Proof that 1/7 is a repeated decimal

Kevin Jacobs

Oct 5, 2018 • 3 min read

Photo by CHUTTERSNAP / Unsplash

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

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!