Year 9 Maths Advanced Working with Numbers

Converting Recurring Decimals to Fractions

Q: How do you convert a recurring decimal to a fraction?

Let x be the decimal. Multiply by 10, 100 or 1000 so the repeating part lines up, subtract the original, then solve for x and simplify. For 0.27 recurring, 100x minus x gives 99x equals 27, so x is 27 over 99, which is 3 over 11.

Q: What do you multiply by when two digits repeat?

Multiply by 100, because 100 shifts the digits two places, which is one whole block. In general, multiply by 10 to the power of the number of repeating digits.

Q: How do you convert a decimal like 0.16 recurring where only the 6 repeats?

Use two multipliers: 100x is 16.666 recurring and 10x is 1.666 recurring. Subtracting gives 90x equals 15, so x is 15 over 90, which is 1 over 6.

20 practice questions 1 video lesson Theory + worked examples

A recurring decimal can always be written as a fraction. In NSW Year 9 Advanced maths, students let \(x\) equal the decimal, multiply by a power of \(10\) and subtract, so the repeating part cancels (NESA outcome MA5‑IND‑P‑02).

You will convert decimals such as \(0.\dot{6}\dot{8}\), \(0.2\dot{3}\) and \(0.5\dot{3}\dot{7}\) to fractions in simplest form, write \(2.\dot{1}\dot{8}\) as an improper fraction, and add and multiply recurring decimals by changing them to fractions first. You will also find the exact value of \(0.\dot{9}\), prove a shortcut for two-digit repeats, and find how many games a netball team won. It is part of Year 9 Advanced Working with Numbers.

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Theory

Every recurring decimal is a fraction. Let \(x\) be the decimal, multiply by \(10\), \(100\) or \(1000\) so the repeating tails line up, and subtract: \(0.\dot{2}\dot{7}\) gives \(99x = 27\), so \(x = \dfrac{3}{11}\). This Year 9 Advanced maths page covers converting recurring decimals to fractions, including digits before the block, whole-number parts and arithmetic with recurring decimals.

A recurring decimal repeats a block of digits forever. The block length is the number of digits that repeat: \(0.\dot{7}\) has a block of \(1\), \(0.\dot{2}\dot{7}\) a block of \(2\), and \(0.\dot{1}0\dot{8}\) a block of \(3\).

The method rests on one idea: multiplying by \(10\) shifts every digit one place left. Multiplying by \(10^k\), where \(k\) is the block length, produces a number with exactly the same repeating tail. Subtracting cancels the tail and leaves a whole number.

Some decimals have digits before the block starts, such as \(0.1\dot{6}\). These need two multipliers, chosen so that both results have the same digits after the point.

Because the method always works, every recurring decimal is a rational number. This is how it can be proved that a recurring decimal is rational, and why irrational numbers can never repeat.

Multiply by \(10^k\), where \(k\) is the number of digits in the block. The repeating tails line up and cancel when you subtract.

When the block starts straight after the point, the fraction is the block over as many \(9\)s as it has digits. Then simplify.

Let \(x\) be the decimal. Multiply by a power of \(10\) so the repeating tails line up, then subtract:

\[k \text{ digits repeat} \Rightarrow 10^k x - x = (10^k - 1)x\]

10^{k} x - x = (10^{k} - 1) x

Shortcut when the block starts straight after the point:

\[0.\dot{a} = \dfrac{a}{9} \qquad 0.\dot{a}\dot{b} = \dfrac{10a + b}{99} \qquad 0.\dot{a}b\dot{c} = \dfrac{100a + 10b + c}{999}\]

0. \dot{a} \dot{b} = \frac{10 a + b}{99}

With \(p\) non-repeating digits first and a block of \(k\) digits, subtract \(10^p x\) from \(10^{p+k} x\). For \(0.3\dot{1}\dot{8}\): \(1000x - 10x = 990x\).

