elementary-number-theory

[wraith oasis]

(This is usually where people start reacting with the emote)

[ionic latch]

Interestingly, it does form a monoid under elementwise addition that's isomorphic to the natural numbers with standard multiplication (or polynomials with addition)

[kindred fulcrum]

isn't this just $\bigoplus_{n \in \bN} \bN$?

[sterile pumice]

ExpertEsquieESQUIE

[west glade]

yes

[ionic latch]

What's that mean

Is $\oplus$ the symbol for prime factorization?

[sterile pumice]

Coolempire93

[west glade]

its basically the set of elements of sequences of natural numbers for which only finitely many entries are nonzero

aka the possible set of sequences of the exponents you could have in the prime factorization of natural numbers

and the addition of those sequences is exactly what you get when multiplying the relevant natural numbers

[sand yarrow]

@sharp shadow thanks. is the goal of a proof like this to convert one form of the equation or inequality into another form? if you can convert the equation or inequality into the final form then you've effectively proven it? I think I'm still weak with proofs. I think I can follow the above logic, but I have no idea how I'd find a uniqueness proof from there. like, somehow I'd have to show that q and r are not equal. q and r exist, but they must also not be equal to each other.

[sharp shadow]

a typical way to prove uniqueness is to assume that there are different values q,r and q',r' that both satisfy the conditions, and then show somehow that q=q' and r=r'

woohoo, I finally proved that! I was silly and didn't notice that there was a useful theorem at the end of the chapter

As a bonus I also proved that there is infinite number of primes of form 4n+3 (actually I started with that, because it doesn't require any additional quadratic theorem)

here are my proofs (somewhat long...):

and another one, Problem 49 from the problem set above, it's about 4n+1 primes:

that was somewhat tricky, happy to have done it!

[unique cypress]

Now do it in general

[sharp shadow]

I bet that must be hard, even Niven and Zuckerman don't prove it in the book, saying that it uses techniques beyond the scope of the book (I assume you mean Dirichlet theorem)

[sharp shadow]

well, let's start from the beginning. The problem says: prove that there exist some numbers q, r satisfying some property. How can we prove that?

We need to just show such numbers, i.e. find them somehow, and also show that they satisfy that property

but how do we find them? We need to start from somewhere, something similar. And we have a similar thing, this "division algorithm" that was described in the book. It says that we can always find numbers q and r satisfying another property, slightly different to ours, but similar

now, the idea is to use that as a starting point, and somehow derive the numbers that we need for our problem from the numbers that are given by "division algorithm"

and that's what I did. I started with q and r that are given to us by the division algorithm, and transformed them to other numbers: q' and r', that satisfy the conditions of this problem

so now, I can just show those found numbers q' and r' to anyone interested, and they will satisfy the criteria specified in the problem, so I solved it, woohoo

the only remaining bit is that the problem asks us to show that those q' and r' not only exist, but are also unique, i.e. that there is no other pair of numbers (say, q'' and r'') that can satisfy the same criteria. Only our previously found q' and r' can

Does that make sense?

(I am off to sleep, so will check tomorrow :)

[ionic latch]

Interesting way to express

Is there anything special about the polynomial part that I mentioned

Sequences of natural numbers for which only finitely many entries are nonzero sounds quite isomorphic to $\ZZ[x]$ (or $\mathcal{P}(\ZZ)$? tbh I don't necessarily know the difference)

[sterile pumice]

Coolempire93

[tall arch]

P(Z) has cardinality equivalent to R

[ionic latch]

Not referring to the powerset

[kindred fulcrum]

Its pretty much the same.

[ionic latch]

Nothing fun or interesting since polynomials have a slightly different idea of what multiplication entails once we move to the ring structure?

[kindred fulcrum]

You are only using the additive structure in what you mentioned

[ionic latch]

That's why I brought up Z[x] I wanted to see if I could bring in any form of multiplication operation

[kindred fulcrum]

I think the structure here is kinda wierd

[west glade]

x^i*x^j=x^(i+j) would translate to p_i*p_j=p_(i+j) which would certainly be weird

I doubt this would be in any way useful

you could try checking what e.g. (1+x+3x^2)*(2x+7x^5) would correspond to

but its probably nonsense

[versed flax]

Well the set of sequences of integers with finitely many non-zero entries is the ring Z[x] if you define the product of 2 sequences the right way

idk if that's what you were looking for tho

[sharp shadow]

What is the Royal Road to Algebraic Number Theory? Currently I see it as this: study rings in a typical algebra introductory course, study ENT to get some idea what it’s all about in the simpler setting, then read some chapters from Ireland/Rosen to revisit NT from a more general rings perspective and maybe read Marcus “Number Fields” or Stewart “ANT” as an intro to ANT proper. Is this reasonable? (I suspect Galois Theory, Complex Analysis and a deeper dive in Commutative Algebra might be handy too, but not strictly necessary?)

And in general: is ANT exciting for you? Are there interesting unexpected results or problems? Or it’s not that fun actually?

[unique cypress]

Yup

[dusty dragon]

This seems reasonable to me. The roadmap that I followed in a curriculum to an intro in ANT is as you say: study ENT and abstract algebra to build the machinery and intuition for studying ANT. You’ll see analogs of many different number theoretic results in different fields; for example, the Chinese Remainder Theorem is studied in context of Z/mZ in number theory but you can extend that to arbitrary rings with ideals. You should generally conclude a first course in algebra with an intro to field theory. Naturally, ANT extends upon this through the study of field extensions. Indeed, solving NT problems with algebra often means looking at field extensions and studying properties of field extensions (norms and traces). You’ll see this in the culmination of ANT through Dirichlet’s unit theorem and how it can be used to obtain solutions to Pell equations

We used Marcus’ Number Fields book and it served as a good intro to the field, so you shouldn’t go wrong with it

Galois theory would be quite doable to do alongside ANT since they cover pretty similar topics but their motivations are largely distinct

[sharp shadow]

Thank you!

[ionic latch]

Yeah either some very hidden insight or useless and likely the second xd

[sterile pumice]

azart

[unique cypress]

hmm

[modern temple]

forget about it, I found the counter-example

[mint stone]

Yeah a=1 and b=1.

[sand yarrow]

I just got a copy of the elementary number theory book by rosen. I think I'm going to try to work through it. it looks cool.

[ionic latch]

That’s the one I learned from

I liked it

I found that the chapters were pretty self-contained as well

[tribal palm]

It has been proven, by James Maynard that there are infinite number gap sizes of g<246 within the prime number sequence.

Many people think that this can be reduced further. I proved by calculation exhaustion that the lowest this bound can come down to is g<54.

There are 4 different types of intervals within the prime number sequence, and each one has a maximum bound. I will list them:

(+,-) g<54

(-,-) g<54

(+,+) g< infinity

(-,+) g< infinity

In other words, moving forward in the prime number sequence, the prime gaps are unbounded. The bounded gaps are when you move backwards because there is a limit on how far you can go backwards while still generating a new prime that is still bigger than the last prime.

This theoretical hard limit of g<54 is less than the smallest proven known gap sizes of g<246 and the technique that I used has nothing to do with sieve theory.

But what I’m saying is this, the gap size of g<54 is the limiting factor for the smallest size of gap that can be “proven” to happen infinitely many times.

While most mathematicians think the twin prime conjecture is true (gap size of 2), I’m saying, even if it is true, you can’t prove it. The gap size of 54 is the lowest bound that can exist. This bound cannot be improved upon. It is the hard limit.

So for gap sizes of g<2 , g<4, g<6, while these gaps probably do happen an infinite number of times, it is impossible to write a proof for them. The best we can do is g<54 and that is it.

Let me know if anyone wants to confirm my result.

[patent acorn]

even if it is true, you can't prove it
which specific theory did you show can't prove it?

[wide snow]

> tagged with undergraduate math
> claims to have improved a result by James Maynard
> without sieve theory

[dusty dragon]

looks like AI slop

[unique cypress]

May I ask what LLM did you use for this?

[versed flax]

no

[tribal palm]

This was a paper I wrote with the assistance of Grok xAI, but the idea should be pretty novel.

When you are generating prime numbers, the next prime that you generate can only go “backwards” in the sequence so far (this is not necessarily true the with forward generating primes) and because of this, these particular types of intervals have a maximum size of gap=54.

I used the AI to brute force the max bound by extensive calculations.

I will attach the paper.

It is AI slop? I don’t it know…it might be, but it might not. If someone can confirm my findings then, the idea in the paper is probably correct.

The power of any tool is based upon the user.

[karmic thorn]

bro is 👍ing his own message

[boreal garden]

it is AI slop

[tribal palm]

This has nothing to do with AI slop or not.

In math and science, when someone proposes an idea, you either:

A: Confirm it to be true.

B: Prove that the claim is false.

In this case, this AI tool was only used to brute force millions of calculations, but the idea is totally mine.

If a person were to play a game of chess with an AI bot, such as Stockfish and were beaten, nobody would ever say “I was beaten by AI slop”.

Instead, they would say, “I was beaten by one of the best artificial players in the world”.

The reason why people use the phrase “AI slop” is because anyone can sit down and generate a result with it, because most people are “mediocre” they produce a work that is of “low quality output”.

But in regards to my paper, I don’t think this is a low quality effort.

[boreal garden]

Well, let me rewrite it for you, it is nonsense.

The reason why we don't say that stockfish is not AI slop is because we can measure how good stockfish is, there's a certain parameter. For llms most of the times it just produces random nonsense.

It takes an expert actively curating, searching and discriminating information produced by a llm to get some good results. And I don't think nobody respectable is using them to solve an impossible problem like Goldbach conjecture.

[tribal palm]

LLMs like ChatGTP is actually solving unanswered math problems such as the Erdos problems.

It’s true that you can get these AIs to spit out nonsense, that is not relevant.

My claim is that there are an infinite number of bounded gaps size 54 and the only matter now is to either prove or disprove the claim. Opinions about LLMs are not related to the claim in the paper.

[boreal garden]

You haven't prove nothing.

