Mathematics - 2017-07-24 (page 1 of 4)

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12:01 AM

@AkivaWeinberger I am not sure that I understand what you mean, but $V$ should be a complex vector space, yes.

12:14 AM

If we have a unitary space $V$ and a normal endomorphism $\varphi$, we can can define the eigenvalues of $\varphi$, then the eigenspaces. For every eigenspace we can build an orthonormal basis with Gram-Schmidt and union them to the basis of $V$. This one will be orthonormal, too, as the eigenvectors of different eigenspaces are perpendicular to each other. Have a got the idea of the spectral theorem right?

and this works, as the sum of the eigenspaces is direct, so the dimension of $V$ is the sum of the dimensions of the eigenspaces, right?

1:06 AM

@Kirill: The crucial point is that the eigenspaces add up to everything, i.e., that for every eigenvalue the geometric multiplicity is equal to the algebraic multiplicity. There's an inductive proof, basically, that guarantees this.

1:16 AM

rehi @MikeM

@TedShifrin on the other side, proving the theorem we see that there is a basis of $V$ consisting of eigenvectors of $\varphi$. That is equivalent to the statement that $\varphi$ is diagonalisable. And that is sufficient for the fact that for every eigenvalue its algebraic multiplicity equals to the geometric one.

Yes, but you need an inductive argument to get the basis of eigenvectors. Precisely.

hi

good work done today

Good for you, Mike.

I probably won't understand details.

@TedShifrin oh you mean the induction by the number of spanning vectors?

1:21 AM

well, no. Induction on the size of the matrix, say.

oops, that was the proof for the Gram-Schmidt algorithm, saying that it produces an orthonormal basis. My fault.

The usual proof doesn't require Gram-Schmidt.

yes, and about the question I asked before? Mr. Shifrin, do you explain the chain rule in "Multivariable Mathematics"? I can borrow it tomorrow from our library. @TedShifrin

Sure. It's a standard theorem in multivariable calculus.

@TedShifrin I looked for it at Wikipedia and found a totally different notation as we use. We use something like $D_f(x)$ and the Wikipedia article is full of differences of differentials $\frac{dt}{dx}$.

1:32 AM

I state it in terms of linear maps.

I am very interested to look in the book then.

Most multivariable calculus students don't know any linear algebra, so you find less conceptual treatments on line.

You can also watch one of the videos if you want, @Kirill.

@TedShifrin some particular one, or in general?

No, the one on the chain rule :P You can watch others if you want.

@TedShifrin does it make sense to you to have a course that teaches calculus 2, multivariate calculus, and linear algebra all at once in roughly 10 weeks? Very confused by a friend of mine describing this class and it sounds intense, especially for someone that isn't majoring in math. Is that normal to your knowledge?

1:33 AM

Just pass to the hyperreals and chain rule becomes actual multiplication!

Jk

also, how is your vacation going?

wonders if Demonark is capable of serious discussion

hello

I tend to fall back on stuff like $\frac{df}{dt}=\sum_k \frac{\partial f}{\partial x_k}\frac{dx_k}{dt}$. That's what I think of when someone says "multivariable chain rule".

totally NOT normal (nor good), Typhon.

1:34 AM

@Daminark um.... what.

i tab into chat, @Daminark says nonsense, par for the course tbh

@TedShifrin The one on the chain rule, but yours?

Maybe it's just a few topics from each for some particular sort of students, Typhon.

@Kirill: There are 112 lectures on my YouTube posting covering all sorts of stuff.

Hi @EricSilva. Sorta what I said.

yup

@TedShifrin thought so. It's for biology students, but it just sounded rough. It's an online course with no textbook. Just horrible videos that are maybe a minute long (they showed me because they wanted help studying for an exam). I let them borrow the textbooks from when I took those classes. At least now I know I'm not overthinking that.

1:35 AM

Hence I always feel a bit lost when I see the matrix notation for the multivariable chain rule.

Typhon, I'm sure they only do a few topics from each. There are similar things for business students.

It just doesn't play on the right reflexes.

oh

I mean, much as I don't anticipate that it's a good idea to think about hyperreals in multivariable calculus, I think that does legitimately check out

fair enough

1:36 AM

Semiclassic, that's too bad. You should intuit that the linear approximation of the composition is the composition of the linear approximations.

@TedShifrin I will look up. I paused after dozen of videos the last time. That was linear algebra.

