Mathematics - 2020-02-05

topologicalmagician

00:00

@TedShifrin yeah, it seems intuitive that its uncountable. Since for any real number you can construct a function $f$: $\{$ 1 $\}$ $\rightarrow \mathbb{R}$ which maps to it.

Ted Shifrin

Not just uncountable. Specifically in bijection with $\Bbb R$.

topologicalmagician

yeah, but how does that help with showing that $B(\{$ 1 $\})$ is equal to $\mathbb{R}$?

they're in one to one correspondence. I agree.

Ted Shifrin

Every function on a finite space is bounded.

Thorgott

$B(\{1\})=\mathbb{R}^1$, but $\mathbb{R}^1\neq\mathbb{R}$

Ted Shifrin

Huh?

topologicalmagician

00:04

what? @Thorgott

Thorgott

Speaking in a very strict set-theoretic sense

Ted Shifrin

Let's not confuse the poor @topologicalmagician

Thorgott

I think this distinction is what the confusion is about, $\mathbb{R}^n$ is the set of all functions from $\{1,...,n\}$ to $\mathbb{R}$

Ted Shifrin

then $\Bbb R^1$ is the set of all functions from $\{1\}$ to $\Bbb R$. So?

That is in obvious bijection with $\Bbb R$, as I said.

Thorgott

So $B(\{1\})=\mathbb{R}^1$, because any such function is bounded.

topologicalmagician

00:08

$\mathbb{R}^n$ $=$ $\{$ $(x_1,x_2...,x_n)$ $:$ $x_i \in \mathbb{R}$ $\}$ Thats the definition the lecturer gave. The one you provided is equal to that one ? @Thorgott

Ted Shifrin

@topologicalmagician: $f(i)=x_i$.

Thorgott

How do you define a tuple? I define it as a function $\{1,...,n\}\rightarrow\mathbb{R}$, so the definitions are equal.

(I'm aware I'm skipping over a slight subtlety here, but I'm doing this intentionally)

topologicalmagician

OK, so what I wrote earlier makes sense.

1

Q: Space of bounded functions and $\mathbb{R}^N$

Let $B(X)$ $=$ $\{$ $f:X\rightarrow \mathbb{R}$ $:$ $f$ is bounded $\}$. Show that if $X=$ $\{$ $1,2...,n$ $\}$. Then $B(X)$ $=$ $\mathbb{R}^n$. My attempt: If $g$ is a bounded function from $X$ to $\mathbb{R}$ then observe that $g(1)=x_1$, $g(2)=x_2$...., $g(n)=x_n$. Hence $g$ coincides with t...

Thorgott

The substantial part is to note that any real-valued function on a finite set is bounded, simply because the range is finite. The rest is just a matter of definitions.

Ted Shifrin

To me the point is that every function on a finite set is bounded (why?), and so we forget about the boundedness criterion.

LOL, we're saying the same thing. I will shaaddup :)

topologicalmagician

00:15

since it maps to finite many points, so it has to be bounded

anakhro

Hey hot cats.

topologicalmagician

thanks guys @TedShifrin @Thorgott

Ted Shifrin

What's up, @anakhro?

anakhro

Just thinking about food.

Dreaming, one might say.

Ted Shifrin

LOL

anakhro

00:18

What are you up to?

Ted Shifrin

Just hanging out for a little bit.

anakhro

Was doing a cute exercise today: f:R->R additive & continuous implies linear.

Ted Shifrin

Yup.

anakhro

Follow up is to drop continuous and add measurable.

Thorgott

The follow-up is significantly harder.

anakhro

00:25

It takes a bit more machinery, but about the same length of proof.

Thorgott

The continuous case is a two-liner. Can you get the measurable case that short?

anakhro

I am curious to hear your two liner. Mine involves several steps proving what f(0) is, f(-x), f(nx) and f(x/n).

And then one line to finish it off.

So 5 lines to be exact.

