4 ms·

As seems typical, this video goes on for a while and is straightforward until it completely loses you all of a sudden. I have no idea what would even motivate t

by kbd 10y ago

As seems typical, this video goes on for a while and is straightforward until it completely loses you all of a sudden. I have no idea what would even motivate the separation of numbers into "rooms" like the video shows, let alone how that explains why an infinite positive sum = -1.

espeed 10y ago
The infinite sum makes sense when using the p-adic number system [1] -- and measuring distance using the p-adic metric -- which forms an ultrametric space [2]. An interesting related concept is the Bruhat–Tits building [3]. For an intro on p-adic numbers, read these two short articles "A first introduction to p-adic numbers" [4] and "A Tutorial on p-adic Arithmetic" [5] or see the short video "Introduction to p-adic Numbers" [6]: [1] https://en.wikipedia.org/wiki/P-adic_number https://en.wikipedia.org/wiki/P-adic_number [2] https://en.wikipedia.org/wiki/Ultrametric_space https://en.wikipedia.org/wiki/Ultrametric_space [3] Bruhat–Tits building https://en.wikipedia.org/wiki/Building_(mathematics) https://en.wikipedia.org/wiki/Building_(mathematics) [4] A first introduction to p-adic numbers http://www.madore.org/~david/math/padics.pdf http://www.madore.org/~david/math/padics.pdf [5] A Tutorial on p-adic Arithmetic https://koclab.cs.ucsb.edu/docs/koc/r09.pdf https://koclab.cs.ucsb.edu/docs/koc/r09.pdf [6] Introduction to p-adic Numbers https://www.youtube.com/watch?v=vdjYiU6skgE https://www.youtube.com/watch?v=vdjYiU6skgE
- cderwin 10y ago
  I think the problem is that the introduction of the p-adic metric was poorly motivated. It's not clear why we would want 1+p+p^2+p^3+... to converge to -1, and introducing the p-adic metric to do so doesn't show why the p-adic numbers are useful in general (all it's really saying is that the partial sums are 2^(n+1)-1). I'm sure you can derive all sorts of weird metrics so that various weird identities are true; that alone fails to make them interesting. Based on this video alone it's not clear that either the identity or the p-adic norm are interesting in any non-trivial sense of the word. The result is that the introduction of the p-adic metric is hard to follow and the resulting identity seems arbitrary, even if you manage to follow the bit about the metric. (And these combined with a lack of rigor where it's needed seem to be recurring problems in 3Blue1Brown videos.)
  - t-ob 10y ago
    > I'm sure you can derive all sorts of weird metrics so that various weird identities are true On the rational numbers, at least, the p-adic metrics are more or less your whole lot, according to Ostrowski's Theorem [1]. There is a kind of cognitive hurdle everyone who studies these numbers has to clear, in that things that should be "large" turn out to be very small indeed, when viewed under a p-adic lens. I think it's more instructive to build up the ring of p-adic integers first [2, chapter 2], and construct the p-adic numbers from there. I can assure you they are very useful, though! A general theme in number theory is to take a "global" problem, defined over the integers, and to translate it into infinitely many "local" ones (over the p-adics, for each prime p). These are sometimes easier to solve and, if you're lucky, offer insight into the global solution you're looking for. [1]: https://en.wikipedia.org/wiki/Ostrowski's_theorem https://en.wikipedia.org/wiki/Ostrowski's_theorem [2]: http://www.springer.com/gb/book/9780387900407 http://www.springer.com/gb/book/9780387900407
  - posterboy 10y ago
    >doesn't show why the p-adic numbers are useful in general well, you certainly are neither more help either ;)
  - stablemap 10y ago
    My short advertisement is that it has proven very useful to study Diophantine equations by reducing mod n. The Chinese remainder theorem [1] tells you to focus on reducing mod p^r, where p is a prime. If r = 1 then you are working in a field but in the ring Z/4, for example, I know 2 ≠ 0 and yet 2·2 = 0. To make the p-adics Z_p I stitch all of these Z/p^r together: an element is a choice of a_r in each Z/p^r and these have to be compatible: a_2 reduces mod p to a_1 and so on. The resulting Z_p has no "zero divisors", and if I allow myself to invert p I get a field Q_p. This is a huge improvement, a foundation on which to build analysis and geometry as we did over R. [1] https://en.wikipedia.org/wiki/Chinese_remainder_theorem https://en.wikipedia.org/wiki/Chinese_remainder_theorem
taneq 10y ago
"Step 2: Draw the rest of the owl."

