[Math] n alternative proof for periodic expansion of decimal fraction

elementary-number-theory

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,

Best Answer

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.

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