Thorgott

It's the same. Additive implies linear over the prime field $\mathbb{Q}$, which is dense in $\mathbb{R}$, so it's $\mathbb{R}$-linear by continuity. You're right that it's a couple more lines if argued in complete detail, but I think the measurable case will be significantly longer still if argued in the same detail.

It's not hard to get linearity once you have local boundedness, but I can't remember how you get that and I think that's the non-trivial part

anakhro

My solution isn't significantly longer.

Thorgott

00:41

Is there a trick?

anakhro

00:52

Definitely.

What's that cipher again people use for leaving hints?

topologicalmagician

Definition: Let $n\in\mathbb{N}$ be fixed. Given a set $X$. An $n-tuple$ of $X$ is **defined** by a function $\textbf{x}:$ $\{$ 1,2....,n $\}$ $\rightarrow X$. The $n-tuple$ is denoted by the following:

$\textbf{x}=(x_1,x_2....,x_n)$ where $x_i$ $=$ $\textbf{x}(i)$

From the above definition ,it is immediate that:
If $X,Y$ are sets. Then

$X\times Y$ $=$ $\{$ $(x,y)$ $:$ $x\in X$ and $Y\in Y$ $\}$ $=$ $\{$ $\textbf{x}$ $:$ $\{$ 1,2 $\}$ $\rightarrow $ $X\cup Y$ : $\textbf{x}(1)\in X$ and $\textbf{x}(2)\in Y$ $\}$

@Thorgott, @TedShifrin what I wrote above is an attempt to organise my thoughts. Is everything correct?

anakhro

Is there anything you have doubts about?

topologicalmagician

after what I wrote, no. @anakhro

Adam

04:20

what is the branch of applied logic that is used for things like developing false flag operations and the like?

Alessandro Codenotti

09:59

Are you around? @Balarka

Balarka Sen

11:44

@Alessandro Am now.

Alessandro Codenotti

I'm looking for "interesting" examples of metric spaces which are not proper

Where by interesting I think I mean that they are unbounded but not infinite dimensional

The examples I usually think of are either infinite dimensional Banach spaces or things like $d'(x,y)=\min\{d(x,y),1\}$ for some metric $d$ on a noncompact metric space

Balarka Sen

I forgot what proper means. Closed balls are compact?

Alessandro Codenotti

@BalarkaSen yes

Balarka Sen

What about infinite valence graphs with length metric

Alessandro Codenotti

It's a strong version of locally compact

Balarka Sen

11:52

Right, makes sense.

Alessandro Codenotti

@BalarkaSen Hm that's interesting

I think something like an infinite plane grid with a complete countable tree attached to each node should be an example of a nonproper unbounded metric space with asymptotic dimension strictly between $1$ and $\infty$

Balarka Sen

It immediately makes the point clear to me regarding why these should not be studied in coarse geometry

Alessandro Codenotti

Yes, properness is pretty much automatically assumed everywhere in coarse geometry

Balarka Sen

@AlessandroCodenotti Hm, what is this attaching procedure? I can attach a monovalent tree to each node and get something from $\Bbb Z^2$ which is very much proper. So I may be misunderstanding.

Alessandro Codenotti

The complete countable tree is the rooted tree with countably many immediate successors for every node (as a set theorist would say, $\omega^\omega$). I'm thinking about the Cayley graph of $\Bbb Z^2$, where every vertex is the root of such a tree

Balarka Sen

11:59

Ah, by countable you mean countably infinite.

Alessandro Codenotti

Oh yes

Balarka Sen

Got the picture.

Infinite valence pops up quite often if you try to take universal cover of complicated looking spaces, like $T^2 \vee \Bbb{RP}^2$, and see what the Bass-Serre description looks like

Alessandro Codenotti

I see

Balarka Sen

Makes me wonder if there's a coarse way to handle these stuff

Alessandro Codenotti

Trying to think about non proper spaces will be part of my thesis hopefully :P

Balarka Sen

12:03

By complicated I mean if one of $X$ and $Y$ has infinite fundamental group the Bass-Serre tree corresponding to $\widetilde{X \vee Y}$ is not finite valence.

