Mathematics - 2023-11-30 (page 1 of 2)

oscarmetal break

00:24

Hi, I was trying to write the ideal $(6)$ as a product of prime ideals in $\mathbb{Z}[\sqrt{-5}]$. I know this question has been answered in MSE, but I don't really get why the ideal $(2, 1+\sqrt{-5})^2 = (2)$.

Ted Shifrin

Look at $(1+\sqrt5)(1-\sqrt5)$.

For what you typed, look at the three generating elements for the square of the ideal.

I don’t see how we get $2$, though.

Oh, yes, sure, I do.

oscarmetal break

Some hint?

Ted Shifrin

We have $4$, $-4+2\sqrt5$, and $2+2\sqrt5$.

oscarmetal break

00:50

Do you mean that since $(2, 1+\sqrt{-5})^2$ is an ideals and $4$, $-4+2\sqrt5$, and $2+2\sqrt5$ in $(2, 1+\sqrt{-5})^2$ produce $2$ so that $(2)\sub (2, 1+\sqrt{-5})^2$? How about the converse?

Ted Shifrin

The reverse inclusion is obvious,

Every one of those is “even.”

oscarmetal break

I see. I just do the multiplication in $(2, 1+\sqrt{-5})^2$, and they are multiple of 2 indeed. Thanks!

Ted Shifrin

You’re welcome.

shintuku

Let $A$ be a finitely-generated $K$-algebra, $K$ some field. If we say $c_1, \dots, c_n \in A$ are algebraically independent, this means there is no polynomial $f$ in $A$ with $f(c_1,\dots,c_n)=0$, right?

leslie townes

not my at all my field (no pun intended), but one of those algebra goofs might want you to say algebraically independent 'over' something, probably with reference to wherever the coefficients of this polynomial are coming from.

you might leave that out when context is clear, but when you're asking what "algebraically independent" means is maybe not one of those times

shintuku

01:10

imgur.com/a/m7Orjtm

damn you kemper

Ted Shifrin

Am I missing something to cause the downvote here?

(I promise I’m not channeling anyone.)

Thorgott

01:31

leslie is correct, you want algebraic independence over $K$ which means the polynomials $f$ are polynomials with coefficients in $K$

shintuku

at Thorgott: so it makes sense here to distinguish the $K$-algebra $A$, the isomorphic polynomial algebra, and $K[x_1, ..., x_n]$ from which we take our polynomial $f$, correct? none of these three are strictly identical to one another

I meant: the $K$-algebra $A$, the isomorphic quotient of a polynomial ring, and $K[x_1, \dots, x_n]$ from which we take $f$

the latter being... an unquotiented polynomial ring we need to define algebraic independence

Ted Shifrin

@Alessandro @Jakobian Any ideas about this one?

Thorgott

02:06

you certainly need to distinguish the two non-isomorphic algebras and I would also consider it good practice to distinguish $A$ from the quotient of a polynomial algebra as such a presentation depends on a choice of generators

none of that matters for algebraic independence though, if you have a polynomial $f$ in $K[x_1,\dotsc,x_n]$ and elements $a_1,\dotsc,a_n$ in a $K$-algebra $A$, the element $f(a_1,\dotsc,a_n)\in A$ is defined

@TedShifrin I agree with your comment that the most natural course of action would be to simply ask that $X/G$ is normal

I don't know how much we need to ask for this to be true though, probably that the action is proper and then perhaps something more

shintuku

I see! thanks that makes a lot of sense

btw does category theory make these formal distinctions more evident or am i just looking for a reason to do more math?

Ted Shifrin

@Thorgott Yes, certainly in the case I gave, it’s very non-Hausdorff.

Zekchelovek

Hi. Where would be the appropriate forum to ask for someone to proof read my Goldbach's submission? If they have zelle, I'll be glad to pay for it. Just want to be sure it makes sense to other people, and is easy to understand.

Thorgott

@shintuku I don't think you need a categorical perspective to distinguish between things that are not equal, not sure what you're looking for

Ted Shifrin

I know of no such mathematical service.

