SPEC N10 • NUMBER
Decimals & Recurring Decimals
Converting between fractions and decimals in both directions. The denominator test that tells you whether a fraction terminates. And the Higher-tier algebraic method that turns any recurring decimal into a fraction.
Key Definitions
Every decimal at GCSE is either terminating or recurring — there is no third option for a rational number.
Terminating Decimal
A decimal with a finite number of decimal places. Can be written exactly.
Examples: 0.75 = 3/4, 0.625 = 5/8, 0.14 = 7/50
Recurring Decimal
A decimal where one or more digits repeat indefinitely. Notation uses a dot over the first and last digit of the repeating block.
Examples: 0.3̇ = 0.333… (only 3 repeats)
0.1̇2̇ = 0.1212… (the block "12" repeats)
0.6̇23̇ = 0.623623… (three-digit block "623" repeats)
Equivalent Fraction
A fraction representing the same value in different form. 3/4 = 6/8 = 9/12. Simplify by dividing top and bottom by their HCF.
Terminating Decimal → Fraction
Place the digits over a power of 10 matching the number of decimal places, then simplify.
Method
Count the decimal places.
Write the digits as numerator; write 10, 100, 1000, … as denominator.
Simplify by dividing by the HCF.
Worked examples
0.625 → 3 decimal places → 625/1000
HCF(625, 1000) = 125 → 625/1000 = 5/8
0.24 → 2 decimal places → 24/100 = 6/25
0.35 → 2 decimal places → 35/100 = 7/20
Fraction → Decimal: Long Division
Divide numerator by denominator. On Paper 1 you'll do this by hand.
The method in three lines
At each step: how many times does the denominator go into the current number? That digit goes on top. The remainder becomes the next "tens" — bring down a zero.
Useful fraction-to-decimal table
1/4 = 0.25 | 1/5 = 0.2 | 1/8 = 0.125 | 1/10 = 0.1
1/3 = 0.3̇ | 1/6 = 0.16̇ | 1/9 = 0.1̇ | 1/11 = 0.0̇9̇
The Denominator Test
How do you know whether a fraction will terminate or recur, without doing the long division? Look at the prime factors of the denominator (after the fraction is simplified).
Denominator has only 2s and 5s as prime factors
→ Terminating decimal ✔
Denominator has any other prime factor (3, 7, 11, …)
→ Recurring decimal ↻
Why it works
Any terminating decimal can be written over 10ⁿ, and 10 = 2 × 5. So a fraction terminates only if it can be rewritten with a pure-10ⁿ denominator — which requires the denominator's primes to already be 2s and 5s.
Quick examples
7/50 → 50 = 2 × 5² → terminates = 0.14
11/24 → 24 = 2³ × 3 → recurs (because of the 3)
1/140 → 140 = 2² × 5 × 7 → recurs (because of the 7)
3/8 → 8 = 2³ → terminates = 0.375
Showing a Fraction Terminates (or Doesn't)
AQA often asks you to show that a fraction is terminating or recurring. The expected method is prime factorisation of the denominator.
A
Show that 7/50 is a terminating decimal 1 MARK
Method 1 — express with power-of-10 denominator:
7/50 = 14/100 = 0.14 ✔ terminates
Method 2 — prime factorise denominator:
50 = 2 × 5² → only 2s and 5s → terminates
B
Show that 11/24 is not terminating 2 MARKS
24 = 2³ × 3
Denominator contains a factor other than 2 or 5 (there is a 3), so 11/24 cannot be written with a power-of-10 denominator. Therefore 11/24 is recurring.
Recurring → Fraction: The Algebraic Method Higher
Most often a 2- or 3-mark "Prove that…" question. You must show the algebra — calculators that convert are not enough.
Set up
Let x equal the recurring decimal. Write out several digits to make the pattern visible.
Multiply
Multiply both sides by 10ⁿ where n is the length of the recurring block.
Subtract
Subtract x from 10ⁿx. The recurring tails cancel, leaving integers on both sides.
Solve & simplify
Divide to isolate x. Simplify the fraction by dividing top and bottom by their HCF.
