Recurring Decimals to Fractions

What you'll learn

How to read recurring decimal dot notation.
How to convert fractions into recurring decimals.
How to prove recurring decimals as fractions using algebra.
How to handle recurring decimals in calculations.

Recurring decimals and dot notation

A decimal can either stop, like 0.25, or carry on forever. If it carries on with a repeating pattern, it is called a recurring decimal.

Definition

Recurring decimal

A recurring decimal is a decimal where a digit or block of digits repeats forever. A dot above one repeated digit means that digit repeats, while dots above the first and last digit of a block mean the whole block repeats, such as $0.2˙16˙=0.216216…0.\dot{2}1\dot{6}=0.216216\ldots$ .

Dot notation shows whether one digit repeats or a whole block of digits repeats.

Example

Reading recurring notation

In $0.72˙0.7\dot{2}$ , the 7 happens once and the 2 repeats.
So $0.72˙=0.7222…0.7\dot{2}=0.7222\ldots$ .
In $0.4˙05˙0.\dot{4}0\dot{5}$ , the recurring block is 405.
So $0.4˙05˙=0.405405405…0.\dot{4}0\dot{5}=0.405405405\ldots$ .

Converting fractions to decimals

In a fraction, the numerator is the top number and the denominator is the bottom number. The fraction bar means “divide”.

For example, $511\frac{5}{11}$ means 5 divided by 11.

Example

Convert a fraction to a recurring decimal

The fraction $511\frac{5}{11}$ means 5 divided by 11.

$The long division for 5\div 11 repeats when the remainder 5 appears again.$

Start dividing: 11 goes into 50 four times, with remainder 6.
Continue: 11 goes into 60 five times, with remainder 5.
The remainder 5 has appeared again, so the digits 45 repeat forever.
Therefore $511=0.4˙5˙\frac{5}{11}=0.\dot{4}\dot{5}$ .

The main algebra trick

To turn a recurring decimal into a fraction, we give it a letter name, usually $x$ .

Definition

Algebraically

To prove something algebraically means using letters and equations rather than just a calculator. Here, $x$ is a temporary name for the recurring decimal.

Key Idea

Make the repeating tails cancel

Set the recurring decimal equal to $x$ , multiply by powers of 10 so the repeating parts line up, then subtract. The infinite recurring tail disappears, leaving a normal equation.

Example

Prove a simple recurring decimal

Prove algebraically that $0.7˙0.\dot{7}$ is $79\frac{7}{9}$ .

Multiplying by 10 lines up the recurring 7s so the infinite tail cancels when subtracting.

Let $x=0.7˙x=0.\dot{7}$ , meaning:
$x=0.777…x=0.777\ldots$
Multiply by 10 because one digit repeats:
$10x=7.777…10x=7.777\ldots$
Subtract the original equation:
$10x=7.777…x=0.777…9x=7\begin{aligned} 10x &= 7.777\ldots\\ x &= 0.777\ldots\\ 9x &= 7 \end{aligned}$
Divide both sides by 9:
$x=79x=\frac{7}{9}$
So $0.7˙=790.\dot{7}=\frac{7}{9}$ .

When the repeat does not start straight away

Sometimes a decimal has some non-recurring digits first. These are digits that happen once before the repeating part begins.

For example, in $0.53˙0.5\dot{3}$ , the 5 happens once, then the 3 repeats.

Example

A fixed digit before the repeat

Write $0.53˙0.5\dot{3}$ as a fraction in its simplest form.

$For 0.5\dot{3}, using 100x and 10x makes the recurring 3s line up exactly.$

Let $x=0.53˙=0.5333…x=0.5\dot{3}=0.5333\ldots$ .
Use $100 x$ and $10 x$ so the recurring tails match:
$100x=53.333…10x=5.333…\begin{aligned} 100x &= 53.333\ldots\\ 10x &= 5.333\ldots \end{aligned}$
Subtract:
$100x−10x=53.333…−5.333…90x=48\begin{aligned} 100x-10x &= 53.333\ldots-5.333\ldots\\ 90x &= 48 \end{aligned}$
Divide by 90 and simplify:
$x=4890=815x=\frac{48}{90}=\frac{8}{15}$
Therefore $0.53˙=8150.5\dot{3}=\frac{8}{15}$ .

