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.
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.
@@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 this is unnecessary. The definition of a prime number is any number p that satisfies the implication p|ab => (p|a or p|b) The definition you gave, as far as I know, is equivalent. But generally it's good to avoid quantifiers like "for all" and "exists" when you can, since it leads to simpler and more operable definitions.
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.
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.
I could never grasp Euclid's Lemma, it was the one step in the whole thing that I just couldn't get my head around - now I got two proofs for it that I could both understand, thanks!
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.
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.
Hey, thanks for the amazing video. I especially liked the examples. The only thing that i would have liked to be added eas explaining why the proof does not hold for the examples given. From what i can tell, the unnatural numbers are not closed under adition or substraction, so there goes part of the proof. While the Q✓5 numbers are not well ordered. So from what i can tell, the reason why it holds for natural numbers is because we have division with remainder. But i would love it if anyone has extra insight.
not only removing elements from natural number set but adding some new ones could also hinder the uniqueness of factorization...it seem like set of natural number is the chosen 1 :)
your euclid's lemma proof...i almost understood everything from beginning till euclid's lemma ...now i think i will have to take piece of paper and write down everything step by step to fill this gap...its very abstract ! btw great video on such a niche topic...and i loved the reimann zeta function proof the most ! can you provide a summary type to remember what to do while proving euclids lemma as there are many steps...
well at 11:19 what is beautiful ? is it the fact that every power of x can be represented as some unique way of either selecting 1 or x from brackets in multiplication to get that power...?
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
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.
@@ProactiveYellowSo 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
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
i have a doubt at 6:37 where you said "6= -2*-3 is no different than 2*3 and you stated the reason that -1 is an unit "...so what if -1 is unit ? i don't got the reason...sorry if it is a silly doubt but i not got it so ... can you explain why ?
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.
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.
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.
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.
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.
I don’t know if it was intentional, but nice joke!
Very true.
We should call every number that isn't an integer an outteger.
What's a teger, though?
@@Duiker36 The superset of integers and outtegers, of course.
A system of numbers that doesn't have unique prime factorization? That's not natural.
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.
@@DontWatchWhileHigh do you have examples?
Quality joke.
@@InputOutput-b2l
The example of the video: Z[√-5]
@@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.
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?
@@DanielJackson6742 this is unnecessary. The definition of a prime number is any number p that satisfies the implication
p|ab => (p|a or p|b)
The definition you gave, as far as I know, is equivalent. But generally it's good to avoid quantifiers like "for all" and "exists" when you can, since it leads to simpler and more operable definitions.
8:48
This is such a cool hypothetical scenario, really caught me with that one!
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.
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.
I could never grasp Euclid's Lemma, it was the one step in the whole thing that I just couldn't get my head around - now I got two proofs for it that I could both understand, thanks!
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.)
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.
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.
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?
Damn why your videos look so attractive and beautiful? This colors are amazing!
Hey, thanks for the amazing video. I especially liked the examples. The only thing that i would have liked to be added eas explaining why the proof does not hold for the examples given. From what i can tell, the unnatural numbers are not closed under adition or substraction, so there goes part of the proof. While the Q✓5 numbers are not well ordered.
So from what i can tell, the reason why it holds for natural numbers is because we have division with remainder. But i would love it if anyone has extra insight.
I just found this channel, amazing content! Keep it up:)
not only removing elements from natural number set but adding some new ones could also hinder the uniqueness of factorization...it seem like set of natural number is the chosen 1 :)
you're damn underrated! keep it up man
Obviously this proof must fail somehow in non-UFD rings. It would be interesting to see exactly where it breaks down.
How do you compare in non-UFDs?
your euclid's lemma proof...i almost understood everything from beginning till euclid's lemma ...now i think i will have to take piece of paper and write down everything step by step to fill this gap...its very abstract ! btw great video on such a niche topic...and i loved the reimann zeta function proof the most ! can you provide a summary type to remember what to do while proving euclids lemma as there are many steps...
Ugh... Those "alien numbers" can't be added and don't even form a group under multiplication (no inverse).
I think they form a monoid, but I agree it wasn’t a great example for the mathematically inclined.
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?
There's an *either* missing in your implication:
p|ab -> either p|a or p|b
Exactly 1 must hold
@@kyay10 no. 3 | 3× 6 -> 3|3 and 3|6. And 3 is a prime too.
@@28aminoacids oops you're absolutely right!
@@28aminoacids yeah according to Wikipedia they specify that p can't be the zero element or a unit
yep, by fiat primes must be non-units
well at 11:19 what is beautiful ? is it the fact that every power of x can be represented as some unique way of either selecting 1 or x from brackets in multiplication to get that power...?
yes after 12:50 i feel this is beautiful before for me it was obvious :)...
why do you call multiplication "times by"
Keep up the fab work 💪🏼💯
babe wake up new polyamath video dropped
11:20 how did Euler looked at Reimann Zeta function? Euler was around a century before Reimann
Well it was just the zeta function but it gets the same point across.
1:42 noooo naturals without 0
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
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.
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.
@@ProactiveYellowSo 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
@@irigima9974He said to you that you only need to test up to sqrt(N)
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
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.
The unnatural numbers aren't closed under addition, so it fails at any stage involving addition.
@@TheArizus What about the actual ring posed as a counterexample?
Please, stop with the "times by" thing. It's "times" or "multiplied by", but not "times by".
i have a doubt at 6:37 where you said "6= -2*-3 is no different than 2*3 and you stated the reason that -1 is an unit "...so what if -1 is unit ? i don't got the reason...sorry if it is a silly doubt but i not got it so ... can you explain why ?
Great video.
You forgot 0, and I'm very vehement about it!
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.
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.
@@fullfungo Makes sense. Thank you
Why not 60 frames per second though? ;_; Great video!
That seems overkill when the video is mostly instances of seconds-long static frames with brief transitions between them
@@Anonymous-df8itOn the contrary, the brief transitions are the aestethics engine of the video, it is the real life videos that don't need the 60fps.
if the middle p were an e, it would describe a proton decay?
interesting that this is called the fundament
yet is on the top of the graph rather than bottom
This helps
Bro try to avoid white background🙃
Thank you for the awesome Video! :-)
It would be great if you could tell me the name of the music at 14:25 :))
@5:08 not either or, p could devide both a and b
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.
Not true, please take a look at the last word in the line above
Watch out with how you use irreducible and prime. They're in general not the same
wgere is my conversations edit!111111!!!!
I still might have it somewhere loll
Your channel name is so similar to polymath lol
Idk the math was cool and all, but still don't see why this theorem is fundamental to arithmetic. What does fundamental even mean in math
So this Manim?
"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)
Vieta jumping
Question 11 lol
"times by"?? "Such integer such that"?? Lol dude
Bro. Stop saying “times by.” It’s just “times.”
But you are times by a number of groups. "By" is useful.
Cope and seethe nerd.
unnatural