@Alessandro Cool!

I'd try to read something like that

Edward Evans

12:19

y0

Rithaniel

Heya

Edward Evans

How’s it going?

Rithaniel

It's going okay. Sitting in my office and reading about Model theory while I wait for my first class. How about you?

Edward Evans

Nise, sitting at home smashing my head against modular forms notes

Rithaniel

This is that complex analysis course?

Edward Evans

12:29

Well it's a modular forms course lol

Rithaniel

Fair enough

Edward Evans

but modular forms are complex theoretic objects so

Rithaniel

The detail of $f(\frac{az+b}{cz+d})=(cz+d)^kf(z)$ is an interesting one

"Why is that there?" I find myself wondering

Like, what is it that makes that property desirable?

Edward Evans

Well, you have an action of $\operatorname{SL}_2(\Bbb Z)$ on meromorphic functions on the upper half plane by $(f|_k M)(z) = (cz + d)^{-k} f(Mz)$ and a modular form of weight $k$ and level 1 is then a function that is invariant under this action

and then idk

you get a fourier expansion for modular forms because the invariance under this action guarantees also 1-periodicity

and then there's interesting number theoretic information stuck inside the fourier coefficients

Rithaniel

You lost me at meromorphic, unfortunately. You're a bit more advanced into this stuff than I could hope to be

Edward Evans

12:36

well it just means the functions are holomorphic on the upper half plane except for some poles

Anyway a nice thing about modular forms is that you can find some nice generators for the space of holomorphic modular forms as a direct sum of copies of $\Bbb C$

and that gives you a nice way of computing the Fourier series for a modular form in terms of the Fourier series for "Eisenstein series"

and then some cool shit happens idk

Prashin Jeevaganth

Does anyone here know whether log(n!) and (log n)! are asymptotically the same?

Edward Evans

what's (log n)!

Rithaniel

12:57

I imagine $\Gamma(\text{log}(n)+1)$

Prashin Jeevaganth

@Rithaniel Hmm for some reason I cant see the Latex for this

Rithaniel

Check out the links in the top right of the screen: tinyurl.com/cfqcvpc

Prashin Jeevaganth

Ah ok I see it

I tried using Stirling's approximation to get the bounds for (log n)!

Let k = log n

lg(k!) = theta(k lg k)
lg((lg n)!) = theta k lg k
(lg n)! = n lg n

1010011010

13:16

HI

Im completely stuck on something super basic

I have shown that for some $n\times n$ matrix $M$, the following holds:
$$
-\log M = \int_0^\infty d\alpha [(M+\alpha)^{-1}-(1+\alpha)^{-1}]
$$

The next question is to show that $\operatorname{Tr}(M\delta\log M)=0$ whenever $\operatorname{Tr}(\delta M)=0$

This seems so basic but I cannot for the life of me derive it

Well ok I've been stuck on it for 4 hours but that seems excessive for something this simple

user193319

Let $x$ be an element in some $C^*$-algebra. What does it mean to say that it is a contraction? I know what it means for a linear operator between normed spaces to be a contraction, but what about an element in an abstract $C^*$-algebra?

Does it just mean that $||x|| \le 1$?

Kemono Chen

14:04

Does there exist a polynomial of integer coefficient such that it is irreducible over every finite fields? I believe it has been asked before but I cannot find a link.

Leaky Nun

14:14

5

Q: Minimal polynomial reducible modulo every prime $p$

Suppose $K = \mathbb{Q}(\alpha)$ with $\alpha = a + b\sqrt{D_1}+c\sqrt{D_2}+d\sqrt{D_1D_2}$ with $D_1,D_2 \in \mathbb{Z}$. Prove that the minimal polynomial $m_\alpha(x)$ for $\alpha$ over $\mathbb{Q}$ is irreducible of degree 4 over $\mathbb{Q}$ but is reducible modulo every prime $p$. In partic...

abstract-algebra galois-theory finite-fields minimal-polynomials

@KemonoChen

Thorgott

In case that was not a typo and you want irreducible mod every prime, then $X$ suffices.