Common Mistake

Subtracting before the tails match

For $0.53˙0.5\dot{3}$ , subtracting $x$ from $10 x$ does not cancel the recurring part neatly. Use $100 x$ and $10 x$ so both numbers have the same recurring tail.

Subtracting before the tails match leaves different decimal tails, so the recurrence does not cancel cleanly.

Repeating blocks of digits

The recurring part might be more than one digit long. If the block has 3 digits, multiplying by 1000 shifts one whole block along.

Tip

Which powers of 10?

If $m$ non-recurring decimal digits come before a recurring block of $n$ digits, use $10^m x$ and $10^{m+n}x$ . For example, one fixed digit then two recurring digits means use $10 x$ and $1000 x$ .

The powers of 10 are chosen from the number of fixed digits and the length of the repeating block.

Example

A three-digit recurring block

Prove algebraically that $0.1˙89˙0.\dot{1}8\dot{9}$ is $737\frac{7}{37}$ .

A three-digit recurring block needs a shift by 1000 to align the repeated blocks.

Let $x=0.1˙89˙=0.189189189…x=0.\dot{1}8\dot{9}=0.189189189\ldots$ .
The recurring block 189 has 3 digits, so multiply by 1000:
$1000x=189.189189…1000x=189.189189\ldots$
Subtract the original equation:
$1000x=189.189189…x=0.189189…999x=189\begin{aligned} 1000x &= 189.189189\ldots\\ x &= 0.189189\ldots\\ 999x &= 189 \end{aligned}$
Divide by 999 and simplify:
$x=189999=737x=\frac{189}{999}=\frac{7}{37}$
So $0.1˙89˙=7370.\dot{1}8\dot{9}=\frac{7}{37}$ .

Numbers bigger than 1

If there is a whole number part, include it in $x$ . The same method still works.

A fraction is in simplest form when the top and bottom have no common factor bigger than 1.

Example

Recurring decimal with a whole number part

Write $2.43˙6˙2.4\dot{3}\dot{6}$ as a fraction in its simplest form.

Let $x=2.43˙6˙=2.43636…x=2.4\dot{3}\dot{6}=2.43636\ldots$ .
There is one fixed decimal digit, 4, then a two-digit recurring block, 36.
Use $1000 x$ and $10 x$ :
$1000x=2436.3636…10x=24.3636…\begin{aligned} 1000x &= 2436.3636\ldots\\ 10x &= 24.3636\ldots \end{aligned}$
Subtract:
$990x=2412x=2412990\begin{aligned} 990x &= 2412\\ x &= \frac{2412}{990} \end{aligned}$
Simplify by dividing top and bottom by 18:
$x=13455x=\frac{134}{55}$
Therefore $2.43˙6˙=134552.4\dot{3}\dot{6}=\frac{134}{55}$ .

Calculations with recurring decimals

For calculations, convert each recurring decimal into a fraction first. Then use normal fraction rules.

For division, multiply by the reciprocal, which means the second fraction flipped upside down.

Example

Work out a division

Work out $0.18˙÷0.6˙0.1\dot{8}\div0.\dot{6}$ .

