# Terminating or not

*Terminating or Not printable sheet*

*A terminating decimal is a decimal which has a finite number of decimal places, such as 0.25, 0.047, or 0.7734*

Take a look at the fractions below.

$$\frac23 \qquad \frac45 \qquad \frac{17}{50} \qquad \frac3{16} $$ $$\frac7{12} \qquad \frac58 \qquad \frac{11}{14} \qquad \frac8{15}$$

Which ones do you think can be written as a terminating decimal?

Once you've made your predictions, convert the fractions to decimals.

Click below to check which ones terminate.

*The remaining four fractions can be written as recurring decimals, with a repeating pattern that goes on forever.*

I wonder whether there is a quick way to decide whether a fraction can be written as a terminating decimal...

Choose some fractions, convert them to decimals, and write down the fractions whose decimals terminate.

What do they have in common?

Can you explain a method you could use to identify fractions which can be written as terminating decimals?

*Next you might like to explore recurring decimals in the problem Repetitiously.*

*You may also be interested in the other problems in our Comparing and Matching Feature.*

What are the prime factors of 10?

What are the prime factors of 100?

What are the prime factors of 1000?...

You could rewrite the eight fractions like this:

$$\frac23 \qquad \frac{2^2}5 \qquad \frac{17}{2\times5^2} \qquad \frac3{2^4} $$ $$\frac7{2^2 \times 3} \qquad \frac5{2^3} \qquad \frac{11}{2\times 7} \qquad \frac{2^3}{3 \times 5}$$

What do the fractions with terminating decimals have in common?

The year 7 mentoring group from Bangkok Patana School in Thailand, Ruby from Loughborough High School and Jayden and Kiefer from Leys Junior School, both in England, worked out which of the fractions in the example are equivalent to terminating decimals. This is Jayden and Kiefer's work:

First, we predicted which fractions would have a terminating decimals, this was quite easy as we could use process of elimination. $\frac 23$ - this is really easy as it is commonly known that this is equivalent to $0.6\dot6$ reoccurring.

Originally, we thought that $\frac{17}{50}$ wouldn't be terminating, but then realised it could be converted into $\frac{34}{100}$, this makes it terminating.

Then, you can eliminate $\frac7{12}$ as there is no possible way to convert it into [a fraction with denominator] that ends with zero (10,100 etc.) apart from numbers in the 120 times table.

$\frac{11}{14}$ is also easy to eliminate for the same reason as the last one.

$\frac{8}{15}$ is a harder one than the rest, but we eventually found out that it also wouldn't convert into 10,100 etc.

This leaves us with: $\frac45, \frac{17}{50}, \frac{3}{16}, \frac58, \frac{17}{50}$

The year 7 mentoring group put the fractions in a table, and focused on the denominators:

We took all the terminating fractions and factorized the denominator. We got the numbers: $5,2$ or both.

Then we did the same with the recurring numbers: $3$; $2,2$ and $3$; $2$ and $7$; $3$ and $5.$

This made us think that to get a terminating decimal, the factors of the denominator should be $2,5$ or both.

W.H., Dashiell from Sequoyah in the USA and Carlos from Kings College Alicante in Spain all found this same rule using unit fractions (numerator = 1). This is some of W.H.'s work:

The fractions that I am going to work out these values for are $\frac12, \frac13, \frac14, \frac15, \frac16, \frac17, \frac18, \frac19$ and $\frac1{10}.$

The only fractions that terminate are $\frac12, \frac14, \frac15, \frac18$ and $\frac1{10}.$

To understand this problem better, I [rewrote] all of the denominator numbers so that they are expressed as a product of their prime factors. All of the terminating ones' denominators contain only prime factors of $2$ or $5$ ($2, 4, 5, 8$ and $10$), and the recurring ones can have $2$ and $5$ as factors, but they also have factors such as $3$ in them.

One possible reason that the terminating fractions have exclusively prime factors of $2$ and $5$ could be to do with methods like percentages. Because a percentage is defined as a fraction with a denominator of $100,$ we can see that $100$ is a power of $10,$ and $10 = 2^1\times5^1.$ This means the only numbers that divide $10$ and return a whole number are $2$ and $5,$ so it logically follows
that only fractions that have denominators with these prime factors are terminating. As all fractions have a fraction $\frac1{10^n}$ that is smaller than them, we can treat all of these cases the same.