Kemono Chen

Thanks to you two, what about polynomials of degree $\ge2$?

Thorgott

$X^4+1$ is irreducible over Q and reducible mod every prime

christmas_light

How can I prove $(C^1([a,b]), \left\lVert \cdot\right\rVert_{\infty} )$ is not a relfexive space?

Thorgott

14:30

It is irreducible over Q by Eisenstein and reducible mod every prime, because you can factor it into two quadratics depending on whether -1,2 or -2 is a square mod p (at least one of which is always the case)

Kemono Chen

Sorry, I meant irreducible over both $\mathbb F_p$ and $\mathbb Z$.

christmas_light

I know $C^1([a,b]), \left\lVert \cdot\right\rVert_{C^1})$ is not reflexive, and $C^1([a,b])$ is not closed in $C^0([a,b])$

Lukas Heger

14:46

@KemonoChen there's no such polynomial of degree $\geq 2$. Let $f \in \Bbb Z[x]$ of degree $\geq 2$. Then $f -1$ and $f+1$ have only finitely many roots (bounded by the degree of $f$), so there's a $n \in \Bbb Z$ such that $f(n) \neq \pm 1$. But then $f(n)$ is divisible by some prime $p$, so $f$ has a root mod $p$ and is not irreducible mod $p$

Edward Evans

Hey @Lukas

Lukas Heger

Hey @Edward

Edward Evans

How's it going ? :P

Thorgott

Neat.

Edward Evans

Hiya @Thorgott too

Lukas Heger

14:48

Learning for the adic spaces exam, the material is pretty tough. @Edward

and yourself?

Edward Evans

Studying modular forms, I have just decomposed the Hecke algebra into a tensor product of the p-primary parts

Thorgott

Hey

Kemono Chen

@LukasHeger That's great! Thanks

Edward Evans

and also getting excited over the ANT seminar :P

Leaky Nun

@LukasHeger wow

Lukas Heger

15:21

@EdwardEvans I actually know Marius Leonhardt from a conference

the guy who does the ANT seminar

Rithaniel

16:41

So, I had this idea. Kind of a generalization of polynomial rings. Given a ring $R$ and a monoid $M$, define $R[M]$ as the set $\{\sum\limits_{m_i\in M}n_im_i\mid n_i\in R\}$ with operations $$\sum\limits_{m_i\in M}n_im_i+\sum\limits_{m_i\in M}x_im_i=\sum\limits_{m_i\in M}(n_i+x_i)m_i$$ and $$(\sum\limits_{m_i\in M}n_im_i)(\sum\limits_{m_i\in M}x_im_i)=\sum\limits_{m_i\in M}p_im_i$$ where $$p_i=\sum\limits_{m_jm_k=m_i}n_jx_k$$

I'm not sure it's necessarily a ring, but showing it to be so (or not) should be fairly straight forward. Is this anything you guys recognize at a glance?

Edward Evans

@Lukas nice, is he cool? :D

Lukas Heger

yeah, he's a cool guy

and knows a lot of NT

Edward Evans

Nice :D

I think next semester is gonna be a good semester for me lol

Also @Lukas this is the paper we're reading from for ANT arxiv.org/pdf/math/0606108.pdf

Lukas Heger

ah, so you're doing Lubin-Tate theory

Edward Evans

yeah

which looks cool I guess

although there doesn't seem to be any explicit mention of topological properties at all

which for some reason I expect

anyway I'm super excited for that and analytic nt and algebra 2 lol

Balarka Sen

16:58

Here's a cool identity you two might like @Edward @Lukas

$$\left ( \sum_{a = 0}^{p-1} \left ( \frac{a}{p} \right ) \zeta_a^p \right )^2 = \left ( \frac{-1}{p} \right ) p$$

Although you probably already know this

Edward Evans

@Lukas didn't we talk about this in some form in L-functions?