$Convert each recurring decimal to a fraction first, then divide by multiplying by the reciprocal.$

Convert $0.18˙0.1\dot{8}$ into a fraction:
$a=0.1888…100a=18.888…10a=1.888…90a=17a=1790\begin{aligned} a &= 0.1888\ldots\\ 100a &= 18.888\ldots\\ 10a &= 1.888\ldots\\ 90a &= 17\\ a &= \frac{17}{90} \end{aligned}$
Convert $0.6˙0.\dot{6}$ :
$0.6˙=69=230.\dot{6}=\frac{6}{9}=\frac{2}{3}$
Divide by multiplying by the reciprocal:
$1790÷23=1790×32\frac{17}{90}\div\frac{2}{3}=\frac{17}{90}\times\frac{3}{2}$
Simplify and multiply:
$1790×32=1760\frac{17}{90}\times\frac{3}{2}=\frac{17}{60}$
The exact answer is $1760\frac{17}{60}$ .

Common Mistake

Recurring 9s

A recurring string of 9s is exactly equal to the next number: $0.9˙=10.\dot{9}=1$ , so $1.29˙=1.31.2\dot{9}=1.3$ . Do not treat it as just less than the next number.

Exam technique

In the exam

Write the decimal as $x = ...$ with a few repeated digits, so the examiner can see the pattern.
Choose powers of 10 that make the recurring tails identical, then subtract the smaller equation from the larger one.
Always simplify the final fraction; for a calculation, convert every recurring decimal to a fraction before multiplying or dividing.

Self review

Check yourself

Can you explain why $0.4˙2˙0.\dot{4}\dot{2}$ has denominator 99 before simplifying?
For $0.37˙0.3\dot{7}$ , which two multiples of $x$ would you subtract?
If a recurring decimal has one non-recurring digit and a two-digit recurring block, which powers of 10 should you use?

Recurring Decimals to Fractions

What you'll learn

How to read recurring decimal dot notation.
How to convert fractions into recurring decimals.
How to prove recurring decimals as fractions using algebra.
How to handle recurring decimals in calculations.

Recurring decimals and dot notation

A decimal can either stop, like 0.25, or carry on forever. If it carries on with a repeating pattern, it is called a recurring decimal.

Definition

Recurring decimal

Dot notation shows whether one digit repeats or a whole block of digits repeats.

Example

Reading recurring notation

In $0.72˙0.7\dot{2}$ , the 7 happens once and the 2 repeats.
So $0.72˙=0.7222…0.7\dot{2}=0.7222\ldots$ .
In $0.4˙05˙0.\dot{4}0\dot{5}$ , the recurring block is 405.
So $0.4˙05˙=0.405405405…0.\dot{4}0\dot{5}=0.405405405\ldots$ .

Converting fractions to decimals

In a fraction, the numerator is the top number and the denominator is the bottom number. The fraction bar means “divide”.

For example, $511\frac{5}{11}$ means 5 divided by 11.

Example

Convert a fraction to a recurring decimal

The fraction $511\frac{5}{11}$ means 5 divided by 11.

$The long division for 5\div 11 repeats when the remainder 5 appears again.$

Start dividing: 11 goes into 50 four times, with remainder 6.
Continue: 11 goes into 60 five times, with remainder 5.
The remainder 5 has appeared again, so the digits 45 repeat forever.
Therefore $511=0.4˙5˙\frac{5}{11}=0.\dot{4}\dot{5}$ .

The main algebra trick

To turn a recurring decimal into a fraction, we give it a letter name, usually $x$ .

Definition

Algebraically

To prove something algebraically means using letters and equations rather than just a calculator. Here, $x$ is a temporary name for the recurring decimal.

Key Idea

Make the repeating tails cancel

Set the recurring decimal equal to $x$ , multiply by powers of 10 so the repeating parts line up, then subtract. The infinite recurring tail disappears, leaving a normal equation.

Example

Prove a simple recurring decimal

Prove algebraically that $0.7˙0.\dot{7}$ is $79\frac{7}{9}$ .

Multiplying by 10 lines up the recurring 7s so the infinite tail cancels when subtracting.