[unique cypress]

Can we keep #elementary-number-theory clean please? The AI slop is ruining it

[warm coral]

imagine a dude microwaving a meal and then proclaiming himself a chef

[patent acorn]

your pdf explicitly says that the "results" are just empirical data, lists the statement that would imply that there are infinitely many gaps of size <= 54 as a conjecture, and concludes with "Although a rigorous proof remains elusive, [...]"

also the fact that you can get AIs to spit out nonsense is relevant, because in the real world, we do not have infinite time to consider every argument that anyone puts forward and so sometimes we have to use crude heuristics to decide what's worth spending time on

[karmic thorn]

also isn't ai slop against the server rules?

iirc there was like a clause in #rules

[unique cypress]

Someone ask a question about solving equations of the form a^2-db^2=p to clean this

[boreal garden]

This was a cool chanel to go to every now and then, but yeah AI slop has ruin it.

[warm coral]

I miss when cranks didn't use ai

[boreal garden]

This week I teach some guys about the case d=-1. And they really loved it

[unique cypress]

based

tfw p splits iff p = 1 mod 4

[kindred fulcrum]

what primes are of the form x^2 + 14y^2 then?

[boreal garden]

They asked me about other cases and I told them that there's a subtle relationship with complex multiplication. But I didn't have too much time to explain it

[unique cypress]

tfw Q(sqrt(-14))

[kindred fulcrum]

👍

[unique cypress]

tfw j invariant is scary

[kindred fulcrum]

it is scary

but I am more scared of modular forms

one day I will get over it xD

[unique cypress]

I assure you modular forms are not scary. Maass forms is what you should be scared of

[kindred fulcrum]

ignorance is bliss

[versed flax]

elementary number theory
modular forms
so elementary

[unique cypress]

I mean it's the 5th operation right (Enpeace reminded me)

[kindred fulcrum]

you summoned enpeace

[warm coral]

I'm always watching

[versed flax]

these days everyone is becoming nt pilled huh

[versed flax]

i know where you stole this line from

[warm coral]

lwk only in it for the geometry

[boreal garden]

Modular forms are elementary.
If you don't know them in elementary school you are wasting your time

[karmic thorn]

[tribal palm]

So the other day, I posted a claim about certain gaps being maxed at 54 as a gap size.

Everyone said “this is AI slop” then called it a day.

After further looking at the problem, it turned out that all 4 gap cases, indeed were unbounded, it’s just that there were some cases that grew so slowly that you have to look at thousands of cases to see the growth rate.

So the whole claim of gap<54 was false, however still when you do good research, you just focus on trying to prove or disprove claims.

Tools are little exactly what they are…. Tools.

The term “AI slop” is referred to, because it’s easy for anyone to spit out large volumes of low effort content.

But if used in the right hands, could produce novel results.

Imagine this, you are a carpenter who builds wooden chairs at your wood shop.

It takes you 1 week to build a single chair, but the quality is really high.

Then another woodworker sets up shop nearby. He can crank out 1 chair per day, however the chairs are low quality.

So one day, the true craftsmanship guy walks over to the newcomer and says “Hey that’s wood slop”.

But at the end of the day, they are both using the same tools. The only reason why the true woodworker says “wood slop” is because his competitor can crank out massive amount of poorly made chairs, which undermines his own high quantity work.

I don’t normally use AI to solve math problems, but for some problems involving heavy computation, such as calculating 5 million primes, it seems somewhat useful.

Also in every field, using a bot, or advanced tool can actually be beneficial.

I solve sokoban puzzles, someone’s I come across sokoban puzzles online and for this one website it will tell you if nobody has solved it before. Sometimes I crack these with Jsoko, a sokoban solving program. Why? Because there is a reason why nobody has solved them on the website….. because they are hard.

[fiery zenith]

AI slop is AI "slop" largely because of the people who use it in the way they do.

[lime trout]

so are we just going to ignore the fact that you called a your own paper, which you wrote, so you should have a clear understanding of its contents, a "proof" when the paper itself says that a rigorous proof remains elusive and all it does it do a check of 5 million primes to see if your result holds?

Is it not low quality slop if the author himself doesn't even make the effort to understand what his own AI generated paper just spat out?

[candid torrent]

Using AI for heavy computation
What is python for?

[lime trout]

you say that as if you are the so-called "right hands" to process the AI's work, yet clearly no processing has been done as evidenced by your language contradicting that of your paper's

[ionic latch]

python is for training ai models

[lime trout]

you are the woodworker who cranks out the low quality chairs btw. enough said

[fiery zenith]

i sure can feel your anger in this rebuke...

[lime trout]

LOL

i am not angry

[ionic latch]

[versed flax]

just use brainpower to do the heavy computations, you dont need anything else fr

[lime trout]

It has been proven, by James Maynard that if it holds for 5 million primes it holds for all primes so im right suck it bitches

[cosmic zenith]

I'm bad at English, but I'd like to know how many people know that Fermat's little theorem is topological.

)

[candid torrent]

I don't, so that's 0 / 1 now

Are there any resources on that btw? It sounds interesting

[cosmic zenith]

I can only send you a short article by Academician Arnold in Russian. In short, it discusses a directed graph mapping a certain set onto itself. In theory, the same idea is in the book "Topology of Numbers."

[candid torrent]

thanks

[boreal garden]

We've been using computers in math for a very long time... You are at least 30 years late to the controversy, yet that doesn't make your work valid. Math requires you to be humble, If you want to learn math, there's a very nice theorem called "prime number theorem" which tells how prime numbers grow, there's a very elementary proof and it would take you around 1 year of learning and hard work to completely understand it but I guarantee that you would learn a lot more than doing whatever you are doing right now.

Like, you can pick a book, read it and post your questions here and we will help you.
But please, stop trying to solve this conjecture, you will not solve it.

[boreal garden]

what do you mean by topological?

[cosmic zenith]

I just sat down to figure it out and here it is
Arnol’d calls Fermat’s little theorem ‘topological’ in the sense that it describes the cycle structure in a monad graph: elements of the multiplicative group under the map x->ax mod p form a closed cycle, which he refers to as a ‘topological circle’. Here, ‘topological’ means a discrete, combinatorial analogue of a topological structure, not classical topology
I'm not really sure how "topological" this is. Well, at least it sounds interesting.

[karmic thorn]

lets use c then

(for computation)

<insert rant about pari macro based whatever tf am i even looking at at this point here>

[ionic latch]

Reminds me of #computing-software who was like can my python run with no C please

[karmic thorn]

I mean, once it's compiled it's no longer c, right?

[boreal garden]

I mean, that's one way to say it. But if just feels wrong to say it is topological.

Is that Arnorld the same "Bunch of hard problems" Arnorld? Like the mathematical physics guy?
Idk, I would say that there's something more on the fourier side than the topological say when it comes to fermat's little theorem.

[cosmic zenith]

Yes, you're right. I didn't expect that topologicality in this sense was simply a term invented by Arnold. On the other hand, I thought, "That's just his style."

[boreal garden]

Are you talking about Vladimir Arnorld?

[cosmic zenith]

Yes

[boreal garden]

It kind of feels like the opposite of what he would say tbh

[cosmic zenith]

Why

In Russian sorry

[cosmic zenith]

Well, here he clearly doesn't mention this fact, but he does in his lecture on Number Hydrodynamics. (That's where I learned about him and started looking for what he wrote back then.) There are also two lectures on this article, if I'm not mistaken.

[boreal garden]

uhh that makes more sense

[tribal palm]

I final finished my power point on the prime number sequence. This is a pdf. Human created.

[lunar rover]

Let $n = a_1a_2a_3$ be a three-digit number (where $a_1, a_2, a_3$ denote its digits).
Define the number
[
N = a_1a_2a_3a_1a_2a_3,
]
obtained by repeating the digits of $n$ twice.
For example, if $n = 504$, then $N = 504504$.

Prove that $N$ is always divisible by $143$.

[sterile pumice]

Rodro

[west glade]

what can you say about the number $a_1a_2a_3000$ ?

[sterile pumice]

Denascite

[lunar rover]

Yeah 1001=143*7

[west glade]

which answers the question

[lilac bronze]

can i check if the following is true

Let $n, k \in \mathbb{N}$ with $k \neq 0$. Then $\forall m \in \mathbb{N}, n | km \iff \frac{n}{\text{gcd}(n, k) } | m$

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[ionic latch]

I'm sure you can

Anyway
$n|km \implies \frac{n}{(n,k)} | \frac{km}{(n,k)} \implies \frac{n}{(n,k)} | m$ is one direction

[sterile pumice]

Coolempire93

[kindred fulcrum]

in the other direction if n/gcd(n,k) divides m then n divides m * gcd(n,k) which divides m * k

[ionic latch]

$\frac{n}{(n,k)}|m \implies n | m\cdot(n,k) | km$
ah yeah

[sterile pumice]

Coolempire93

[ionic latch]

Nice

[lilac bronze]

The second implication uses Euclid’s lemma right

[ionic latch]

Idk what euclid's lemma is 😅 but I use the fact that if a and b are coprime, then $a | bc \implies a | c$

[sterile pumice]

Coolempire93

[lilac bronze]

That’s Euclid’s lemma i believe

[ionic latch]

Ah

There you go then 🙂

[lilac bronze]

This will make an appearance in my next article

[ionic latch]

Youre writing an article about it but wasn't sure how to prove it? Or wanted to see what way was easiest

[lilac bronze]

The latter

My next article will be baby Yoneda 4, on Adjunctions

[ionic latch]

Oh lord

[lilac bronze]

And this seems like a cool example of an adjunction to use

I hope it will be approachable though

[ionic latch]

This is a nice example

[wind ridge]

why does euler's totient function work?

[ionic latch]

Our book (Rosen ENT) developed it really nicely

A whole chapter on it starting with multiplicative functions

If no one explains by the time I come back I'll run through it

[lilac bronze]

let $a, b, c \in \mathbb{N}$

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[lilac bronze]

it's true that $a \leq b + c \iff \max(a - b, 0) \leq c$, right?

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[kindred fulcrum]

yes

[lilac bronze]

cool cool

[kindred fulcrum]

just split into cases a <= b and a > b

[lilac bronze]

mhm mhm, makes sense

now, for $a + b \leq c$..

