To clarify, prime elements must also be non-zero and non-units and it is possible for both p|a and p|b to hold. (So I shouldn't have included the word "either"). Sorry for the confusion.

@chriswebster24

13 күн бұрын

It’s ok. Just don’t let it happen again, please. Thanks 🙏🏿

The more you know about pure mathematics, the less fundamental and trivial some previously trivial-looking theorems look, and this is a prime example.

@brodymiller9299

13 күн бұрын

I don’t know if it was intentional, but nice joke!

A system of numbers that doesn't have unique prime factorization? That's not natural.

@DontWatchWhileHigh

13 күн бұрын

Considering that almost all number systems don't have unique prime factorization, i'd say the one we have is the "unnatural" one, depending on how you define natural ofcourse.

@user-ng3kf4cs4d

13 күн бұрын

​@@DontWatchWhileHigh do you have examples?

@MiroslawHorbal

13 күн бұрын

Quality joke.

@MiroslawHorbal

13 күн бұрын

​@@user-ng3kf4cs4d The example of the video: Z[√-5]

@user-ng3kf4cs4d

12 күн бұрын

@@DontWatchWhileHigh nvm I just watched the whole video, I forgot about restrictions (but they all still seem secondary/derived from the natural number sistem) yet they have less propreties as if adding things only ruins how perfect the original one was

Ohhhhh man, that was good. That way of defining primes never clicked before. Since, like 4|36 and the statement: "4|3 or 4|12" is a true statement, but "4|6 or 4|6" is a false statement.

@DanielJackson6742

13 күн бұрын

would it be better to say if p|k, for all a,b s.t. ab=k, either p|a or p|b instead then?

We should call every number that isn't an integer an outteger.

8:48 This is such a cool hypothetical scenario, really caught me with that one!

Nice video! FYI, there's a small typo at 20:45: "kb < 2p" should say "ka < 2p". (You said it right out loud, of course.)

Excellen video. One remark though: Euler has had this insight (and many other insights) BEFORE Riemann constructed the analytic continuation. In a sense, you are talking about the Euler Zeta function, with domain positive real numbers greater than 1. There is absolutely no need for the extension of the domain to complex numbers or for analytic continuation.

@kyay10

13 күн бұрын

TIL! It makes sense though because both infinite products and sums should be respected the same by derivates, and so analytic continuation on the sum formula Vs the product formula must produce the same result.

Bruh I thought you had more likes and subs :0 This channel is criminally underrated given the quality and amount of information it gives

Nice video. Easy to follow.

I just found this channel, amazing content! Keep it up:)

you're damn underrated! keep it up man

Keep up the fab work 💪🏼💯

Damn why your videos look so attractive and beautiful? This colors are amazing!

That case of "unnatural" numbers seems weird because, for example, addition in that system never works as a binary operation (no sum will be within unnatural numbers), and thus it's not a ring and multiplication has no real meaning as well. On the other hand, if you say that 1 + 11 = 21 in unnaturals, then it means that those numbers are just natural numbers in disguise and you may just re-label 11 as 2, 21 as 3, and so on, and the math works fine again. For illustrative purposes it's fine, but it's much worse than Z[sqrt(5)] for example which is a legit ring that shows how it can have non-unique factorization.

@andrewkarsten5268

12 күн бұрын

You’re right about the ring issue, and it would’ve been nice for him to be more explicit about that, however the idea is about factoring an element into a product. Since you are familiar with rings, I assume you are familiar with groups as well. You can define “multiplication” without defining “addition” in a group structure (though it’s not a group either, I think it’s a monoid?) and you can have irreducibles in groups as well. There’s an area called representation theory which focuses on group representations and breaking those down into irreducible representations, and it has a similar flavor.

how have i never seen an explanation of how the two different riemann zeta functions are the same?

I have a question. At 10:56, doesn't this expansion only reach choices with an infinite tail of ones at the end? Like, if we choose x^n each time, this would be some kind of infinite power of x that doesn't appear in the expansion. Thanks.

@fullfungo

14 күн бұрын

The final expansion does not include x^∞ and here is why. The expression (1+x)•(1+x^2)•(1+x^4)•… is not a shorthand for an infinite expression with infinitely many brackets, because all well-formed formulas in most logical systems are limited to finite strings of text. Instead, this expression is a shorthand for a limit of the following finite expressions: (1+x) (1+x)•(1+x^2) (1+x)•(1+x^2)•(1+x^4) … Of course this can all be compacted into lim_{n->∞} Π_0^n (1+x^(2^n)) which is once again a finite formula. Either way, if we expand every step individually, we get: 1+x 1+x+x^2+x^3 1+x+x^2+x^3+x^4+x^5+x^6+x^7 … So the resulting expression 1+x+x^2+… just means the limit of the expressions defined above. If we want, we can compact it into lim_{n->∞} Σ_0^n x^n which is a finite expression. In either case, the term x^∞ does not actually appear.

