I must have missed p-adics in school. This was a fun, interesting, and clear read.

They are typically not taught in school, at least in the US but probably elsewhere too. I don’t even know of a commonly taken undergrad-level course that I’d expect to cover them.

They're often -- but not always -- touched on in advanced undergraduate classes, e.g. after real analysis and abstract algebra. Not every math undergraduate takes number theory these days.

Which classes are you thinking of?

My honors analysis class covered it. Although it the professor that taught the class worked in p-adic analysis so he snuck p-adics into the first quarter of the analysis sequence.

there must be a rule that bans using “marvelous, magic, mysterious, never seen before, …” words used in math posts. There’s nothing marvelous about any of this—it’s just math. Imo it’s better to let people face math head on than try to sugar coat it and click bait them into liking math.

I will continue to marvel at it, though.

> just math

I don't know what this is supposed to convey. Modern mathematics is one of the most amazing, marvelous, beautiful etc. achievements of the mind. I wish more people could appreciate that.

The marvel of Math is in its non-obviousness. Obvious truths are not marvelous but non-obvious ones are.

"Magic" is also a word we can use about Math because we know no true magic exists. Therefore when something is said to be "magic" we know it means something "looks like magic".

For the Magician (/Mathematician) it of course doesn't look like magic because they clearly understand and see how it works.

So, you are perhaps the Mathematician/Magician for whom it doesn't look like magic but for the rest of many of us it does. We are even willing to pay for the ticket to see magicians perform their "magic" and to marvel at it.

> Mathematician/Magician

Or mathemagician. ;-)

Great word

Yeah, I admit the title is a bit click-baity. But I got the word “marvelous” from the Audre Lourde poem, so it’s not completely gratuitous.

For another accessible introduction to p-adic numbers, check out Eric Rowland's overview [1].

[1]: https://www.youtube.com/watch?v=3gyHKCDq1YA

It's also not a full introduction, but for those who want to have a taste of p-adic numbers, see 3b1b's video https://www.youtube.com/watch?v=XFDM1ip5HdU which motivates 2-adic numbers. There's also a neat connection between twos complement and 2-adic numbers.

Shameless promotion of a blog post I made a few years ago connecting 2-adic numbers with computationally fast exponentiation: https://kevinventullo.com/2020/12/21/2-adic-logarithms-and-f...

Oh hey, I enjoyed your hashing multisets article and wrote my own take on the topic here: https://jeremykun.com/2021/10/14/group-actions-and-hashing-u...

Thanks for writing it and clueing me into the idea!

> Likewise, if you start with a positive integer that ends in 6 and repeatedly raise it to the fifth power, you converge digit by digit toward the strange number

> b = ···743740081787109376

No you don't. Trying with 6, 16, 26, any number, you don't converge.

6^5 ends in ...76

6^25 ends in ...376

6^125 ends in ...9376

6^625 ends in ...09376

and so on

I was hoping for something on metric spaces, I was completely wrong but not disappointed. Great post.

Edit: Arguably there is a weird kind of metric in that, so not completely wrong I guess...

I like the subject matter, but I find the writing much too long-winded for my taste, with all the “you” and “we” and “let’s” and so on. That conversational style might work in a video, but as written text I find it exhausting.

> It seems a + b = 1.

I've seen this before, and it makes no sense. Is it common to ignore that you are carrying a one?

There isn't some state where you are "carrying the one" and the number is different. Carrying is just an algorithm to canonicalize a number w.r.t. to a radix.

consider the natural number 9. it is always equal to (1+1+1+1+1+1+1+1+1+0)

but the radix is what determines allowable digits in the representation.

in any radix >9 you simply write 9, now consider radix=7

you will compute the fact 9 = 7^1*1 + 1^1*2, which implies the numeral is 12. because 9 = 9 + 0 = 7 + (9 - 7) = 7 + 2 = 10_7 + 2_7 = 12_7

computing coefficients doesn't change a number or a sum.

Given that

    a = ···256259918212890625

and

    b = ···743740081787109376

already they "make no sense" as natural numbers. (And they aren't.) We can define p-adic addition more formally, and "following the usual rule for adding numbers" (giving c = ···000000000000000001) is just serving as useful motivation in the meantime (that's why the author says "it seems"). You can look up the formal addition rule for p-adic integers, but meanwhile if you accept that the sum c=a+b is also going to be a p-adic integer, then you can think about: if c is not ···000000000000000001, what else could it be? (If you pick a certain position, say the 100 trillionth digit from the right, what is that digit of c going to be? Same for any other position other than the rightmost one.)

(Another analogy: If you consider that 1.0000 - 0.3333 = 0.6667, and 1.00000 - 0.33333 = 0.66667, etc, when you say 1.00000… - 0.33333… = 0.66666…, do you wonder about “where did the 7” go? Footnote 5 in the post discusses this.)