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[lilac bronze]

if $b > c$ this is impossible

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[lilac bronze]

and if $b \leq c$ then it's just $a + b \leq c \iff a \leq c - b$

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[kindred fulcrum]

yeah... I don't think there is more to this

[lilac bronze]

cool cool

i've got some more examples for my next article now

[kindred fulcrum]

xD

[unique cypress]

why did I expect pseudo to ask a question about something related to local cohomology

[lilac bronze]

sorry what

[boreal garden]

What are you writting about anyway? I'm curious

[lilac bronze]

this is going in the next article in the baby yoneda series

on adjunctions

[boreal garden]

And how does your questions relate to this?

[lilac bronze]

so, this says that addition by $b$ on $\mathbb{N}$ has a left adjoint

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[lilac bronze]

namely, taking the "predecessor" b times

except when you hit 0, you stay at 0

[lilac bronze]

whereas this shows you don't get a right adjoint

[boreal garden]

I think I see it

[lilac bronze]

there's also a multiplicative analog of this

$n | km \iff \frac{n}{\text{gcd}(n, k) } |m$

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[unique cypress]

nvm

[boreal garden]

Can I read your article once you finished it?

[lilac bronze]

yeah ofc, it'll go up on my blog after all#

https://pseudonium.github.io/

[red yacht]

how does, 5 | x , 2 | x imply 10 | x because gcd(5,2)=1?

how to prove
5 | x and 2 | x => 10 | x

maybe, because gcd(5,2) = 1 then we can use bezouts identity

1 = 5a + 2b

10 = 50a + 20b

(a,b) = (1,-2)

[kindred fulcrum]

x = 5ax + 2bx

both 5ax and 2bx are divisable by 10

so x is divisable by 10

[red yacht]

what

[red yacht]

you are jumping to conclusions

[kindred fulcrum]

am I? what do you not understand?

[red yacht]

after here how you continue

[kindred fulcrum]

.

x = 2y = 5z for some y,z and then you have 5ax = 10ay, 2bx=10bz

[red yacht]

how

[lilac bronze]

The easiest way is by considering prime factorisations

[red yacht]

care to elaborate

[patent acorn]

"x = 2y for some y" is just the definition of 2 | x, and similarly x = 5z for some z because 5 | x

or an alternative proof: 2 | x implies x = 2k for some k, so 5 | x implies 5 | 2k, and because gcd(5, 2) = 1, this implies 5 | k

the last step uses a generalisation of euclid's lemma: if a | bc and gcd(a, b) = 1, then a | c

(you can prove it using bezout's lemma: given ax + by = 1, we have a | axc + bcy = (ax+by)c = c, so a | c)

[ionic latch]

I didn't get to it yesterday but I'm here today, what aspect of it are you referring to by 'work'

[wind ridge]

why is it true? how does the function actually compute the amount of numbers relatively prime to it?

[ionic latch]

Ah

This'll be fun because the book presents it in the opposite direction

No matter

All we need is totient multiplicative on the primes and it all comes together

[wind ridge]

alr

[ionic latch]

So first let me lay out the formula so that we have the same starting point

Where $n = p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ is the prime-power factorization of $n$, $$\phi(n) = n(1 - \frac{1}{p_1})(1 - \frac{1}{p_2})\cdots(1 - \frac{1}{p_k})$$

[sterile pumice]

Coolempire93

[wind ridge]

yeah

[ionic latch]

Great

So let's break it down a bit

\begin{align*}
\phi(n) &= n(1 - \frac{1}{p_1})(1 - \frac{1}{p_2})\cdots(1 - \frac{1}{p_k}) \
&= p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}(1 - \frac{1}{p_1})(1 - \frac{1}{p_2})\cdots(1 - \frac{1}{p_k}) \
&= (p_1^{a_1} - p_1^{a_1 - 1})(p_2^{a_2} - p_2^{a_2 - 1})\cdots(p_k^{a_k} - p_k^{a_k - 1})
\end{align*}

[sterile pumice]

Coolempire93

[wind ridge]

hmm

interesting

[ionic latch]

Do you see how I got there

[karmic thorn]

isn't $\phi(pq) = (p-1)(q-1)$

[ionic latch]

Yes that's what's coming next

[wind ridge]

yes

[sterile pumice]

xtea418

[ionic latch]

So now comes how this becomes the number of numbers<n relatively prime to n

\begin{theorem}
If $p$ is a prime, then $\phi(p) = p-1$
\end{theorem}

[sterile pumice]

Coolempire93

[ionic latch]

Do you see why this is true based on the formula

[wind ridge]

well, this is intuitive

yes, i understand

[ionic latch]

Okay, how about $\phi(p^a) = p^a - p^{a-1}$

[sterile pumice]

Coolempire93

[ionic latch]

Do you see also how that comes from the formula

[wind ridge]

gimme a sec

wait yeah

i see it

[ionic latch]

Now let's take multiple primes, say p and q
What's left to show is that $\phi(pq) = \phi(p)\phi(q)$

[sterile pumice]

Coolempire93

[ionic latch]

This is the multiplicative property (when extended to any relatively prime integers anyway)

[wind ridge]

this, again, i can get why is true

[ionic latch]

Ah, well if you understand that, then you are essentially done

Going back to the theorem

[ionic latch]

This is definitely the number<p of relative primes to p

[wind ridge]

wait, if p=q, this is false, right?

[ionic latch]

They have to be relatively prime, so yes, different primes

[wind ridge]

oh wait ye

[ionic latch]

Let's consider why this is the number of numbers<p^a relatively prime to p^a

Well, p is the only divisor of p^a

So the only ones not relatively prime are multiples of p

[wind ridge]

mhm

[ionic latch]

There are p^a numbers <= p^a so all we need to do is subtract from that however many multiples of p there are <= p^a

[ionic latch]

Based on this, there should be p^{a-1}, can you prove it? 👀

[wind ridge]

p^(a-1) numbers relatively prime?

oh wait ye

[ionic latch]

p^{a-1} multiples of p less than or equal to p^a (so not relatively prime)

[wind ridge]

so p^1 up to p^(a-1)

[ionic latch]

p^0 is relatively prime since it's 1

[wind ridge]

mkay, changed it

[ionic latch]

The multiples of p are all the numbers of the form k*p
So
p, 2p, 3p, 4p, etc.
There are p of these before p^2
There are p^2 of them before p^3
etc.
So p^(a-1) such before p^a

So yes yu are correct

[wind ridge]

but that's only a numbers

[ionic latch]

That's p^(a-1) numbers

1p, 2p, 3p, ..., p^(a-1)p = p^a and we stop

[wind ridge]

wait, p,2p,3p...p^(a-1)*p

[ionic latch]

Are all the multiples of p

[wind ridge]

mhm

[ionic latch]

Less than or equal to p^a

Of which there are p^(a-1) of them

[wind ridge]

yeah

[ionic latch]

👌

So that means

\begin{theorem} $\phi(p^a) = p^a - p^{a-1} = $ the number of numbers$<=p^a$ relatively prime to $p^a$ \end{theorem}

[sterile pumice]

Coolempire93

[ionic latch]

Great 👍

So we have that Euler's totient counts the number of relatively primes for prime powers

Next comes the final step, why we can multiply them

Consider $n = p^aq^b$

[sterile pumice]

Coolempire93

[wind ridge]

mkay...

[ionic latch]

(With p and q different primes)

We want to count the number of relatively primes to n

Which is p^aq^b - (multiples of p or q)

We already know how this goes though

[wind ridge]

so if you have pq numbers, pq-q-p+1 (from PIE) is the number of them coprime to pq?

[ionic latch]

That's essentially what we plan to do ✅

[wind ridge]

ok alr

so we can apply multiplicativity repeatedly

[ionic latch]

Yes, as we did before

I'm trying to come up with a way to express the overlap in a table so it's easier to visualize

But we know already as before

For higher powers, it looks like

$#{1 \leq k \leq p^{a-1}q^b} = p^{a-1}q^b$ is the number of multiples of p less than $p^aq^b$ (we can write in the form $kp$)

Similarly, $p^aq^{b-1}$ is the number of multiples of q less than $p^aq^b$

[sterile pumice]

Coolempire93

[wind ridge]

pq-q-p+1=(p-1)(q-1)

so...

wait

phi(p)phi(q)=phi(pq) then

[ionic latch]

The overlap will be $p^{a-1}q^{b-1}$ such that
\begin{align*}
\text{relatively primes to } p^aq^b &= p^aq^b - p^{a-1}q^b - p^aq^{b-1} + p^{a-1}q^{b-1} \
&= (p^a - p^{a-1})(q^a - q^{a-1}) \
&= \phi(p^a)\phi(q^b) \
&= \phi(p^aq^b) = \phi(n)
\end{align*}

[ionic latch]

✅

[wind ridge]

so it's basically proved now?

[ionic latch]

Yes, this is essentially how the proof goes

[wind ridge]

oh ok, great!

[sterile pumice]

Coolempire93

[wind ridge]

thank you : )

[ionic latch]

I remember being like wtf this crazy formula does what the first time I saw it too

But yeah essentially you can show that

1. Crazy formula is multiplicative for prime powers
2. phi = relative primes for primes (and then prime powers)
3. Since every number has a prime-power factorization, show that by PIE, the formula works for the individual parts, and factors into each multiplicative term of phi

[wind ridge]

yep

[ionic latch]

It's easier from the other direction (start by supposing phi counts the number of relatively primes, and then show that it is equal to the totient formula) I will say

This direction requires more combinatorics

[wind ridge]

i'm not the best at it 😅

thx tho

[ionic latch]

Nobody is 😔 but that's the fun of it! 💃

[ionic latch]

No problem 🙂

[ionic latch]

Since you spend a lot of time in the calculus channel you may find interesting https://categorified.net/CombinatorialCalculus.pdf

[wind ridge]

ty

[versed flax]

btw since you are talking about euler's totient function, here is something nice related to it: \For any positive integer $n$, the following holds:$$\sum_{d\mid n}\varphi(d)=n$$

[sterile pumice]

ali yassine

[wind ridge]

thank u

[versed flax]

np, there are other functions which have nice properties too if you want to look them up

an example is something called the mobius function

you can check many nice properties of such functions, called arithmetic functions, in something like apostol's "introduction to analytic number theory" if you want

this starts in chapter 2, if you get more interested then ig continue with the book

[ionic latch]

Yeah this is what I was really asking about as the other possibility for saying the totient 'works'

This one has just as interesting a proof in our book

Apostol is probably a better resource than my recommendation of chapter 7 of the ENT book

Which covers the big multiplicative functions but then suddenly detours to combinatorial number theory

And generating functions

[versed flax]

does the proof go like this: you partition the set ${1,2,...,n}$ in the following way: let $A_d={k\in {1,2,...,n}\mid (k,n)=d}={k\in{1,2,..,n}\mid (k/d,n/d)=1}$, then the sets $A_d$ form a partition of ${1,2,..,n}$ and you get $\sum_{d\mid n}\varphi(d)=\sum_{\frac nd\mid n}\varphi(\frac nd)=1$?

