Number Systems • Topic 4 of 4

Decimal Expansions of Rational Numbers

What are the decimal expansions of rational numbers? Every rational number written as $\frac{p}{q}$ in lowest terms ($q \ne 0$) has a decimal expansion that is one of two kinds:

  • Terminating — it ends after finitely many digits (e.g. $\frac{1}{4} = 0.25$, $\frac{3}{5} = 0.6$).
  • Non-terminating but repeating — it runs forever, repeating a fixed block of digits (e.g. $\frac{1}{3} = 0.\overline{3}$, $\frac{2}{7} = 0.\overline{285714}$).

A rational number can never give a non-terminating, non-repeating decimal — that behaviour belongs to irrational numbers only.

The terminating test. Put $\frac{p}{q}$ in lowest terms and factorise the denominator. Then:

  • If $q = 2^{m}\times 5^{n}$ (only the primes 2 and 5, any whole-number powers), the decimal terminates.
  • If $q$ has any other prime factor (3, 7, 11, …), the decimal is non-terminating repeating.

Why the test works. A decimal terminates exactly when the fraction can be rewritten with a denominator that is a power of 10. Since $10 = 2 \times 5$, a power of 10 contains only the primes 2 and 5. If $q = 2^{m}\times 5^{n}$, multiply top and bottom by the missing factors to balance the powers of 2 and 5. For example, $\frac{7}{20} = \frac{7}{2^{2}\times 5} = \frac{7 \times 5}{2^{2}\times 5^{2}} = \frac{35}{100} = 0.35$. The decimal terminates after $\max(m, n)$ places.

Examples of the test:

  • $\frac{7}{20}$: $20 = 2^{2}\times 5$ → terminates.
  • $\frac{13}{3125}$: $3125 = 5^{5}$ → terminates (after 5 places).
  • $\frac{5}{6}$: $6 = 2\times 3$ → the factor 3 forces it to repeat.
  • $\frac{64}{455}$: $455 = 5\times 7\times 13$ → the factors 7 and 13 force it to repeat.

Repeating-decimal notation. A bar is placed over the repeating block: $0.333\ldots = 0.\overline{3}$ and $0.142857142857\ldots = 0.\overline{142857}$. A terminating decimal can be seen as repeating with zeros, e.g. $0.5 = 0.5\overline{0}$.

Converting a repeating decimal back to a fraction. Let $x$ equal the decimal, multiply by a suitable power of 10 to line up the repeating blocks, then subtract to eliminate the repeating tail. For example, with $x = 0.\overline{63}$: $100x = 63.\overline{63}$, so $100x - x = 63$, giving $99x = 63$ and $x = \frac{63}{99} = \frac{7}{11}$.

Common mistakes to avoid:

  • Forgetting to reduce to lowest terms first. $\frac{6}{15}$ looks like it has a factor 3, but $\frac{6}{15} = \frac{2}{5}$ terminates.
  • Thinking the numerator matters — only the denominator (in lowest terms) decides terminating vs repeating.
  • Mis-reading $0.\overline{0}$-type cases: a terminating decimal is still rational.
  • When converting a repeating decimal, multiplying by the wrong power of 10 so the repeating blocks do not align before subtracting.
Flowchart deciding whether a rational p/q gives a terminating or repeating decimalTerminating or repeating? Test the denominatorp/q in lowest termsIs q = 2^m · 5^n(only 2s and 5s)?YESTERMINATING7/20, 20 = 2^2·5 → 0.35NONON-TERMINATING5/6, 6 = 2·3 → 0.8̅3Examples classifying fractions as terminating or repeating by their denominatorReading the denominatorFractionDenominatorDecimal type3/88 = 2^3Terminating9/1010 = 2 × 5Terminating2/77 (prime ≠ 2, 5)RepeatingOnly the denominator (in lowest terms) decides — the numerator is irrelevant.
Solved Examples
1
Worked Example
Without doing the division, decide whether $\frac{7}{80}$ has a terminating or repeating decimal.
Solution
  1. Step 1: $\frac{7}{80}$ is already in lowest terms since $\mathrm{HCF}(7, 80) = 1$.
  2. Step 2: Factorise the denominator: $80 = 2^{4} \times 5$.
  3. Step 3: It contains only the primes 2 and 5, so the decimal terminates.

