Special cases are $m=0$ (then $V(m\mathbb{Z})=V(\{0\})=Spec(\mathbb{Z})$) and $m=1$ (then $V(m\mathbb{Z})=V(\mathbb{Z})=\emptyset$)

Can one be good at pedagogy but bad at teaching it

So... the closed sets of $Spec(\mathbb{Z})$ are the finite sets $\{(p_1),\dots,(p_n)\}$ for some prime numbers $p_1,\dots,p_n$ and the whole space?

actually I think I got it now^^

@lattice Yes, you do

anyone got any clues regarding this?

I.e. the Zariski topology is the cofinite topology on $Spec (\mathbb{Z})$

@Semiclassical I think my conjecture holds

I was thinking that a ring shaped vector times a ring shaped vector will result in a sphere or ellipse.

for any integer equal to or greater than 2

I think that the dimension of an ellipsoidal matrix will be the combination of the 3 radii.

if n is odd $\frac{n+1}{2}$

if n is even $\frac{3n}{2}$

@Dodsy prove it

please

its unprovable

it's not trivial.

it's literally the collatz conjecture.

if you can prove the collatz conjecture you can prove the Dodsy conjecture.

They're equivalent?

@Dodsy then you don't believe it. One does not believe statements they cannot prove.

:-)

@AkivaWeinberger he just means that the conjecture's are fundamentally so similar that proving one provides the means to prove the other.

@AkivaWeinberger no not equivilent

Typhon

Oh OK

?

up there you told semi that appararently the problem is as you said "trivial"

Which I find strange.

it really isn't.

@Dodsy the guy who responded claimed to have a proof almost instantly.

that is... trivial.

Oh Zee?

His proof wasn't even real :P

@Krijn I just somehow mixed up $0$ as an element of $\mathbb{Z}$ and $(0)$ as an element of $Spec(\mathbb{Z})$ when trying to show that $0$ is a generic point...

in short, I don't really know what I did there^^

D:

i thought he was solving the case for even integers

No, this is not solved.

@Dodsy proving it works for all odd integers would prove it is true

because: all factors of two are changes to three

The reason the Collatz conjecture is hard, I think, is that it seems to hint at some relation between the prime factorization of $x$ and the prime factorization of $x+1$

And we don't know anything about that

True.

actually

Well, between $x$ and $3x+1$, but same deal

we do know something

x and 3x+1 share no factors

morningh

because 3x+1 - 3x = 1

"if 1 is an integer linear combination of a and b, then the GCD of a and b is 1"

so we can say that they share no factors

For any integer $\ge2$, if the integer $n$ is odd then; $\frac{n+1}{2}$ if the integer $n$ is even then $\frac{3n}{2}$ all integers will eventually terminate to 2.

It's hard to prove anything like this.

@Dodsy yes, but we can restate it as: "Change all factors of 2 into 3's, then add one and divide by 2. Repeating this infinite times will result in 2 or 3."

infinite?

it will terminate in a finite amount of time.

@Dodsy I'm thinking of significantly large numbers.

Also if it terminates to 3

it will terminate to 2.

yes, even a significantly large number will terminate in a finite number of steps.

I meant that one can repeat it any number of times and all successive passes will be either 2 or 3

3 is odd

add 1

divide by 2

@Dodsy true, but as one continues to increase the integer, the steps will increase without bound. Hence "infinite'

you get 2.

YES I KNOW THAT

well that's the problem with these problems, you can't check every number, numbers are infinite.

I'm saying that after so many passes all successive steps will only produce either a 2 or 3.

@Dodsy that's the point of induction

I don't even know what that means.

anyways, I'm happy

I have a conjecture now.

@Dodsy yeah. If you iterate on 2, you get 3. If you iterate on 3, you get 2. Easier to build the conjecture as 2 or 3, since then one might be able to avoid 2.

I was merely saying that 2 doesn't map to 2

"termination" makes me think that the numbers stop changing

2*3/2 = 3

as in the case of 1

3+1/2 = 2

@Dodsy dude no

2*3/2 = 3

"change all factors of 2 into 3"

2 -> 3

so it still terminates to 2.

I tend to think the conjecture is plausible but well---not a proof

no

termination would mean 2 -> 2

@Semiclassical its not proven

it terminates at the set {2,3}

When typhon told you it was "trivial"

he was wrong.

@Semiclassical Zee posted a troll proof and fooled us.

No, all integers equal to or greater than 2 will terminate to 2.

:p

@Dodsy termination means 2 -> 2

you do not need to mention 3.

termination means a sequence ends

Troll proof assumes he doesn't believe it himself

Right...

what is your point?

@Dodsy the sequence doesn't end. It becomes the sequence 3,2,3,2,3,2,3,2,3,2...

what...

the sequence would terminate

once you hit 2

you wouldn't continue goin!

I think what Typhon wants is: the relevant condition is not that the sequence necessarily terminates, but that the only cycle is 2->3->2

cmon man, give your head a shake

@Semiclassical 1 is also a cycle.

Point

So in the collatz conjecture

we would say that he sequence doesn't terminate at 1

@Dodsy umm... your formula indicates that you keep going forever and that it just hits 2 and stops changing.