[sterile pumice]

ali yassine

[ionic latch]

Yep

[versed flax]

ohhh i see, yea thats a nice proof

i remember that i found it really nice when i saw it for the first time

[ionic latch]

As a combinatorics enjoyer I find it particularly nice

A partition into easily countable sets horrible work

[versed flax]

have you seen stuff with the mobius function too?

its a very nice function

like mobius inversion etc..

[ionic latch]

Yeah

Like I said the book covers the major multiplicative functions

Although I didn't really get to get super into it

I'll probably add it to my list to reread this week and then @ you if I have anything interesting to say

[versed flax]

alright feel free to ping me whenever you want

[fleet sluice]

Let $n \in \N^{\times}$ and $(r_k)_{k > n}$ be a sequence contained in $[0, 1]$. Consider the function $\beta:\Z/n\Z \to \N \cup {\infty}$ given by

$$\beta(s) = \inf \left{k \in \N_{>n} : r_k + \frac{ks}{n} \not \in \Z + \frac12 \right}.$$

Does $|\text{Im}(\beta)| \le 3$ for every sequence $(r_k)$?

[sterile pumice]

apolus_

[ionic latch]

lord

[boreal garden]

wdym by N^x?

[fleet sluice]

It's {1,2,3,...}

The set of positive integers

[boreal garden]

This is phrase in a really weird way, but consider n=8 and r_k=1/8 (the constant sequence) then beta(1)=4

[nova sequoia]

it's not "for all x in im(beta), x <= 3" but rather whether the cardinality of im(beta) is <= 3

[boreal garden]

There's an inf right there

But sure. This should work to disprove that as well.

Just take n=16 and r_k the same sequence

I'm just thinking of beta as a map from N to N. I have no clue why do they take Z/nZ

[nova sequoia]

there must be a reason innit

[fleet sluice]

I think this example doesn't work

[boreal garden]

Not really

[boreal garden]

why not?

[nova sequoia]

how do you know that

[boreal garden]

The quotient is not well defined

[fleet sluice]

Technically you need 4 different s_1, s_2, s_3, s_4 with β(s_j) ≠ β(s_i) for i ≠ j

Which s_1, s_2, s_3 and s_4 you think should work?

Taking n = 16 and r_k constant = 1/8

[boreal garden]

s=2,4,8,16

well

s=1,2,4,8

[nova sequoia]

they're just taking $[x] \in Z/nZ (\text{for }x < n) \mapsto x$ and then composing them implicitly, it's no big deal

[sterile pumice]

sigmund

[boreal garden]

uhh, not really

I'm too drunk rn to think of a specific reason

[nova sequoia]

thats fine

[boreal garden]

But, this is just phrase really weirdly

[nova sequoia]

it explain things, really

[fleet sluice]

β(1) = 17 and β(2) = 17 for this example bro

[nova sequoia]

man your bolts are just misaligned, get less drunk before trying to help ppl out mate

[boreal garden]

What?

[fleet sluice]

Yep, the least possible value for β(s) for n is n+1, I'm restricting to k \in N_{>n}

[nova sequoia]

https://tenor.com/view/kingbluprint-gif-21436316

my man helping with maths

[boreal garden]

Let me work out this example. Take rk=1/13, and n=13 then
s=1 we have
{2/13, 3/13,...,} so beta(1)=12
s=2
{ 3/13, 5/13, 7/13, 9/13, 11/13, 1} so beta(2)=6
s=3
{ 4/13, 7/13, 10/ 13, 1} so beta(3)=4
s=4
{5/13, 9/13, 1} so beta(4)=3
this is a counterexample

[fleet sluice]

β(s) is always greater than the n you're taking, it cannot be 6, 4 and 3

If you're considering n = 13

[boreal garden]

then take n=26

Idk what are you working on

But that excercise is weird as hell

[boreal garden]

I'm still better than you in your best day.

[nova sequoia]

dont need to be mean my dude

https://tenor.com/view/microsoft-clippy-gif-27340116

[boreal garden]

It is not mean to tell the truth.

[nova sequoia]

alr alr

well in my best days i can usually provide with actual working examples

doesnt seem to be the case for you right now innit

[boreal garden]

Still, that doesn't change that the original post is phrase in a confusing way.

And my idea still works.

[nova sequoia]

what in heavens do you have against the letter d omg just say PHRASED in a confusing wayyy

[boreal garden]

So I'm calling it a day.

[nova sequoia]

dkeopfkpweoikfoiwp

[boreal garden]

Yeah, english is my fourth language. I have no respect for this language

[nova sequoia]

and english is my fifth language, smh

[boreal garden]

👍

[boreal garden]

This is not true btw. Since even if you have taken k bigger than some n. The effect on the denominator is still the same.

[nova sequoia]

\beta(s) is greater than n by definition tho

man i'm much better than that in my good days

[boreal garden]

The idea of chossing 13 is choosing some element n such that n-1 has a lot of divisors then the divisors of k of n-1 should kill the denominator at n-1/k

[boreal garden]

Take n=2

[fleet sluice]

bro it's the infimum restricted to {n+1, n+2,...}

[nova sequoia]

lemme snatch it away

[boreal garden]

Still, this are different values for beta

What are we even talking about?

[nova sequoia]

skamdkamaaff

GO REST

like really

you should go rest

[boreal garden]

Sure

I mean, what is this problem for?

[wild mesa]

It's a problem that came up in his research

[nova sequoia]

yeah it's like a minor problem but it's got its importance in its own right

not a major thing whatsoever anyway

[ionic latch]

I probably don't need to ping a third time

I guess I do

<@&268886789983436800>

because the other two disappeare dbut not that one

[nova sequoia]

what happened?

[ionic latch]

spam

scam spam

[nova sequoia]

i see

[smoky cargo]

a^n + b^n = c^n

Do you fear it?

Is it even true

[kindred fulcrum]

No

[kindred fulcrum]

Yes

[versed flax]

no, it fears me instead

[smoky cargo]

WAIT...now that i think abt it smth powered smth plus smth powered smth will always result in smth powered smth damn

Unless c is specified to be smth unique

[ionic latch]

You should see if you can write a proof of it

[smoky cargo]

Bruh ik the story of some japanese mathematician developing a whole branch of mathematics just to prove that

And I am just a mere grim reaper

[loud maple]

How would one go about simplifying the expression $\sum\limits_{m\mid n} \frac{2^{\omega(m)}}{m}$, where $\omega(m)$ is the number of distinct prime factors of m?

[sterile pumice]

KnightWatch

[kindred moss]

maybe try sone small examples to see how it behaves

[loud maple]

Yeah I've worked it out now. Thanks

[loud maple]

I just had to recall how to sum a multiplicative function over the divisors of a number. Hadn't done that in quite some time

[kindred fulcrum]

That is Shinichi Mochizuki and his "proof" of the abc-conjecture

Its a bit of a rabbit hole though

[wide snow]

Love the scare quotes

abc is only a theorem in Kyoto, as I often joke

[sharp shadow]

No, it won’t. For example 2^2 + 3^2 =13 , and 13 is not “something powered”

[queen sphinx]

well it could be 13 powered 1 :^)

[sharp shadow]

Well, but then we can just say that he discovered a mind boggling fact that a + b = c :) which is hardly worth a “WAIT” exclamation:)

[queen sphinx]

come on now, this is only elementary number theory

[smoky cargo]

Oh yea lol i got the wrong idea i forgot n is the same everywhere

[queen sphinx]

sorry ill go back to bashing my head on primitive roots

[sharp shadow]

The Last Fermat theorem says that “there are no integer solutions to that equation for all n > 2”, not that every combo of a, b has a corresponding c

[smoky cargo]

Oh

Ty for correcting me :)

I just forgot it straight out of nowhere 😭(the fact abt n)

[smoky cargo]

Btw is number theory only abt numbers?or does it hold surprises like algebra

You there?

[sharp shadow]

Yes, it progresses to algebraic structures

That are generalisations of integer numbers

[smoky cargo]

Ahh i see

So like there isn't smth crazy like transformation of *space type stuff right?

[smoky cargo]

Nvm i just googled that and I see RINGS

[sharp shadow]

Not sure what you mean by “bending space”, but yeah, space transformations are not a subject of number theory)

For that you likely need to go to linear algebra, topology and some geometry

[smoky cargo]

pheww

That's great

Ty man :)

I feel that I was a bit clumsy mid convo but nvm

[sharp shadow]

No worries

In addition to rings, algebraic number theory also heavily uses fields

[smoky cargo]

Fields?

[sharp shadow]

And there are also some surprising connections to complex analysis

[sharp shadow]

Another algebraic structure

[stiff osprey]

yeah , it simplifies things heavily there

[smoky cargo]

Ohh got it

[smoky cargo]

I see

Anyways that means that i got a lot to learn then

Let's see how my journey unfolds

[sharp shadow]

Ring is basically like a number that supports three operations: +,-,*, and a field also supports /

You take the essential axioms that capture the essence of those operations and voila! You’ve got a generalised structure that behaves like a number

For example a polynomial: it can be added, can be subtracted, can be multiplied

So it’s an example of a ring

[smoky cargo]

Ohh

[sharp shadow]

It can’t be divided nicely though, so it’s not a field

[smoky cargo]

So any ring which can also be divided, can be called a field?

