A 10-adic number that is a zero divisor

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 x and y such that xy=0.

Today Foxmaths! wrote about a 10-adic number (although not using that terminology)

F=\ldots4106619977392256259918212890625

such that F^2=F (in other words F is an idempotent element of the 10-adic numbers). Of course, with a little algebra we can turn this into 0=F^2-F=F(F-1). In other words, both F and F-1 are zero divisors in the set of 10-adic numbers.

In the post Foxmaths discusses how to generate other numbers of this form. In particular, assume F is such a number and a_n is the integer which corresponds to the first n+1 digits of F. Foxmaths goes on to write:

The n-th digit of a_n is given by the n-th digit of a_{n-1}^2 . Which I think is a strikingly simple result.

Observe. Starting at a_0=5, we have 5*5=25. So the 1-th digit of a_1 is given by the 1-th digit of 25, the 2. Thus a_1=25.

Carrying on, we see that 25*25=625, so the 2-th digit of a_2 is given by the 2-th digit of 625, the 6. a_2=625.

Next, 625*625=390625. So the 3-th digit of a_3 is a 0. This means that a_3=625 in value.

But then we have again, 625*625=390625, and the 4-th digit of a_4 is given by the 4-th digit of 390625, the 9. Thus, a_4=90625.

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 a determines the rest of the digits of a. 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 a_0^2=a_0 (mod 10). In particular, there must be only four idempotent 10-adic integers (two of which are 0 and 1).] Very cool.

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.

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5 Comments

  1. Fox says:

    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!

  2. Of course! Thank you for pointing that out. I fixed the post accordingly.

  3. Michael Lugo says:

    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.

  4. Brian Dean says:

    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?

  5. Fibonacci says:

    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.

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