anyway cool

Lukas Heger

yeah I like it

Balarka Sen

This can be used to prove $\Bbb Q(\sqrt{p})/\Bbb Q$ is abelian lol

Edward Evans

lmfao

Balarka Sen

Kronecker-Weber for $\Bbb Q$ with a bunch of square-roots adjoined

Lukas Heger

17:02

@EdwardEvans yes, this Gauss sum appears in the $\varepsilon$-factor for the Dirichlet $L$-function associated to the Dirichlet character $(\Bbb Z/p)^\times \to \Bbb C^\times, a \mapsto \left ( \frac{a}{p} \right )$

Edward Evans

right

nice

Rithaniel

(The thing I just posted is apparently already a studied thing, called a Monoid Ring)

Balarka Sen

@EdwardEvans Recommend good depressive suicidal black metal

Edward Evans

Hvis lyset tar oss

heh heh

Oh I know a band you might like

Balarka Sen

Oh Burzum I have heard this of course

Edward Evans

17:03

although idk if I'd call it DSBM

Lukas Heger

@Rithaniel yes, but it's great that you came up with this generalization on your own! for example if $R$ is a field and $M$ is a group, then this ring is heavily used in representation theory

Edward Evans

@Balarka Have you heard of Lurker of Chalice?

Balarka Sen

Nope

Gonna check this out

Edward Evans

it's the guy from Leviathan

Balarka Sen

Ah

Edward Evans

17:04

Lurker Of Chalice - Lurker Of Chalice (Full Album)

►

very cool album, but a bit more doomy than DSBM

Balarka Sen

Thanks!

Shall look into it

Edward Evans

Granite is fookin amazing

both the mineral and the song

Balarka Sen

Lol

Thorgott

What are your top 5 minerals?

Edward Evans

Granite, granite, granite, granite, evian

Thorgott

17:14

acquired taste

Edward Evans

lol

I don't know much mineralogy

@Balarka you know wolves in the throne room already right?

Balarka Sen

@EdwardEvans Yeah they are great

Which album from them is your fav?

Alessandro Codenotti

Speaking of doom do you listen to SubRosa?

Wolves in the throne room are fantastic

Archer

17:32

Is (a,b,c) (i.e. a+bx + ce^x mapped to (a,b,c)) a valid isomorphism from $\mathbb{Span(1,x,e^x)}$ to $\mathbb {R^3}$?

Specifically, I don't see the need to write the expansion of e^x.

Thorgott

Indeed, there's no need

Just counting dimensions is enough

Edward Evans

@Balarka Diadem of 12 Stars is great hehe

@Alessandro approve

Don't know SubRosa

oh and Black Cascade is a great album too

Archer

@Thorgott Thanks.

Alessandro Codenotti

18:12

@EdwardEvans try "More Constant than the Gods"

Edward Evans

18:24

@Alessandro taking a look now

Edward Evans

18:38

@Lukas ANT 2 is gonna be held in the Wintersemester 20/21, I guess alongside algebraic geometry lol

which is fun

Alessandro Codenotti

@EdwardEvans let me know what you think

Edward Evans

it was cool :D

long ass songs

Also @Balarka and @Alessandro: Check out Darkspace if you like inaccessible black metal where the songs consist of 15 minutes of blast beating

based on space and alien horror movies

hahaha

Alessandro Codenotti

Uhm I'm not sure whether I like that

Edward Evans

hahaha

Alessandro Codenotti

I'll check it out though

topologicalmagician

18:50

Let $B(V,V)$ be the space of bounded linear maps from $V\rightarrow V$. Show that the operator norm (defined in terms of infimum) is a norm: Is the infimum the minimum in this case?

