I am looking for the name of the theorem that says that a number $x$ is rational if and only if its fractional part terminates or repeats (where "fractional part" refers to the representation of $x$ with respect to some integral base $b \ge 2$, not necessarily $10$).
EDIT: I know how to prove this theorem, though my proof is tedious. In the original version of this question I asked for a proof of the theorem in the hope of finding a quicker, slicker argument. Afterwards I realized that all I really need is the name of the theorem, so that I can refer to it in another proof.
$\sum_{i=1}^{\infty}\frac{1}{10^{ni}}, n>0 = \frac{1}{1-10^{-n}}-1,$ which is rational. These are the numbers whose digits are a periodic 1 in the fractional part. Every number with a fractional part which repeats after a finite number of steps can be expressed as a linear combination of finitely many of these numbers with integer coefficients, which is therefore rational.
Conversely, consider the base 10 expansion of a fraction $\frac{q}{p} \lt 1$. This can be performed by the following steps, for $a$ set initially to $10*q$, $n$ to $1$:
Set digit $n$ (past the decimal) to the number of times $p$ divides $a$. Note that $\frac{a}{p}\ge10 \iff \frac{a}{10} \ge p$. This should not happen at the first step, since $\frac{q}{p} \lt 1$, and you will see it should not happen on recursion.
Set $a$ to its remainder under $p$. Now $\frac{a}{q} \lt 1$.
Multiply $a$ by 10, add 1 to $n$, and recurse.
Now, you'll notice that if the same remainder ever occurs, the same digit will occur in the next step, which will look like the same step it was first produced in, but with a greater value for $n$. Hence, it will repeat. However, since $p$ is finite, there are a finite number of remainders $a$ can have. Thus, the digits must repeat if the $q, p$ are integers and thus $\frac{q}{p}$ has digits derivable through this process of integer arithmetic.