Real numbers and rationals - Decimal Expansion

Question

How would one endeavor to show that A real number is rational if and only if its decimal expression ends in recurring digits?

Answer 1

A recurring decimal expansion is a geometric series with common ratio $10^{-\tau}$, where $\tau$ is the period in number of decimal places. Thus, through the relationship $\frac{a}{1-r} = a\,(1+r+r^2+\cdots);\,r<1$ it is a number of the form $\frac{p}{q};\,p,\,q\in\mathbb{N}$.

To infer that rationals have finite or periodic $N-ary$ expansions: if a number is rational, then the $N-ary$ expansion is computed by iterated integer division on $\frac{p}{q};\,p,\,q\in\mathbb{N}$. At a given $j^{th}$ iteration of this algorithm, the input to the next iteration is the integer $N^x\,R_{n-1}$ where $R_{n-1}$ is the remainder after the former iteration and $x$ is the least integer such that $N^x\,R_{n-1}>q$. The next digit in the expansion is $N^x\,R_{n-1}\mathrm{div}\, q$ and $R_n = 10^x\,R_{n-1}\mod q$ is the input to the next iteration. There are only a finite number of input states to each iteration, to wit, the numbers $N^{x_j}\,j;\,j\in0\cdots\,q-1$ and so the input state after $q$ iterations must be the same as the state at a former iteration so that the process either ends ($R_n=0$) or is periodic.