@dani-rybe

14 күн бұрын

@@fullfungo Makes sense. Thank you

Thank you for the awesome Video! :-) It would be great if you could tell me the name of the music at 14:25 :))

why do you call multiplication "times by"

Ugh... Those "alien numbers" can't be added and don't even form a group under multiplication (no inverse).

@andrewkarsten5268

12 күн бұрын

I think they form a monoid, but I agree it wasn’t a great example for the mathematically inclined.

This helps

p|ab -> p|a or p|b. This definition also works for p = 1. So, do we have to say that p has to be a non-unit?

@kyay10

13 күн бұрын

There's an *either* missing in your implication: p|ab -> either p|a or p|b Exactly 1 must hold

@28aminoacids

13 күн бұрын

@@kyay10 no. 3 | 3× 6 -> 3|3 and 3|6. And 3 is a prime too.

@kyay10

13 күн бұрын

@@28aminoacids oops you're absolutely right!

@kyay10

13 күн бұрын

@@28aminoacids yeah according to Wikipedia they specify that p can't be the zero element or a unit

@GaborRevesz_kittenhuffer

13 күн бұрын

yep, by fiat primes must be non-units

Very informative video. Is anyone aware of the pattern to how all integers (N) are factored?? The only issue is that all P needs to be tested up to N, so not really a major breakthrough, but there is a pattern which definitely continues to inf.

@ProactiveYellow

14 күн бұрын

Actually, you only need to test primes P≤√n for a number n. The "pattern" for factoring is one of the Hard Problems involved in the P vs NP problem.

@irigima9974

14 күн бұрын

@@ProactiveYellow​​⁠So for example, if you gave me any N, I could instantly tell you if any P was part of the factor of N, and how. Unfortunately, in this case - all P

@henokvanni3831

13 күн бұрын

⁠@@irigima9974He said to you that you only need to test up to sqrt(N)

@kyay10

13 күн бұрын

You mean checking if P divides N? As in if P/N is a whole number? An easy way would be to just run Euclidean algorithm to compute gcd(N, P) and check if it's == P. The most naive form of that algorithm only takes O(N/P) steps. There's probably much better ways though. I'd guess just calculating N mod P would also give you the answer pretty well, as would N/P and calculating the factional part out of it

babe wake up new polyamath video dropped

I wish you'd shown why Euclid's Lemma cannot be applied to the unnatural alien numbers. It is not clear which step in the proof does not apply.

@TheArizus

6 күн бұрын

The unnatural numbers aren't closed under addition, so it fails at any stage involving addition.

You forgot 0, and I'm very vehement about it!

@5:08 not either or, p could devide both a and b

@DeJay7

13 күн бұрын

On the screen it says "p|a or p|b", which does includes the possibility of p|a AND p|b, it's just that only one NEEDS to be true, both is just a valid possibility.

@dig_dus

12 күн бұрын

Not true, please take a look at the last word in the line above

1:42 noooo naturals without 0

Your channel name is so similar to polymath lol

Obviously this proof must fail somehow in non-UFD rings. It would be interesting to see exactly where it breaks down.

Vieta jumping

So this Manim?

Real numbers including the 2 complex solutions to z³=1 Gaussian integers too (z⁴=1) 10:53 huh i guess that's equivalent to the geometric series? I don't really get the point of irreducibility yet, we'll see 11:42 oh my god that's genius 16:30 well I'm not proving anything myself but this proof makes sense I guess I see where you're going Infinite descent

Watch out with how you use irreducible and prime. They're in general not the same

Please, stop with the "times by" thing. It's "times" or "multiplied by", but not "times by".

"most people would say six is two times by three"... I don't know anyone in person who would lol. (Most people would say six is two times three)

"times by"?? "Such integer such that"?? Lol dude

Bro. Stop saying “times by.” It’s just “times.”

@ExistenceUniversity

12 күн бұрын

But you are times by a number of groups. "By" is useful.

@hambonesmithsonian8085

12 күн бұрын

Cope and seethe nerd.

unnatural