This is somewhere in the middle of the first set ... probably around 30.

Meh. Truth is I just don't do that kind of thing enough for it to be a problem.

@Daminark I think it does work out, iirc Terry Tao likes nonstandard analysis a weird amount and has some expository posts about it somewhere on his blog, seems weird to me though

@Semi I always think of chain rule in terms of composition of linear maps, so the matrix form makes the most sense to me

The classic notation can really mess people up, @Semiclassic. Try the proof of Euler's theorem on homogeneous functions. A disaster.

1:37 AM

iirc Terry Tao's take on nonstandard analysis is that it's a way to automate epsilon-management.

I think in logic, there are other contexts in which these sorts of hyperconstructions, the idea being to encode certain things as arithmetic

interesting

Which is, to say nothing else, nifty

@ted Yeah, I have vague recollections of how annoying that can be in the usual notation.

it's amazing how much notation affects how easy or hard math is

1:39 AM

Not really that amazing when you think about it.

@AkivaWeinberger been looking at modular arithmetic and I realized that aside from very specific examples the only time equivalent mod x (where x is what is appended) make sense for the arbitrary case of quadratic rings is when x and x' (the other solution) are both dividing each other (making the modular arithmetic of each equivalent). I think this only occurs when b divides a and x^2 + ax + b = 0, but I'm not sure. I know x and x' must divide a to divide each other. I think it has to do...

with showing that irrational algebraic integers only divide an integer if their norm divides it.

I mean, it's like a trivial statement, but I think it's generally pretty amazing that notation (or language) can influence thought.

Not real sure. Any thoughts?

@AkivaWeinberger On another note, I found out that a weaker version of legendres conjecture was once proven. Apparently there exists either a prime or the product of two primes between any two squares.

@EricSilva as hieroglyphs are based on pictures, Chinese surely implies a different mind work as me doing algebra now. On the other side, I still think different in different languages, because sometimes the grammar doesn't allow me to think in a right way.

Akiva Weinberger

@EricSilva Maybe it's a side-effect of the fact that when you learn a notation you learn a notion

1:51 AM

Linguistic relativity

The principle of linguistic relativity holds that the structure of a language affects its speakers' world view or cognition. Popularly known as the Sapir–Whorf hypothesis, or Whorfianism, the principle is often defined to include two versions. The strong version says that language determines thought, and that linguistic categories limit and determine cognitive categories, whereas the weak version says that linguistic categories and usage only influence thought and decisions. The term "Sapir–Whorf hypothesis" is considered a misnomer by linguists for several reasons: Edward Sapir and Benjamin Lee...

Relevant

Akiva Weinberger

and it's the notions that change how easy or hard it is

DogAteMy, I think you're truly addicted. :P

Akiva Weinberger

@AreaMan TL;DR The Sapir–Worf hypothesis says that one's language defines how they perceive the world. This was popular in the 19th century due to nationalism and the like, I think. Everyone essentially agrees it's wrong. However, there are various weaker forms that are still up for debate.

@AkivaWeinberger I'm no linguist, but I think you are describing the "Strong" Sapir Worf. (See the link.) I don't think it is true that every agrees it is wrong, though I know I hear that a lot from linguistics majors, I've heard also heard it contradicted. What is your source?

Akiva Weinberger

So, in terms of how they let you see the world, all natural languages are essentially equivalent. There's no reason that that sort of thing should apply to math, though, since its notations are very far from "natural language"

@AreaMan Yeah, you're right, that is the strong version…

1:55 AM

Is it though? Formal grammers and notations have a lot in common.

Akiva Weinberger

Well, for one, it needs to be explicitly taught, unlike most actual language which you kinda just learn as a baby

That is just my opinion, to be clear (about math notations being far from natural language)

The main example people point to about influential changes to notation, I think, is Gauss's $\equiv$ for modular arithmetic, right?

I don't know what Gauss used.

Akiva Weinberger

I thought he invented $\equiv$.

One of the arguments for the existence of a natural grammar is that the complexity of the system learned vs. the amount of information taken in implies a great deal of redundancy / structure. There is a neat debate between Chomsky and Foucault were Chomsky makes the same point about morality. I think it's not too far a leap to make the same point about mathematics. Especially since it would explain something about why some things are obvious once understood but very hard to explain...

I have no idea, DogAteMy.

Akiva Weinberger

2:00 AM

@AreaMan Ooh, that sounds very interesting. Could you point to resources about that debate?

