What is a recurring decimal in Mathematics?

A recurring decimal, also known as a repeating decimal, is a decimal number in which a digit or a group of digits repeats infinitely. For example, 0.333... and 1.666... are recurring decimals. In the IB MYP Mathematics curriculum, understanding recurring decimals is essential for grasping the concept of rational numbers.

How can I convert a recurring decimal into a fraction?

To convert a recurring decimal into a fraction, you can use algebraic methods. For example, if you have a recurring decimal like 0.666..., set it as x = 0.666... Multiply both sides by 10 to shift the decimal point: 10x = 6.666... Subtract the original equation from this new equation: 10x - x = 6.666... - 0.666..., which simplifies to 9x = 6. Solving for x gives x = 6/9, which simplifies to 2/3. This method is commonly taught in the IB MYP Mathematics curriculum.

Why do some decimals repeat while others do not?

Decimals repeat when they represent rational numbers that cannot be expressed as a finite decimal. This occurs because the division of the numerator by the denominator results in a remainder that eventually repeats. Non-repeating decimals are typically irrational numbers, which cannot be expressed as a simple fraction. Understanding the difference between repeating and non-repeating decimals is a key concept in the IB MYP Mathematics program.

How do I identify the repeating part of a recurring decimal?

To identify the repeating part of a recurring decimal, look for the sequence of digits that repeats. This sequence is often indicated by a bar over the digits or parentheses. For example, in 0.142857142857..., the repeating part is '142857'. Recognizing this pattern is important for solving problems related to recurring decimals in the IB MYP Mathematics curriculum.

What is the significance of recurring decimals in the IB MYP Mathematics curriculum?

Recurring decimals are significant in the IB MYP Mathematics curriculum because they help students understand the nature of rational numbers and their decimal representations. They provide a foundation for more advanced topics such as number theory and algebra. Mastery of recurring decimals is crucial for developing problem-solving skills and mathematical reasoning in the IB MYP framework.

Why is "For $0.\overline{54}$, multiplying by 10 instead of 100." a common mistake in Recurring Decimals?

Why it happens: Not counting the block length. How to avoid it: Multiply by 10^n where n is the LENGTH of the repeating block. 0.54 has a 2-digit block → multiply by 100.

Why is "Leaving $\tfrac{54}{99}$ unsimplified." a common mistake in Recurring Decimals?

Why it happens: Stopping at the first fraction. How to avoid it: Always simplify. 5499 has HCF 9, so simplify to 611.

Why is "Diving into the calculation without naming the variable." a common mistake in Recurring Decimals?

Why it happens: Trying to save time. How to avoid it: Always start with 'Let x = ...'. The criterion-C marker wants to see your structured working.

NumberIB MYP Maths Extended Level

Recurring Decimals

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Recurring Decimals — IB MYP Mathematics (Extended): converting repeating decimals into exact fractions

Decimals that repeat forever are still rational numbers. This MYP Mathematics Extended note covers the dot notation, the conversion technique, and why this proves $0.\overline{9} = 1$ .

What you’ll learn

Mapped to the IB MYP Mathematics subject guide (2026 onwards).

MYP Mathematics A — Recognise recurring decimals.
MYP Mathematics A — Convert recurring decimals to fractions.
MYP Mathematics C — Use dot notation correctly.
MYP Mathematics D — Discuss the link between rationals and decimals.

What is a recurring decimal?

A digit or block that repeats forever.

Some fractions give terminating decimals — $\tfrac{1}{4} = 0.25$ . Others go on forever in a pattern — $\tfrac{1}{3} = 0.333\ldots$

A recurring decimal has at least one digit (or block) that repeats indefinitely. Common examples:

Fraction	Decimal	Notation
$\tfrac{1}{3}$	$0.333\ldots$	$0.\overline{3}$ or $0.\dot{3}$
$\tfrac{2}{3}$	$0.666\ldots$	$0.\overline{6}$
$\tfrac{1}{7}$	$0.142857142857\ldots$	$0.\overline{142857}$
$\tfrac{5}{6}$	$0.8333\ldots$	$0.8\overline{3}$
$\tfrac{1}{11}$	$0.090909\ldots$	$0.\overline{09}$

The vinculum (bar) goes over the digits that repeat. So $0.8\overline{3}$ means $0.8333\ldots$ — the 8 appears once, then the 3 repeats forever.

In some notations, dots are placed above the FIRST and LAST digit of the repeating block: $0.\dot{1}\dot{4}$ means $0.141414\ldots$ .

