# Proof that 1/7 is a repeated decimal

This blog post serves as an exercise and solution to the following question:

$\frac{1}{7}=0.\overline{142857}$

In plain English: is the fraction $\frac{1}{7}$ a repeated decimal (0.142857142857142857…)?

## Recurring pattern

How can we tackle such a problem? First note that we have to deal with a recurring pattern: a pattern that refers to itself. One useful concept to model such problems is the concept of sequences. We can model the repeated decimal by the following sequence:

$0.\overline{142857}=x_n$

$x_{n+1}=0.142857 + 10^{-6} x_n$

With $x_0=0$

How does that work? How does this sequence approximate the repeated fraction? First, take a look at $x_1$. $x_1=0.142857+10^{-6}\cdot x_0=0.142857$. Now, take a look at the following item in the sequence: $x_2=0.142857+10^{-6}\cdot x_1=0.142857142857$. So, it concatenates “142857” and the recurring pattern constructed so far. By that property we know that the sequence $x_n$ equals the repeated decimal if $n$ approaches $\infty$.

## The proof

We would like to proof that $\frac{1}{7}=0.\overline{142857}$. This is equal to proving that $1=7 \cdot 0.\overline{142857}$. This equals $1 – 7 \cdot 0.\overline{142857}$ and this is equal to proving that $1 – 7 \cdot \lim\limits_{n \rightarrow \infty} x_n = 0$.

### Another sequence

Lets introduce another sequence:

$a_n = 1 – 7 \cdot x_n$

Notice that the statement we need to proof is the same as the following:

$1 – 7 \cdot \lim\limits_{n \rightarrow \infty} x_n = 0 \iff \lim\limits_{n \rightarrow \infty} a_n$

Now notice that:

$a_{n+1}=1-7 \cdot x_{n+1}=1 – 7 \cdot (x_n \cdot 10^{-6} + 0.142857)$

$a_{n+1}=1 – 10^{-6} \cdot 7 \cdot x_n – 0.999999$

$a_{n+1}=0.000001 – 10^{-6}\cdot 7 \cdot x_n$

$a_{n+1}=10^{-6}-10^{-6} \cdot 7 \cdot x_n$

$a_{n+1}=10^{-6}\cdot(1 – 7 \cdot x_n)$

Thus:

$a_{n+1}=10^{-6} \cdot a_n$

And thus we know that:

$a_{n+k}=10^{-6k} \cdot a_n$

Thus (for $n = 0$):

$a_k=10^{-6k} \cdot a_0$

From this, we get that:

$\lim\limits_{k \rightarrow \infty} a_k = a_0 \cdot \lim\limits_{k \rightarrow \infty} 10^{-6k} = 0$

And from this we can deduce the earlier statement, thus:

$\frac{1}{7}=0.\overline{142857}$

Thus, $\frac{1}{7}$ is indeed equal to the repeated fraction $0.\overline{142857}$! If you have any questions or suggestions, feel free to post them below!

#### Kevin Jacobs

Kevin Jacobs is a certified Data Scientist and blog writer for Data Blogger. He is passionate about any project that involves large amounts of data and statistical data analysis. Kevin can be reached using Twitter (@kmjjacobs), LinkedIn or via e-mail: kevin8080nl@gmail.com. Want to write for our website? Then check out our write for us page!