

$(2x+1)\cdot \frac{5}{2}=5x + \frac{5}{2}$.... :-/ Wait

It's an application of $\pi_2(S^2) = \Bbb Z$ and the homotopy long exact sequence

[Random]

@Mr.Xcoder 5/2 isn't an integer

Alternatively Pointryagin-Thom and compute $\pi_1(SO(2))$

@LeakyNun Hence the Wait :P

Well I know that $4x+2$ is divisible by $2x+1$

So $x+17$ must be divisible by $2x+1$

yes

And IDK further...

can you use Euclidean algorithm to compute the GCD of $x+17$ and $2x+1$?

Given a predicate P that is non constructive, what criteria is needed for P to have a constructive proof that there exists no constructive proof for P?

@Secret constructively show that a counter-example exists in the constructible universe :P

@BalarkaSen I'm sure you mean $\pi_3(S^3) = \Bbb Z$

Right.

one thing that intrigue me about axiom of choice is whether there exists a constructive proof that no examples that involve the axiom of choice is constructive

so I guess I should start looking at the constructive universe for answers...

@LeakyNun We weren't told about the Euclidean Algorithm. I'll try to though

@Mr.Xcoder alright then

so $x+17$ is divisible by $2x+1$, and so is $(2x+1)-(x+17) = x-16$

can you say anything from this?

$x-16$ divisible by $2x+1$...

yes, that's what I said

if $x+17$ and $x-16$ are both divisible by $2x+1$, then what do you know?

brain is heavily malfunctional

That $x+17 + x-16$ is divisible by $2x+1$

so $2x+1$ is divisible by itself

Yup

good

That's not terribly useful!

well at least we haven't bork arithmetic

@LeakyNun What was this for?

@LeakyNun Haha fair enough

I'd say the fact that divisibility is a preorder is quite useful

lol

@Mr.Xcoder come on, try again :P

ಠ_ಠ Mr. Xcoder

$x+17-x+16=33$

yes

So $(2x+1)|33$

@MatheiBoulomenos $420 = 2^2 \times 3 \times 5 \times 7$, so by Sylow $n_7=1$ or $n_7=15$. If $n_7=1$, then by Sylow we are done since the $7$-subgroup must be normal. If $n_7=15$, then the normalizer of $P_7$ the $7$-subgroup has $28$ elements. Take $\varphi:N(P_7)\to\operatorname{Aut}(H)$ be the group action by conjugation of the normalizer onto $H$. Since $|H|=7$, $H$ must be cyclic. Therefore, $\operatorname{Aut}(P_7) = \Bbb Z_7^* = \Bbb Z_6$. Elements in $H$ must map to the identity.

Then I've been stuck for like an hour

@Mr.Xcoder therefore?

@BalarkaSen what is the generator of $\pi_2(S^2)$?

Hence... $2x+1$ is among the divisors of 33. Thanks!!! So $2x+1\in\{1,3,11,33}\implies x\in\{0,1,5,16\}, but x\in\Bbb{Z}^{*}$ so x\in\{1,5,16\}$

The map with winding number $1$? @BalarkaSen

the identity map

@BalarkaSen what?

@LeakyNun good, so far. Is there something else in the kernel?

@LeakyNun ?

@MatheiBoulomenos well everything is either in the kernel or mapped to $3$

$S^2 \to S^2$ the identity map generates the group

@LeakyNun okay, what can you say about the size of the kernel?

@jublikon wie bekommst du $e^1/1!$?

@MatheiBoulomenos 14 or 28

nvm

@Mr.Xcoder no, 2x+1 in {1,3,11,33} isn't everything

Wait, rewriting

33 has 8 divisors

Wait, rewrititng

@LeakyNun okay, know try to show that there's an element of order $14$

and you still need to check if each of them works

@LeakyNun ich bin darauf wegen der definition hier gekommen:

@jublikon ja, und was ist $a$ hier?

$0$?

ja, denn was ist $f'(a)$?

$2x+1\in\{±1,±3,±11,±33\}$

And it's easy checking from now on... Thanks a lot @LeakyNun!

@MatheiBoulomenos well I can only say that there's a subgroup of order $14$...

that's not really helpful

but you know that there is an element of order $2$ in the centralizer!

that's even better than in the normalizer

oh, the centralizer is the kernel of the thing

@jublikon erst ableiten, dann einsetzen

hm okay.. danke@MatheiBo @LeakyNun

Viel Deutsch...

For people wondering, the above is in fact a reply to a message posted in another room. (Probably.)

How does exponentiation work with complex numbers?

Namely, I was wondering how $(2+5i)^2 = (21+20i)$

@Mr.Xcoder try to expand it

and use the fact that $i^2=-1$

cantorsattic.info/Zero

@LeakyNun think about the definition of the centralizer

I am also one of those people who classify 0 as a limit ordinal because it is the only ordinal where adding itself arbitrary number of times is a fixed point

@MatheiBoulomenos it's abelian hence cyclic :P

@Secret how is that related to limit ordinal?

You cannot reach one by any combination of 0 (except maybe $0^0$)

@Secret Also, $f:x \mapsto x \omega$ has quite a lot of fixed points...

@LeakyNun the centralizer itself is not necessarily abelian

Umm, where did I go wrong?

@MatheiBoulomenos but they commute with every element in $N(H)$, including themselves

@Mr.Xcoder you didn't

the answer given is wrong

the centralizer of $H$ is not the same thing as the center of $N(H)$

Oh....

@LeakyNun In fact I have mistaken when copy-pasting. It was $(5+2i)^2$. Makes sense anyway. Thanks!