A lot of what I know about linguistics is from this YouTube channel LingSpace (which may or may not dock my credibility a bit, I dunno), and they're big fans of universal grammar

(It's an ongoing series of like ten-minute lectures)

Actually I was mistaken, the interview I'm remembering is a reflection on the debate between Foucault and Chomsky: youtube.com/watch?v=i63_kAw3WmE (at least, I think it is this interview)

@AkivaWeinberge At about 4:50

ncbi.nlm.nih.gov/pubmed/28447839

Akiva Weinberger

Relevant: youtu.be/u6eXw0AAKZ8

2:25 AM

The more I read scifi and physics literature about time, the more nonlinear and jump around my perception of time gets. Interestingly, time is still relatively normal in my dreams, but not space (which is always 2x larger), which is opposite to what happens in my waking life

heh. through detective work, I have ascertained John Baez upvoted one of my posts.

I find this funnier than I should.

@AkivaWeinberger the programming languages course I had last semester would say that written languages and notations are mathematically the same kind of structures. They are literally the same thing until you go to compile them.

@AkivaWeinberger think of it like number theory and integers versus quadratic rings. The integers are the trivial spoken language to which most properties apply. Everything else is some other variation.

smbc-comics.com

@Secret L

M

A

O

> Always study things for their own sake, before thinking about how to use it, unless it is obvious how it is used

2:34 AM

My response: "does it matter?"

Indeed

nothing is useless

Akiva Weinberger

What's that quote that Dami referenced before?

Jul 19 at 20:28, by Daminark

One of the golden quotes from atop: "We can do calculus on R^n, but we want more. Now, we should know why we want more, for funding reasons. For instance, some people care about something called physics"

but even Von Neumann found a purpose for nothing... we call them integers.

Akiva Weinberger

As a side note, I always found it strange that cavemen are depicted as speaking with broken grammar.

^^^

Akiva Weinberger

They obviously didn't speak English, and they spoke with perfect grammar of whatever ancient language they did speak.

2:37 AM

yeah. Man hasn't increased in intelligence. Cavemen were just as civil and intellectual.

Hello

Akiva Weinberger

Hi

they were intellectually challenged by things such as why leaves are green and why fire hurts.

I'm working on a problem where I'm trying to compute the product xAy' where A is skew symmetric, A' = -A, and x and y are vectors

there was an age where "flame theory" was the pinnacle of science

2:39 AM

I'm trying to write it as a sum over i from 1 to N and j from 1 to N, where N is the dimension of the matrix

TBH, modern man is probably more unintelligent than cavemen.

Consider how many children on Reddit act like bafoons.

I know a_ii is zero for every i, but I'm not sure if it reduces more. I also know that a_ij = -a_ji since A is skew symmetric, but I'm not sure how to use this

Or 4chan?

Akiva Weinberger

To be fair, we're still intellectually challenged by why leaves are green

Black would be more efficient

o, and don't forget water, it will weird you out despite we drink it on a daily basis

2:41 AM

Fair enough

I mean, define intelligent

Akiva Weinberger

There are several competing theories for why plants use *chlorophyl (which is green) rather than a potentially more efficient black chemical, but no consensus

Idk

In terms of how much stuff we know, it's clearly grown

In other respects it's not likely that much has changed

@AkivaWeinberger I meant why the leaves look green though. Answer: chlorophyll

Akiva Weinberger

2:43 AM

Dammit, I knew I used the wrong word, corrected

Like, we tend to look upon the past with some sense of "Wow, look at the degenerates we've become!"

Hmm, how do we write skew symmetric inner products and quadratic forms again?
$\langle x,-Ay\rangle = \langle Ax,y\rangle$ ?

@Daminark I know. I was making a dog towards people who act dumb on the internet.

Akiva Weinberger

It just shifts the question to why chlorophyl

I would say, the only reason ancient man didn't advance to as far as we are now, is due to a lack of resources.

In reality, it's mostly a product of how you can see all of it thanks to the internet, and when presumed degeneracy happens, it often takes the spotlight. So you're just exposed to more of it

@Akiva I mean... Over time, they kinda did

Akiva Weinberger

2:44 AM

Who cares if they were all Einsteins if they have to hunt and gather all day?

Hi @Secret, thanks for the comment, but I'm not sure how to use it to write the sum :(

@Daminark you're taking it too seriously, mate.

