Converting recurring decimals to fractions GCSE questions use one reliable algebraic method: let the decimal equal x, multiply by a power of 10 that shifts the repeating block, then subtract the original equation so the repeating part cancels out. The result is always a fraction in the form $\frac{\text{integer}}{\text{integer}}$.
What is a recurring decimal?
A recurring decimal has one or more digits that repeat forever after the decimal point. It is written with a dot above the repeating digit, or dots above the first and last digits of a repeating block. For example, $0.\dot{3}$ means $0.333333...$, and $0.\dot{1}\dot{6}$ means $0.161616...$
Every recurring decimal is a rational number, which means it can always be written as an exact fraction — this is what distinguishes it from an irrational number like $\pi$, whose digits never settle into a repeating pattern. GCSE exam boards test the algebraic method for finding that exact fraction, rather than accepting a decimal approximation.
What is the algebraic method for a single repeating digit?
- Let $x$ equal the recurring decimal.
- Multiply both sides by 10 (for one repeating digit) so the decimal parts line up.
- Subtract the original equation from the new one — the repeating digits cancel.
- Solve the resulting equation for $x$ and simplify the fraction.
Worked example: Write $0.\dot{3}$ as a fraction.
Let $x = 0.333333...$
Multiply by 10: $10x = 3.333333...$
Subtract the first equation from the second:
$$10x - x = 3.333333... - 0.333333...$$ $$9x = 3$$ $$x = \frac{3}{9} = \frac{1}{3}$$
So $0.\dot{3} = \frac{1}{3}$. Subtracting cancels every repeating digit after the decimal point, leaving a whole number on the right-hand side.
How does the method change for two repeating digits?
When two digits repeat as a block, multiply by 100 instead of 10, so that the repeating block shifts by exactly two places and lines up for subtraction.
Worked example: Write $0.\dot{1}\dot{6}$ as a fraction.
Let $x = 0.161616...$
Multiply by 100: $100x = 16.161616...$
Subtract:
$$100x - x = 16.161616... - 0.161616...$$ $$99x = 16$$ $$x = \frac{16}{99}$$
The fraction $\frac{16}{99}$ does not simplify further, since 16 and 99 share no common factors. The general rule is: multiply by $10^n$, where $n$ is the number of digits in the repeating block.
How do you handle a recurring decimal with a non-repeating part first?
Some numbers, like $0.41\dot{6}$, have digits before the repeat begins. Here you need two multiplication steps to isolate the repeating block correctly.
Worked example: Write $0.41\dot{6}$ as a fraction, where only the 6 repeats.
- Let $x = 0.41666...$
- Multiply by 100 to move the decimal point past the non-repeating digits: $100x = 41.666...$
- Multiply by 1000 to move it past one more repeating digit: $1000x = 416.666...$
- Subtract: $1000x - 100x = 416.666... - 41.666...$, giving $900x = 375$.
- Simplify: $x = \frac{375}{900} = \frac{5}{12}$.
Checking the answer by dividing 5 by 12 on a calculator gives $0.41\overline{6}$, confirming the fraction is correct.
What is the fastest method for common recurring decimal patterns?
| Decimal | Fraction | Number of repeating digits |
|---|---|---|
| $0.\dot{1}$ | $\frac{1}{9}$ | 1 |
| $0.\dot{2}$ | $\frac{2}{9}$ | 1 |
| $0.\dot{3}$ | $\frac{1}{3}$ | 1 |
| $0.\dot{1}\dot{2}$ | $\frac{12}{99} = \frac{4}{33}$ | 2 |
| $0.\dot{1}\dot{4}\dot{2}\dot{8}\dot{5}\dot{7}$ | $\frac{1}{7}$ | 6 |
A useful shortcut for single-digit repeats: the recurring digit sits directly over a 9. For two-digit blocks, it sits over 99, and for three-digit blocks, over 999. Recognising this pattern lets you check an algebraic answer quickly, though exam mark schemes still expect the full working shown.
How do you check your answer is correct?
Divide the numerator of your final fraction by the denominator using long division or a calculator, and confirm the decimal matches the original recurring decimal you started with. If the digits do not repeat in the same pattern, an arithmetic slip has occurred somewhere in the subtraction step, and it is worth re-checking that the correct power of 10 was used for the number of repeating digits.
Exam questions sometimes ask you to prove that a given recurring decimal equals a stated fraction, rather than find the fraction from scratch. The method is identical: set up the algebra, subtract, and show the working leads to the given fraction, simplified fully.
Frequently asked questions
Why do you multiply by 10, 100, or 1000 in this method?
Multiplying by a power of 10 shifts the decimal point exactly far enough to align the repeating digits in both equations. Subtracting one equation from the other then cancels the infinitely repeating part, leaving a finite equation that can be solved normally for x.
What is 0.3 recurring as a fraction?
$0.\dot{3}$, meaning 0.333333... forever, equals $\frac{1}{3}$. Using the algebraic method: let $x = 0.\dot{3}$, so $10x = 3.\dot{3}$, and subtracting gives $9x = 3$, so $x = \frac{3}{9} = \frac{1}{3}$.
Can every recurring decimal be written as a fraction?
Yes. Every recurring decimal represents a rational number, and rational numbers are defined as numbers that can be expressed as a fraction of two integers. This is different from a non-repeating, non-terminating decimal such as $\sqrt{2}$, which is irrational and cannot be written as an exact fraction.
What if the whole decimal repeats, with no leading non-repeating digits?
Use a single multiplication by $10^n$, where $n$ is the number of repeating digits, then subtract the original equation once. This is the simplest case and needs no extra steps to strip out non-repeating digits first, unlike a decimal such as $0.41\dot{6}$.
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