How to choose the multiplier
1 recurring digit → multiply by 10. 2 recurring digits → by 100. 3 recurring digits → by 1000. The denominator of the final fraction will always be 9, 99, 999, … (before simplifying).
Worked Example — 0.2̇7̇ as a Fraction Higher
01Write 0.2̇7̇ as a fraction in simplest form 3 MARKS
- 1Set up: Let x = 0.272727…
- 2Multiply: The recurring block "27" has length 2, so multiply by 100.100x = 27.272727…
- 3Subtract:100x = 27.272727…−x = 0.272727…99x = 27
- 4Solve & simplify:x = 27/99HCF(27, 99) = 9, so divide top and bottom by 9:x = 3/11
Answer: 0.2̇7̇ = 3/11
MCE Worked Example 3 MARKSHigher02
Show that 0.4̇5̇ = 5/11 . You must show your working.
•
MMethod. Let x equal the decimal. Multiply by 100 (two-digit recurring block). Subtract to eliminate the recurrence. Solve and simplify.
•
CCalculate.
Let x = 0.454545…
100x = 45.454545…
100x − x = 45.454545… − 0.454545… = 45
99x = 45
x = 45/99
Simplify by dividing top and bottom by HCF(45, 99) = 9:
x = 5/11
•
EEvaluate.Therefore 0.4̇5̇ = 5/11 as required. ✓
Mark scheme: M1 for setting up the equation and multiplying by the correct power of 10; M1 for the subtraction step; A1 for the final simplified fraction. Missing the subtraction line means no method marks — it is the critical step.
Mixed Recurring Decimals Higher
Sometimes only part of the decimal recurs. The method is essentially the same — but the final fraction won't have a clean 9s denominator.
Example: 0.47̇3̇ — the "4" doesn't recur, only "73" does
Let n = 0.473737…
100n = 47.373737…
Subtract: 100n − n = 47.373737… − 0.473737…
99n = 46.9n = 46.9 / 99
To clear the decimal in the numerator, multiply top and bottom by 10:
n = 469/990
Key technique
If subtracting leaves a decimal in the numerator (like 46.9 here), multiply top and bottom by a power of 10 to make it a whole number. Then check whether you can simplify further.
Weak vs Strong: 0.3̇
One of the cleanest examples of the difference between treating a recurring decimal like a terminating one and doing it properly.
Write 0.3̇ as a fraction in simplest form.
Weak student
"0.3 as a fraction is 3/10. Dot on top doesn't matter."
0.3̇ = 3/10 ✘
Treats the recurring dot as decoration. 3/10 = 0.3 exactly, not 0.333…
Strong student
"Set up equation, multiply by 10, subtract."
x = 0.333…
10x = 3.333…
9x = 3 → x = 1/3 ✔
Quick sanity check
Convert your answer back: does 3/10 = 0.3 exactly? Yes — but the question's 0.3̇ is larger (it keeps adding). So 3/10 cannot be the answer.
Examiner Alert: Subtraction is the Critical Step
What examiners report
Students write x = 0.27…, then 100x = 27.27…, and then try to divide the two equations or just solve 100x = 27.27… directly. Neither works — because the recurring tail is still there.
What to write instead
The critical step is subtracting x from 100x. This is what cancels the infinite decimals — leaving a clean integer equation you can solve.
100x − x = 99x on the left; 27.27… − 0.27… = 27 on the right.
Mark scheme insight — Jun23 P1 Q23
AQA explicitly credits the subtraction step as M1. "Subtraction signs must be seen or the results correct." Students who skip showing the subtraction — even if they somehow land on the right fraction — can lose the method mark.
Examiner Alert: Use the Right Multiplier
What examiners see
On 0.1̇2̇ (two-digit recurring block), students multiply by 10 only. Then after subtraction the recurring parts don't align — they still have a decimal tail.
Match the multiplier to the block length
1 digit recurs (e.g. 0.3̇) → multiply by 10
2 digits recur (e.g. 0.1̇2̇) → multiply by 100
3 digits recur (e.g. 0.3̇51̇) → multiply by 1000
The multiplier shifts the decimal by the length of the block, so the recurring tails line up and cancel perfectly on subtraction.
