# Comments on Why does the decimal expansion of $1/(10n - 1)$ have this neat pattern?

## Parent

# Why does the decimal expansion of $1/(10n - 1)$ have this neat pattern?

I was playing around with the reciprocals of some positive integers and found these interesting patterns:

$$ \frac{1}{19} = 0.\overline{052631578947368421} $$

Now, this repeating decimal can also be obtained by "concatenating" the powers of $2$ successively to the left as follows:

\begin{align}
1& \\
21& \\
421& \\
8421& \\
{\color{red}{1}}68421 \\
{\color{red}{3}}{\color{green}{3}}68421 \\
\vdots\quad &
\end{align}
Notice that here I think of the red $\color{red}{1}$ of $16$ and $\color{red}{3}$ of $32$ as being "carried over", so the units digit of $32$ and the tens digit of $16$ add to give me the green $\color{green}{3}$. The sequence gets increasingly complicated as more terms are concatenated in this manner, but it definitely looks like it's converging to $\dotsc 052631578947368421$, where the ellipsis indicates that the pattern repeats forever towards the *left*. This is a bit surprising because (informally) this would mean that if I perform this pattern infinitely far out to the right of the decimal point, then I get the decimal expansion of $1/19$.

How can I go about making this idea precise?

Some more data for this pattern: \begin{align} \vdots\quad & \\ {\color{red}{3}}{\color{green}{3}}68421 \\ {\color{red}{6}}{\color{green}{7}}368421 \\ {\color{red}{1}}{\color{green}{34}}7368421 \\ {\color{red}{2}}{\color{green}{69}}47368421 \\ {\color{red}{5}}{\color{green}{38}}947368421 \\ {\color{red}{10}}{\color{green}{77}}8947368421 \\ \vdots\quad & \end{align}

In this pattern, one can see that at each stage one new digit from the required sequence is correctly added, despite all the carrying-over.

I have also observed this to be the case for the decimal expansion of $1/29$: $$ \frac{1}{29} = 0.\overline{0344827586206896551724137931} $$ In this case, we concatenate the powers of $3$ in a similar fashion.

Is there any significance to the fact that $19 = {\color{red}{2}}*10 - 1$ and $29 = {\color{red}{3}}*10 - 1$ to the use of powers of $2$ and $3$, respectively, in these computations?

I also checked for $1/39$, and though the repeating part of its decimal expansion is not as impressively long as for the other two cases, the pattern still holds up: $$ \frac{1}{39} = 0.\overline{025641} $$ and concatenating the powers of $4$ indeed gives $\dotsc 025641$.

The last case I checked was $1/49$, and it also obeys this pattern: $$ \frac{1}{49} = 0.\overline{020408163265306122448979591836734693877551} $$ and concatenating the powers of $5$ gives us this very repeating set of digits.

Of course, the pattern for $1/49$ makes it obvious that there is a concatenation that works in the *forward* direction: namely, the powers of $2$ concatenated in bunches of two digits. But this just means that
$$
\frac{1}{49} = \frac{2}{100} + \frac{2^2}{100^2} + \frac{2^3}{100^3} + \dotsb
$$
which is true by the formula for the sum of an infinite geometric series. So, "forward concatenating" patterns can be explained in this manner, but I'm stumped regarding how to explain the "backward concatenating" ones.

If anyone can throw some light on this, it would be greatly appreciated!

## Post

First of all, there *is indeed* a pattern.

To figure out why, there's an easier example than what you've found: $$0.\bar{1} = \frac{1}{9}.$$

How does that work? $1$ is the only power of $1$: $1^n = 1$.

Now, how do we *prove* this? if $x = 0.\bar{1}$, then $10x = 1.\bar{1}$ and so, $9x=1$.

Hidden in that proof lies the answer for the other cases - you have some repeating number and in order to show what fraction that repeating number is equal to, here $1/\left(10n-1\right)$, you multiply it by $10n$ and subtract the number (again, I'll just call it $x$).

To take your example of $1/19$, we have that the fraction is $x = 0.\overline{052631578947368421}$, so we multiply by $20$ to get $20x = 1.\overline{052631578947368421}$, subtract $x$ and $19x = 1$. Crucially, the number left on the RHS after subtracting is exactly $1$, as everything after the decimal point *cancels*. It must do this, because the fraction is some integer divided by some other integer. That is, if $x=n/m$, then $mx=n$.

As we're in a decimal system, multiplying by $10$ is shifting one digit to the left. So, in doing this multiplication by $10n$, you're shifting the digit to the left, then multiplying by $n$, which is exactly the procedure outlined in the question.