[sharp shadow]

Essentially yes. There are some technicalities, but they don’t matter much

[smoky cargo]

I see*

[sharp shadow]

Fun fact: field is called “body” in some languages, for example in German

[smoky cargo]

Lol

[sharp shadow]

But anyway, thus channel is about Number Theory so I should stop rambling:)

[smoky cargo]

Yea lol but no problems tho ig

I just don't know like there is so much math everywhere and I am liking every math for some reason

And you felt like a very serious person at first lol,but turns out I was wrong:)

[sharp shadow]

I am very serious, why!

[queen sphinx]

ehh, calling + and - different operations feels weird, additive inverses and multiplicative inverses arent exactly a different operation

[smoky cargo]

Wait like are you admitting or asking?💭

Anyways now you feel like a fun person:)

[sharp shadow]

Thanks :)

[smoky cargo]

But i don't think there's any harm to that

[sharp shadow]

Yeah, I see where you are coming from. But if we just look at it from an elementary perspective I think this is a good explanation

[smoky cargo]

I mean it's understandable

[queen sphinx]

the interesting thing about going from a ring to field isnt the concept of being able to divide

[sharp shadow]

Why not?

Even Wikipedia highlights this idea

I mean, of course you can look at all those boring axioms, inverses and stuff

[queen sphinx]

sorry, yeah, that is the main part about fields. i misspoke. i guess ive never really abstracted a ring to 3 operations like that, but it works

[sharp shadow]

But I think what I said captures the essence well, and for technicalities one can look at definition and axioms

[sharp shadow]

You see, so you learnt something new too ;)

[queen sphinx]

😄

[keen quartz]

So quick question (I just want to clarify that my knowledge is correct) on how we describe GCDs and LCMs of any integers a and b in terms of p-valuations--
Since we can clearly express a and b in terms of factorizations of a prime p--
we can express the GCD as the integer we get from \pm \prod_p{p^{\min(v_p(a), v_p(b))}}. This is because, considering the p-valuations for all common divisors of a and b referred as d\prime and how the gcd d implies that d | d\prime, we get v_p(d) \le v_p(d\prime).

(I apologize in advance if the latex comes out bad btw)

[carmine tide]

In the future, please put $ on each side of your latex so it actually renders.

Because d is the greatest common divisor, you should have that d’|d not d|d’.

[keen quartz]

Ohhh, oops yeah made a bit of a mistake in my notation. Thanks for the heads up on how the latex bot works, appreciate it.

[lilac bronze]

are the following manipulations correct

$a | m \text{ and } b | m \iff b | m \text{ and } a | b (\frac mb) \iff b | m \text{ and } \frac{a}{\text{gcd}(a, b)} | \frac mb \iff \frac{ab}{\text{gcd}(a, b)} | m$

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[sharp shadow]

Why does second <=> hold? Can you elaborate on that?

Other <=>’s are trivial

[lilac bronze]

So it’s a general fact that $n | km \iff \frac{n}{\text{gcd}(n, k)} | m$

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[lilac bronze]

The proof is as follows

$n | km \iff \frac{n}{\text{gcd}(n, k)} | \frac{k}{\text{gcd}(n, k)} m$ first, by cancelling common factors

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[lilac bronze]

But $\frac{n}{\text{gcd}(n, k)}$ and $\frac{k}{\text{gcd}(n, k)}$ are coprime

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[lilac bronze]

So we can apply Euclid’s lemma to deduce that $\frac{n}{\text{gcd}(n, k)} | \frac{k}{\text{gcd}(n, k)} m \iff \frac{n}{\text{gcd}(n, k)} | m$

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[sharp shadow]

Yep, sounds good to me

[lilac bronze]

I suppose this result generalises Euclid’s lemma

Cause that applies for gcd(n, k) = 1

[lilac bronze]

Anyway I was just trying to prove $\text{lcm}(a, b) = \frac{ab}{\text{gcd}(a, b)}$

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[sharp shadow]

Yeah, but you also proved that all common multiples of a and b are also multiples of lcm(a, b)

Which might be useful

[lilac bronze]

Oh that was the definition I was taking

https://pseudonium.github.io/2026/01/18/Products_Categorically.html

[sharp shadow]

Ah, right. If we take a definition that is the smallest multiple, then this property becomes something to prove

But if we take the definition that LCM is a multiple that divides all other multiples, then your whole chain of arguments is not really needed?

[lilac bronze]

well my chain of arguments it to show that lcm(a, b) actually exists

[sharp shadow]

Well, it’s needed for proving relationship to gcd

[lilac bronze]

and is equal to ab/gcd(a, b)

[sharp shadow]

Yes

[lilac bronze]

i may try to include this in my next article

[sharp shadow]

I need to explore your blog more closely

Looks interesting

[lilac bronze]

i have been on this "baby yoneda" series for a while

it's a simplified version of the yoneda lemma that requires you to learn a lot less category theory

but still has plenty of interesting consequences

[sharp shadow]

Cool!

[lilac bronze]

https://pseudonium.github.io/2026/01/22/The_Baby_Yoneda_Lemma.html

[wide snow]

Oh you wrote the category theoretic argument for why preimages are so good!

You must be stopped /j

(To expand on the joke: I think this is a good gateway into category theory, but not so helpful if you just want to know why preimages are so nice. The best explanation is "preimages preserve information in a way that forward images do not, unless they are injective". But I appreciate that may not have been your aim!)

[lilac bronze]

My point is understanding why that statement is true

Not just “that” it is true

Like, the point of the article is to explain why preimages preserve information in a way that forward images don’t

[wide snow]

Maybe I am too far removed from when I didn't understand it, but it seems like it just follows from the definition, and a few examples of forward images having "collisions"

[lilac bronze]

To me, “preimages are precomposition and subset operations are postcomposition” is what made it click

[wide snow]

That absolutely tracks with your username lmao

[lilac bronze]

I mean is that not intuitive

Forward images don’t have nearly as nice a description in terms of just composition

[wide snow]

I agree that it is if you are leaning towards a category-theoretic framework anyway. I haven't, traditionally, but maybe I should

[lilac bronze]

Function composition isn’t even intrinsically category theoretic

So I’m not sure what you mean

Category theory studies composition, sure, but that’s hardly unique to category theory

[wide snow]

Not intrinsically, no, but the diagrams fall into what I'd call a "category-theoretic paradigm"

[lilac bronze]

I mean I guess?

[wide snow]

As a mentor of mine once said off the cuff (!), "The category-theoretic paradigm has had trouble gaining a toehold in the undergraduate zeitgeist, for a range of reasons."

[lilac bronze]

They’re not really more complicated than flowcharts

[wide snow]

Well not to you or I, no - I am wrapped up in thinking about how to communicate these ideas to undergrads when I teach discrete

[lilac bronze]

I suppose my point is that it is misguided to equate “category-theoretic paradigm” with “difficult or abstract”

The concept of a flowchart is not something that requires an undergrad math degree to understand

[wide snow]

No, but I think the computer science students in my discrete class would lead a revolt if I sent them the article unfortunately 😭

(They should think more about composition but that culture change is beyond my ability to instigate, alas)

[lilac bronze]

How come?

(I mean this sincerely, my exposition is by no means perfect and I’m always open to ways to improve)

[wide snow]

Possibly because they are mostly data science-brained, so this level of abstraction (which, yes, to a math major, wouldn't be much!) is not easily received. I think they are just not the target audience for your exposition, which I would peg as "for the category-theoretically curious"

[lilac bronze]

Hm, by “this level of abstraction” - could you be more precise? Do you mean using arrows to represent functions, for example?

[wide snow]

No that's fine - I will have to comb through and think about where I'd lose them

[wide snow]

I think mainly this would Look Scary, if they got that far

[lilac bronze]

Interesting

Would you say that’s because the diagram appears elaborate?

Like, that there just happen to be lots of points and lines?

Because that might be more to do with the way I present the concepts than the underlying difficulty

[charred roost]

I think this diagram in particular would lose a lot of people. It's not at all clear to someone not familiar with commutative diagrams that this is equivalent to (a o f, b o f) = (a, b) o f

[lilac bronze]

As a comparison, here is the diagram for the construction of a regular pentagon with ruler and compass:

[wide snow]

I mean I just keep coming back to this. It's embedded in the definition that preimages preserve information while forward images lose it. I would show two of the definition chasing proofs and note how similar they are, to highlight that something deeper is going on. Then I'd do a few counterexamples of images behaving badly, to contrast with what's going on with preimages. Then the "information preserving" idea should be grokkable.

The category-theoretic framework is neat, but in a class that doesn't otherwise make use of it feels like lipstick on a proof, so to speak.

[lilac bronze]

Mhm - is there anything I could do to make that more clear?

[wide snow]

I just realized what board we're in (oops!) so I will just say that I think the exposition is great, I just think we disagree about who its target audience is

[lilac bronze]

I mean you say this but, in my undergrad, I remember seeing how the preimage proofs were “definition-chasing” and similar, but overall felt unenlightened

Those kinds of “symbol-chasing” proofs were the ones I disliked the most, I guess

[wide snow]

I think "sets as predicates" is a really good insight, and could be used to explain why the definition chasing proofs all look so similar without further diagrams

[lilac bronze]

Sure, but I guess in my mind that’s a stone’s throw away from “preimage is precomposition, subset operations are postcomposition”

[charred roost]

I don't have any specific suggestion, but I think it should be possible to explain simply. And also maybe give an explicit definition of (a, b) : X -> {0, 1}^X, because a beginner might not be able to fill in that definition automatically

[lilac bronze]

I don’t see much of the utility in viewing subsets as predicates if you don’t also cast preimage in that light

[lilac bronze]

If you mean the “packaging”, that’s covered earlier on

Hence why the double arrow is labelled “package”

[charred roost]

How about coloring the double lines to highlight them, then just write something like "we want to prove that package o compose = compose o package"?