Fact: A fraction $\tfrac{p}{q}$ (in lowest terms) gives a TERMINATING decimal if and only if the denominator $q$ has no prime factors other than 2 or 5. Otherwise it gives a RECURRING decimal.

Recurring decimal: a digit or block that repeats forever.
Notation: bar over the block, or dots above first/last.
Fraction terminates iff denominator's only prime factors are 2 and 5.

Converting a recurring decimal to a fraction

Multiply, subtract, solve.

This is a classic Extended-level technique. Use the variable $x$ to stand for the decimal, then multiply by a power of 10 to SHIFT the repeating part so it lines up.

Worked example 1. Convert $0.\overline{3}$ to a fraction.

Let $x = 0.\overline{3} = 0.3333\ldots$

Multiply both sides by 10 (the period of the repeat is 1): $10x = 3.3333\ldots$

Subtract the first equation from the second — the repeating decimals CANCEL: $10x - x = 3.3333\ldots - 0.3333\ldots$ $9x = 3$ $x = \frac{3}{9} = \frac{1}{3}.$

Worked example 2. Convert $0.\overline{27}$ to a fraction.

Let $x = 0.27272727\ldots$ . The repeating block has 2 digits, so multiply by $10^2 = 100$ : $100x = 27.272727\ldots$ $100x - x = 27.272727\ldots - 0.272727\ldots$ $99x = 27$ $x = \frac{27}{99} = \frac{3}{11}.$

Worked example 3 (mixed terminating + recurring). Convert $0.1\overline{6}$ to a fraction.

$x = 0.16666\ldots$ . The 1 is not part of the recurring block. Multiply by 10 to push past the 1: $10x = 1.6666\ldots$

Multiply by another factor of 10 to shift the repeating block: $100x = 16.6666\ldots$

Subtract: $100x - 10x = 16.6666\ldots - 1.6666\ldots$ $90x = 15$ $x = \frac{15}{90} = \frac{1}{6}.$

General rule: if the repeating block has length $n$ , multiply by $10^n$ . If there are also some non-repeating digits after the decimal point, use TWO multiplications: one to push past the non-repeating digits, one to shift the repeating block.

Let $x$ = the decimal.
Multiply by $10^n$ where $n$ = length of repeating block.
Subtract original $x$ ; the repeating tail cancels.
Solve for $x$ — a fraction.

$0.\overline{9} = 1$ — really

The most counter-intuitive equality in school maths.

Apply the conversion technique to $0.\overline{9}$ :

Let $x = 0.\overline{9} = 0.9999\ldots$ $10x = 9.9999\ldots$ $10x - x = 9.9999\ldots - 0.9999\ldots$ $9x = 9$ $x = 1.$

So $0.\overline{9} = 1$ , EXACTLY. Not 'very close to'. Equal.

Why is this counter-intuitive? Because we imagine $0.999\ldots$ as a number 'just below' 1. But there is no number between $0.999\ldots$ and 1 — they have ZERO gap, so they must be the same number.

Other ways to see it:

$\tfrac{1}{3} = 0.\overline{3}$ . Multiply both sides by 3: $1 = 0.\overline{9}$ .
Any terminating decimal has an alternative recurring form: $0.5 = 0.4\overline{9}$ , $7 = 6.\overline{9}$ .

This is a great example for criterion D — discussing what NUMBERS really are, and how different notations can describe the same thing.

$0.\overline{9} = 1$ exactly.
Confirmed by the conversion technique.
Also follows from $3 \times \tfrac{1}{3} = 1$ .

How it’s examined

Criterion A: convert a recurring decimal to a fraction. Criterion C: show all subtraction steps clearly.

Sources: IB MYP Mathematics subject guide (IBO, official). Last reviewed 2026-05-25.

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Step-by-step worked examples — Recurring Decimals

Step-by-step solutions to past-paper-style questions on recurring decimals, written exactly the way a tutor would explain them at the board.

