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

Question

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..

Please don't close this.

Answer 1

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.

Answer 2

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.