Number and ratio · GCSE Maths
Recurring decimals to fractions
GCSE Maths recurring decimals to fractions: set x equal to the decimal, multiply to shift the repeating block, subtract, and simplify, including one-digit and two-digit repeats.
Let x be the recurring decimal. Multiply to move one full repeat past the point, subtract, and the repeating tail cancels. Then divide by 9, 99 or 999 according to the length of the block.
The important bits
What you need to know
- 1
A recurring decimal has a block of digits that repeats forever. 0.3̇ = 0.333… and 0.1̇6̇ = 0.161616…. A dot over the first and last repeating digits is the UK notation.
- 2
Every recurring decimal is a rational number: it can be written as a fraction in lowest terms. Terminating decimals are the special case whose repeat is 0.
- 3
Method: let x = 0.777… . Then 10x = 7.777… . Subtract: 9x = 7, so x = 7/9. One repeating digit produced a 9 in the denominator.
- 4
Two repeating digits produce 99: x = 0.151515… , 100x = 15.1515… , subtract, 99x = 15, x = 15/99 = 5/33. Three repeating digits produce 999.
- 5
If some digits after the point do not repeat, multiply first to park the non-repeating part to the left of the point, then multiply again to shift one full repeat, then subtract those two equations.
- 6
Example: 0.16̇ (only 6 repeats). Let x = 0.1666… . 10x = 1.666… and 100x = 16.666… . Subtract: 90x = 15, so x = 15/90 = 1/6.
- 7
Always simplify the fraction. 0.3̇ is 1/3, not just 3/9. Check by dividing on the calculator: 1 ÷ 6 should return 0.1666… .
- 8
0.9̇ = 1, which surprises students and is a useful check of the method: x = 0.999… , 10x = 9.999… , 9x = 9, x = 1. The algebra is allowed to finish at a terminating value.
Quotations worth analysing
Short evidence. Real method.
“Let x = 0.1̇6̇”
The method mark is on defining x and writing a multiplied version. A jump to 16/99 without x rarely scores if the arithmetic then slips.
“Multiply by 10, 100 or 1000 to shift the repeating block”
The multiplier is 10^k where k is the length of the repeating block (after any non-repeating digits have been parked). Subtract, then simplify.
“Give your answer as a fraction in its simplest form”
15/99 is not finished. Divide numerator and denominator by 3 to get 5/33. The accuracy mark sits on the simplified fraction.
Go deeper
One repeating digit is a nine in the denominator
The algebra is a cancellation of infinite tails. If x = 0.777…, then 10x = 7.777… . The decimal parts are identical, so 10x − x = 7.777… − 0.777… = 7. Hence 9x = 7 and x = 7/9. The same pattern gives 0.1̇ = 1/9, 0.2̇ = 2/9, …, 0.8̇ = 8/9, and 0.9̇ = 9/9 = 1. You do not need a new trick for each digit; you need the subtract-to-cancel move. On a non-calculator paper write all four lines: x = …, 10x = …, subtract, simplify. Examiners follow those lines. A memorised “one digit means over 9” is fine as a check, but it fails as soon as a non-repeating digit appears, which is why the algebra is the method you actually learn.
Go deeper
Two-digit repeats and mixed repeats
For 0.1̇8̇, the block “18” has two digits, so multiply by 100: x = 0.181818… , 100x = 18.1818… , 99x = 18, x = 18/99 = 2/11. For 0.1̇6̇ the block is “16”, so 99x = 16, x = 16/99. Do not confuse 0.1̇6̇ (16 repeating) with 0.16̇ (only 6 repeating). The dots tell you the block. Mixed example: 0.53̇, meaning 0.5333… . Let x = 0.5333… . Then 10x = 5.333… and 100x = 53.333… . Subtract the first from the second: 90x = 48, x = 48/90 = 8/15. You subtracted 10x from 100x because those two lines have matching repeating tails. Choosing 10x and x would not cancel, because 0.5333… and 5.333… do not share the same tail after the point in the same way.
Go deeper
Why subtraction is legal, and how to check
You are subtracting two genuine numbers, so the infinite matching decimals cancel just as 8.333 − 0.333 leaves 8. That is not a fudge. After simplifying, check by short division or a calculator: 8 ÷ 15 = 0.5333…, and 2 ÷ 11 = 0.1818… . If the calculator shows a terminating decimal, you either simplified wrongly or misread the repeating block. Keep the fraction improper only if it is greater than 1; most GCSE recurring questions sit between 0 and 1 and want a proper fraction in lowest terms. If a question starts 0.2̇7̇ × 0.1̇, convert each to a fraction first, then multiply the fractions — do not try to multiply the recurring decimals digit by digit.
See the idea in action
Write 0.1̇6̇ as a fraction in its simplest form. (Both 1 and 6 repeat, so 0.161616… .) Step 1: Let x = 0.161616… Step 2: The repeating block has 2 digits, so multiply by 100: 100x = 16.161616… Step 3: Subtract: 100x − x = 16.161616… − 0.161616… Step 4: 99x = 16, so x = 16/99. Step 5: 16 and 99 have no common factor greater than 1, so 16/99 is in simplest form. Check: 16 ÷ 99 = 0.161616… on a calculator. Contrast: if the decimal had been 0.16̇ (only 6 repeating), the fraction would be 1/6, not 16/99.
Exam technique
Turn knowledge into marks
Count the repeating digits before you choose 10, 100 or 1000. Write let x = … as line 1. Simplify at the end; 16/99 left unsimplified is a dropped accuracy mark when it can cancel.
Common mistakes
Do not give these marks away
- 01
Treating 0.1̇6̇ as 0.16̇, so only one digit is shifted, and the fraction becomes 1/6 instead of 16/99.
- 02
Multiplying by 10 when two digits repeat, so the tails do not match and subtraction does not cancel.
- 03
Leaving 18/99 unsimplified instead of 2/11, against “simplest form” on the answer line.
0.2̇ as a fraction in its simplest form is
A2/9
B2/99
C1/5
D2/10
Show the answer
2/9. Let x = 0.222… . Then 10x = 2.222… , so 9x = 2 and x = 2/9. 2/99 would be a two-digit repeat such as 0.02̇. 1/5 is 0.2 terminating. 2/10 is 0.2 terminating, not 0.222… .
Quick questions
If this is the bit you searched
How do you convert a recurring decimal to a fraction?
Let x equal the decimal, multiply by 10, 100 or 1000 to shift one full repeating block, subtract to cancel the tail, then simplify the fraction.
When do I multiply by 100 instead of 10?
When two digits repeat, as in 0.1̇8̇. Three repeating digits need × 1000. The multiplier is 10 to the power of the length of the repeating block.
How do I handle a decimal where only some digits repeat?
Write two multiplied versions so that their repeating tails match, subtract those two equations, then simplify. For 0.16̇, use 10x and 100x.
Is 0.9̇ really equal to 1?
Yes. Let x = 0.999… , 10x = 9.999… , 9x = 9, x = 1. It is the same method, and it is the reason terminating decimals can be written with a recurring 9.