Now, each particular digit (after this multiplication procedure) must cancel with the digit to the left so at this point, they must be equal and we have the exact behaviour outlined in your question.

To explain this generally and in more detail (taking $x<1$ wlog.) I'll use a fraction $$x = \frac{m}{10n-1}.$$

When written in decimal format, this is $x = 0.x_1x_2\ldots x_k\ldots$, with $x_{k+r}=x_k$ for some $r$, where each $x_k$ is a single digit (i.e. an integer in the range $0-9$ as we're in a decimal basis).

Now, we know that $\left(10n-1\right)x = m$, or $10nx - x = m$. However, m is an integer, so all the digits after the decimal point (the 'fractional part', in other words) are $0$.

This means that the digits in the fractional part of $10nx - x$ *must also* be $0$.

So, we look at the digits $x_r$ and $x_1\ldots x_{r-1}$.

The fractional part of $10nx -$ the fractional part of $x$ equals $0$, or the fractional part of $10nx$ equals the fractional part of $x$. For this to be the case, *all the digits must be equal*.

So, we take the digit $x_r$. To calculate $x_{r-1}$, we use the properties of the fractional parts being equal and multiplication by 10 as shifting the digit to the left to give that the fractional part of $10n\left(0.x_1\ldots x_r\right) - 0.x_1\ldots x_r = 0.0\ldots x_r$ (where the $0$s after the point are repeated $r-1$ times).

This is *exactly* what's written in the question, just in a different way: $nx_r \mod 10 = x_{r-1}$. Or, the digit to the left of $x_{r-1}$ is $nx_r$, with change. Accounting for this change gives that $x_{r-j} = n^jx_r$, which is where the powers of $n$ come from, the change being what needs to be added to digits further to the left.

To explain what I mean by phrases such as "Accounting for this change", let's take the initial stream of digits again and write them out more explicitly, where the number in brackets to the right of each line represents the number of times the 0 is repeated after the decimal point:

\begin{align*}x &= 0.0\ldots x_k \qquad(k-1)\\ &+ 0.0\ldots x_{k-1} \qquad(k-2)\\ &+ 0.0\ldots x_{k-2} \qquad(k-3) + \ldots\end{align*}

Now, we apply this procedure of multiplying by $10n$. It is a *requirement* that, because the fractional part of $x$ *must* equal the fractional part of $10nx$, $10nx-m=x$, that is

\begin{align*}10nx - m &= 0.0\ldots x_k \qquad(k-1)\\ &+ 0.0\ldots x_{k-1} \qquad(k-2)\\ &+ 0.0\ldots x_{k-2} \qquad(k-3) + \ldots\end{align*}

Taking this line by line (where the term in brackets on the LHS has $j-1$ $0$s after the decimal point and the superscripts denote that this series arises from the $j^{th}$ term in the similar expansion of $x$) gives that

\begin{align*}10n\left(0.0\ldots x_j\right) &= 0.0\ldots x_{k-1}^{(j)} \qquad(k-2)\\ &+ 0.0\ldots x_{k-2}^{(j)} \qquad(k-3)\\ &+ 0.0\ldots x_{k-3}^{(j)} \qquad(k-4) + \ldots\end{align*}

We can add this to the next term in the expansion of $x$ to get that (unfortunately at this point, notation gets very confusing but the first term on the left hand side has $j-1$ $0$s and the second, $j-2$) \begin{align*}10n\left(0.0\ldots x_j + 0.0\ldots x_{j-1}\right) &= 0.0\ldots x_{k-1}^{(j)} \qquad(k-2)\\ &+ 0.0\ldots x_{k-2}^{(j)} \qquad(k-3)\\ &+ 0.0\ldots x_{k-2}^{(j-1)} \qquad(k-3)\\ &+ 0.0\ldots x_{k-3}^{(j)} \qquad(k-4)\\ &+ 0.0\ldots x_{k-3}^{(j-1)} \qquad(k-4) + \ldots\end{align*}

Now, having done this, the $k^{th}$ term is exactly $$\sum_{j>k}0.0\ldots x_k^{(j)} \qquad(k-1),$$

which is the sum over $10n\times$ all the digits to the right of the $k^{th}$ digit, where you're only summing the resulting $k^{th}$ digit.

However, you can recursively apply this to get that the $k^{th}$ digit is the sum over $\left(10n\right)^p\times$ all the digits, $p$ positions to the right of the $k^{th}$ digit, where you're only summing the resulting $k^{th}$ digit.

This is precisely what's done in the question, although written in a different way and **shows that the behaviour arises because the terms are required to be sums of powers of $10n$ of previous terms because this is a fraction with a denominator of $10n-1$**.

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