Let $x=0.7˙x=0.\dot{7}$ , meaning:
$x=0.777…x=0.777\ldots$
Multiply by 10 because one digit repeats:
$10x=7.777…10x=7.777\ldots$
Subtract the original equation:
$10x=7.777…x=0.777…9x=7\begin{aligned} 10x &= 7.777\ldots\\ x &= 0.777\ldots\\ 9x &= 7 \end{aligned}$
Divide both sides by 9:
$x=79x=\frac{7}{9}$
So $0.7˙=790.\dot{7}=\frac{7}{9}$ .

When the repeat does not start straight away

Sometimes a decimal has some non-recurring digits first. These are digits that happen once before the repeating part begins.

For example, in $0.53˙0.5\dot{3}$ , the 5 happens once, then the 3 repeats.

Example

A fixed digit before the repeat

Write $0.53˙0.5\dot{3}$ as a fraction in its simplest form.

$For 0.5\dot{3}, using 100x and 10x makes the recurring 3s line up exactly.$

Let $x=0.53˙=0.5333…x=0.5\dot{3}=0.5333\ldots$ .
Use $100 x$ and $10 x$ so the recurring tails match:
$100x=53.333…10x=5.333…\begin{aligned} 100x &= 53.333\ldots\\ 10x &= 5.333\ldots \end{aligned}$
Subtract:
$100x−10x=53.333…−5.333…90x=48\begin{aligned} 100x-10x &= 53.333\ldots-5.333\ldots\\ 90x &= 48 \end{aligned}$
Divide by 90 and simplify:
$x=4890=815x=\frac{48}{90}=\frac{8}{15}$
Therefore $0.53˙=8150.5\dot{3}=\frac{8}{15}$ .

Common Mistake

Subtracting before the tails match

For $0.53˙0.5\dot{3}$ , subtracting $x$ from $10 x$ does not cancel the recurring part neatly. Use $100 x$ and $10 x$ so both numbers have the same recurring tail.

Subtracting before the tails match leaves different decimal tails, so the recurrence does not cancel cleanly.

Repeating blocks of digits

The recurring part might be more than one digit long. If the block has 3 digits, multiplying by 1000 shifts one whole block along.

Tip

Which powers of 10?

If $m$ non-recurring decimal digits come before a recurring block of $n$ digits, use $10^m x$ and $10^{m+n}x$ . For example, one fixed digit then two recurring digits means use $10 x$ and $1000 x$ .

The powers of 10 are chosen from the number of fixed digits and the length of the repeating block.

Example

A three-digit recurring block

Prove algebraically that $0.1˙89˙0.\dot{1}8\dot{9}$ is $737\frac{7}{37}$ .

A three-digit recurring block needs a shift by 1000 to align the repeated blocks.

Let $x=0.1˙89˙=0.189189189…x=0.\dot{1}8\dot{9}=0.189189189\ldots$ .
The recurring block 189 has 3 digits, so multiply by 1000:
$1000x=189.189189…1000x=189.189189\ldots$
Subtract the original equation:
$1000x=189.189189…x=0.189189…999x=189\begin{aligned} 1000x &= 189.189189\ldots\\ x &= 0.189189\ldots\\ 999x &= 189 \end{aligned}$
Divide by 999 and simplify:
$x=189999=737x=\frac{189}{999}=\frac{7}{37}$
So $0.1˙89˙=7370.\dot{1}8\dot{9}=\frac{7}{37}$ .

Numbers bigger than 1

If there is a whole number part, include it in $x$ . The same method still works.

A fraction is in simplest form when the top and bottom have no common factor bigger than 1.

Example

Recurring decimal with a whole number part

Write $2.43˙6˙2.4\dot{3}\dot{6}$ as a fraction in its simplest form.