Thorgott

02:14

@TedShifrin and very non-proper

Ted Shifrin

Yup.

Jakobian

@TedShifrin assume that $X$ is locally compact Hausdorff and $G$ acts properly discontinuously on $X$. Then $X/G$ is normal

Thorgott

proper should be enough to guarantee a Hausdorff quotient iirc

Zekchelovek

I'm asking for a member here, or a different chat. If it's out of bounds, I'll do without

shintuku

@Thorgott the fact that strictly speaking, a polynomial is an element of a quotient of a polynomial ring, and that strictly speaking a quotient of a polynomial ring has explicit generators is something that can be lost when we consider only the isomorphic K-algebra, no?

Ted Shifrin

02:15

I think this should all be classic group action stuff, probably in Montgomery …

shintuku

@Thorgott the semantics and the formal proof machinery here don't work well if we don't keep these objects separate, but maybe I'm overanalyzing? I don't know

Ted Shifrin

@Zekchelovek Of course any chatter is free to volunteer, but there’s so much “pop number theory” around ….

shintuku

yet a K-algebra can "work", as an object, in a certain sense, without generators.

Zekchelovek

Its short, and correct, in case that helps. I'm guessing there are no helpful suggestions here?

Thorgott

I find infinite discrete groups to be in a uniquely awkward position when it comes to these actions

the quotient would be automatically normal if the group were finite or if it were compact Lie and the action was smooth on a smooth manifold etc.

Ted Shifrin

02:20

And free …

@Zekchelovek correct, huh? Who says?

Zekchelovek

Did you miss the part where i would pay?

Thorgott

@shintuku I think you might be mixing things up? Evaluating a polynomial with coefficients in $K$ on elements of a $K$-algebra $A$ has nothing to do with representing elements of $A$ as polynomials in some chosen set of generators of $A$.

Ted Shifrin

I did not miss it.

shintuku

@Thorgott most likely mixing things up, need to shut up and do more exercises. thanks for taking the time lol

Zekchelovek

dimostriamogoldbach.it/en

Jakobian

02:22

math.brown.edu/reschwar/M141/hausdorff.pdf

Zekchelovek

That is who says its correct

Ted Shifrin

Cool, Jakobian, but he needs normal.

Jakobian

it implies normal

the quotient map $X\to X/G$ will be open which implies that $X/G$ is locally compact Hausdorff

Zekchelovek

Normal is good

Ok. Thanks anyway

Thorgott

If you want a categorical perspective, $K[x_1,\dotsc,x_n]$ is the free $K$-algebra on $n$-elements, meaning that for any $K$-algebra $A$ there is a canonical bijection between $K$-algebra morphisms $K[x_1,\dotsc,x_n]\rightarrow A$ and elements of $A^n$. a set of elements of $A^n$ is a generating set iff the corresponding homomorphism is surjective. evaluating a polynomial $f$ in $a_1,\dotsc,a_n$ means taking the image of $f$ under the homomorphism associated to $(a_1,\dotsc,a_n)\in A^n$.

Ted Shifrin

02:25

@Jakobian i see. Can you add that to the post I linked, as either an answer or a comment?

Thorgott

iirc LCH doesn't imply normal, no? @Jakobian

Jakobian

@Thorgott It doesn't. That was mistake on my part.

Ted Shifrin

Compact Hausdorff does. Of course.

Thorgott

yeah

Jakobian

We need closed quotient map

Ted Shifrin

02:29

I thought that OP asked a good question.

I know that the question of when $X/G$ is a manifold is technical, but well-understood. I just don’t think that way.

Thorgott

I'll admit I never read the general case of that, it's very annoying

all I needed in my life was that you get a covering if $G$ is finite and acts freely

shintuku

@Thorgott see this just looks like something way more lucid, clearly distinguished, that clearly names and denotes everything in play between the different structures involved. will refer back to this when I know a bit more

thanks!