?

okay but we hit 1 in the collatz conjecture

Hello chat denizens.

are we going to multuply it by 3

add 1

@Dodsy collatz conjecture of 1 is 1, correct?

and divide by 2?

Wrong.

well then it terminates to {1,2,4}

that's stupid

Depends how you formulate the Collatz map

Once you hit the number you would stop computing.

I'm not saying it terminates at

I'm saying it will terminate to.

it doesn't mean it's impossible to keep going.

I think the word 'terminate' is unnecessary here

perhaps you're right, it might be the language.

The statement is "every number eventually gets to 2" (except 1)

Better to say

"the process will eventually reach 2"

for $n\ge2$

I have a feeling that whoever solves the dodsy conjecture

@Dodsy That is better phrasing to me. I know what you want. I'm just commenting on the clarity of the exposition. As in... to me "terminate" sounds like that if I kept going I wouldn't go anywhere. The function stops changing.

will solve the collatz conjecture.

hence why I suggested "terminated at" {2,3}

@Typhon fair enough, better to just say it will eventually reach 2.

Right. That said, there is a legit role for the word terminate here; namely, reaching 2 is the condition for anyone running the process to stop

Because once it gets there you know what will happen from then on

So it's the break condition in a while loop

true.

You know how I came up with the dodsy conjecture?

I just moved the +1 over to the other side

and flipped the even and odds.

@Semiclassical No. It is the termination condition. A break statement is bad coding practices.

Am I genius?

Maybe.

@Dodsy nah. But you did create an intriguing thought. I would ask that you ask the question since it your conjecture that spawned it.

@Dodsy please go ask the following question: "is there a way to express a function without recursion that takes all factors of some integer and replaces those factors with an equal number of multiples of some other integer?"

that would be pretty intriguing.

Ugh. I think I'm going to hate myself over the next 5-6 weeks though.

@Typhon I wondered if I was using it wrong. Oh well

@Semiclassical It's ok. You're a physics person. I can give you the benefit of the doubt on programming terminology.

Now if you claimed to be studying computer science... I would be obligated to kiddingly say something like "you should know better. XD"

lol

@Semiclassical you ever write any large papers before?

I've been part of some.

None by myself.

how large?

hmm. nothing too long, I'd say

5000 words would you say?

Lemme find some for reference.

I forget how long 5000 words is tbh

at least 15-16 pages

I haven't saved any papers from undergrad, for better or worse, so I have no idea how that would compare.

But this is the first paper I was part of as a grad student: arxiv.org/pdf/1303.6386.pdf

i just thought maybe you'd have a ballpark estimate

Not really. I've never had writing assignments that were in terms of word count.

me neither

but I just thought maybe you looked. XD

lol, not in a while.

I tend to err on the side of "long"

And as I said I haven't held onto any undergrad papers, so I don't remember how long those would've been.

It's just that I was trying to think of how to describe this thingy I have to do to you.

@Semiclassical I know I wrote a 50 page geometry paper but to be fair "You didn't need to do this much."

lol

There's an old line whose origin is unclear

To say I enjoyed it is an understatement.

"I made this letter longer only because I have not had the leisure to make it shorter"

Which is to say, a lot of the work of writing is editing and getting rid of stuff that's not needed.

@Semiclassical Actually, I ended up describing the axiomatic concept of Euclidean geometry, the history of non-Euclidean geometry's development, a general (brief) overview of what I imagine would be prospective surfaces satisfying the different negation combinations of the 5 postulates, a pretty in-depth description of analytical (I prefer the term vectoral) Euclidean geometry, a long and complete description of almost all of spherical geometry, and a very tiny section describing hyperbolic geometry.

^^the last one is because I ran out of time.

oh and I also talked about projective geometry in a build up to hyperbolic.

I'd argue that while I could've shrunk down a little... most of the size was the sheer amount of content and not wordiness.

@Semiclassical true, but that letter probably isn't describing 3 different geometries. XD

Hey!

@Daminark wassup?

Not much, how about you?

@Daminark I have to start writing the documentation for my game. Ugh.

(Or rather, the documentation describing how it all works)

Oh, not the most fun part, it seems

Documentation is always the most tedious part.

@Semiclassical not referring to that kind of documentation. I'm referring to the document describing the math behind everything

note: a high school teacher is going to have to read it eventually

(one whose level of CS knowledge could be anywhere between 0 and Dr. of CS)

                ^ I feel like the initial result is wrong and I tried using tools from trigonometry

@Semiclassical the game is technically a project a kid at another school somewhere is supposed to do over the summer. They are allowed to have anyone help.. so they asked me. It was posed as "make a video game over the summer". Needless to say, I think between the 3D surfaces and the memory allocation, I have a lot of stuff to try to explain. Otherwise, I doubt anyone else is going to be able to read the code. It ain't trivial.

and to be frank... I don't know how much experience this person has had. They might just be a high school math teacher. XD

@Semiclassical Plus, if I start writing anything of significant size I will become tempted to start writing something to rival that geometry paper. I'm not yet prepared to lose myself to that rabbit hole. Btw, the tree grabber worked nicely. Thanks for that formula.

honestly, I should probably just dive into it now. Still have a few months. It will give me time to be patient with it.

im gonna head out