Period length of a decimal expansion

Question

I would like some guidance for this problem: Suppose $a/b$ is a rational number. Prove that, if $2$ and $5$ are not divisors of $b$, the period length of the decimal expansion is the smallest integer $k$ such that $10^k \equiv 1 \mod b$.

I can find an explanation related here, but without proof: https://math.stackexchange.com/a/377696/737799 Thanks in advance!

Answer 1

Here are a few pointers:

• If the decimal expansion of $\ \frac{a}{b}\ $ is non-terminating, its period depends only on its fractional part. It therefore simplifies matters to assume that you've already got rid of the integer part, and so $\ 0\le a<b\ $.
• The decimal expansion of the fraction $\ \frac{a}{b}<1\ $ is periodic with (smallest) period $\ k\ $ if and only if \begin{align} \frac{a}{b}&=c\sum_{i=1}^\infty10^{-ik}\\ &=\frac{c10^{-k}}{1-10^{-k}}\\ &=\frac{c}{10^{k}-1}\ , \end{align} where $\ c\ $ is an integer less than $\ 10^k\ $ which is not itself periodic with a period smaller than $\ k\ $.
• A positive integer $\ c<10^k\ $ is periodic with period $\ j<k\ $ if and only if $\ k=mj\ $ for some positive integer $\ m\ $, and there exists a positive integer $\ d<10^j\ $ such that \begin{align} c&=d\sum_{i=0}^{m-1}10^{ij}\\ &=\frac{\left(10^m-1\right)d}{10^j-1} \end{align}
• Given the above observations, my strategy for proving the result would be:
  • Let $\ k\ $ be the smallest positive integer such that $\ 10^k\equiv 1\pmod{b}\ $, and let $$ c=\frac{a\left(10^k-1\right)}{b}\ , $$ noting that this is always possible if neither $2$ nor $5$ is a divisor of $\ b\ $.
  • Show that $\ c<10^k\ $ and that it satisfies the equations given in the second of the above dot points.
  • Show that if $\ c\ $ were itself periodic, with a period $\ j< k\ $ then $\ 10^j-1\ $ would be evenly divisible by $\ b\ $, thus contradicting the fact that $\ k\ $ was the smallest such positive integer. It's this step where you need to use the supposition that $\ \gcd(a,b)=1\ $.