[charred roost]

I think a bit of color could go a long way to help a beginner keep track of difficult diagrams

[lilac bronze]

I’ll keep that in mind

[charred roost]

Hmm, I don't know why my eyes skipped over that part 😅 but I think it's a sign that it maybe isn't highlighted clearly enough

[lilac bronze]

I’ll keep that in mind too - sometimes I have a tendency to go into long prose

[charred roost]

Wait, where do you define (a, b)? This is the first mention I can find, but you only give the domain and codomain

[lilac bronze]

Hm, I thought I had given a link to products, categorically

In that case I will edit the article at some point to give the actual definition of packaging

Perhaps after finishing the baby Yoneda series

[lilac bronze]

It shouldn’t be hard, it’s just $(\alpha, \beta)(x) = (\alpha(x), \beta(x))$

[sterile pumice]

Pseudo (Cat theory #1 Fan)

[charred roost]

I'm sure this could be a great article for explaining preimages, I think it just needs a few modifications for it to be easily digestible for a beginner. Your article on products was great btw, no notes 💪

[lilac bronze]

Thank you! I’m writing down all these suggestions so I don’t forget when it comes time to actually writing them

[charred roost]

Yep, it's not hard, but someone who's learning about preimages might not have seen that exact notation before

[lilac bronze]

I also want to rework the way my images are done

Currently I just make them in quiver, take screenshots and upload

[frosty ridge]

If $F(n)= \sum_{d\mid n}f(n)$ then for any positive integer $m $, $$\sum_{i=1}^m F(i)= \sum_{k=1}^m f(k)\lfloor \frac{m}{k} \rfloor$$ I am not sure how to do this

like neither i am able to understand what is happening

[sterile pumice]

AlienX

[frosty ridge]

i mean u can think about tau as f for reference but i amnot able to do that either

i can imagine some form of matrix but cnt get it completelly

[kindred fulcrum]

Count how many times each f(i) appears in the double sum

[digital warren]

Hey, I hope this is the right channel for this. How would one approach a problem like this one?

In my discrete mathematics class, we've gotten into the properties of integers and divisibility. I'm trying to tackle some problems similar to this one, but I kind of just stare at the screen and my mind blanks, so I'm looking for some starting points.

[ionic latch]

In general I would say a starting point is the definition

For example, if my definition of divisibility is $$a | b \text{ if there exists an integer } k \text{ such that } ak = b$$

[sterile pumice]

Coolempire93

[ionic latch]

Then I might write

What I have
a | b
c | d

What I want to show
ab | cd

So what this means to me is
There is an i such that a*i = b
There is a j such that c*j = d

What I want to show is
There is a k such that ab*k = cd

Which would by my starting point for a proof, to see if I can manipulate these two equations to come up with one that looks like the last one

[ionic latch]

Even before this though what I really should begin with is considering some examples

Since it says the statement may or may not be true

I want to gain an intuition/understanding of what I'm working with

So let's choose integers a,b,c,d such that a | b and c | d

Say, a = 2, b = 4 and c = 3, d = 6

We can see that a | b is true and c | d is true

But ab = 8 does not divide cd = 18 and we have a counterexample

Really most numbers for this will give you a counterexample so if you just try a few examples you'd end up at one quickly hopefully

[ionic latch]

So yeah that would be my process

1. Try some examples to get a feel for the problem
2. If none of the examples give you a counterexample, try to construct one (based on what knowledge you have gained about the problem, are there any integers you think that the statement might not work for? If not, skip this step)
3. Set up to prove, beginning by writing your assumptions, and using the definitions to get them to a form you can work with, and then write your final goal (as you get more familiar with proofs, you will be able to start skipping this step)
4. Attempt to prove the statement; if you can get from your assumptions to your conclusion you are done; if not, see where in the proof failed - can you come up with a counterexample based on this?

As an example, say I had tried to prove it

I know $a\cdot i = b$ and $c \cdot j = d$

So $ab = a^2 \cdot i$ and $cd = c^2 \cdot j$

I want to show that for some integer $k$, $ab \cdot k = cd$

Which would mean $a^2 \cdot i \cdot k = c^2\cdot j$, or in other words $$k = \frac{c^2 \cdot j}{a^2 \cdot i}$$ must be an integer. Now I can just choose an example carefully, let $a = 2$ and $i = 2$, $c = 3$ and $j = 3$. Of course, $k = \frac{9}{4}$ is not an integer in this case, so $a = 2, b = a \cdot i = 4$, $c = 3, d = c \cdot j = 9$ is a counterexample.

[sterile pumice]

Coolempire93

[ionic latch]

I believe I've done an example of every step of the process now

Hope this helps!

[digital warren]

ok thank you so much for that. I'll try some problems (after I finish lunch hehe) and I'll come back here if I get stuck again

[ionic latch]

Sounds good 🙂

[digital warren]

Is there a specific notation for "does not divide"? For example, if I said 4|6, that's false. How would one go about writing that?

As a related question, when we say a|b, is it sort of implied that it's saying "a divides b as long as the result is also an integer?" Because in my head, 4 does divide 6, with the result being 1.5 . (I hope that makes sense, I have a very specific way of thinking about things sometimes)

[ionic latch]

Yeah

$a\not| b$

Why is it so spaced

[sterile pumice]

Coolempire93

[ionic latch]

$a \nmid b$

[sterile pumice]

Coolempire93

[ionic latch]

There we go

[digital warren]

thank you

[ionic latch]

As a related question, when we say a|b, is it sort of implied that it's saying "a divides b as long as the result is also an integer?" Because in my head, 4 does divide 6, with the result being 1.5 . (I hope that makes sense, I have a very specific way of thinking about things sometimes)
Yes, again the definition is
a | b if there exists integer k such that ak = b
That is b/a = k is an integer

[ionic latch]

The idea I used here

Selecting numbers so that cd/ab is not an integer

So yes, you are correct

[digital warren]

thank you

[exotic herald]

How do I count again?

[tiny hazel]

https://tenor.com/view/ragebait-i-will-not-interact-gif-9380134741995340701

[wide snow]

You just did

[keen quartz]

Oh my god, thank you. I've been trying to figure out a way to make my latex notation more clean whenever I talk about divisibility.

[junior swallow]

Can somebody please give me a hint as to why 2r and r^2 - Ds^2 are in Z implies that r + s \sqrt{D} is in Z[\sqrt{D}]?

[ionic latch]

I think it's as a result of this if you have something like this anywhere

Oh wait I read your question wrong
Disregard what I just said

Hmmmmm

[ionic latch]

I think this notation is unfamiliar to me

[junior swallow]

What part of the notation confuses you?

[ionic latch]

I don't think I've seen S[number] before

Is that just polynomials using power of number with coefficients in S

[versed flax]

are you trying to show the hint given by the author?

if yes then the first direction, ie r+s\sqrt D is an algebraic integer implies 2r and r^2-Ds^2\in Z follows from exercise 5

for the other direction, if 2r and r^2-Ds^2 are in Z, what can you say about the polynomial x^2-2rx+r^2-Ds^2?

[versed flax]

yes

S[a] is the set of polynomials in a with coefficients in S (assuming that S is a ring)

you can generalize this in the following way: let A be a subring of a ring B and let S be any subset of B, then A[S] is the set of all not necessarily commutative polynomials in elements of S with coefficients in A. "not necessarily commutative" because the elements of S may not commute with each other

tho if you take B (and hence B, A and S because A,S\subset B) to be commutative, you can drop the "not necessarily commutative" part

[ionic latch]

Okay then the new suggestion i was going to give would have been righf

I stopped before I did it because I was annoyed I whiffed the first and wasn't completely certain

[versed flax]

ohhh, what was your suggestion

[vale pawn]

How trivial is it to prove the bezout identity? If I haven't proved it in 2 hours should I keep trying?

[kindred moss]

have you seen the euclidean algorithm?

you basically apply quotient remainder theorem repeatedly

[vale pawn]

Yeah the quotient one i came up with a proof that was maybe a bit not 100% rigorous but this can't figure out

Ai is telling me its more complicated then euclid algorithm is that not true?

[kindred moss]

like, you can do something called the extended euclidean algorithm that exactly gives the identity

[versed flax]

let a,b be non-zero integers and consider the set S={ax+by| x,y in Z and ax+by>0}, this set has a least element by the well ordering principle which you can denote by d. Now write a=qd+r for some integers q and r with 0=< |r|<d. What can you say about r? similarly write b=q_1 d+r_1 for some integers q_1 and r_1 with 0=< |r_1|< d, what can you say about r_1?
Finally let c be a common divisor of a and b, prove that c|d

also i dont recommend spending this much time on a problem before asking for a hint or looking at a solution

because sometimes you would be missing something important for the proof and you wouldnt be able to proceed if you dont look it up

that being said, this doesnt mean that you should ask for a solution/hint after thinking about the problem for a few secs

like give it a good thought for a few mins, if you still cant proceed then ask for a hint or something like that

[vale pawn]

Oh wow that fast after a few min?

[vale pawn]

BTW are you a ring theorist?

What if you still can't figure out with a hint

[versed flax]

no, i am a 2nd year undergrad student. What made you think that i am ring theorist tho

[versed flax]

you ask for another one or you just search for the solution

[vale pawn]

I googled your name that's what I found

[versed flax]

i see

there are many people called ali yassine in this world

[vale pawn]

Is it fine if when I read the various proofs I dont fully follow parts? Like is it ok if it takes me a few hours to get it?

[abstract ferry]

yeah lol there is no issue

it's usual to get struggling with understanding/following stuff

[vale pawn]

Ok ai told me bezout is much harder then eucluds lemma not sure if its true or not. Should we prove the statements in text ourselves or just the exercises should be fine?