An unknown digit. If \(0.\dot{a}\dot{2} = \dfrac{8}{11}\), write both over \(99\): \(\dfrac{10a + 2}{99} = \dfrac{72}{99}\), so \(10a + 2 = 72\) and \(a = 7\).

Making a whole number. \(n \times \dfrac{p}{q}\) is a whole number (with \(\dfrac{p}{q}\) in simplest form) exactly when \(n\) is a multiple of \(q\). So convert first, simplify, and read off the denominator.

How to convert a recurring decimal to a fraction

Let \(x\) equal the decimal and write out a few repeats.
Multiply by powers of \(10\) to make two numbers with identical repeating tails: \(10^k\) for a block of \(k\) digits, and an extra \(10^p\) on both if \(p\) digits come first.
Subtract the smaller from the larger. The tails cancel, leaving a whole number (or a terminating decimal).
Solve for \(x\). Clear any decimal in the numerator by multiplying top and bottom by \(10\).
Simplify with the HCF, and check by dividing on a calculator.

Example 1 — The block starts at the point

Write as fractions in simplest form:
(i) \(0.\dot{7}\)
(ii) \(0.\dot{2}\dot{7}\)
(iii) \(0.\dot{1}0\dot{8}\)

Solution

Multiply by \(10\), \(100\) or \(1000\) to match the length of the block, then subtract.

(i) One digit repeats.

\(10x - x\)	\(=\)	\(7.777\ldots - 0.777\ldots\)
\(9x\)	\(=\)	\(7\)
\(x\)	\(=\)	\(\mathbf{\dfrac{7}{9}}\)

9 x = 7, x = \frac{7}{9}

(ii) Two digits repeat.

\(100x - x\)	\(=\)	\(27.2727\ldots - 0.2727\ldots\)
\(99x\)	\(=\)	\(27\)
\(x\)	\(=\)	\(\dfrac{27}{99} = \mathbf{\dfrac{3}{11}}\)

99 x = 27, x = \frac{3}{11}

(iii) Three digits repeat, including the \(0\).

\(1000x - x\)	\(=\)	\(108.108\ldots - 0.108\ldots\)
\(999x\)	\(=\)	\(108\)
\(x\)	\(=\)	\(\dfrac{108}{999} = \mathbf{\dfrac{4}{37}}\)

999 x = 108, x = \frac{4}{37}

\(108\) and \(999\) share the factor \(27\).

Example 2 — A digit before the block

Write as fractions in simplest form:
(i) \(0.1\dot{6}\)
(ii) \(0.3\dot{1}\dot{8}\)

Solution

Make two numbers whose decimal parts are identical, then subtract. One multiplier moves the first block to just after the point; the other moves a whole block further.

(i) \(x = 0.1666\ldots\)

\(100x\)	\(=\)	\(16.666\ldots\)
\(10x\)	\(=\)	\(1.666\ldots\)
\(90x\)	\(=\)	\(15\)
\(x\)	\(=\)	\(\dfrac{15}{90} = \mathbf{\dfrac{1}{6}}\)

90 x = 15, x = \frac{1}{6}

(ii) \(x = 0.31818\ldots\)

\(1000x\)	\(=\)	\(318.1818\ldots\)
\(10x\)	\(=\)	\(3.1818\ldots\)
\(990x\)	\(=\)	\(315\)
\(x\)	\(=\)	\(\dfrac{315}{990} = \mathbf{\dfrac{7}{22}}\)

990 x = 315, x = \frac{7}{22}

Check: \(7 \div 22 = 0.31818\ldots\)

Example 3 — Whole-number part and arithmetic

(i) Write \(3.\dot{4}\dot{5}\) as an improper fraction.
(ii) Write \(0.\dot{6} - 0.\dot{1}\dot{2}\) as a fraction in simplest form.

Solution

(i) Two digits repeat, so use \(100x - x\). The whole number comes along.