It was a joke

Akiva Weinberger

I dunno, first figure out what each entry of Ay' is

and then use that to find xAy'

I will argue however that warfare and such has grown worse over time if only due to obvious scale increases.

my guess is that it is sum_{i = 1 to N} sum_{j = i+1 to N} a_ij (x_iy_j - x_jx_i) ?

2:46 AM

Eh, that was inevitable

@AkivaWeinberger did you see my number theory post in chat?

Akiva Weinberger

About modular arithmetic? You said you needed $x$ and $\bar x$ to divide each other for some reason but I forget why

skew symmetric matrices are consists of a upper triangular with diagonal of zeros add to the minus of a corresponding lower triangular with diagonals of zeros, not sure how that helps

Hmm, let $T$ be the upper triangular entries of $A$, then $-T^T$ is the lower triangular part and thus $A=T-T^T$?

it will definitely make the computation of $Ay$ a bit more tractable since it is easy to multiply a vector with a triangular matrix

@Secret Hmm, I see, I think that helps, thank you

In other news, I am still annoyed that I cannot visualise any arbitrary basis free linear operator as a glyph yet we have no problem visualising vectors in $\Bbb{R}^{whatever}$ as arrows

2:57 AM

@AkivaWeinberger I want to prove that x and $\bar x$ for x^2 + ax + b = 0 divide each other only when b divides a. The other stuff was just context. :-)

It seems we humans not only don't have the ability to perceive the 4th spatial dimension (if any), we also don't have the ability to visualise an entity living "in many directions at once"

@Secret the phrase "in many directions at once" confuses me.

there are many directions in space.

Well, take a matrix, it is a linear map which maps basis vectors to basis vectors, thus in a sense, for every unit of the e.g. x unit vector, it has a "weighting" of some other vector

@Secret umm what?

so its like every unit in the x direction it has a "size" that is a vector v

3:02 AM

@Secret what are you referring to?

take a 2x2 matrix as an example. The 1st and 2nd column of the matrix corresponds to the x and y direction (or in general any pair of direction for a given basis set)

ok...?

such that when you multiply a matrix A=(a,b;c,d) to a vector v =(e,f), you get e(a,b)+f(c,d)

you're describing coordinate systems as I view them.

now how does this relate to something we cannot perceive?

The issue is, can we ever visualise A=(a,b;c,d) on its own, like how we visualise any two vectors without using a coordinate system as two arrows pointing in some direction

Akiva Weinberger

3:05 AM

@Typhon Can you give an example for $b\ne\pm1$?

@AkivaWeinberger sure

x^2 + 5x + 5 = 0

Akiva Weinberger

Also, Wikipedia hole has led me to this Albanian folk song

@AkivaWeinberger note that the norm of x is b, so b dividing a shows that x divides a + x = x'. The reverse is what I wish to prove to show equivalency.

I wish to show that x and x' dividing a and therefore dividing each other shows that b divides a.

@AkivaWeinberger in turn, x and x' dividing each other is the condition that makes n mod x and n mod x' the same equivalency relations. The moldular arithmetics become one and the same. That provides an interesting situation I am writing about (for fun).

Akiva Weinberger

And $b=-x\bar x$, right?

@AkivaWeinberger no

$b = x\bar x$

Akiva Weinberger

3:10 AM

Oh yeah you're right sorry

cccccccccccccccccccccccccccccccccccccccccccccccccccccc

just do n^2 + an + b = (n-x)(n-x')

sorry

Akiva Weinberger

You OK

that provides the relations I'm using.

interestingly

all x^2 + b = 0 are values fulfilling my requirements.

since then the conjugates are negatives of each other

3:12 AM

It also does not help that this intuitive thing we call direction, is mathematically really a complicated entity that is represented by how the components of a vector are related to each other under any basis set

@Secret ugh fair enough. I still don't know what you meant by multi directional. We live in multi directional.

I can move up and right.

Akiva Weinberger

I successfully have gotten that Albanian folk song stuck in my head

Are you saying a situation where there is a space where directions exist and then a different space where another direction can exist?

Akiva Weinberger

which is harder than it looks because the tune for the chorus is ever so slightly different from the verse, so it was hard to see exactly what the difference was

@AkivaWeinberger I'm thinking that the norm of irrational numbers must divide any integers the irrational numbers divide.

Akiva Weinberger

3:15 AM

Is that equivalent?

no. it is stronger

Akiva Weinberger

Ah

consider sqrt{3}

it divides 27

does 3 divide it? Yes, it does.