[abstract ferry]

well anything is okay

you can prove the theorem first, and do the exercises

but similarly, you can do the reverse

struggling with exercise and sometimes that gives you the idea

which allows to understand the proof easier sometimes

[vale pawn]

What i mean is it ok to read the proof if you will be doing the proofs in the exercises? Whats the difference btw the two

Like proofs in the text provided vs proofs that are exercises

[abstract ferry]

well if you are caring about proofs that are left as exercises then you should try to get an idea at least, by trying to make examples and play around with them

it's still okay that you end up with proving nothing but at least playing around examples helps a lot

[vale pawn]

Not sure i understand what you mean. My question is basically what's the difference btw trying to prove the statements proved in the text vs the proofs that are in the end of the section left as exercises if any

[abstract ferry]

basically reading the proof for that theorem first and trying to do proof exercise is like there is no point, kinda

[vale pawn]

Are the proofs in the text vs proofs in the exercises different in sone way just trying to get how they are the same or different

[abstract ferry]

i dont get what you mean lol

[sharp shadow]

I guess proofs that authors give directly in the text are considered something that you are not meant to come up with yourself

[vale pawn]

Ok in the actual chapter there are theorems stated and proved as part of the text. At the end of each section/chapter there are exercises many of which are proofs (usually no solution provided). I am trying to understand what's the difference in doing the proofs in the text on your own vs those in the exercise if they are roughly the same or not. If doing 1 without the other us sufficient

[sharp shadow]

You just learn by reading them and trying to understand

And the proofs requested in the exercises is something that you are expected to be able to prove after reading the relevant chapter

Of course there might be easier or harder exercises, and it is also possible to try to prove things in the main text yourself before reading the author’s proof

It all depends on your style, concrete book and other factors

[vale pawn]

Are the proofs the authors put harder/easier and different then the exercises?

[sharp shadow]

It depends

There are books with devilishly hard exercises

Sometimes they are rated

[sharp shadow]

I guess you can just try and see for yourself:)

Next time you see a theorem in your book: do NOT read the proof

And try to prove it yourself

Then compare with the author’s proof

Note the difficulty

[ionic latch]

is what I do almost always
Usually either the author did it in a different and insightful way or they omitted stuff that I would liked to have seen in my proof

[sharp shadow]

Then do the same for 5 exercises

Compare and report here :)

[vale pawn]

I see so just depends on the textbook

[vale pawn]

I wasn't asking now about my vs the authority proof though I always wonder about that. I mean difference in difficulty of exercises vs proofs in the actual text

[sharp shadow]

Yeah, I sometimes do this too. At least it makes me understand the theorem statement well and be alert to various subtleties, even if I don’t manage the proof myself

I don’t spend a lot of time on this, just seeing if I can spot the idea of the proof

[sharp shadow]

Yes. Did you really expect that is a general convention on this for all math textbooks?

[ionic latch]

I think I'm the opposite, I basically write every proof possible by hand as long as I have time in my schedule my method of taking notes

[vale pawn]

Does it often take you guys long to prove stuff and even just understand given proofs?

[sharp shadow]

I do that for exercises

And only occasionally for proofs from the text

If a theorem and/or proof is particularly puzzling :)

[sharp shadow]

Yes, sometimes it takes days

And then I come here and ask for hints :)

[ionic latch]

Just depends on how novel the field is for me I suppose
Like linear algebra proofs either feel like magic or ingenuity to me

But analysis proofs so far have been really routine

[vale pawn]

I guess I mean more on more beginner stuff like elementary number theory

You guys are undergrads?

[sharp shadow]

Bezout is cool, I think that was the first unexpected result in Elementary number theory for me

[ionic latch]

ENT wasn't too bad for me, I had to take it independent study (and I hadn't joined this server yet), so I basically learned it all on my own

[ionic latch]

This

[vale pawn]

So it didn't take you hours to understand or prove anything?

[sharp shadow]

Don’t compare yourself to random people on Internet, maybe I am a future Fields medalist

[vale pawn]

Yeah I just feel dumb now cause I tried to prove it for 2 hours didnt get far been looking at proves dont get it yet

[sharp shadow]

There will always be someone significantly smarter than you, no need to worry about that

[ionic latch]

Some things did, especially the theorem that led to bezout

[vale pawn]

Euclids lemma seemed way easier

[ionic latch]

oh looks like I left some things unproven

hensel's lemma, chinese remainder theorem

[sharp shadow]

What’s your proof for Bezout?

[ionic latch]

mine or michael's

[sharp shadow]

Is it just plug and chug in Euclid’s algorithm or the proof based on sets of integer linear combinations?

[sharp shadow]

Both :) but mainly Michael

Because I think you are already sorted with your proof long time ago

[versed flax]

you could use the well-ordering principle

[sharp shadow]

Time to get to work!

[sharp shadow]

Yeah, that’s the other approach that I meant

But just to get it done piggybacking on Euclid is much easier

Good old Euclid, what would happen to us all without you

[ionic latch]

Yes

[vale pawn]

The ai gave me 7/10 on euclid saying I need to improve rigor. What do you mean by piggybacking

[ionic latch]

forgot to spoiler

[sharp shadow]

AI grading for proofs - sus!

[unique cypress]

[sharp shadow]

AI is so rubbish at this…

[sharp shadow]

Asked today: give me conjugacy classes of A_4

And both Claude and ChatGPT gave me wrong answers, very persuasively, as usual

[ionic latch]

yeah I basically only use gemini now

[vale pawn]

Wait but it can solve the easy erdos problems and imo

Surely it can check elementary number theory proofs

As those are much easier in comparison

Also I check on that halmos website made by mathematicians

Hallmos

[ionic latch]

Sadly when it comes to AI that is a logical fallacy 😔

[kindred fulcrum]

Guys lets not talk about ai

[ionic latch]

Surely it can do a harder thing does not imply it can do an 'easier' thing from a human perspective

Because the AI doesn't reason or build on easy to get to hard

[sharp shadow]

Well, it can’t do elementary algebra proofs unfortunately, it gave me 7 wrong proofs in a row and then said that I spent all my free tokens

[ionic latch]

glad I've never gotten an incorrect proof from gemini, I would use szl.ai for an ENT proof though

[vale pawn]

Depends on which model too

You are saying it can do imo but not way easier stuff you sure??

[ionic latch]

Anyway here is that bezout proof, now with spoiler

[unique cypress]

If your bar is imo then there is something that has gone deeply wrong in your studies

[sharp shadow]

Nice handwriting!

[ionic latch]

oh I included the corollary

Well the next corollary was bezout

But it's on the next page

[vale pawn]

What do you mean? I said if it can do imo surely it should be easily able to handle basic proifs?

[ionic latch]

Again

Fallacious logic because an LLM doesn't learn in that manner

[vale pawn]

Hmm not sure lol if one day it proves the remaining hypothesis you will say it can't do college math

[ionic latch]

If one day it proves the riemann hypothesis I would be very interested in the training data

[sharp shadow]

I’d recommend posting your proofs here instead

If you are unsure about them and want feedback

Maybe we will give you 10/10

[ionic latch]

[vale pawn]

Yeah I should

[worldly zenith]

https://youtu.be/pO3rDM3USeM

This person claims collatz is solved lol

21 hours ago?

https://zenodo.org/records/18626114

Here is the linked proof by the way?

Didnt read it yet but thought it would be funny to post here

[stoic current]

is that real or is it AI slop

[worldly zenith]

I have no clue

LKMOAOOOO

akgadkgaodga

[stoic current]

It's AI slop

[worldly zenith]

tragic

[carmine tide]

https://scholar.google.com/citations?user=B0kvFSEAAAAJ&hl=en

all of the citations of his work are him citing himself

also the publish dates are wild

yup ignore

[stoic current]

"why is my paper getting ignored? 😭 " - bro who generated his whole paper with claude after generating the last 3 with claude as well

[worldly zenith]

WHERE DOES IT SAY THAT plagplaeg

or does it not actaully say "why is my paper getting ignored"

[carmine tide]

I think they said that in response to my comment about all of his citations being from himself

[worldly zenith]

oooh ok got it

least ai generated text ever:

[abstract ferry]

well anyone claiming they solved collatz is literally false unless it's Terence Tao

[silent rock]

Has anyone yet noticed that 11^37 is very close to 2^128, admitting a relatively compact representation of a (non-trivial prime power) finite field with a non-power of 2 order?

Would such a thing be useful for anything, practically speaking?

(Thinking perhaps PRNG stuff or fields for polynomials or elliptic curves for cryptography or other similar uses)

[ionic latch]

I believe my advisor mentioned something about something like this when going over RS and BCH codes but I can't remember 🫠

[carmine dune]

Hey guys, I am currently doing some reading on class field theory and I would like to get an intuitive overview of how the main results were arrived. Hence, I am working through many different sources but have mostly landed on weil's basic number theory focusong on the explicit central simple algebra approach (as right now, the cohomological approach seems entirely mysterious to me)

However, even for this, I do not really understand the motivation for introducing these objects. My understanding for the local case is as follows: for a finite extension of local fields L/K and each character chi: Gal(L/K) --> C, of course the image of Gal(L/K) is a group of roots of unity because Gal(L/K) is finite, and groups of unity are cyclic, so the kernel of chi gives a normal subgroup H s.t. the fixed field L^H is a cyclic extension over K. To this cyclic extension L^H/K and to an element a in Kx, we can choose a generator g of Gal(L^H/K) define a certain central simple algebra over K that embeds L by introducing an element y which implements g as an inner automorphism: yly^{-1} = g(l) and satisfies y^n = a. Then, we have a function inv on CSAs --> Q/Z which we can exponentiate to give a root of unity, and define a pairing of the characters on Gal(L/K) and Kx by this. My question is simply: why consider this construction of a CSA this way, why is it interesting, why does this give the correct correspondence thta in the global case is the artin symbol. What's the intuition for coming up with it and working with it in the first place

[abstract ferry]

if you are doing something similar/higher than algebraic number theory or analytic number theory, like these better to go to #advanced-number-theory

[unique cypress]

this but also I can answer

[unique cypress]

Have you gone through the motivation of "why class field theory"?

fwiw look at #book-recommendations message

[carmine dune]

Thanks, I’ll move the convo there

[eager silo]

I realised something interesting, the binomial formula works for rising and falling powers too

Is this the umbral calculus they talk about?

