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# 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:

$a_{n+1}=1-7 \cdot x_{n+1}=1 - 7 \cdot (x_n \cdot 10^{-6} + 0.142857)$ $a_{n+1}=1 - 10^{-6} \cdot 7 \cdot x_n - 0.999999$ $a_{n+1}=0.000001 - 10^{-6}\cdot 7 \cdot x_n$  $a_{n+1}=10^{-6}\cdot(1 - 7 \cdot x_n)$

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!