Is there an alternative proof for periodic expansion of decimal fraction?

Question

I'm currently reading Elementary Number Theory and Its Applications by Kenneth H. Rosen. In chapter 12 - Decimal Fraction, he provided a proof about the period length of the base $b$ is $\mathrm{ord}_n b$. and pre-period length. However, I found this proof so confusing, and it's not short, so I'm looking for an alternative proof for this theorem. Any reference or information would be greatly appreciated.

This is the original theorem from Rosen's book.
Theorem 12.4 - page 474

Let $b$ a positive integer. Then $a$ periodic base $b$ expansion represents a rational number. Conversely, the base $b$ expansion of a rational number either terminates or is periodic. Futher, if $0 < \alpha < 1$, $\alpha = r/s$, where $r$ and $s$ are relatively prime positive integers, and $s = TU$, where every prime factor of $T$ divides $b$ and $(U,b) = 1$, then the period length of base $b$ expansion of $\alpha$ is $\mathrm{ord}_U b$, and the pre-period length is $N$, where $N$ is the smallest positive integer such that $T|b^N$.

Furthermore, is there an alternative presentation of this theorem? I was really confused about what this theorem is all about.

Thanks,

Answer 1

Here is something about period length but nothing about pre-period length.

---

The first question is whether a fraction is periodic or not. If it's not periodic it's of the form $x/b^r$ for example 36.542 = 36542/1000 = 18271/500. The contrapositive of this tells us that if a fraction is of the form $x/y$ with $b \not | y$ then the expansion is periodic, for example in base 10: $10 \not | 73$ so 15/73 is periodic: $0.2054794520547945\ldots$.

It suffices to consider only fractions between 0 and 1 so let's do that. How do we compute the base $b$ expansion of a fraction? For example how do we compute $0.14285714285714285\ldots$ from $1/7 = p/q$ multiply both sides by 10 and throw away the remainder to get the first digit $d = \text{floor}(\frac{b p}{q})$. To get the rest of the digits we just do the same process on $r/q$, where $r$ is the remainder of the division $\frac{b p}{q}$. Here is an example of this

p q d r
1 7 1 3
3 7 4 2
4 7 2 6
8 7 8 4
4 7 5 5
5 7 7 1
...

and if you read the digits column we have the correct digits.

So how can we use this idea of repeatedly applying the division algorithm to find the period length? It seems like a difficult problem so for a warm up lets just find out when the period has length 1, for example $1/3 = 0.333\ldots$ has that property. Why? Well it's obvious from the table

p q d r
1 3 3 1
1 3 3 1
...

It's got period 1 because $p = r$, the remainder of dividing $bp$ by $q$ is $p$, $bp \equiv p \pmod q$. In fact it turns out this solves the problem completely - 1/7 is periodic with period 6 in base 10 because it is periodic with period 1 in base 10^6! So what is the smallest $r$ such that $b^r p \equiv p \pmod q$? It's just that ord thing that was mentioned.

Answer 2

For the pre-period part, perhaps it is easier if we specialize to base 10. Then $T=2^a5^b$ has all the factors of $2$ and $5$ in $s$, and $U$ has all the other factors. It says you need to find the smallest $N$ such that $T$ divides $10^N$, which is just the maximum of $a$ and $b$. The repeat comes about because you are finding the smallest number of the form $999\ldots000\ldots$ that $s$ divides into. The number of zeros is just $N$ and the number of nines is enough to take care of the other factors of the number. You can see this with $\frac{1}{3}=0.33333, \frac{1}{6}=0.1666666, \frac{1}{12}=0.083333$, where $\frac{1}{6}$ has one digit after the decimal point before the repeat starts (the pre-period) and one factor of $2$, and $\frac{1}{12}$ has two digits before the repeat starts and two factors of $2$. I'd suggest trying some more examples.

Answer 3

A simple proof: if $\rm\,\alpha\in \mathbb R\cap(0,1)\,$ is periodic in radix $\rm\,b\,$ with preperiod,$\,$ period lengths $\rm\,n,m\,$ then (as below or here) multiplying or dividing by $\rm\,b^n\,$ and $\rm\,b^k-1\,$ corresponds to shifting left or right by the preperiod and period, so it follows that $\rm\,\alpha\,$ is rational, expressible with denominator $\rm\,b^n (b^k-1),\, $ so its least denominator $\rm\, s\ |\ b^n (b^k-1),\,$ so $\rm\ s = tu,\ t\ |\ b^n,\ u\ |\ b^n-1\,$ by Euclid's Lemma or unique factorization, by using $\rm\, (b^n,\,b^k-1) = 1.\ $ In a bit more detail $$\begin{array}{rrl} &\alpha\!\!\! &\rm =\,\ 0\:.\overbrace{a_1a_2\cdots a_n^{\phantom{l}}}^{\textstyle a}\ \overbrace{\overline{c_1c_2\cdots c_k}}^{\textstyle \overline c}\ \ \text{ in radix } b\\[.2em] \iff& \beta \!\!&\!:= \rm b^n\, \alpha\! -\! a = 0\:.\overline{c_1c_2\cdots c_k}\\[.2em] \iff&\rm b^k\: \beta \!\!&=\ \rm c + \beta\\[.2em] \iff&\rm (b^k-1)\ \beta \!\!&=\ \rm c\\[.2em] \iff&\!\!\!\! \rm (b^k-1)\ b^n\: \alpha \!\!&\in\ \Bbb Z \end{array}\qquad\qquad$$