# Period of the decimal expression for the rational number $\frac{1}{n}$ is at most $n-1$

Proposition: Show that for an integer $n\geq 2$, the period of the decimal expression for the rational number $\frac{1}{n}$ is at most $n-1$. I'm unsure of where to start. This is my first class on proofs. Do I state:

$\frac{1}{n}=a_n=a_1a_2...a_nb_1b_2...b_n$ with $b$ referring to the repeating part of the expression. I've looked at several other examples but am more confused than aided. I'm unsure how to prove the $n-1$ part. Any help would be appreciated.

When you carry out long division of $1$ by $n$, either the process terminates and you have a finite decimal, or you obtain a sequence of remainders among $1, 2, \dots, n-1$. Once a remainder is repeated, the decimals must start repeating too. Since there are only $n-1$ possible remainders, the repetition must occur by the $n$th decimal place at the latest. The period is then the distance between this and the previous occurrence of the same remainder, which must be at most $n-1$ decimal places.

Clearly, it suffices to take $$n$$ such that $$n$$ is co-prime with $$10$$. This is because the periods of $$\frac{1}{5k},\frac{1}{2k}$$ and $$\frac{1}{k}$$ are all the same.

So suppose $$n>1$$ is an integer coprime to $$n$$.

The period of $$n$$ is equal to the smallest positive integer $$a$$ such that $$10^a-1$$ is a multiple of $$n$$ (because each of the blocks in the expansion of $$\frac{1}{n}$$ must yield a sequence of nines when multiplied by $$n$$).

Notice that $$n$$ divides $$10^{\varphi(a)}-1$$. Since $$\varphi(a)\leq a-1$$ we are done.

Notice that equality holds if and only if $$n$$ is primes and $$10$$ is a primitive root $$\bmod n$$.

• short answer: the period of $n$ is equal to the order of $10\bmod m$ where $m=n/(10,n)$. And this period is a divisor of $\varphi(m)<n$. Feb 12, 2017 at 0:51