Alessandro Codenotti

How is the operator norm defined?

topologicalmagician

$inf\{$ $k\geq 0$ $:$ $k$ is a bound for $T$ $\}$

$=$ $||T||$

Alessandro Codenotti

what is a bound for $T$ then?

topologicalmagician

$V$ is a normed space. The operator norm is denoted by $||T||_{op}$ = $inf $\{$ $k\geq 0$ $:$ $k$ is a bound for $T$ $\}$ where $||T(x)||\leq k||x||$ for all $x$

Alessandro Codenotti

Do you see why testing all $x$ with $\|x\|=1$ is equivalent?

topologicalmagician

18:57

@AlessandroCodenotti do you mean for all non-zero x?

Alessandro Codenotti

I mean do you see why asking $\|Tx\|\leq k\|x\|$ for all $x$ is equivalent to asking it for all $x$ such that $\|x\|=1$?

Balarka Sen

19:11

@EdwardEvans Thanks! I will look into it

Listening to Lurker of Chalice now

Edward Evans

niiiice

Balarka Sen

19:22

Darn this is good

@Edward Have you listened to Enslaved btw

Edward Evans

19:44

@Balarka I definitely have

Striker is my favourite song of theirs

Balarka Sen

That's a good one!

Alessandro Codenotti

20:34

Hi @Ted

urgh my laptop keeps showing people as online while they're not, I greet them, the page refreshes and they're gone

Balarka Sen

I'm still around but leaving to get a cigarette

I'll come back in 5

@EdwardEvans Diadem of 12 Stars is absolutely amazing

I finished, and then put it back on

Edward Evans

Truly is

Alessandro Codenotti

20:58

Two Hunters is also really good

Balarka Sen

Yeah that's one of the albums by them I have listened to

Leaky Nun

21:25

@BalarkaSen what kind of chess do they play in India?

Jacksoja

21:37

@LeakyNun what kind of question is that? chess is chess

Leaky Nun

Xiangqi

Xiangqi (Chinese: 象棋; pinyin: xiàngqí; English: ), also called Chinese chess, is a strategy board game for two players. It is one of the most popular board games in China, and is in the same family as Western (or international) chess, chaturanga, shogi, Indian chess and janggi. Besides China and areas with significant ethnic Chinese communities, xiangqi is also a popular pastime in Vietnam, where it is known as cờ tướng. The game represents a battle between two armies, with the object of capturing the enemy's general (king). Distinctive features of xiangqi include the cannon (pao), which must jump...

Jacksoja

@LeakyNun wow , i knew of many variations of chess

but not as totally different game

Edward Evans

21:48

@Balarka @Alessandro Black cascade is also a good album by them hehe

Ted Shifrin

Demonic @Alessandro: Well, I was here briefly. Now here briefly once again.

Alessandro Codenotti

I see, maybe my laptop isn't completely drunk then

Jacksoja

@TedShifrin Hi Ted

Ted Shifrin

I can pour your laptop a martini in a few hours, @Alessandro.

hi @Jacksoja.

Jacksoja

What does that mean Ted?

pour martini on laptop?

Ted Shifrin

21:52

No, it can drink it.

Jacksoja

that is not good, for anyone ..

Okay haha

@TedShifrin I have a question about what course does one take in order to learn more about surfaces

I want to study projective geometry

and algebraic geometry in the future

Ted Shifrin

Most schools no longer teach projective geometry, sadly.

OK, I have to leave for an hour, but I shall return eventually.

skullpatrol

hi @TedShifrin did you enjoy the Austrian open men's final?

::dang:: missed him, again

Jacksoja

22:11

@skullpatrol it is very common to miss Ted , happens to me alot of times too

skullpatrol

yeah, that's why I said "again" :-)

Jacksoja

there is another Ted that you could try to reach him if you want .

Yes I know, I just pointed the fact that this occur way to often

am just horsing around :)

skullpatrol

:)

Adam

history is a weird subject

skullpatrol

and, as they say, it likes to repeat

Adam

22:19

Well it's impossible to know for events that occur in your lifetime as to what exactly happened

But I'm sure it repeats itself alot

skullpatrol

~~Austrian~~ Australian open @TedShifrin

Jacksoja

22:34

@LukasHeger Hi Lukas

Lukas Heger

Hi @Jacksoja

Jacksoja

I have written a new version of the proof of finite abelian group problem

want to check it and see if it is good?

if you are not busy ofcourse

Lukas Heger

go ahead

Jacksoja

am not sure how to upload it here

but you do not mind if i just type it?