\(100x\)	\(=\)	\(345.4545\ldots\)
\(x\)	\(=\)	\(3.4545\ldots\)
\(99x\)	\(=\)	\(342\)
\(x\)	\(=\)	\(\dfrac{342}{99} = \mathbf{\dfrac{38}{11}}\)

99 x = 342, x = \frac{38}{11}

Or convert only the decimal part: \(0.\dot{4}\dot{5} = \dfrac{5}{11}\), so \(3.\dot{4}\dot{5} = 3\dfrac{5}{11} = \dfrac{38}{11}\).

(ii) Convert each decimal, then subtract the fractions.

\(0.\dot{6}\)	\(=\)	\(\dfrac{6}{9} = \dfrac{66}{99}\)
\(0.\dot{1}\dot{2}\)	\(=\)	\(\dfrac{12}{99}\)
\(\dfrac{66}{99} - \dfrac{12}{99}\)	\(=\)	\(\dfrac{54}{99} = \mathbf{\dfrac{6}{11}}\)

\frac{66}{99} - \frac{12}{99} = \frac{6}{11}

Example 4 — Using the fraction

In a survey, \(0.\dot{4}\dot{8}\) of the people asked said yes. What is the smallest number of people who could have been asked, and how many said yes?

Solution

Convert to a fraction in simplest form. The number asked must be a multiple of the denominator.

\(100x - x\)	\(=\)	\(48.4848\ldots - 0.4848\ldots\)
\(99x\)	\(=\)	\(48\)
\(x\)	\(=\)	\(\dfrac{48}{99} = \dfrac{16}{33}\)

x = \frac{48}{99} = \frac{16}{33}

The smallest group is \(\mathbf{33}\) people, with \(\mathbf{16}\) saying yes. Check: \(16 \div 33 = 0.4848\ldots\)

Using \(\dfrac{48}{99}\) unsimplified would give \(99\) people, which is possible but not the smallest.

Common pitfalls

Match the multiplier to the block. For \(0.\dot{2}\dot{7}\), multiplying by \(10\) gives \(2.7272\ldots\), whose tail does not line up with \(0.2727\ldots\). A block of two digits needs \(100\).

Watch out for digits before the block. \(0.1\dot{6}\) is not \(\dfrac{16}{99}\) or \(\dfrac{16}{90}\). Use \(100x - 10x\) to get \(\dfrac{15}{90}\).

Clear decimals and simplify. \(9x = 1.5\) gives \(x = \dfrac{1.5}{9} = \dfrac{15}{90} = \dfrac{1}{6}\). Leave no decimal in a fraction, and always reduce with the HCF.

A zero inside the block still counts. \(0.\dot{1}0\dot{8}\) has a three-digit block \(108\), so it is \(\dfrac{108}{999}\), not \(\dfrac{18}{99}\).

Frequently asked questions

How do you convert a recurring decimal to a fraction?

Let \(x\) be the decimal. Multiply by \(10\), \(100\) or \(1000\) so the repeating part lines up, subtract the original, then solve for \(x\) and simplify. For \(0.\dot{2}\dot{7}\): \(99x = 27\), so \(x = \dfrac{27}{99} = \dfrac{3}{11}\).

What do you multiply by when two digits repeat?

Multiply by \(100\), because \(100\) shifts the digits two places, which is one whole block. In general, multiply by \(10^k\) for a block of \(k\) digits.

How do you convert a decimal like 0.16 recurring where only the 6 repeats?

Use two multipliers: \(100x = 16.666\ldots\) and \(10x = 1.666\ldots\). Subtracting gives \(90x = 15\), so \(x = \dfrac{15}{90} = \dfrac{1}{6}\).

Is there a shortcut for recurring decimals?

When the block starts straight after the point, the fraction is the block over the same number of \(9\)s: \(0.\dot{1}0\dot{8} = \dfrac{108}{999} = \dfrac{4}{37}\).

Are recurring decimals rational?

Yes. The multiply-and-subtract method turns every recurring decimal into a fraction of two integers, so every recurring decimal is rational.

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Converting Recurring Decimals to Fractions

📖 Theory

How to convert a recurring decimal to a fraction

Common pitfalls

Frequently asked questions

Theory