How to Convert Recurring Decimals to Fractions
This page explains how to convert recurring decimals into exact fractions for GCSE and IGCSE Mathematics. You will learn how to identify the repeating part, set up equations correctly, and use subtraction to eliminate the recurring digits. You will also learn to predict, without dividing, whether a fraction terminates or recurs. Worked examples are written in clear exam style to help secure full method marks, followed by five rooms of free, auto-marked practice questions.
\(0.\dot{3}\) means the 3 repeats forever. \(0.\dot{8}\dot{1}\) means the block 81 repeats, and \(0.\dot{4}1\dot{2}\) means the block 412 repeats. Some books write the same thing with bars, like \(0.\bar{3}\).
Which fractions give terminating decimals?
You can predict whether a fraction terminates without doing any division. Our decimal system is built on 10, and \(10 = 2 \times 5\), so:
- Write the fraction in its simplest form. This step matters: \(\dfrac{15}{24}\) looks like it should recur, but it simplifies to \(\dfrac{5}{8}\).
- Look at the denominator's prime factors. Only 2s and 5s (like 8, 10, 16, 20, 25, 40): the decimal terminates. Any other prime factor (3, 7, 11, ...): the decimal recurs.
For example \(\dfrac{9}{40}\) terminates because \(40 = 2^3 \times 5\), while \(\dfrac{7}{60}\) recurs because \(60 = 2^2 \times 3 \times 5\) keeps a factor of 3 no matter what. Room 1 of the practice below drills exactly this decision.
How to convert a recurring decimal to a fraction
- Identify the repeating block. Count how many digits repeat. One digit repeating means \(\times 10\); two digits repeating means \(\times 100\); three digits means \(\times 1000\), and so on.
- Set up an equation. Let \(x\) equal the recurring decimal.
- Multiply both sides by the matching power of 10 so the repeating tail lines up to the right of the decimal point.
- Subtract the original equation from the multiplied one. The repeating tails cancel, leaving a clean equation in \(x\) with no decimals.
- Solve for \(x\) and simplify the fraction to lowest terms (using HCF if needed).
For mixed decimals (a non-repeating part followed by a repeating part), you multiply twice: once to move past the non-repeating digits, then again to line up one full repeat block.
Worked examples
💡 Example 1: a single repeating digit
Step 1. Let \(x\) equal the decimal.
\[ x = 0.\dot{6} \]Step 2. Multiply by 10 (one digit repeats).
\[ 10x = 6.\dot{6} \]Step 3. Subtract \(x\) from \(10x\) to cancel the recurring tail.
\[ \begin{array}{r@{\;}l} 10x & = 6.666\ldots \\ -\,x & = 0.666\ldots \\ \hline 9x & = 6 \end{array} \]Step 4. Solve and simplify.
\[ \begin{aligned} x &= \frac{6}{9} \\ x &= \frac{2}{3} \end{aligned} \]What's happening?
- Naming the decimal \(x\) gives us something to manipulate algebraically.
- Multiplying by 10 shifts the decimal one place. Both \(10x\) and \(x\) now end in the same \(.666\ldots\) tail.
- Subtracting cancels that infinite tail completely, leaving a nice whole number on the right.
- Cancel \(\frac{6}{9}\) by dividing top and bottom by 3 to get \(\frac{2}{3}\).
💡 Example 2: a two-digit repeating block
Step 1. Let \(x\) equal the decimal.
\[ x = 0.\dot{8}\dot{1} \]Step 2. Multiply by 100 (two digits repeat).
\[ 100x = 81.\dot{8}\dot{1} \]Step 3. Subtract \(x\) from \(100x\).
\[ \begin{array}{r@{\;}l} 100x & = 81.8181\ldots \\ -\,x & = \phantom{00}0.8181\ldots \\ \hline 99x & = 81 \end{array} \]Step 4. Solve and simplify (HCF of 81 and 99 is 9).
\[ \begin{aligned} x &= \frac{81}{99} \\ x &= \frac{9}{11} \end{aligned} \]What's happening?
- Two digits repeat, so we multiply by 100 instead of 10. This shifts the decimal point past both repeating digits in one go.
- The repeating tails of \(100x\) and \(x\) match exactly, so subtraction wipes them out.
- The denominator is always made of 9s: one repeating digit gives 9, two gives 99, three gives 999.
- To simplify, find the HCF of numerator and denominator and divide both by it.
💡 Example 3: a non-repeating start then a repeating block
Step 1. Let \(x\) equal the decimal.
\[ x = 0.12\dot{4}\dot{5} \]Step 2. Multiply by 100 to move past the non-repeating "12".
\[ 100x = 12.\dot{4}\dot{5} \]Step 3. Multiply again by 100 (since two digits repeat) to line up one full repeat block.
\[ 10000x = 1245.\dot{4}\dot{5} \]Step 4. Subtract the smaller equation from the larger so tails cancel.
\[ \begin{array}{r@{\;}l} 10000x & = 1245.4545\ldots \\ -\,100x & = \phantom{0}12.4545\ldots \\ \hline 9900x & = 1233 \end{array} \]Step 5. Solve and simplify (HCF of 1233 and 9900 is 9).
\[ \begin{aligned} x &= \frac{1233}{9900} \\ x &= \frac{137}{1100} \end{aligned} \]What's happening?
- The decimal has two parts: a non-repeating start (12) and a repeating block (45). The trick is two multiplications instead of one.
- The first \(\times 100\) moves past the non-repeating part so the decimal looks like \(12.\dot{4}\dot{5}\).
- The second \(\times 100\) (i.e. \(\times 10000\) overall) lines up one whole repeat block before subtracting.
- Subtracting the two equations cancels the repeating tail, leaving a clean integer equation.
- Always finish by simplifying. Here \(1233 \div 9 = 137\) and \(9900 \div 9 = 1100\).
🔑 Key points
- Let \(x\) equal the recurring decimal, then multiply by a power of 10 to shift the repeating block.
- Match the multiplier to the repeat length: 1 digit repeating gives \(\times 10\); 2 digits gives \(\times 100\); 3 digits gives \(\times 1000\).
- Subtraction is the step that cancels the infinite tail. Without it, you can't escape the recurring decimal.
- For mixed decimals, multiply twice: once to skip the non-repeating start, then again to line up one full repeat block.
- Always simplify the final fraction using the HCF of numerator and denominator.
- A fraction in simplest form terminates exactly when its denominator has no prime factor other than 2 and 5.
- The method even shows that \(0.\dot{9} = 1\) exactly: if \(x = 0.\dot{9}\) then \(9x = 9\), so \(x = 1\). A tail of recurring 9s is just another way of writing the number above it.
- One equals per line, vertically aligned, earns method marks.
⚠️ Common pitfalls
- Using the wrong power of 10 (e.g. multiplying by 10 when two digits repeat). Count the digits in the repeating block first.
- Skipping the subtraction step. Just multiplying does not remove the recurring tail; you need to subtract.
- Not multiplying both sides. Whatever you do to one side, do to the other.
- Stopping at the unsimplified fraction. \(\frac{6}{9}\) is correct working but the final answer must be \(\frac{2}{3}\).
- For mixed decimals, multiplying only once. You need both multiplications: skip the non-repeat, then line up the repeat.