Generalisation of integers for infinite length?

Question

As most people know, an integer can only be finite length when expressed in the form of a series of digits in some base. However, real numbers in general can be infinite length, so long as there is a "first digit".

That is, if we express an "infinite integer" in the form $$ \sum_{n=0}^\infty a_n 10^n $$ where $a_n\in\mathbb{Z}_{10}$, then addition and multiplication should both be well-defined on this sequence. Indeed, I suspect that such a system may actually form a field, as additive and multiplicative inverses should exist (consider that $\bar{9}+1=0$ where $\bar{9}$ is the number for which $a_n=9$ for all $n$, and $\overline{285714}3\times7=1$, where $\overline{285714}3$ has $a_0=3$ and then $a_{6n+1}=4$, $a_{6n+2}=1$, and so on).

Has such a set with addition and multiplication been investigated? Does it have a name? Does it actually form a field, or does it fail one of the required properties of a field? Is the system functionally the same irrespective of chosen base, or do different bases change some properties of the system?

Answer 1

You've rediscovered the $10$-adic integers. Such "infinite decimal expansions of integers" form a ring using the usual rules for adding and multiplying decimals. However, they do not form a field. A quick and easy way to see this is that $10$ has no inverse (since if you multiply $10$ by anything, then the last digit of the product will be $0$).

Even worse, however, this ring has zero divisors: there are nonzero elements $x$ and $y$ such that $xy=0$. These take a bit of work to construct, but here's the idea. We'll construct the digits of $x$ and $y$ one at a time, starting from the end. Start by saying the last digit of $x$ is $2$ and the last digit of $y$ is $5$, so the last digit of $xy$ is $0$. Then choose the preceding digit of $x$ so that $x$ is divisible by $4$, and the the preceding digit of $y$ so that $y$ is divisible by $25$, so their product will be divisible by $100$ and end in two zeroes (for instance, $x$ might end in $12$ and $y$ might end in $25$). Continue choosing digits of $x$ and $y$ one at a time so that $x$ is divisible by every power of $2$ and $y$ is divisible by every power of $5$. Every digit of the product $xy$ will end up being $0$.

You can, of course, do the same thing with a different base instead of $10$, giving you the $b$-adic integers for any integer $b>1$. Unlike the case of finite base expansions, however, you get a genuinely different number system for different values of $b$! (To be precise, it turns out that the number system you get depends only on the set of prime factors of $b$, so for instance the $10$-adic integers are isomorphic to the $50$-adic integers.)

It turns out that whenever $b$ has two different prime factors, the $b$-adic integers will have zero divisors, as in the case $b=10$. When $b$ is a prime (say $b=p$), however, the $p$-adic integers are an integral domain (they have no zero divisors). The $p$-adic integers aren't quite a field, though, because $p$ has no inverse. If you adjoin an inverse to $p$, you do get a field, called the $p$-adic numbers and written $\mathbb{Q}_p$. Elements of $\mathbb{Q}_p$ are "base $p$ expansions that can be infinite on the left": that is, formal sums $$\sum_{n=k}^\infty a_np^n$$ where each $a_n$ is an integer between $0$ and $p-1$ and $k\in\mathbb{Z}$ (so $k$ can be negative, allowing finitely many negative powers of $p$ in the sum).

There is a much larger story here: $p$-adic numbers play a huge role in modern number theory, and have a very rich structure. Since it is easily possible to write an entire book on the topic, I'll end my answer here though. You can read a bit more about them on the Wikipedia page I linked at the beginning.

Answer 2

Yes. What you've defined is called the 10-adic numbers. They don't form a field for two reasons, one of which is that $10$ is not invertible; the other reason is left as an exercise.

Most books will only talk about the $p$-adic numbers for $p$ prime, and this is because the general case turns out to reduce to the prime case, essentially because of the Chinese remainder theorem.

Answer 3

Another fun way that the rings differ depending on the base used is that in base 5, $-1$ (i.e. $\bar{4}$) has a square root while $2$ doesn't, but in base 7 it's the other way around (and $-1$ is of course $\bar{6}$ in that base).

I should note that I haven't proved those, but my messing around with the idea pointed strongly that way. If you like programming it's a fun exercise to write up a class that implements these numbers. I did it in Python, as its yield/generator system made it easy to model these numbers as a potentially infinite list that produced the digits from units place up.