[swift fox]

yes

modern umbral calculus is basically about treating polynomial sequences that behave like powers under translation as if they were powers

sakura a sequence ${p_n(x)}$ is called of binomial type if
[
p_n(x+y)= sum_{k=0}^n binom nkp_k(x)p_{n-k}(y)
]

[sterile pumice]

Oléagineux Distilliànus VIVII

[swift fox]

falling factorials are the natural basis for the finite difference operator

[eager silo]

I know what you are talking about, is this related to the whole binomial thing though?

[mortal osprey]

I wonder what kind of "umbral" calculus leads to $e^t = \sum_{z = 0}^{\infty} \frac{t^z}{z!} \Leftrightarrow \tilde{\Gamma}(z) = \int_0^{\infty} t^ze^{-t} dt$, with big quotations around the equivalence sign

[sterile pumice]

PKThoron

[mortal osprey]

like screw bound variables and discrete vs. continuous, we're making this work!

[limber ferry]

I have just started studying elementary no. Theory by David m bruton and feel a little lost on the preliminaries section concerned with the well ordering theorem and induction I read the proofs I think I get em but minutes later I feel like i don't understand anything what to do

Ping me please if u can help me out in any way

[ionic latch]

Honestly I felt the same way in the preliminaries of my ENT class

[limber ferry]

Then what should I do

Like im doing it from self just using books

Should I just stick to it?

Like keep rawdogging it?

[ionic latch]

I think having someone on here guide you through it will be helpful (since WOP proofs are special) but yeah I had to independent study mine as well (no class, just exams) and the rest of the book was honestly easier

Than getting familiar with the new technique

Having someone outline the process for you will probably be good for you though

I would do it but I have to run

[limber ferry]

For instance I uh learnt the principle behind mathematical induction that if for a relation A, we have a set of natural no.s till n then it must also hold true for n+1 thus proving the statement or relation for all positive integers but I tend to forget it again and again

[limber ferry]

Lmao okok

I think I will raw dog it and if I get stuck on trying to prove SMTH myself I will just create a help channel

Thanks tho

[wraith sage]

Speaking for myself, teaching myself, I copy out the worked examples from the textbook, making sure I fully understand each step going from one line of working to the next. Then I do the exercises, modelling my solutions on the way the worked examples went.

There is just no substitute for getting the practice in from the exercises

[limber ferry]

ohkk

[wraith sage]

Hi Cohle, is there an induction question you are stuck on?

I am new here. Hi! I taught Maths and tutored Maths at high school and college. I am retired now

I am currently teaching myself Multivariable Calculus

[limber ferry]

not really a question but when i read proofs i think im just rote learning instead of actually understanding

[wraith sage]

Cohle, I think I could help you but we need to work on examples together, discuss, and attempt questions

[limber ferry]

ohkayy

well im planning to study today ill ping u if i need help

[slow birch]

Honestly for me, I learnt from the very same book.
The preliminaries were quite easy.

It's literally Induction and Binomial Theorem

Learn WOP. Then, we prove various properties that follow from WOP, like Archimedean Property, First Principle of Finite Induction, Second Principle of Finite Induction, do some problems, and yeah. If you are trying to prove anything by Well-Ordering Principle, try proof by contradiction by taking a subset of the non-negatives, taking a least element, trying to find a element less than the least element, and by contradiction of the principle, proving the original statement.

For Binomial Coefficients, learn the expansion. Pascal's rule follows simply. The Proof of Binomial Theorem follows from Induction.

The Book explains it very nicely in my opinion.

Whenever the book explains hard stuff, the book usually brushes it over in an intuitive manner. So its a more intuitive book than a rigorous book. For example Dirichlet's theorem. It offers no proof of that.

For example, here.

[limber ferry]

thats the problem whenever i read or reread i feel as if im just memorising stuff instead of learning it

[slow birch]

I mean memorizing is reading without understanding

If you understand the proof, if you can do it on your own, whenever asked, and if you genuinely understand the various paths we follow to prove theorems related to WOP, then you are good.

[limber ferry]

how to do that

[slow birch]

Try to prove the theorems yourself, before reading the book.

And use the theorems on examples

Also, do a LOT of problems. You already have exercise problems

Try them.

[limber ferry]

okay

[slow birch]

But the book does a very GOOD job if you read it

Its very intuitive.

After Chapter 10, you have full choice to do chapters in any ORDER.

[sharp shadow]

Think of induction as a ladder: if you can get to the first step and can climb from one step to the next, then you can get to any step

It’s hardly possible to forget that

[limber ferry]

nice analogy

[limber ferry]

ill lock in then, your making me pretty excited lmao

[fresh sand]

I don’t know if this counts as number theory but can someone help me with proving that the first n digits of an irrational number r, multiplied by 3, are determined by the first f(n) digits of r

I have a homework assignment about proving that multiplication by 3 is computable on irrationals and it boils down to this being true

[mint stone]

In order to determine the first n digits of 3r you need to know the carry into the n-th digit, and that carry can be only 0,1,2. The carry is uniquely determined if the (n+1)-st digit is 0,1,2 (giving carry 0), 4,5 (carry 1), 7,8,9 (carry 2). If the (n+1)-st digit is 3 or 6, then you must examine the (n+2)-nd digit in order to determine the correct carry into the n-th digit. The process then continues in the same way. Therefore, I believe your function f(n) should be defined as f(n)=n+1+t(n), where t(n) is the length of the consecutive block of 3s or 6s starting at position n+1. Say, if r=0.3336623337... we get the sequence f(n)=5,5,5,7,7,... for n=1,2,3,4,5,... (n=1 corresponds to digit 0)
@fresh sand

[sharp shadow]

It’s interesting that Aluffi likes to prove things using Well Ordering Principle as opposed to induction. In his book “Algebra. Notes from the Underground”, he writes: “Several proofs in what follows could be handled by induction or interchangeably by an appeal to the well-ordering principle. Which to use is essentially a matter of taste. I will often insist on using the well-ordering principle, to stress that we are really using a specific feature of Z. Also, I often find it somewhat easier to write a lean, rigorous argument by directly invoking the well-ordering principle rather than induction”.

[fresh sand]

And since the number is irrational, the sequence of $3$s or $6$s has to be finite. $f(n)$ even is partial recursive (but not primitive recursive). It seems then I can extend the result to rationals that are not of the type $\overline{a}.\overline{b}(c)$ where $c \in 3, 6$

[sterile pumice]

Ted

[mint stone]

Yes, I think so

[hearty valley]

Whoever created number theory deserves a Darwin

If I were allowed to swear, I would do that to number theory.

[unique cypress]

https://tenor.com/view/he-made-a-statement-he-made-a-statement-so-ass-the-dog-left-this-planet-gif-268224778921118459

[midnight cloak]

LMFAO

[sharp shadow]

What’s up with this unlove for number theory, didn’t Gauss said that it’s the Queen of math

[kindred fulcrum]

it is the queen

[unique cypress]

It's just people learning number theory the wrong way

[abstract ferry]

true, they need to start from weil conjectures

or modular forms, CFT, or whatever

[slow birch]

For problem practice, they should lean towards Goldbach conjecture. Really strengthens your number theory skills.

[warped tendon]

The best way to get in to something is to begin with the best it has to offer

Thats why your first number theory book should be Modular Curves and the Eisenstein Ideal by Barry Mazur

[rugged pasture]

Stifel's relation ☝️

[lilac bronze]

<@&268886789983436800> spam

[lilac bronze]

<@&268886789983436800> spam

[warped tendon]

No see the difference is that the Eisenstein ideal paper is actually good

[sharp shadow]

He is also the youngest Fields medalist ever (at 27), so must know how to write to appeal to younger students of mathematics

[lunar laurel]

Hi, is there an intuitive way of understanding equivalence classes?

[sharp shadow]

It’s like equality in a way: equivalence classes are subsets of elements that are “equal” (equivalent) in some sense. Those classes also partition the original set in disjoint parts

But I’m not sure what exactly you mean by intuitive understanding, what is your current issue with understanding of equivalence classes?

Why is it not intuitive?

[lunar laurel]

I mean, i find it as a concept that I have to memorize but it's not "tangible" nor easy for me to think about it

[lilac bronze]

https://pseudonium.github.io/2025/09/16/Three_Perspectives_on_Equivalence_Relations.html

there's a third perspective that might be helpful

[sharp shadow]

what about the partition perspective then? I think it's quite tangible. Basically, you can just partition any set into disjoint subsets and declare them equivalance classes, and it will work for some relation, namely "x and y are equivalent if and only if they are in the same subset"

and this "naming function" that @lilac bronze writes about is also a very nice way to think about them, I like how it was neatly combined with Poincare's quote about mathematics and thus raised to a meta-level :)

[lilac bronze]

one nice thing is that a lot of equivalence relations in practice arise from such naming functions!

e.g. the equivalence relation for modular arithmetic comes from the function sending an integer to its remainder modulo n

[lilac bronze]

in this case, your naming function sends x to the subset of the partition it lies in

[sharp shadow]

yeah, after all they just alternative ways of looking at equivalence. We can also say that we partition objects in such a way that they have the same "name" or property (remainder mod N, in this case)

I just feel that the parition is the most "tangible" way, because it literally takes a set and splits it into parts, what can be more intuitive than that? We are doing it with sets of apples or whatever since we were kids :)

naming is also intuitive though

the only problem with all of this is that it was already intuitive for us, but not for @lunar laurel :)

so let's see what he thinks

[lunar laurel]

mmm

im still thinking about it

I'll say something once i finish thinking xD

https://tenor.com/view/marcus-aurelius-marcus-aurelius-marc-aurèle-rome-gif-14213982345192092969

[lunar laurel]

I mean, i see that the concept is easy, but its not intuitive as other things😅 , atleast for me

in any case, i'll study it more and see if I can get to some point

thanks to both of you

[sharp shadow]

It would help if you can actually discuss it, ie say what questions you have, challenge the required properties (like why on earth does equivalence relation has to be transitive? What happens if it’s not?) or rephrase the definition in your own words or restate that blog post or something like that. Instead of simply saying that “it’s not intuitive”, because this gives almost zero feedback to us

[lunar laurel]

I mean, I understand the properties, but I find it hard to imagine what is a class of equivalence, I know that a class of equivalence is a subset of equivalence relation, but, I don't see any easy example to imagine it, I just see a definition