Let $x=2.43˙6˙=2.43636…x=2.4\dot{3}\dot{6}=2.43636\ldots$ .
There is one fixed decimal digit, 4, then a two-digit recurring block, 36.
Use $1000 x$ and $10 x$ :
$1000x=2436.3636…10x=24.3636…\begin{aligned} 1000x &= 2436.3636\ldots\\ 10x &= 24.3636\ldots \end{aligned}$
Subtract:
$990x=2412x=2412990\begin{aligned} 990x &= 2412\\ x &= \frac{2412}{990} \end{aligned}$
Simplify by dividing top and bottom by 18:
$x=13455x=\frac{134}{55}$
Therefore $2.43˙6˙=134552.4\dot{3}\dot{6}=\frac{134}{55}$ .

Calculations with recurring decimals

For calculations, convert each recurring decimal into a fraction first. Then use normal fraction rules.

For division, multiply by the reciprocal, which means the second fraction flipped upside down.

Example

Work out a division

Work out $0.18˙÷0.6˙0.1\dot{8}\div0.\dot{6}$ .

$Convert each recurring decimal to a fraction first, then divide by multiplying by the reciprocal.$

Convert $0.18˙0.1\dot{8}$ into a fraction:
$a=0.1888…100a=18.888…10a=1.888…90a=17a=1790\begin{aligned} a &= 0.1888\ldots\\ 100a &= 18.888\ldots\\ 10a &= 1.888\ldots\\ 90a &= 17\\ a &= \frac{17}{90} \end{aligned}$
Convert $0.6˙0.\dot{6}$ :
$0.6˙=69=230.\dot{6}=\frac{6}{9}=\frac{2}{3}$
Divide by multiplying by the reciprocal:
$1790÷23=1790×32\frac{17}{90}\div\frac{2}{3}=\frac{17}{90}\times\frac{3}{2}$
Simplify and multiply:
$1790×32=1760\frac{17}{90}\times\frac{3}{2}=\frac{17}{60}$
The exact answer is $1760\frac{17}{60}$ .

Common Mistake

Recurring 9s

A recurring string of 9s is exactly equal to the next number: $0.9˙=10.\dot{9}=1$ , so $1.29˙=1.31.2\dot{9}=1.3$ . Do not treat it as just less than the next number.

Exam technique

In the exam

Write the decimal as $x = ...$ with a few repeated digits, so the examiner can see the pattern.
Choose powers of 10 that make the recurring tails identical, then subtract the smaller equation from the larger one.
Always simplify the final fraction; for a calculation, convert every recurring decimal to a fraction before multiplying or dividing.

Self review

Check yourself

Can you explain why $0.4˙2˙0.\dot{4}\dot{2}$ has denominator 99 before simplifying?
For $0.37˙0.3\dot{7}$ , which two multiples of $x$ would you subtract?
If a recurring decimal has one non-recurring digit and a two-digit recurring block, which powers of 10 should you use?

Recurring Decimals to Fractions

What you'll learn

Recurring decimals and dot notation

Converting fractions to decimals

The main algebra trick

When the repeat does not start straight away

Repeating blocks of digits

Numbers bigger than 1

Calculations with recurring decimals

Recap questions

You've reached the end

Recurring Decimals to Fractions

What you'll learn

Recurring decimals and dot notation

Converting fractions to decimals

The main algebra trick

When the repeat does not start straight away

Repeating blocks of digits

Numbers bigger than 1

Calculations with recurring decimals

Recap questions

You've reached the end

Recurring Decimals to Fractions

What you'll learn

Recurring decimals and dot notation

Converting fractions to decimals

The main algebra trick

When the repeat does not start straight away

Repeating blocks of digits

Numbers bigger than 1

Calculations with recurring decimals

Recap questions

You've reached the end

Recurring Decimals to Fractions Revision Guide

Recurring Decimals to Fractions

What you'll learn

Recurring decimals and dot notation

Converting fractions to decimals

The main algebra trick

When the repeat does not start straight away

Repeating blocks of digits

Numbers bigger than 1

Calculations with recurring decimals

Recap questions

You've reached the end

Recurring Decimals to Fractions Revision Guide