# Tiny Nines

## Problem

*This problem follows on from Terminating or Not and accompanies Repetitiously*

In the video below, watch as three calculations are performed on the calculator.

The fourth calculation is not completed. Can you predict what the result will be?

*This video has no sound.*

*If you can't see the video, click below.*

Work out these answers, and then use them to predict the answer to $1 \div 9999$.

The decimal representations of $\frac19$, $\frac{1}{99}$, $\frac{1}{999}$ and $\frac{1}{9999}$ can be used to help you work out the decimal representations of other families of fractions.

Can you use what you now know to make predictions about the decimal representations of these and other fractions?

- $\frac13$, $\frac{1}{33}$, $\frac{1}{333}$...

- $\frac1{11}$, $\frac1{111}$, $\frac1{1111}$...

- $\frac{23}{99}$, $\frac{37}{99}$, $\frac{52}{99}$, $\frac{n}{99}$...

Can you show that the recurring decimals in your predictions are equivalent to the fractions that they are supposed to represent?

*Calculator used in video: https://www.eeweb.com/tools/online-scientific-calculator/*

## Getting Started

Take a look at the problem Repetitiously to explore a method for converting recurring decimals into fractions.

## Student Solutions

Sanika from PSBBMS in India, ThÃ´ng-sahge from Tiffin School in the UK, Zofia from the Netherlands, Mahdi from Mahatma Gandhi International School in India and Turbo from Bangkok Patana School in Thailand all predicted the same decimal for $1\div9999$. This is some of Turbo's explanation:

Since $\frac19$ is $0.1111111111111”¦$ and $\frac{1}{99}$ is $0.01010101010101”¦$ and $\frac1{999}$ is $0.001001001001001”¦$ there is a pattern that follows these rules:

- For every $1$, there is a certain amount of zeros following it (with the only exception of $\frac19.$
- Depending on how many numerals in the denominator of the factor you wish to convert, the amount of numerals is one more than the number of zeroes you put before the one.

E.g.

$0.111111111111111”¦$ is $\frac19$ and since $\frac19$ has one [digit in its] denominator, you minus one away from the number one (the [number of digits in the] denominator) to get $0,$ therefore, there are no zeroes before the one.

$0.01010101010101”¦$ is $\frac1{99}$ and since $\frac1{99}$ has two [digits in its] denominator (the two $9$s), you minus one away from the $2$ (the [number of digits in the] denominator) to get one, therefore, there is $1$ zero before each one.

Using this method, we can get $\frac{1}{9999}$ equal $0.00010001000100010001”¦$ since there are four [digits in the denominator] from which we minus one to get the result of $3.$ So, there are $3$ $0$s before the one.

Turbo wanted to generalise algebraically. Turbo wrote:

$\frac1a = 0.b1b1b1b1b1b1”¦ $

$a$ is how many denominator $9$ digits

$b$ is $a-1$ amount of $0$s.

Having read the rest of Turbo's working, we understand what Turbo means. Mahdi also wanted to generalise algebraically. What Mahdi wrote is easier to understand even without seeing the rest of Mahdi's working:

$$\frac1{\underbrace{99...00}_k} = 0.\underbrace{\dot{0}00...\dot{1}}_k$$

Sanika, Zofia and Mahdi found the decimals equivalent to the other fractions. This is some of Sanika's work:

Part A

$\begin{align}\tfrac13 &= \tfrac19\times 3\\ &=0.1111...\times3\\ &= 0.3333...\end{align}$

$\begin{align}\tfrac1{33} &= \tfrac1{99}\times 3\\ &=0.010101...\times3\\ &= 0.030303...\end{align}$

$\begin{align}\tfrac1{333} &= \tfrac1{999}\times 3\\ &=0.001001001...\times3\\ &= 0.003003003...\end{align}$

Part B

Similarly,

$\begin{align}\tfrac1{11} &= \tfrac1{99}\times 9\\ &=0.010101...\times9\\ &= 0.090909...\end{align}$

$\begin{align}\tfrac1{111} &= \tfrac1{999}\times 9\\ &=0.001001001...\times9\\ &= 0.009009009...\end{align}$

$\begin{align}\tfrac1{1111} &= \tfrac1{9999}\times 9\\ &=0.00010001...\times9\\ &= 0.00090009...\end{align}$

This is some of Zofia's work:

Sanika explained how to extend this to any integer $n$:

$\tfrac n{99} = \tfrac1{99}\times n = 0.010101...\times n$

- if $n$ has one digit, replace all $1$s in $0.01010101...$ with $n$
- if $n$ has two digits then replace the zeros with the digit in the tens place and the $1$s with the digit in the units place
- if $n\gt99$ then do the following:

$\left[\left(n \text{ (mod }99)\right)\times 0.010101... \right] + \left[n - \left(n \text{ (mod } 99)\right)\div9\right]$*(Where $n \text{ (mod }99)$ means the remainder when $n$ is divided by $99$)*

Mahdi gave an example which shows what Sanika means for the cases where $n\gt99$:

$\begin{align}\tfrac{320}{99} &=3+\tfrac{23}{99}\\ &=3+0.\dot2\dot3\\ &=3.\dot2\dot3\end{align}$

Mahdi also extended this to other similar fractions:

$\frac{n}{\underbrace{99..99}_k} = n\times0.\underbrace{\dot000...\dot1}_k$

SHORT CUT: Make $n$ [contain $k$ digits] by adding zeros in front of $n$

$\frac{n}{\underbrace{99..99}_k} = 0.\underbrace{\dot{n}}_k$

To prove that these recurring decimals are equivalent to the fractions that they are supposed to represent, look at the problem Repetitously.

## Teachers' Resources

### Why do this problem

There are fascinating patterns to be found in recurring decimals. This problem explores the relationship between fraction and decimal representations. It's a great opportunity to practise converting fractions to decimals with and without a calculator.

### Possible approach

This problem could be explored alongside Repetitiously.

Show students the video in the problem, or ask them to find the decimal representations of $\frac19$, $\frac1{99}$, and $\frac1{999}$ for themselves. Invite them to predict what $\frac1{9999}$ will be as a decimal.

Challenge students to convince themselves, and convince you, that the decimal representations really do go on forever. They may do this by performing a division calculation by hand and considering the remainders, or converting the recurring decimals back into a fraction (as demonstrated in Repetitiously). Then they could explore other related fractions such as those suggested in the problem:

- $\frac13$, $\frac{1}{33}$, $\frac{1}{333}$...

- $\frac1{11}$, $\frac1{111}$, $\frac1{1111}$...

- $\frac{23}{99}$, $\frac{37}{99}$, $\frac{52}{99}$, $\frac{n}{99}$

### Key questions

How can a fraction be turned into a decimal representation?

Without using a calculator?

If you know that $\frac19=0.\dot{1}$, how can you work out $\frac13$ as a decimal?

If you know that $\frac1{99}=0.0\dot{1}$, how can you work out $\frac1{11}$ as a decimal?

### Possible support

If they haven't already done so, students could start by exploring Terminating or Not.

### Possible extension

Challenge students to prove that the patterns they have noticed will continue.