@Mr.Xcoder ok

@MatheiBoulomenos elements in the centralizer commute with every element of $H$

yes, that's correct

@BalarkaSen one thing I don't understand about the $H^*(X; G) \cong [X, K(G, -)]$ thing: we only get a ring structure on $H^*(X;G)$ when we take $G$ as (the additive group of) a commutative ring. What kind of extra structure does $G$ being a ring give $K(G,-)$ so that we can define multiplication by smash product?

@MatheiBoulomenos It gives $K(G, -)$ a structure of a spectrum.

It's the Eilenberg-Maclane spectrum.

Or at least I think so

Well, I should rather say that it makes it a "ring spectrum". Spectra are dual to cohomology theories, ringed spectra are dual to multiplicative cohomology theories.

I see

"ring spectrum" sound oddly familiar

Nothing to do with $\text{Spec} \, R$ I'm afraid :)

@MatheiBoulomenos I'm lost though this sounds like an easy question

@LeakyNun on the 420 problem?

@MatheiBoulomenos yes

do you understand how to get an element of order $14$ ?

in the normalizer

I don't

Okay, take an element of order $2$ in the centralizer

consider the subgroup generated by that thing and the $7$-Sylow

hint: if the generators of a group commute, the whole group is Abelian

@MatheiBoulomenos :o

what

how have I not thought of this

that's nothing to be surprised about...

@BalarkaSen ... except my stupidity

I think you can finish from here

I really need to step up in group theory lol

I really can't visualize those things

you don't need to visualize everything

@MatheiBoulomenos but I can't even come up with that argument above

everyone has those moments. I confused the roots of sine and cosine in my advanced analysis exam ...

@MatheiBoulomenos sure that's just careless

but I still can't see it now

I mean, I understand the argument

but I cannot "see" it

tbh, I'm giving you quite difficult numbers so that you can see a bunch of different tricks, so you shouldn't feel too bad.

@MatheiBoulomenos I still can't see that if the centralizer of a vierzehnergruppe has order 7 then it is abelian

those two facts seem very unrelated to me

I mean, I don't have a good intuition for centralizer and normalizer

that's not quite what I said

btw vierzehnergruppe is much more compact than "a group with 14 elements" lol

@MatheiBoulomenos hmm?

oh lol

I reversed the two things haven't I

yes, you have

I mean, if the centralizer of a siebenergruppe has order 14 then it is abelian

(technically, we only have a subgroup of the centralizer in this case, but that's not important)

how do you "see" it

Well, I think about it like this: we have a group of order 7 and we add an additional element of order 2. If we choose the element of order 2 such that it commutes with the elements already there, then the whole thing is abelian

so you don't "look" at the centralizer?

I don't really have a visualization of a centralizer in an abstract setting, no

you can also explain this with semi-direct products which turn out to be direct, but that's overkill

@MatheiBoulomenos how?

btw I still can't continue lol

Well, we have that our 14 element group has a normal group of order 7 and a subgroup of order 2 which intersect trivially and generate the whole group, thus it is isomorphic to a semidirect product $\Bbb Z _7 \rtimes \Bbb Z_2$, but the multiplication of a semidirect product is determined by a homomorphism $\Bbb Z_2 \to \operatorname{Aut}(\Bbb Z_7)$. This homomorphism is a restriction of the same homomorphism that we considered before, i.e. action by conjugation

but we chose our element of order $2$ from the kernel of that homomorphism

Now a semidirect product is direct iff the homomorphism is trivial

but that's not really insightful, it's just a bunch of abstract language to say the same thing

@LeakyNun you have an element of order $14$. What did we do in previous cases when we had an element of a specific order in the normalizer?

@MatheiBoulomenos eyebrow raising over 9000

@BalarkaSen I said it's overkill and mostly language, but I would be lying if I said I didn't think about it like this

I think in terms of short exact sequences so I don't have any excuse either

:)

well, semidirect products are not that far from short exact sequences

Sure, they are the middle group of the split short ones

@MatheiBoulomenos you argue with its embedding in $S_{15}$, but here it embeds perfectly into a 14-cycle...

@LeakyNun is embedding into $S_{15}$ the best we can do?

@MatheiBoulomenos $A_{15}$ :P

it cannot be 14 as it would be even

it cannot be 2n+7 either

wait, so $A_{15}$ has no element with order $14$?

Sure, 2+2+7

@MatheiBoulomenos oh, nvm

but no element that fixes only one letter

right

@LeakyNun leaky :D

hi

hi :)

@LeakyNun gonna do group actions today, and sylow theorem =p I kinna need ur help with the latter after i do my readings :D

ok

@MatheiBoulomenos certainly knows more than me though

Hm, I am a little confused about somewhat basic linear algebra. Suppose $A$ is an $n \times n$ real, hyperbolic matrix, that is, none of it's eigenvalues have real part $0$. I want to prove $\Bbb R^n = V^+ \oplus V^-$ where $V^+$ and $V^-$ are invariant subspaces of $A$ (and in fact $A$ should "dialate" on one subspace and "contract" on the other).

So I look at $A$ as a complex $n \times n$ matrix and write down it's Jordan form

I think $\Bbb C^n$ decomposes as $W^+ \oplus W^-$ where $W^+$ and $W^-$ are the span of the generalized eigenvectors of $A$ with eigenvalues of real part greater than $0$ or less than $0$ respectively.

I'd take the generalized Jordan form (en.wikipedia.org/wiki/Jordan_normal_form#Real_matrices) and then just order the blocks according to the real part of the corresponding eigenvalue.

that seems less complex

Oh interesting

then you make two big blocks, one for positive real part, one for negative, this corresponds to to the decomposition of $\Bbb R^n$ that you want

Right, this is very nice. Thanks a bunch!

np

it's not difficult to derive this real normal form from the complex case, too. (Though I prefer to derive it for general fields to begin with, which won't surprise you)

Is this for the ODE class?