# Using the pigeonhole principle prove that the decimal expansion of $1/k$ for any $k\in\Bbb N_{>0}$ becomes cyclic at some point [duplicate]

Prove if $k$ is any positive integer, the decimal expansion of $\frac{1}{k}$ eventually gets into a repeating cycle.

I'm stuck here , i've thought about applying the pigeonhole principle and induction here but to no avail...

My answer is asking for a proof through pigeonhole principle or induction.. the one linked does not show those..

## marked as duplicate by Mees de Vries, Jean Marie, Henry, Marc van Leeuwen, user223391 Mar 10 '17 at 7:11

• @MeesdeVries Can u check my edit. – bigfocalchord Mar 8 '17 at 8:57
• @dydxx, actually the second answer in the linked question provides a solution that uses the pidgeonhole principle. – Mees de Vries Mar 8 '17 at 9:12

If you divide $1$ by $k$ in the way everyone learnt it in school, you calculate $10$ times a number divided by $k$ with remainder multiple times.

The possible remainders are $0,1,\cdots ,k-1$, so there are $k$ possible remainders.

If we have determined $k+1$ remainders, the pigeonhole-principle guarantees some duplication.

So, there must be two positions with equal remainders.

The earliest point for which this occurs is the point where the decimal expansion gets periodic.

Perform long division:

To find the decimal expansion for $1/k$, let $a_0=1$. Then $10a_0=kb_1+a_1$, and we continue so that $10a_1=kb_2+a_2$, and so on. From this, we see that $b_1,b_2,...$ are indeed the digits in the decimal expansion.

by the pigeonhole principle: we know that $a_1 \in \{0,1,..k-1\}$ because it is a remainder, so eventually $a_i=a_j$ for some $i,j \leq k$. But then $b_{i+1}=b_{j+1}$ and so forth after this, so the decimal expansion is periodic.