Posted by: Dave Richeson | December 29, 2008

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)


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.



  1. 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. 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. 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

    0000000000012 (4*3)

    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:


    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. 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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