But for a rank 2 tensor, which can be wrote in some matrix representation e.g. $$\sum_{i,j=1}^n a_{ij} e_i \otimes e_j$$

we have the trouble that its basis element $e_i \otimes e_j$ is some bizarre thing which is like a grid but each vertical line of the grid has each point of the line that is a vector, while the horizontal line has each point that is some other vector

it doesn't divide 2

because 3 doesn't divide 2

but is that.... sufficient?

Akiva Weinberger

3:17 AM

So you want to prove or disprove that $x|a\implies N(x)|a$, for $a\in\Bbb Z$

indeed

for the quadratic integers only

In attempt to illustrate, naively it seemed to look something like this:

Akiva Weinberger

Hm. Consider the set of all integers that are multiples of $x$

alrighty?

Akiva Weinberger

If $a$ and $b$ are both multiples of $x$, then so are $a+b$ and $a-b$

3:19 AM

hmm

true

linear combinations

Akiva Weinberger

This implies that the set of integers that are multiples of $x$ is equal to $\langle n\rangle$ (which means the set of multiples of $n$) for some $n$

how?

i think there was a leap of logic there

Picture this: For an ordinary grid, we move up y horizontal lines if we want to walk north by y units, and we move to the right x vertical lines i we want to walk east by x units.

But matrices does something strange, it's like for every unit north we walked e.g. 6 units east west and every unit east we walked e.g. 7 units south

Akiva Weinberger

Proof: Look at the smallest positive number in the set. Call it $n$. We want to show that everything in that is a multiple of $n$.

hi @Daminark

Akiva Weinberger

3:21 AM

Suppose not; then there is some integer $m$ in the set that isn't a multiple of $n$.

ok

whatever matrices and multilinear maps live in, is an intuitively very bizarre space where every unit of xyz direction actually count towards walking in some other direction

Akiva Weinberger

By the division algorithm for the integers, there exists $q$ and $k$ such that $m=nq+k$ where $0<k<n$

(obviously $k\ne0$ since we assumed it's not a multiple of $n$)

Then, by the property from before, $m-nq=m-n-n-\dotsb-n=k$ is in the set

but that contradicts the fact that $n$ was the smallest positive integer in the set.

This demonstrates a property of the integers: They form a principle ideal domain.

so it is like one is walking in two different directions at the same time since you have something like e.g. 6 units east west / 1 unit north

@AkivaWeinberger I.e. that they have unique prime factorization?

@Secret I got it.

Akiva Weinberger

3:24 AM

An ideal in a ring is a set of numbers that is closed under addition and subtraction, as well as multiplication by anything in the ring

@Secret Walking 6 units west and 1 unit north is just an angled direction....

Akiva Weinberger

The set of even numbers is an ideal, for example.

ooooh

so the subset we are looking at is an ideal?

Akiva Weinberger

Yep

Hey @Adeek!

Akiva Weinberger

3:25 AM

A principle ideal is something of the form $\langle n\rangle$, that is, the set of multiples of a single element

A principle ideal domain is one in which every ideal is principle.

@Typhon No, it works like this. If I have a geographical map such that every vertical unit north is the same as 6 units east west, then walking 4 units north actually translates to 24 units east west

@AkivaWeinberger well with that condition it must be "b". want to know why?

Akiva Weinberger

Why?

simple

It is not walking 6 units west and then 1 unit north

3:26 AM

the norm of two multiples is the multiple of the norms

but more like every unit walking to the north trades into 4 units east west

so the norm of anything multiplied times x is b times that thing

therefore, the only way we could produce a thing smaller than b is if.... there was a multiplicative way to produce integers without norms.

@AkivaWeinberger I guess I don't. All we did was go in a circle. XD

Akiva Weinberger

Isn't the norm of an integer its square?

yaw

norm of an irrational, though.

So the weird thing is that you might saw me walking 6 units north, but if you actually measure the displacement, I am now 24 unit east west away from you

3:30 AM

so you walk in one direction yet you end up in a totally different direction afterwards?

yes

this sounds like discontinuous motion....

i'd refer to portals.

it really weird, but that pretty much what is happening when you do a matrix multiplication and translate that into walking in some space

uuuh

nah

the problem is that you are not considering the motion that corresponds to that matrix multiplication

if we are dealing with isometries then they are rigid motions. Those are just the same as sliding a thing on a table.

affine transformations are scalings as well

but scaling the entire universe or even an object is just growth

other matrices would relate more to funky mirroring.

i think the flaw is more in how you think of geometry.

and matrices

@AkivaWeinberger Let's call N the minimal ordinary integer multiple of some quadratic integer. We know for the integers x that this is... x. We wish to show that for irrationals x such that x^2 + ax + b = 0, that n = b. Any ideas?