1Convert $0.\overline{4}$ to a fraction
Getting started• conversion
Question
Express $0.\overline{4}$ as a fraction in simplest form.
Step-by-step solution
1. Step 1
  Let $x = 0.\overline{4} = 0.4444\ldots$ Repeating block length is 1.
2. Step 2
  Multiply by 10: $10x = 4.4444\ldots$
3. Step 3
  Subtract: $10x - x = 4.4444\ldots - 0.4444\ldots = 4$ . So $9x = 4$ .
4. Step 4
  $x = \tfrac{4}{9}$ . Already in simplest form (HCF $(4,9) = 1$ ).
Answer
$\dfrac{4}{9}$
2Two-digit block
Building confidence• two-digit block
Question
Express $0.\overline{54}$ as a fraction.
Step-by-step solution
1. Step 1
  Let $x = 0.\overline{54} = 0.5454\ldots$ Repeating block length is 2.
2. Step 2
  Multiply by 100: $100x = 54.5454\ldots$
3. Step 3
  Subtract: $100x - x = 54$ . So $99x = 54$ .
4. Step 4
  $x = \tfrac{54}{99} = \tfrac{6}{11}$ (divide top and bottom by 9).
Answer
$\dfrac{6}{11}$
3Mixed terminating + recurring
Stretch• mixed
Question
Express $0.2\overline{3}$ as a fraction.
Step-by-step solution
1. Step 1
  Let $x = 0.2333\ldots$ The 2 is fixed; the 3 repeats.
2. Step 2
  Multiply by 10 to push past the 2: $10x = 2.333\ldots$
3. Step 3
  Multiply by 100 to shift the recurring 3: $100x = 23.333\ldots$
4. Step 4
  Subtract: $100x - 10x = 23.333\ldots - 2.333\ldots = 21$ . So $90x = 21$ .
5. Step 5
  $x = \tfrac{21}{90} = \tfrac{7}{30}$ (divide top and bottom by 3).
Answer
$\dfrac{7}{30}$
4Prove $0.\overline{9} = 1$
Stretch• theory
Question
Use the multiply-and-subtract technique to prove that $0.\overline{9} = 1$ .
Step-by-step solution
1. Step 1
  Let $x = 0.9999\ldots$
2. Step 2
  $10x = 9.9999\ldots$
3. Step 3
  $10x - x = 9.9999\ldots - 0.9999\ldots = 9$ . So $9x = 9$ .
4. Step 4
  $x = 1$ . Therefore $0.\overline{9} = 1$ exactly.
Answer
$0.\overline{9} = 1$ .

Key Definitions and Keywords — Recurring Decimals

Definitions to memorise and the exact keywords mark schemes credit for recurring decimals answers — sharpened from recent examiner reports for the 2026 IB MYP Mathematics (Extended) sitting.

Recurring decimal
Examiner keyword
A decimal in which a digit or sequence of digits repeats forever.
Rational number
Examiner keyword
A number expressible as $\tfrac{p}{q}$ with $p, q \in \mathbb{Z}$ , $q \ne 0$ . ALL recurring decimals are rational.
Terminating decimal
A decimal with finitely many digits, e.g. $0.625$ .

Common Mistakes and Misconceptions — Recurring Decimals

The traps other students keep falling into on recurring decimals questions — taken from recent IB MYP Mathematics (Extended) examiner reports and mark schemes — and how to avoid them.

✕For $0.\overline{54}$ , multiplying by 10 instead of 100.
Why it happens
Not counting the block length.
How to avoid it
Multiply by $10^n$ where $n$ is the LENGTH of the repeating block. $0.\overline{54}$ has a 2-digit block → multiply by 100.
✕Leaving $\tfrac{54}{99}$ unsimplified.
Why it happens
Stopping at the first fraction.
How to avoid it
Always simplify. $\tfrac{54}{99}$ has HCF 9, so simplify to $\tfrac{6}{11}$ .
✕Diving into the calculation without naming the variable.
Why it happens
Trying to save time.
How to avoid it
Always start with 'Let $x =$ ...'. The criterion-C marker wants to see your structured working.

Practice questions

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Past paper style quiz

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Recurring Decimals — frequently asked questions

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Recurring Decimals

Short Study Notes in the form of Slides

Short Study Notes — Recurring Decimals

Detailed Notes

What you’ll learn

How it’s examined

1Convert 0.4‾0.\overline{4}0.4 to a fraction

2Two-digit block

3Mixed terminating + recurring

4Prove 0.9‾=10.\overline{9} = 10.9=1

Recurring decimal

Rational number

Terminating decimal

✕For 0.54‾0.\overline{54}0.54, multiplying by 10 instead of 100.

✕Leaving 5499\tfrac{54}{99}9954​ unsimplified.

✕Diving into the calculation without naming the variable.

Practice questions

Past paper style quiz

Assess your understanding

Recurring Decimals — frequently asked questions

1Convert $0.\overline{4}$ to a fraction

4Prove $0.\overline{9} = 1$

✕For $0.\overline{54}$ , multiplying by 10 instead of 100.

✕Leaving $\tfrac{54}{99}$ unsimplified.