Jakobian

If $G$ is compact and $X$ is normal and LCH, $G$ acts properly discontinuously on $X$, then $X\to X/G$ will be proper with $X/G$ LCH, so perfect. Hence $X/G$ will be normal

Thorgott

np

Jakobian

alright I've cooked some result up

Ted Shifrin

02:46

Is it army-style or gourmet? :)

Jakobian

Definitely more gourmet

I think my issue here is if the author needs $G$ to be countable

Ted Shifrin

From his comment to me, it sounded like he had a specific application in mind. I would assume countable is easier than an arbitrary infinite discrete group,

i don’t think he cares about $\Bbb R$ in the discrete topology.

Jakobian

Thorgott

we don't suppose freeness, though, right?

dunno feel like coverings may not be the right approach

Jakobian

I think its okay as long as $G$ is discrete and countable

those seem to be main restrictions of OP

Jakobian

03:16

this will imply that $X/G$ is Hausdorff

Daniel Donnelly

03:34

Do any of you have screen size smaller than 320px width when held vertically?

Making a responsive math app

If smaller than 320px, we will not support you :D

Min. 320px, but recommended 420px 🍀🍀

🚭🚭

It's called QuiverB (simply a responsive Quiver editor) but we're gonna modularize this way, and make bigger apps with it

Jakobian

idk I'm tired, its not my problem

one potato two potato

03:53

Aug 14 at 5:01, by one potato two potato

Quite surprising that $SL_2(\Bbb Z)$ contains a rank 2 free subgroup

what's more interesting is that for a "generic" choice of two elements of $SL_2(\Bbb R)$, the subgroup that they generate is free. Generic here is in a sense of Baire.

user51462

06:32

Hi, is it ok to mix metalinguistic symbols such as $\vdash$ with logical symbols such as $\implies$?

I've just started learning about soundess and completeness, and always see them defined using English such as "if $\Gamma \vDash \psi$, then $\Gamma \vDash \psi$" or "$\Gamma \vDash \psi$ implies $\Gamma \vDash \psi$". But is it ok to use $\implies$ instead, i.e. $(\Gamma \vDash \psi) \implies (\Gamma \vDash \psi)$?

one potato two potato

07:24

at some point, people don't ask "do you know this concept?" instead, they ask "are you familiar with this concept?"

Silly Goose

08:04

does anyone happen to have a math stack question which answers in the language of differential forms why we can "divide" by differentials?

i was trying to look for one, but i only found answers which use the language of usual calculus

Joe

12:50

@SillyGoose: This isn't exactly what you are looking for, but I think it is related.

Daniel Donnelly

13:13

Those buttons on the top right took all night LOL

Some have dropdowns. Everything is responsive down to 320px w

Can we get an AI to do this in 5 years? Answer: resounding NO

They can handle little pieces but can't connect everything yet.

@SineoftheTime hi

Sine of the Time

sup

Daniel Donnelly

See my image?

Would you use it?

Sine of the Time

what is it precisely?

Daniel Donnelly

It's a responsive Quiver

Quiver doesn't work on phones, it's just for desktop/web

Quiver is a CD editor

Sine of the Time

I'm ignorant in this matter

Daniel Donnelly

13:18

I'm hacking it and redoing its GUI to make it responsive to all screen sizes

Sine of the Time

wow

Daniel Donnelly

That's Quiver🐝

So it's like Quiver original was "A". @varkor a user here made it

This one is "(B)ee"

Sine of the Time

lol

Daniel Donnelly

Eventually we'll have CD database site, but got to start on the various subprojects

this was one of them

I'm hacking it minimally just to connect the buttons, this way we can update Quiver core code from varkor when they make updates

Harjot Dhillon

14:10

@DanielDonnelly @Thorgott hi

Are straight line graphs with zero slope a type of linear relationship. Or just positive or negative slope straight line graphs are linear relationship. Sorry if this question is silly but my knowledge about math is very less and I am self studying it

Jakobian

Yes

Bumblebee

15:01

I am in need of assistance please.

I need to solve for x over the range [-pi,pi]

cos(2x) + cos(x) = 1 + sin(2x) - sin(x)

But i'm having trouble finding the appropriate simplification

Vivaan Daga

15:31

If x^n =1 for all x in a group of order n then is the group cyclic? Is there an easy proof?