Dashiell found the same rule, but described it in a very different way:

I tried a lot of different rules and things [the fractions with terminating decimals] they had in common. Then I realized that they all went into powers of ten

$\begin{align}&\frac12: 2\times5=10 \hspace{1cm} &\frac14:4\times25=100\hspace{1cm} &\frac15:5\times20=100\\ &\frac18: 8\times125=1000 \hspace{1cm} &\frac{1}{10}: 10\times1=10 \hspace{1cm} &\frac1{16}:16\times625=10000\\ &\frac1{20}:20\times5 = 100\end{align}$

If the denominator goes into [a power] of two [it will also have a terminating decimal, because] all powers of two go into powers of ten. I used this [for] numbers like $\frac1{16}.$ It's easier to figure out that $16$ is a power of two than that $16\times625=10000.$

Then I realized that this worked [only] if the fraction was simplified. Otherwise fractions like $\frac3{12}$ wouldn't fit into any of the categories, but the equivalent decimal would be finite or in this case $0.25.$

So if the denominator is a factor of a power of ten then the equivalent decimal is finite.

W.H. explained how the reasoning for fractions $\frac1n$ can be extended to simplified fractions with numerator $>1.$ Click to see this explanation.

The reason that the method still works for these non $\frac1n$ fractions is that we know a terminating fraction in the form $\frac1n$ can't have any multiples that are recurring, meaning that if the first fraction is terminating, then all fractions that are multiples of the first one are also terminating.

However, this process does not flow exactly the same for recurring numbers, as fractions like $\frac6{12}$, which should apparently be recurring according to this logic, are actually terminating. However, we can use the previous simplifying method to show that the reasoning is still sound.

Ahan from Tanglin Trust School in Singapore, Sanika P from PSBBMS in India, Thomas from Lakenheath American High School in England, Homare from Wimbledon High School in the UK, Edward from Worthington Hooker School in the USA, Tiger and Utkarsh and Kaishin from Bangkok Patana School in Thailand, Mahdi from Mahatma Gandhi International School in India, John from Vaels International School in India, John from Royal Latin School in England and An from Loughborough High School in the UK all got the same rule - that fractions whose denominator is a factor of a power of $10$ (when simplified) are equivalent to terminating decimals.

John from Vaels International School wrote the rule using algebra:

If a fractions is in the form $\dfrac{xy}{(x)(5^c)(2^d)}$ it is terminating.

Joseph sent in this method for testing whether a fraction's decimal equivalent will terminate. Can you see how Joseph's method uses the same rule?

To decide whether a fraction will result in a terminating decimal, follow these steps:

1. Simplify the fraction. If the fraction can't be simplified any further e.g. $\frac78$ can't be simplified any further, we do nothing for this step.

2.Look at the denominator.

3. $x=$denominator

4.If $x$ ends in $0$ or $5,$ divide $x$ by $5.$ Repeat until $x$ doesn't end in $0$ or $5.$

5.If $x$ ends in $0$ or $2$ or $4$ or $6$ or $8,$ divide $x$ by $2.$ Repeat until $x$ is an odd number.

6.If $x=1,$ the fraction will result in a terminating decimal.

Otherwise, the fraction will not result in a terminating decimal.

Tiger and Utkarsh, John from Royal Latin School, An and Kaishin all said that fractions whose denominator is a factor of a power of $10$ (when simplified) are equivalent to terminating decimals because of the way we write numbers. Kaishin wrote:

If the denominator's prime factors are 2 or 5 or a combination of both, [then] the denominator will always be able to be converted to 10,100,10000... etc. . We use a base 10 number system, which means that if the denominator can be converted into 10,100,10000... etc., the decimal will always be terminal.

Edward 1 and his twin brother Edward 2 used this idea to prove the rule:

If a fraction, $f = \frac pq$ terminates, then it can be explicitly written as: $f = \dfrac {n_1}{10} + \dfrac{n_2}{10^2} + \dfrac{n_3}{10^3} + \dfrac{n_4}{10^4} + ..... + \dfrac{n_k}{10^k,}$ for some finite $k,$ & where $n$ is some arbitrary placeholder.*(The $n_i$ are the digits of the terminating decimal - $f=0.n_1n_1n_3n_4...n_k$ (because we write numbers in base $10$))*

