H0w t0 prove that periodic decimal numbers are rational? $a_1...a_k(b_1b_2..b_l)={m \over n}$

Given $a_1...a_k(b_1b_2..b_l)={m \over n}$ how can I prove that periodic decimal numbers are rational?

Where do I even begin?

Consider the decimal $r=0.b_1b_2\ldots b_nb_1b_2\ldots$, that is, a simple decimal of period $n$. Multiply $r$ by $10^n$:

$$10^n r = b_1b_2\ldots b_n.b_1b_2\ldots b_n \ldots$$

or

$$(10^n-1) r = b_1b_2\ldots b_n$$

Remember that $b_1b_2\ldots b_n$ is an integer, so that $r$ is a rational number after reduction to simplest form.

For a more general decimal $0.a_1a_2 \ldots a_m b_1b_2\ldots b_nb_1b_2 \ldots$, note that $0.a_1a_2 \ldots a_m$ is a rational number in and of itself, leaving the remainder as a rational number ($10^{-m}$) times the repeating decimal, so that the more general case also leaves a rational number.

I will just consider the case where $a = \frac{m}{n} \in \mathbb{Q}$ is between $0$ and $1$. This is a trick taught by my high school teacher:

Rather than simply prove simply existence, I will supply a constructive method.

Suppose we have $$a = .a_1a_2a_3\ldots a_na_1a_2$$so that the period is $n$. Then we may multiply to get $$10^n \cdot a = a_1a_2\ldots a_n.a_1\ldots$$

Then try subtracting these two to get $$10^n*a - a = a_1a_2 \ldots a_n.00000\ldots$$ And divide to get $$a = \frac{a_1\ldots a_n}{10^n-1}$$You may reduce this fraction (because it need not be in lowest terms), and voila, you are done!