The $10$-adic Numbers

It's a familiar fact that every real number can be written in the form

$$d_n\ldots d_1 d_0.d_{-1}d_{-2}\ldots = d_n 10^{n} + \cdots + d_1 10 + d_0 + d_{-1} 10^{-1} + d_{-2} 10^{-2} + \cdots$$

where each digit $d_i$ is between 0 and $9$, and the sequence can continue indefinitely to the right.

The $10$-adic numbers also have decimal expansions, but everything is backward! To get a feeling for why this might be the case, we consider Euler's nonsensical series

$$\sum_{n=1}^{\infty} (-1)^{n+1}n! = 1! - 2! + 3! - 4! + 5! - 6! + \cdots.$$

One can prove (see Exercise 55) that this series converges in $\mathbf{Q}_{10}$ to some element $\alpha\in\mathbf{Q}_{10}$.

What is $\alpha$? How can we write it down? First note that for all $M\geq 5$, the terms of the sum are divisible by $10$, so the difference between $\alpha$ and $1! - 2! + 3! - 4!$ is divisible by $10$. Thus we can compute $\alpha$ modulo $10$ by computing $1! - 2! + 3! - 4!$ modulo $10$. Likewise, we can compute $\alpha$ modulo $100$ by compute $1! - 2! + \cdots + 9! - 10!$, etc. We obtain the following table:

$\alpha$	$\mod 10^r$
$1$	$\mod 10$
$81$	$\mod 10^2$
$981$	$\mod 10^3$
$2981$	$\mod 10^4$
$22981$	$\mod 10^5$
$422981$	$\mod 10^6$

Continuing we see that

$$1! - 2! + 3! - 4! + \cdots = \ldots 637838364422981$$   in $Q_10$ !

Here's another example. Reducing $1/7$ modulo larger and larger powers of $10$ we see that

$$\frac{1}{7} = \ldots857142857143$$   in $Q_10$$$.$$

Here's another example, but with a decimal point.

$$\frac{1}{70} = \frac{1}{10}\cdot \frac{1}{7} = \ldots85714285714.3$$

We have

$$\frac{1}{3} + \frac{1}{7} = \ldots66667 + \ldots57143 = \frac{10}{21} = \ldots 23810,$$

which illustrates that addition with carrying works as usual.