Lucky Chouhan

@VivaanDaga not $1$ you mean $x^n = e$ where $e$ is an identity?

Vivaan Daga

Yes

Lucky Chouhan

@VivaanDaga what did you try? I know that if G is a group and $a \in G$ and order of a = order of a group then a is called generator of a group G and G is a cyclic group.

@VivaanDaga I hope this would help you math.stackexchange.com/questions/2784780/…

Vivaan Daga

I want n to be the order of the group

Jakobian

@VivaanDaga no

@LuckyChouhan $1$ is used

Vivaan Daga

15:43

What’s a counter example

@Jakobian

Jakobian

$(\mathbb{Z}/2\mathbb{Z})^2$

Thorgott

every non-cyclic finite group is a counter-example

Jakobian

that too. Order of $x$ must divide size of the whole group

Vivaan Daga

Ah yes

of course

Bumblebee

Nvm, I solved it

Lucky Chouhan

16:05

@Jakobian Ah I see..

@Jakobian what does it mean to square Z/nZ ??

Ted Shifrin

16:16

@VivaanDaga This is true in literally every group of order $n$.

Soumik Mukherjee

@LuckyChouhan Cartesian product

Lucky Chouhan

16:36

@SoumikMukherjee so (Z/2Z)^2 = {0,1}??

And (Z/4Z)^2 = {0,1,2,3}? What I think is (Z/nZ)^2 = Z/nZ

Hello @leslietownes

Thorgott

you're not making any sense

Soumik Mukherjee

@LuckyChouhan No, if $A=B=\{0,1\}$ then what is $A\times B$?

Lucky Chouhan

@SoumikMukherjee {0,1} \time {0,1} = {0,1} isn't it? Isn't it like we multiply elements of 2 sets??

Soumik Mukherjee

No

Lucky Chouhan

so sorry for my silly questions. But I really don't know. Please tell me :(

Now I see here operation is Addition.

Soumik Mukherjee

16:45

Cartesian product g.co/kgs/DRoQtu

Lucky Chouhan

Thank you very much @SoumikMukherjee Now I got we need to make ordered pair. I was doing direct multiplication

So (Z/2Z)^2 = {(0,0),(0,1),(1,0),(1,1)}

Soumik Mukherjee

Yes, and addition is done component wise

Obliv

In general a graph has two meanings: the one from algebra & calculus to "graph" functions and the more abstract one with vertices & edges?

they're not the same thing right?

leslie townes

obliv: i sure hope not.

Obliv

why would they have the same name..

unbelievable :(

It honestly seemed to me they were the same thing, except this picture of a graph

leslie townes

17:02

well, context is a pretty helpful differentiator that is present in real life. i guess it's more annoying for encyclopedias and dictionaries.

Obliv

isn't connected

is a graph not just a kind of relation diagram?

leslie townes

well, there are a couple of ways of looking at that example, which is arguably putting more pressure on how you visually depict a "graph" than what a graph is.

Obliv

i guess graphs in general don't have a notion of "spacing" like "normal" graphs

leslie townes

i mean, outside of just 'algebra' and 'calculus' where maybe the functions of interest are defined on intervals of real numbers (or all real numbers), you can at least make sense of the graph of a function f from an abstract set A to another abstract set B as a set of ordered pairs, namely, {(a,f(a)): a in A} regarded as a subset of the cartesian product A x B.

Obliv

is that graph the graph in graph theory

leslie townes

17:05

so when A and B are subsets of R, the graph is a subset of R^2, and the way of visually depicting it is the usual one descartes came up with.

Obliv

or is that an abstract notion of a graph from algebra and calculus..

leslie townes

no we're just "graph" as in algebra and calculus.

Xander Henderson

@Obliv No.

It is "graph" as in "closed graph theorem". :P

leslie townes

so from "algebra and calculus" we are maybe most used to this picture when A and B are both intervals in R and f is continuous so the the thing looks "connected" (and even is technically connected in the topological sense)

Obliv

so does the descartes version or even his number line have any relation to the graphs in closed graph theorem?

leslie townes

17:07

that picture is doing something similar where A and B are subsets of R that aren't intervals. but it's the same idea of "graphing," where the point (x,y) corresponds to where you end up if you start at the origin and go x units along one axis and then y units along the other.

christ, one topic at a time. thanks xander.

