Year 12 Maths Advanced (2027) Sequences and series

Recurring Decimals as Infinite GPs

Q: How do you turn a recurring decimal into a fraction?

A recurring decimal is an infinite geometric series. Put the repeating block over as many nines as its length: one digit over 9, two digits over 99, then simplify.

Q: What is 0.6 recurring as a fraction?

It is a geometric series with first term 0.6 and ratio 0.1, so the sum is 0.6 divided by 0.9, which simplifies to two thirds.

Q: Is 0.9 recurring equal to 1?

Yes. As a fraction 0.9 recurring is 9 over 9, which equals exactly 1.

Q: What is 0.27 recurring as a fraction?

The two-digit block 27 goes over 99, giving 27 over 99, which simplifies to 3 over 11.

20 practice questions 0 video lessons Theory + worked examples

Learn to write recurring decimals as infinite GPs for NSW Year 12 Mathematics Advanced (NESA outcome MAV-12-03). A decimal such as \(0.272727\ldots\) is really a sum of ever-smaller pieces that form a geometric series, so its exact value can be found with a limiting sum.

You will learn to split a recurring decimal into its repeating blocks, identify the first term and common ratio, apply the limiting sum and simplify the answer to an exact fraction — a clear example of an infinite process giving a finite, exact result.

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Theory

Every recurring decimal is an infinite geometric series, so the limiting sum turns it into an exact fraction. This Year 12 Mathematics Advanced topic (MAV-12-03) converts recurring decimals to fractions.

A recurring decimal repeats a block of digits forever, which is a geometric series with \(|r|<1\). Summing it gives an exact fraction.

Shortcut: one repeating digit is over \(9\) (\(0.\dot{d}=\dfrac{d}{9}\)); a two-digit block is over \(99\); a three-digit block over \(999\). A useful check is \(0.\dot{9}=\dfrac{9}{9}=1\).

\(0.\dot{6}=0.6+0.06+0.006+\cdots\), a geometric series.

\[0.\dot{d}=\dfrac{d}{9},\qquad 0.\dot{a}\dot{b}=\dfrac{ab}{99}\]

0 point d recurring equals d over 9; 0 point a b recurring equals a b over 99

Method

Identify the repeating block and its length.
Write it over \(9\), \(99\), \(999\), \(\ldots\) (matching the block length), or sum it as a GP.
Simplify the fraction.

Example 1 — One digit

Write \(0.\dot{4}\) as a fraction.

Solution

\(0.\dot{4}\)

\(=\)

\(\dfrac{4}{9}\)

Example 2 — Two digits

Write \(0.\dot{2}\dot{7}\) in simplest form.

Solution

\(0.\dot{2}\dot{7}\)

\(=\)

\(\dfrac{27}{99}=\dfrac{3}{11}\)

Example 3 — Via the GP sum

Use \(S_\infty\) to write \(0.\dot{6}\) as a fraction.

Solution

\(0.\dot{6}\)	\(=\)	\(\dfrac{0.6}{1-0.1}\)
	\(=\)	\(\dfrac{0.6}{0.9}=\dfrac{2}{3}\)

Example 4 — Two conversions

Write \(0.\dot{1}\dot{2}\) and \(0.\dot{9}\) as fractions.

Solution

\(0.\dot{1}\dot{2}\)	\(=\)	\(\dfrac{12}{99}=\dfrac{4}{33}\)
\(0.\dot{9}\)	\(=\)	\(\dfrac{9}{9}=1\)

Common pitfalls

Match block length to the denominator. One digit over \(9\); two digits over \(99\).

Only the repeating part is the GP. Handle any non-repeating start separately.

Always simplify (e.g. \(\dfrac{27}{99}=\dfrac{3}{11}\)).

Frequently asked questions

How do you turn a recurring decimal into a fraction?

A recurring decimal is an infinite geometric series. Put the repeating block over as many nines as its length: one digit over 9, two digits over 99, then simplify.

What is 0.6 recurring as a fraction?

It is a geometric series with first term 0.6 and ratio 0.1, so the sum is 0.6 divided by 0.9, which simplifies to two thirds.

Is 0.9 recurring equal to 1?

Yes. As a fraction 0.9 recurring is 9 over 9, which equals exactly 1.

What is 0.27 recurring as a fraction?

The two-digit block 27 goes over 99, giving 27 over 99, which simplifies to 3 over 11.

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