Akiva Weinberger

To be clear, when we say that $N$, is a multiple of $x$, we mean that $N=x\cdot y$ where $y\in\Bbb Z[x]$, right?

3:38 AM

yes

Akiva Weinberger

So, $y=mx+n$ for some integers $m,n$

so $N=x(mx+n)$

mmhm

Akiva Weinberger

or $mx^2+nx-N=0$

and we also know that $x^2+ax+b=0$

heh

Akiva Weinberger

which means $x^2=-ax-b$

so $m(-ax-b)+nx-N=0$

I feel like I might have made a mistake somewhere?

But I'll continue

3:40 AM

no mistake so far

you haven't used the fact that N is the smallest integer

Akiva Weinberger

Right

but I suppose that showing it is divisible by b would be sufficient as well

since my claim is merely that b divides all integer multiples.

Akiva Weinberger

Hm. We can rearrange that to $b=(-a-x)x$

wait

that is a mistake

x + x' = a

Akiva Weinberger

No, $-a$, pretty sure

3:42 AM

x' = a - x

Akiva Weinberger

because of the minus signs in Viete's formula

(a-x)x = b

then you have a contradiction somewhere

Akiva Weinberger

@Typhon This is wrong. $x+x'=-a$

oh yeah

you're right

darnit

I keep misusing that

Akiva Weinberger

So $b=(-x-a)x$ and $N=(mx+n)x$

3:44 AM

ok

Akiva Weinberger

and I guess if we show that $|b|\le|N|$ then we're good?

mhmmm

Akiva Weinberger

Because then if $N$ is the smallest then it would have to equal $b$ (or its negative, whatever)

or optionally just show N | b

Akiva Weinberger

You mean b | N

3:47 AM

is that the right notation?

i can never remember if it is "divided by" or "divides"

Akiva Weinberger

$3|6$, not the other way around

It's confusing, yeah

ok

Akiva Weinberger

Hm. $N+mb=(mx+n)x+(-mx-ma)x=(n-ma)x$

The left-hand side is an integer

The right-hand side is $x$ times an integer

let me digest for a moment

Akiva Weinberger

Dividing gives us $x=\dfrac{N+mb}{n-ma}$, unless $n-ma=0$

which is impossible because $x$ is irrational and that's the ratio of two integers

3:49 AM

ok hold on

Akiva Weinberger

5 mins ago, by Akiva Weinberger

So $b=(-x-a)x$ and $N=(mx+n)x$

i see

Akiva Weinberger

^Repost for convenience

So $n-ma=0$, or $n=ma$, or $N=(mx+ma)x=n(x+a)x=-nb$ so $N$ is a multiple of $b$ and we're done

ah

clever

thanks

and by thart

Akiva Weinberger

Yeah that was more confusing than I thought it would be

3:52 AM

therefore

Akiva Weinberger

The set of integers that are multiples of $x$ equals $\langle b\rangle$

and an integer is a multiple of $x$ iff it's a multiple of $b$.

x divides x' and vice versa if and only if a divides b

Akiva Weinberger

(Assuming $x$ is irrational, of course.)

of course

@AkivaWeinberger assuming irrational x is fundamental to this subject. It goes without saying. After all, if x isn't irrational then Z[x] is just Z and well... we're not studying Z are we?

thing is

I think the whole x divides x' applies to integers as well

but it isn't as relevant

thanks @AkivaWeinberger

Akiva Weinberger

I think you meant "b divided a" there. Like, b is a factor of a

Right?

3:55 AM

ya

x divides x' and vice versa if and only if b divides a

Akiva Weinberger

Right, 'cause $x$ divides $x'$ implies $x$ divides $-x-x'=a$ implies (by what we just proved) $b$ divides $a$

and for the converse, $b$ divides $a$ implies $xx'$ divides $-x-x'$ implies $x$ divides $-x-x'$ implies $x$ divides $x'$

Hm. Does that mean that $x|x'$ is equivalent to $x'|x$? @Typhon

Since they're both equivalent to $b|a$

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