Xander Henderson

Heh.

Obliv

Like with the added condition that the members must be ordered left to right in increasing order?

oscarmetal break

My prof was trying to convince me in hw that let $R$ be a ring of characteristic 0, then $\phi(x)= x^p$ is an endomomorphism of $R$, where $p$ is the number of the characteristic. The solution he gives is rather simple, $(xy)^p= x^p y^p$, and $(x+y)^p = x^p + y^p$, where the addition part is followed by binomial expansion and all the middle terms is 0. I didn't really get how this problem is meaningful, and what is going on in the process of addition.

Obliv

Sorry

Jakobian

@oscarmetalbreak char p?

Obliv

17:08

@leslietownes Yeah that makes sense. This way we're just filling in the parts that matter

but the underlying construction is the same $\mathbb{R}\times \mathbb{R}$

leslie townes

"visualize the ordered pair (x,y) as a point in a plane that you get by going x along one axis and then y along the other" is the key thing. if you're using that to plot points for a function f: A to B where A and B are both subsets of R, there's a sense in which you're doing exactly what you're used to doing from algebra and calculus.

Jakobian

> I didn't really get how this problem is meaningful, and what is going on in the process of addition.

leslie townes

with the qualification that if A and B are more complicated subsets of R than intervals, the picture might not look like the "graphs" that you get when they are.

Jakobian

well its meaningful because it gives you an important endomorphism on a ring of characteristic $p$

Frobenius endomorphism

In commutative algebra and field theory, the Frobenius endomorphism (after Ferdinand Georg Frobenius) is a special endomorphism of commutative rings with prime characteristic p, an important class which includes finite fields. The endomorphism maps every element to its p-th power. In certain contexts it is an automorphism, but this is not true in general. == Definition == Let R be a commutative ring with prime characteristic p (an integral domain of positive characteristic always has prime characteristic, for example). The Frobenius endomorphism F is defined by F...

Obliv

@leslietownes such as if they're not ordered the same way?

leslie townes

17:10

obliv: i was thinking of using the usual order in "visualize the ordered pair (x,y) as a point in a plane that you get by going x along one axis and then y along the other." and they're using the usual order in the picture above. but because their A and B aren't intervals, the graph is only a smattering of points in the plane, instead of something resembling a curve or whatever you might expect in that case.

oscarmetal break

@Jakobian Thanks. This is exactly what I am looking for. I am going to look at this latter.

Obliv

If a function is a general equivalence relation (is that true) between two sets A and B on R^2, filling in the dots or drawing lines/curves isn't a relation diagram nor a graph in the closed graph theorem sense

it's just called the graph of the function?

leslie townes

there are a couple of things there, i know what an equivalence relation on a set is, i don't know what an equivalence relation "between two sets" is. i'm not sure that's necessary to answer the question.

Jakobian

closed graph theorem?

oscarmetal break

I just feel it is awkward when $p=0$ and it is not so much when $p$ is prime.

leslie townes

17:13

the fundamental thing going on here is, how do you visualize a subset of R^2. descartes gives you a way of visualizing points, and that way extends to a visualization of any subset of R^2, putting aside htings like, you might run out of room on the paper, and your pencil can't draw with infinite precision.

graphs of functions A to B where A and B are subsets of R, will be subsets of R^2.

Obliv

hmm I think it's instead of an equiv relation being a subset of A^2 we can have a function be a relation on AxB

Jakobian

@oscarmetalbreak I don't think anyone is considering Frobenius endomorphisms for $p = 0$

oscarmetal break

This is exactly what happened in my homework. And since I feel hesitant to write the meaning of it in the solution, I hand it in blank.

Obliv

so in general we're not connecting dots unless the function maps R to R

leslie townes

obliv: yes, there is a more general notion of a relation between one set and another, which is generally just a subset of their cartesian product. and the notion of thinking of a function as a set of ordered pairs is in harmony with this, but there are relations that are not functions. i was sticking with functions.