I factorised this such that $f = \dfrac1{10^k}\times\left(\dfrac{n_1}{10^{1-k}} + \dfrac{n_2}{10^{2-k}} +

..... + \dfrac{n_{k-1}}{10^{-1}} + n_k\right).$*(And, since $k$ is greater than any of $1, 2, 3, ..., (k-1)$, the powers $(1-k), (2-k), ... -1$ are all negative, so $f = \dfrac1{10^k}\left(n_1\times10^{k-1}+n_2\times10^{k-2}+...+n_{k-1}\times10^{1}+n_k\right)$*

I then let whatever is inside the brackets above be equal to $j$ *(where $j$ is a whole number, as seen above, and in fact $j$ also has digits $n_1, ... n_k$ - when $j$ is written out, $j$ is written as $n_1n_2n_3...n_k$)*. I re-write $10^k$ as $2^k\times5^k,$ then $f = \dfrac{j}{2^k\times5^k}.$

This fraction can be simplified by factoring out all the common multiples of $2$ and $5$ in the numerator. Therefore, $f=\dfrac{j'}{2^y+5^z}$ for some new integers $j', y, z.$ Clearly, this form shows that the denominator consists purely of $2$s and $5$s.

Thomas used these ideas to describe how the terminating decimal can be found:

Example: [if a] fraction can be simplified down to $\frac38$, the denominator has a $2$-to-$5$ factor ratio of $3:0$*(Thomas means that the prime factorisation of $8 = 2\times2\times2$ contains $3$ $2$s and $0$ $5$s)*

Bringing that ratio back to $3:3$ ($2\times2\times2\times5\times5\times5$) gives us a power of $10$ in the denominator (namely, $1000$). [So we need to multiply numerator and denominator] by $125$ ($5\times5\times5$) and we can thus see that the decimal form is $0.375.$

This method can be used in every case where the denominator in simplest form of the fraction can be factored into purely $2$s and $5$s.

### Why do this problem?

This problem offers an excellent opportunity for students to practise converting fractions into decimals, while also investigating a wider question that connects their knowledge of prime factors and place value.

### Possible approach

Start by writing a list on the board of the following sequence of fractions:

$$\frac1{40}=$$ $$\frac2{40}=$$ $$\frac3{40}=$$... up to $$\frac{20}{40}=$$

"Do you know how to write any of these fractions as decimals?"

Give students a little time to figure out which fractions they recognise, perhaps using equivalent fractions as an intermediate step. Then fill in the decimal equivalents on the board, inviting students to share the thinking they did to work out the decimal forms. For example:*"I know that $\frac{10}{40}$ is $\frac14$ which can also be written as 25% or $\frac{25}{100}$ so it's $0.25$."**"If $\frac4{40}$ is $0.1$, then $\frac2{40}$ must be $0.05$ because it's half as big."*

Now introduce the main problem. Write up or display these eight fractions:

$$\frac23 \qquad \frac45 \qquad \frac{17}{50} \qquad \frac3{16} $$ $$\frac7{12} \qquad \frac58 \qquad \frac{11}{14} \qquad \frac8{15}$$

"Which ones do you think can be written as a terminating decimal, and which ones do you think have to be written as a recurring decimal?" (If students have not yet met the idea of terminating and recurring, clarify the meanings.)

Give students a bit of time to make their predictions, and then invite them to work out the decimal equivalents and see if they are right. They might do this by using equivalent fractions, a written division calculation, or using a calculator. Once everyone has worked out the decimal equivalents, take time to discuss whether students' predictions were correct and whether there were any surprises.

"In a while, I am going to give you some fractions. Your challenge is to devise a method for working out straight away whether a fraction is equivalent to a terminating or recurring decimal."

Give students some time to try some examples of their own to test any conjectures that they make. You could collect examples on the board in two columns: terminating and recurring.

In the last few minutes of the lesson draw together the insights and methods that have emerged, and test students' methods with some carefully chosen examples.

### Key questions

Think about the denominators of fractions that you know will terminate. What do they have in common?

Why is the prime factorisation of the denominator important?

### Possible support

When you collect examples on the board in two columns (terminating and recurring) consider writing the fractions in their lowest terms and then writing the denominators as a product of their prime factors.

### Possible extension

Students could go on to explore recurring decimals in Tiny Nines and Repetitiously.

For a challenging extension, some students may wish to consider the idea of terminating and recurring representations in other number bases.