Recurring decimal as a fraction

This problem explores how to convert a recurring decimal into a rational number, a fundamental skill in the HSC Mathematics Extension 1 sequences and series topic. By viewing the decimal as an infinite geometric series, you’ll see a powerful connection between decimal representations and limits. Working through this example will help you tackle any purely repeating decimal with confidence.

Problem Statement

Express $0.\overline{23}$ as a rational number.

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Solutions

The repeating decimal $0.\overline{23}$ can be written immediately as a fraction with denominator $99$ because the repeating block has length $2$. So we have

$$0.\overline{23}=\frac{23}{99}.$$

To see why this works, we can express the decimal as an infinite geometric series:

$$0.23+0.0023+0.000023+\cdots$$

with first term $a = \frac{23}{100}$ and common ratio $r = \frac{1}{100}$.

Since $|r| = \frac{1}{100} < 1$, the series converges, so we can apply the sum-to-infinity formula $S_\infty = \frac{a}{1-r}$:

$$S_\infty = \frac{23/100}{1-1/100} = \frac{23/100}{99/100} = \frac{23}{99},$$

matching the direct result.

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Takeaways

• Recurring decimals can be expressed as rational numbers by interpreting them as infinite geometric series.
• A repeating block of length $n$ corresponds to a denominator of $10^n - 1$ when the decimal is purely repeating.
• The sum-to-infinity formula $\frac{a}{1-r}$ applies only when $|r| < 1$, which always holds for such expansions.

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Further Readings

HSC Integrals, HSC Polys Ext 1, HSC Collections, HSC Mechanics