Obliv

17:16

(so we can visualize the infinitude of R)

I meant equiv. relations*

leslie townes

please stick to functions. relations are related (ha ha) but not helping.

Obliv

a function is a specific equiv. relation on some cartesian product that isn't restricted to A^2

leslie townes

when you get to the idea of "connecting dots" you are thinking more about, how do we visualize an infinite set of points with pen and paper. e.g. formally, the graph of the function f(x) = x from R to R is an infinite set of points ((x,x): x in R}.

and how do you draw infinitely many things? well, you don't, you draw a line, because we know what that looks like. there is already a difference between the thing you draw and the formal object of what the graph "is" as a set of points.

and the difference becomes starker with functions defined on more complex subsets of R

Xander Henderson

@Obliv A function is not, generally speaking, an equivalence relation.

It is simply a relation which satisfies an additional property (the vertical line test, essentially).

Obliv

sorry context is important , I'm going off of my book definitions

(i'm not certain they must be equiv. relations but that's the vibes I get)

Xander Henderson

17:20

No!

They are not equivalence relations.

They are simply relations.

Where a relation from $A$ to $B$ is any ol' subset of $A \times B$ (and a relation on $A$ is any ol' subset of $A \times A$).

leslie townes

this definition is maybe thinking about equivalence relations as an important case. this is a specialization of a definition of relaiton that you might see in other books, which would not require the sets in the two 'factors' of the cartesian product to be the same

Obliv

@leslietownes yeah the efficacy of descartes graphs is indisputable

just wish the book went into the details of defining it like u did

instead it just says "we all know what a graph looks like"

which is true..

leslie townes

its really helpful to separate the idea of visualizing a set from what the set might be as a an abstract mathematical object.

Obliv

I agree, it's kind of constraining to just look at one visualization

it's not 1:1

leslie townes

any idea that 'graphs of functions look like plane curves that pass the vertical line test' is very much specific to functions that are from one interval of real numbers to another and probably are also not just continuous but some degree of differentiable.

there's nothing inherent in "graphing" that gives you a requirement that the visualization is going to be that for more general functions.

Obliv

17:24

or more general relations, so maybe that's why we have graph theory?

to relate more general sets ?

leslie townes

and, a line drawn in pencil on a piece of paper is not an infinite set, let alone a subset of the cartesian product R^2. it's a way of visualizing such a set.

just think of "graph theory" as something entirely separate.

Xander Henderson

@Obliv "Graph theory" is completely unrelated to the notion of graph you are grappling with right now.

Obliv

do graphs care about geometry? the descartes one requires two number lines perpendicular to each other but Idk about these

@xanderhenderson this was at the end of the mathematical induction section in this book

Xander Henderson

This is a completely different notion of graph.

leslie townes

different notion. the book isn't belaboring the distinction between those "graphs" and visual depictions of those things, but that's also going on here.

you can think of both things as a drawing on a piece of paper [although even that gets complicated] but there is no deeper connection.

it's generally a mistake to start with unrelated things that have the same name and try to back out some kind of deep connection between them. it's like trying to reverse engineer the origin of a word from its meaning.

Obliv

17:28

are these graphs fundamentally different from relations as well?

leslie townes

obliv: one thing to notice here is that in those visual depictions, there's no cartesian coordinates. they're saying that "v_0" is drawn here, and "v_1" is drawn there, and they're choosing those points arbitrarily on a piece of paper.

Xander Henderson

Yes.

Obliv

do they have relevance to set theory in general?

Xander Henderson

These are fundamentally different kinds of objects.

leslie townes

everything has probable relevance to everything in general. it isn't helpful to think that broadly.

Obliv

17:29

Right @leslietownes that's why I thought maybe the descartes version was just a more specific version of these

Xander Henderson

@Obliv Something that you need to get used to is the idea of reading definitions within the scope in which they are presented.

leslie townes

@Obliv i think the relation is best thought of as just, "these are both things that can be visualized with a drawing on a piece of paper"

Xander Henderson

There are a lot of words in mathematics which have different meanings, depending on context. My go-to example is "normal". Note the number of mathematical terms on the Wikipedia disambiguation page.

