Fraction of a recurring decimal

Many HSC students first meet recurring decimals in primary school, but the Extension 1 syllabus deepens this into an elegant application of limiting sums. Here we use the infinite geometric series formula to turn $0.\dot{1}\dot{2}\dot{3}$ into an exact fraction — and you’ll see that every recurring decimal is simply a sum of infinitely many shrinking terms.

Problem Statement

Find the fraction for $0.\dot{1}\dot{2}\dot{3}$?

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Hints

Use $a = \frac{123}{1000}$ and $r = \frac{1}{1000}$ in your $S_\infty$ formula.

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Solutions

We can think of $0.\dot{1}\dot{2}\dot{3}$ as $0.123\,123\,123\ldots$, which is built from the repeating block “123” shifted three decimal places each time. That makes it an infinite geometric series: the first term is the block over $1000$, and each subsequent term is another $\frac{1}{1000}$ of the previous one. So with $a = \frac{123}{1000}$ and $r = \frac{1}{1000}$, we apply the sum-to-infinity formula:

$$S_\infty = \frac{\frac{123}{1000}}{1 - \frac{1}{1000}} = \frac{123}{999} = \frac{41}{333}.$$

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Takeaways

• Every recurring decimal is a geometric series sum $S_\infty$.
• The first term and common ratio are identified by the length of the repeating block: here $a = \frac{123}{1000}$ and $r = \frac{1}{1000}$.
• Using the sum to infinity formula with $|r|<1$ converts the decimal to a fraction in simplest form, $\frac{41}{333}$, confirming that every purely periodic decimal is rational.

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Further Readings

HSC Trigonometry, HSC Inequalities, HSC Last Resorts, HSC Distributions