## What you need to know

A recurring decimal is a decimal number which has a pattern than repeats over and over after the decimal place. Every recurring decimal can also be written as a fraction. In this topic, we’ll look at how to go from a recurring decimal to and fraction and vice versa.

Two examples of recurring decimals, one quite common and one less common, are

\dfrac{1}{3} = 0.\dot{3} = 0.33333...

\dfrac{6}{11} = 0.\dot{5}\dot{4} = 0.54545454...

Notice the dot on top of some of the digits; this tells us what is repeated. The first dot denotes the start of the repeated section, and the second dot denotes the end of it.

**Convert Fractions to Recurring Decimals**

To convert a fraction to a recurring decimal we must treat the fraction like it is a division and use some method of division to divide the numerator by the denominator. Here we will use short division, also known as the “bus stop method” (I highly recommend this method of division for all purposes). In this example, we’ll see how it soon becomes obvious that the result of your division is a recurring decimal.

**Example: **Write \dfrac{7}{33} as a decimal.

So, we’re going to try dividing 7 by 33. When setting up the bus stop method, you should put in a whole lot of zeros after the decimal place – chances are you’ll only need a few, but it’s better to put more than you need. So, the set up short division should look like

Doing the short division (which you can brush up on here (multiplying dividing revision), we get

Since the last remainder was 7, we looked for how many times 33 goes into 70, but we already did that once. The answer is twice, and then the remainder is 4, which means we’re now going to be asking the question of how many times 33 goes into 40, but we’ve already done that, too. Carrying this on, the pattern becomes obvious.

Therefore, we must have that \frac{7}{33} = 0.\dot{2}\dot{1}. Practice a few of these and you’ll quickly get used to spotting when they start repeating.

**Convert Recurring Decimals to Fractions**

This is really what this topic is about. Let’s dive into an example to see how it goes.

**Example: **Write 0.\dot{1}\dot{4} as a fraction.

Firstly, set x = 0.\dot{1}\dot{4}, the thing we want to convert to a fraction. Then,

100x = 14.\dot{1}\dot{4}.

Now that we have two numbers, x and 100x, with the same digits after the decimal point, if we subtract one from another, the numbers after the decimal point will cancel.

100x - x = 99x = 14.\dot{1}\dot{4} - 0.\dot{1}\dot{4} = 14

Removing the working out steps from this line, we have

99x = 14

Then, if we divide both sides of this by 99, we get

x = \dfrac{14}{99}.

Okay, so how does this method work? Firstly, always assign the thing your converting to be x. Once you’ve done this, the aim is to end up with __two numbers__ (both will be some multiple of ten times by x) that have __exactly the same recurring digits after the decimal point.__ This really is key, because then when you subtract one from the other (in the main step of the process), the digits after the decimal point will cancel and you’ll be left with a nice whole number.

At that point, it’s just a case of solving a straightforward linear equation by doing one division, and you’ve got your answer.

Let’s see another example – remember, the aim is to get two multiples of x both with the same thing after the decimal place.

**Example: **Write 0.8\dot{3} as a fraction.

Firstly, notice how the 8 has no dot above it so isn’t repeating. This will make a difference.

So, let x = 0.8\dot{3}. This time, if we multiply this by 10, 100, 1,000 etc, we’re not going to end up with something that has the same digits after the decimal point as x.

However, if instead we take 10x = 8.\dot{3} and 100x = 83.\dot{3}, then we have two multiplies of x that do have the same digits after the decimal point.

So, subtracting one from the other, we get

100x - 10x = 90x = 83.\dot{3} - 8.\dot{3} = 83 - 8 = 75

Removing the working out steps from this line, we have

90x = 75

Dividing both sides by 90, we get that

x = \dfrac{75}{90} = \dfrac{5}{6}.

This time we were unable to subtract x from anything to get the desired outcome, so we had to be clever and multiply x by 10 *and *100 before doing the subtraction. You will need to do something like this anytime there’s a decimal digit in your number that isn’t involved in the recurring part.

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### Example Questions

1) Write \dfrac{10}{11} as a recurring decimal.

Treating this fraction like a division, we will use short division (or otherwise) to find the result of dividing 10 by 11. The result of the short division should look like

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2) Write 0.\dot{3}9\dot{0} as a fraction.

Set x = 0.\dot{3}9\dot{0}. Then, we get

1,000x = 390.\dot{3}9\dot{0}

Both of x and 1,000x have the same thing after the decimal place, so they are suitable to subtract from one another. Doing so, we get

1,000x - x = 999x = 390.\dot{3}9\dot{0} - 0.\dot{3}9\dot{0} = 390

Removing the working-out steps, we have

999x = 390

Then finally, divide both sides by 999 to get

x = \dfrac{390}{999}.

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3) Write 1.5\dot{4} as a fraction. (HINT: don’t be put off by the fact that it’s a number bigger than 1 – carry out the method as usual and you should be fine)

Set x = 1.5\dot{4}. Then, we get

10x = 15.\dot{4}

This does not have the same thing after the decimal place as x so we cannot subtract yet. Instead, consider

100x = 154.\dot{4}

Then, of 10x and 100x have the same thing after the decimal place, so they are suitable to subtract from one another. Doing so, we get

100x - 10x = 90x = 154.\dot{4} - 15.\dot{4} = 154 - 15 = 139

Removing the working-out steps, we have

90x = 139

Then finally, divide both sides by 90 to get

x = \dfrac{139}{90}.

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### Worksheets and Exam Questions

### Videos

#### Fractions and Recurring Decimals Q1

GCSE MATHS#### Fractions and Recurring Decimals Q2

GCSE MATHS#### Fractions and Recurring Decimals Q3

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