by kbd 10y ago

espeed 10y ago

The infinite sum makes sense when using the p-adic number system [1] -- and measuring distance using the p-adic metric -- which forms an ultrametric space [2]. An interesting related concept is the Bruhat–Tits building [3]. For an intro on p-adic numbers, read these two short articles "A first introduction to p-adic numbers" [4] and "A Tutorial on p-adic Arithmetic" [5] or see the short video "Introduction to p-adic Numbers" [6]: [1] https://en.wikipedia.org/wiki/P-adic_number https://en.wikipedia.org/wiki/P-adic_number [2] https://en.wikipedia.org/wiki/Ultrametric_space https://en.wikipedia.org/wiki/Ultrametric_space [3] Bruhat–Tits building https://en.wikipedia.org/wiki/Building_(mathematics) https://en.wikipedia.org/wiki/Building_(mathematics) [4] A first introduction to p-adic numbers http://www.madore.org/~david/math/padics.pdf http://www.madore.org/~david/math/padics.pdf [5] A Tutorial on p-adic Arithmetic https://koclab.cs.ucsb.edu/docs/koc/r09.pdf https://koclab.cs.ucsb.edu/docs/koc/r09.pdf [6] Introduction to p-adic Numbers https://www.youtube.com/watch?v=vdjYiU6skgE https://www.youtube.com/watch?v=vdjYiU6skgE

cderwin 10y ago
I think the problem is that the introduction of the p-adic metric was poorly motivated. It's not clear why we would want 1+p+p^2+p^3+... to converge to -1, and introducing the p-adic metric to do so doesn't show why the p-adic numbers are useful in general (all it's really saying is that the partial sums are 2^(n+1)-1). I'm sure you can derive all sorts of weird metrics so that various weird identities are true; that alone fails to make them interesting. Based on this video alone it's not clear that either the identity or the p-adic norm are interesting in any non-trivial sense of the word. The result is that the introduction of the p-adic metric is hard to follow and the resulting identity seems arbitrary, even if you manage to follow the bit about the metric. (And these combined with a lack of rigor where it's needed seem to be recurring problems in 3Blue1Brown videos.)
- t-ob 10y ago
  > I'm sure you can derive all sorts of weird metrics so that various weird identities are true On the rational numbers, at least, the p-adic metrics are more or less your whole lot, according to Ostrowski's Theorem [1]. There is a kind of cognitive hurdle everyone who studies these numbers has to clear, in that things that should be "large" turn out to be very small indeed, when viewed under a p-adic lens. I think it's more instructive to build up the ring of p-adic integers first [2, chapter 2], and construct the p-adic numbers from there. I can assure you they are very useful, though! A general theme in number theory is to take a "global" problem, defined over the integers, and to translate it into infinitely many "local" ones (over the p-adics, for each prime p). These are sometimes easier to solve and, if you're lucky, offer insight into the global solution you're looking for. [1]: https://en.wikipedia.org/wiki/Ostrowski's_theorem https://en.wikipedia.org/wiki/Ostrowski's_theorem [2]: http://www.springer.com/gb/book/9780387900407 http://www.springer.com/gb/book/9780387900407
- posterboy 10y ago
  >doesn't show why the p-adic numbers are useful in general well, you certainly are neither more help either ;)
- stablemap 10y ago
  My short advertisement is that it has proven very useful to study Diophantine equations by reducing mod n. The Chinese remainder theorem [1] tells you to focus on reducing mod p^r, where p is a prime. If r = 1 then you are working in a field but in the ring Z/4, for example, I know 2 ≠ 0 and yet 2·2 = 0. To make the p-adics Z_p I stitch all of these Z/p^r together: an element is a choice of a_r in each Z/p^r and these have to be compatible: a_2 reduces mod p to a_1 and so on. The resulting Z_p has no "zero divisors", and if I allow myself to invert p I get a field Q_p. This is a huge improvement, a foundation on which to build analysis and geometry as we did over R. [1] https://en.wikipedia.org/wiki/Chinese_remainder_theorem https://en.wikipedia.org/wiki/Chinese_remainder_theorem

taneq 10y ago

"Step 2: Draw the rest of the owl."