Lukas Heger

sure, just type it

Jacksoja

22:37

okay so the idea is the same as before

we want to find two elements of relatively prime order such that their order will be larger than that of maximal element

only difference I did here now

let a = p^i (PRODUCT primes different from p )

that what i called "terms " last time

but to make it more mathematically sound

and b = a^(p^j) * ( PRODUCT primes different from p )

need not be the same ofcourse

imagine a product sign there haha

and then the argument would follow the same way but by using the uniqueness of prime factorization

@LukasHeger is it clear ? if yes is it good?

Lukas Heger

what is a and what is b?

Jacksoja

oh nooo

that is not supposed to be like that

one moment

order of a = p^i (PRODUCT primes different from p )

order of b = (p^j) * ( PRODUCT primes different from p )

Ok all good now

geocalc33

What do you mean when you say prime factorization?

Lukas Heger

@Jacksoja My question remains. Surely you can't choose any two elements

What is a and what is b?

Jacksoja

it is not really a needed statement @geocalc33 but just to say that we took out all the p's from the product

@LukasHeger yes sorry, a is element of maximum order

and b is any element

geocalc33

22:45

oh okay

Jacksoja

that is what makes this work

Lukas Heger

any element? why is j>i, then?

Jacksoja

coprime order!

:)

Lukas Heger

I don't follow

Thorgott

if you prove j>i, you're done

Jacksoja

22:47

p is dividing the order of b to higher power than order of a

that is not what I prove thor

Lukas Heger

@Jacksoja but why is it possible to find a b with that property?

Jacksoja

I prove under these circumstances it is not possible to find such b

yes Lukas

since the order of ab would be larger than that of a

since the group is abelian

Lukas Heger

so the orders of a and b are assumed to be coprime?

Jacksoja

@LukasHeger I would send you a picture of how i written it if I could , hard to communicate math without latex

NO ! not that omg

Lukas Heger

but if the orders of a and b are coprime, then either i=0 or j=0

Jacksoja

22:51

No no am totally lost , apologies

a^(p^i) and b ^ ( product in the order of b )

those are our elements of coprime order

order of a^(p^i) = the first product in the order of a

and order of b ^ ( product in the order of b ) is p^j

Thorgott

I think this would be more concise and comprehensible if you just gave those products names instead of writing these awfully long expressions

Jacksoja

I agree

Lukas Heger

okay, but you want to prove that the order of every element divides the order of a. How does that follow from what you've done? you're close, but you still need an argument to finish

Jacksoja

Yes i argued that the same situation has to happen with other primes

that is why i mentioned unique factorization property

so we must have that no exponent in the order of the prime factorization of b

can be larger than that of order of a

@LukasHeger My question is ! why did Heirstein wrote is his book

while posting this problem

as he does not know of anyone who solved this problem with elementary arguments in group theory

Did I use something not allowed in this proof?

Lukas Heger

I'm not sure what you're allowed to use, but your proof looks fine now

Jacksoja

23:00

Thanks!

@LukasHeger If you have the book topics in algebra by heirstein page 48

problem 26

Lukas Heger

I don't have the book

Jacksoja

I see , am not sure if am allowed to post a picture on main

In any case, do you know if there are theories about how to do arithmatic mod p ?

I mean not basic things

Stuff like taking square root or n roots

or given g and h in G

you find x such that g^x = h

without the obvious way of testing

Lukas Heger

I don't know of such a theory

Jacksoja

you dont know or there is no such thing ? haha

Lukas Heger

I'm not aware of it, but it might exist

Jacksoja

23:06

Ok ! thanks

Edward Evans

23:38

waddup

Bob

23:52

good evening

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$Mathematics$ Mathematics