A few weeks ago I wrote about p-adic numbers. I mentioned that if p is not prime, then the p-adic numbers can have zero divisors; that is, there are nonzero numbers 
Today Foxmaths! wrote about a 10-adic number (although not using that terminology)
such that 
In the post Foxmaths discusses how to generate other numbers of this form. In particular, assume 
The n-th digit of
. Which I think is a strikingly simple result.
Observe. Starting at
, we have
. So the 1-th digit of
is given by the 1-th digit of 25, the 2. Thus
.
Carrying on, we see that
, so the 2-th digit of
is given by the 2-th digit of 625, the 6.
.
Next,
. So the 3-th digit of
is a 0. This means that
in value.
But then we have again,
is given by the 4-th digit of 390625, the 9. Thus,
.
And so it goes. We can use this to deterministically calculate as many digits of F as we like.
If I am reading his derivation correctly, then it appears the right-most digit of 
In particular, there must only be ten idempotent 10-adic integers (two of which are 0 and 1). [Update: Moreover, this digit can only be 0, 1, 5, or 6 because we need 
Disclaimer: I do not know much more about p-adic numbers than what I’ve written on my blog. Presumably this fact is well-known among experts, or maybe I’m misinterpreting this and I’m completely wrong.
David Richeson: Division by Zero 
But but but!
The rightmost digit determines everything, but what determines the rightmost digit?
It must be that a*a = a (mod 10).
0, 1, 2, 3, 4, 5, 6, 7, 8, 9 square to give
0, 1, 4, 9, 6, 5, 6, 9, 4, 1 (mod 10)
Thus the rightmost digit can only be 0, 1, 5, 6.
So, this means there are only four?
Which is interesting, since 4 has no obvious connection to 10. Hmm!
Of course! Thank you for pointing that out. I fixed the post accordingly.
The connection between 4 and 10 is that 4 is the number of roots of x^2 – x (mod 10). This is the product of the number of roots of x^2 – x (mod 2) and x^2 – x (mod 5), each of which are 2.
In general, the number of solutions of x^2 – x (mod n) is 2^q(n), where q(n) is the number of distinct primes dividing n. This is because x^2 – x (mod p^j) for prime powers p^j has exactly two solutions; then apply the Chinese Remainder Theorem.
So it seems that the number of idempotent n-adic integers is 2^q(n), and there are “nontrivial” idempotent n-adics (i. e. idempotent n-adics other than 0 and 1) if and only if n is composite.
I have a question about 10-adic numbers.
I know that 1/2 doesn’t exist in 10-adic because when you try to divide it, you can’t get 2 times anything to get 1. However, if you look at 2/4, and try to divide that, you get
0000000000002
0000000000012 (4*3)
_____________
9999999999990
So the first digit is 3, but the algorithm breaks down after that because you can’t get 4 times anything to be 9.
If you go with higher powers of 2 (for example 512/1024) you will find that you can go even further before the algorithm breaks down.
This leads me to believe that although 1/2 doesn’t exist in 10-adic, there does exist a limit as n approaches infinity of (2^n)/(2^(n+1))
Actually, there are two answers. They are:
…43441122120343012014132000331301242113113403320313
and
…56558877879656987985867999668698757886886596679688
If you add those two numbers, you get 1.
My question is, is this a natural result of the 10-adics having zero divisors or did I stumble on something more interesting?
But isn’t 1/2 = 0,5 in 10-adics too?
As for zero divisors, as I pointed out in Foxmaths’ blog, is G = 1 – F = …607743740081787109376. This, also, is completely determined by the rightmost digit by a similar procedure – except that you subtract each new nonzero digit from 10 before attaching it to the left.