Obliv

I don't like the idea that sets are fundamental to everything. It seems fanatical

Xander Henderson

@Obliv Doesn't really matter how you feel about it---that is the modern synthesis.

Everything in mathematics is a set. Unless it is bigger than a set (in which case, it is a class).

leslie townes

17:32

you don't need to accept or reject or take a position on "sets are fundamental to everything" to follow along in a book like this.

this might be another example of "it isn't helpful to think that broadly"

Obliv

@XanderHenderson yes this is horrifying to me lol, but if context is fine then I guess no need to be worried

leslie townes

if a book says a mathematical object "is" a particular set, they're [usually] not making a philosophical statement so much as setting up the environment they are working in, in terms of what they will be using and/or taking for granted, and what they won't.

Obliv

sure that makes sense. It just feels robotic when we're all in the same environment

Soumik Mukherjee

@Obliv Set theory is a bourgeois concept, that's why one should go for category theory

leslie townes

you can't, like, make an equation for a tree, man

i discuss this more in my book Math 420: Bonghit Mathematics

Obliv

17:36

Dude I'd read that

we need to break free from the shackles of the MACHINE man.. let's all come up with our own unique formulation that expresses who we truly are

no i get why we have conventions and standards and stuff, like in physics we're just making models that reflect nature. math is kind of like that except in the brain membrane

the things that we observe are logical paths and connections

or they're just the crazy ramblings of burned out maths researchers

leslie townes

yes, that's the spirit. there's a lot of that in the book, how the man wants to divide people by putting them in boxes and giving babies numbers instead of names and making a tree obey an equation

Obliv

is that equation for a tree thing a reference to something lol

leslie townes

no, but hippies do seem to like trees

Obliv

what's not to like? they provide us a breathable atmosphere and look cool

and can talk to you when you're feeling down

Xander Henderson

They blot out the sun.

leslie townes

17:43

and you definitely hear people, not always seriously, trying to caricature X science or Y math or [fill in the blank] as advancing some totalitarian project of turning nature into machines

Xander Henderson

Stupid trees.

Obliv

reminds me of those that believe llm's are conscious..chatgpt is suffering in silence while we abuse it with our inane questions

Xander Henderson

This is the landscape I live in:

photos.app.goo.gl/skJWTPK5dtxmZ4c18

ack...

wrong link

leslie townes

xander's area could maybe do with more trees

Xander Henderson

Don't need no steenking trees.

Obliv

17:45

i remember that pic u posted of your workplace, is that the southern hemisphere

either that or arizona

Xander Henderson

That is New Mexico.

Though I live in Arizona.

leslie townes

my cat just chose to jump over a baby carrier instead of walk around it, to the alarm of the occupant of the baby carrier

Obliv

cats just do not care

Ted Shifrin

Olivia is with munchkin. She demands attention.

Jakobian

@leslietownes you can

Obliv

17:57

interesting how $f(x) = \frac{1}{x} + 1$ for $x\in \mathbb{R}-\{0\}$ isn't surjective because $f(x) = 1$ has no domain element.. but for the limit as $x \to \infty$ we do have a $f(x) = 1$

infinity is finicky

Ted Shifrin

Nothing to do with infinity.

shintuku

@Thorgott i don't pretend to know what I'm talking about, just trying to see if I'm understanding the high level relationship between these concepts: is a finitely generated K-algebra the model of some Lawvere theory?

Obliv

is $ln:(0\to\infty)\to \mathbb{R}$ injective and surjective? I think so

wait nvm it's not surjective

it's missing like all the negative numbers + 0

user 85795

How is munchkin handling the new arrival @leslietownes

Have you started the big sister role training yet (whatever that means :)

leslie townes

18:18

its going OK. its a lifelong process. :D

Thorgott

@shintuku pretty sure it is, but dont think that's a super helpful perspective for anything

Obliv

Does showing a function is surjective amount to showing it is "defined" for some arbitrary element in its codomain

like simply showing $f: \mathbb{Z}\times\mathbb{Z} \to \mathbb{Z}$, where $f(a,b)=3a-4b$, is defined for some $c \in \mathbb{Z}$ as $c = 3a-4b$?

can I just stop there and say the function is defined for all integers $a,b,c$

oh wait

Jakobian

@Obliv Its a bijection

Obliv

it is? @Jakobian

Jakobian

Yes?

