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# Terminating or Not

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### Matching Fractions, Decimals and Percentages

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Age 11 to 14

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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.

Say that we have a fraction such as $\frac58,$ which can't be simplified as $5$ and $8$ are something called coprime, meaning their only common factor is one. However, all of these coprime fractions can be expressed as a multiple of another fraction; in this case $\frac58 = \frac18\times5.$ Because this can be shown in terms of one of the simpler fractions, we can
still apply this method.

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.

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.

Can you match pairs of fractions, decimals and percentages, and beat your previous scores?