Common Errors
Students often writeCorrectWhy it matters
0.3̇ = 3/100.3̇ = 1/3Recurring ≠ terminating — the dot changes the value
Multiply 0.1̇2̇ by 10Multiply by 100 (2-digit block)Multiplier must match block length
Skip subtraction stepAlways show 100x − x = …M1 depends on subtraction being visible
Leave answer as 27/993/11 (HCF = 9)"Simplest form" required
Claim 1/140 terminates140 = 2²×5×7 → recurs (has 7)Denominator test: any prime ≠ 2, 5 → recurs
Practice Question 1
Show that 1/140 cannot be written as a terminating decimal.2 MARKS
1
Prime factorise the denominator:
140 = 2² × 5 × 7
2
The denominator contains a prime factor other than 2 or 5 (it has a 7).
3
Therefore 1/140 cannot be written with a power-of-10 denominator, so it must be a recurring decimal.
Conclusion: 1/140 recurs (does not terminate). ✓
Mark scheme: M1 for 140 = 2² × 5 × 7 (or equivalent showing the 7 is a factor); A1 for the conclusion that this prevents termination.
Practice Question 2
(a) Show that 2/9 = 0.2̇1 MARK
(b) Hence, or otherwise, write 0.7̇ as a fraction in simplest form.2 MARKS
a
Long division: 2 ÷ 9 = 0.222… (9 into 2 is 0 r2; into 20 is 2 r2; repeats).
With the recurring-dot notation: 2/9 = 0.2̇ ✔
b
Let x = 0.7̇ = 0.777…
•
Multiply by 10: 10x = 7.777…
Subtract: 10x − x = 7.777… − 0.777… = 7
9x = 7
•
x = 7/9 (HCF(7, 9) = 1 — already simplest form)
Answer: 0.7̇ = 7/9
Shortcut from part (a): Since 2/9 = 0.2̇, by pattern 7/9 = 0.7̇. Any single-digit recurring decimal 0.ṅ equals n/9.
Practice Question 3 Higher
Prove that the recurring decimal 0.2̇4̇ has the value 8/33 .2 MARKS
(Modelled on AQA Higher revision guide style — full algebraic method required.)
1
Let n = 0.2̇4̇ = 0.242424…
2
Two-digit recurring block, so multiply by 100:
100n = 24.242424…
3
Subtract:
100n − n = 24.242424… − 0.242424…
99n = 24
4
Divide: n = 24/99
HCF(24, 99) = 3, so simplify:
n = 8/33 ✓
Therefore 0.2̇4̇ = 8/33 as required.
Mark scheme (AQA style): M1 for correct setup and multiplication by 100 (with subtraction shown); A1 for simplified fraction 8/33. No working = 0/2 on a "prove" question.
Summary
Terminating ↔ Recurring
Terminating: finite decimal places. Denominator has only 2s and 5s as prime factors.
Recurring: digits repeat indefinitely. Denominator has any other prime factor.
Notation: dot over first and last digit of the repeating block.
Conversions
Terminating → fraction: digits over 10ⁿ, simplify ( 0.625 = 625/1000 = 5/8 )
Fraction → decimal: long division
Recurring → fraction (Higher): set x, multiply by 10ⁿ (n = block length), subtract, solve, simplify
Key Facts to Remember
0.3̇ = 1/3 | 0.6̇ = 2/3 | 0.1̇ = 1/90.2̇7̇ = 3/11 | 0.4̇5̇ = 5/11 | 0.0̇9̇ = 1/11
The Big Two Alerts
Subtraction is the critical step. Without it you haven't used the algebraic method — M1 not awarded.
Match multiplier to block length. 1 digit → ×10; 2 digits → ×100; 3 digits → ×1000.
Ready for 1.07
Next: Percentages as Operators — the multiplier method, percentage increase and decrease in a single step, and compound percentage change for Paper 2/3 problems.
Next: Key Definitions