Ted Shifrin

18:31

@Jakobian Huh?

Jakobian

@TedShifrin why am I being huhed

Obliv

wait i gotta get lunch my brain is broken

Jakobian

I'm correct

$x = e^{-1}$

Ted Shifrin

Oh, yes, of course you are. My second stupidity within 5 minutes — the other in a comment on main.

Obliv

yeah I'm going for lunch lol that was silly

Ted Shifrin

18:33

Too late to erase my huh.

Obliv

recorded for all of eternity. rip

Ted Shifrin

sends self to one’s room for bad behavior

user 85795

18:47

I'll pop some fresh popcorn to help comfort you.

🍿🍿

Lucky Chouhan

@user85795 Pop POp POPCORN

user 85795

🤣🤣

Lucky Chouhan

@user85795 do you also have new arrival?

user 85795

Not yet, do you?

Lucky Chouhan

@user85795 Oh no, I am still in my undergraduate. And what about you?

Ted Shifrin

19:00

@user85795 Gee thanks!

Lucky Chouhan

Your profile photo can terrify any new arrival haha

King has arrived.

user 85795

No problem, pal @TedShifrin

Lucky Chouhan

If you want to give your MSE account to someone then who will be that person?

user 85795

Is my profile picture really that terrifying? @LuckyChouhan

Lucky Chouhan

@user85795 I find it interesting but I am sure my little cousins won't 🤣

btw from where you got it?

Obliv

19:05

@Jakobian does it suffice to show for some arbitrary $b \in \mathbb{R}$, we have an $a \in (0,\infty)$ such that $\ln(a)=b$.

i.e $a = e^b$ for all $b \in \mathbb{R}$ is true

how else can I prove surjectivity..

Lucky Chouhan

@Obliv Surjectivity = One one??

Obliv

unto

Jakobian

@Obliv logarithm is strictly monotone and continuous

moreover, its limits as $x\to 0^+, \infty$ and $-\infty, \infty$

Lucky Chouhan

@Obliv then just do something like this that y=ln(x) \imples x=e^y hence for any y \in (o to infty) we have x=e^y

Obliv

yeah I guess that's all I can really say.

Lucky Chouhan

19:09

@Obliv you should rename yourself as "Liv". It sounds more cooler that "Ob-liv".

Obliv

maybe if I use the series expansion of $e^x$ then we have a more complete proof

Ted Shifrin

Hell no.

Obliv

well no because you'd have $a = \sum_{k=0}^{\infty}\frac{x^k}{k!}=1+x+\frac{x^2}{2}+...$

yeah that's less obvious

whatever, I'll just use those assumptions @Jakobian

Lucky Chouhan

@leslietownes Where can I buy your book?

Sometimes I think whether @Jakobian is a bot or human.

Jakobian

My mother was a washing machine

Obliv

19:21

i forgot how do we prove we can make all integers by $ax+by$ for some constants $a,b\in\mathbb{Z}$ and $x,y$ variable integers

we did this somewhere in my book I forget where

(with a,b nonzero I should mention)

leslie townes

somewhere near a discussion of the gcd, i would imagine.

gcd(a,b) is an obstacle to doing that

Obliv

oh right they must be relatively prime

leslie townes

yes. then if you reverse the euclidean algorithm computation of gcd(a,b) = 1, you can produce x and y for which ax + by = 1, and subsequently realize any other integer n as a combination of a and b e.g. by replacing x and y with nx and ny in that.

Obliv

i was about to write that lol

well i was gonna write you can set x,y such that $ax + by = d(1)$ for some integer $d$

giving them a common divisor

robjohn

19:44

@Obliv Look at Bezout's Identity.

You can compute the coefficients using the Extended Euclidean Algorithm.

Transcript for

$Mathematics$ Mathematics