Answer: Terminating

2
Worked Example
Express $\frac{1}{8}$ as a decimal and classify it.
Solution
  1. Step 1: $8 = 2^{3}$, only the prime 2, so it should terminate.
  2. Step 2: Divide: $1 \div 8 = 0.125$.
  3. Step 3: It ends after 3 places.

Answer: $0.125$ (terminating)

3
Worked Example
Without dividing, state whether $\frac{17}{30}$ terminates.
Solution
  1. Step 1: Check lowest terms: $\mathrm{HCF}(17, 30) = 1$, so it is reduced.
  2. Step 2: Factorise: $30 = 2 \times 3 \times 5$.
  3. Step 3: The prime 3 is present, so the decimal does not terminate.

Answer: Non-terminating repeating

4
Worked Example
State (with reason) whether $\frac{6}{15}$ has a terminating decimal.
Solution
  1. Step 1: Reduce first: $\frac{6}{15} = \frac{2}{5}$.
  2. Step 2: Denominator $5 = 5^{1}$, only the prime 5.
  3. Step 3: So it terminates: $\frac{2}{5} = 0.4$.

Answer: Terminating ($= 0.4$)

5
Worked Example
Is $\frac{64}{455}$ terminating? Give a reason.
Solution
  1. Step 1: $\mathrm{HCF}(64, 455) = 1$, so the fraction is reduced.
  2. Step 2: Factorise: $455 = 5 \times 7 \times 13$.
  3. Step 3: The primes 7 and 13 (other than 2 and 5) are present.

Answer: Non-terminating repeating

6
Worked Example
After how many decimal places does $\frac{13}{2^{2}\times 5^{3}}$ terminate?
Solution
  1. Step 1: Denominator $= 2^{2} \times 5^{3}$ has only primes 2 and 5, so it terminates.
  2. Step 2: The number of places is $\max(2, 3) = 3$.
  3. Step 3: Indeed $\frac{13}{500} = \frac{13 \times 2}{1000} = \frac{26}{1000} = 0.026$.

Answer: After 3 places ($= 0.026$)

7
Worked Example
Write the decimal expansion of $\frac{17}{8}$.
Solution
  1. Step 1: $8 = 2^{3}$, so the decimal terminates after 3 places.
  2. Step 2: Make the denominator a power of 10: $\frac{17}{8} = \frac{17 \times 125}{1000} = \frac{2125}{1000}$.
  3. Step 3: So $\frac{17}{8} = 2.125$.

Answer: $2.125$

8
Worked Example
Write the decimal expansion of $\frac{13}{3125}$.
Solution
  1. Step 1: $3125 = 5^{5}$, so it terminates after 5 places.
  2. Step 2: Balance the 2s: $\frac{13}{5^{5}} = \frac{13 \times 2^{5}}{10^{5}} = \frac{13 \times 32}{100000} = \frac{416}{100000}$.
  3. Step 3: So $\frac{13}{3125} = 0.00416$.

Answer: $0.00416$

9
Worked Example
Classify each as terminating or non-terminating: $\frac{13}{3125},\ \frac{17}{8},\ \frac{64}{455},\ \frac{15}{1600}$.
Solution
  1. Step 1: $\frac{13}{3125}$: $3125 = 5^{5}$ → terminating.
  2. Step 2: $\frac{17}{8}$: $8 = 2^{3}$ → terminating.
  3. Step 3: $\frac{64}{455}$: $455 = 5\times 7\times 13$ → non-terminating.
  4. Step 4: $\frac{15}{1600} = \frac{3}{320}$, and $320 = 2^{6}\times 5$ → terminating.

Answer: Terminating, terminating, non-terminating, terminating

10
Worked Example
Convert the repeating decimal $0.4\overline{6}$ (that is $0.4666\ldots$) into a fraction $\frac{p}{q}$.
Solution
  1. Step 1: Let $x = 0.4666\ldots$ Multiply by 10 to shift past the non-repeating digit: $10x = 4.666\ldots$
  2. Step 2: Multiply by 100 to align the repeating block: $100x = 46.666\ldots$
  3. Step 3: Subtract: $100x - 10x = 46.666\ldots - 4.666\ldots = 42$, so $90x = 42$.
  4. Step 4: $x = \frac{42}{90} = \frac{7}{15}$.

Answer: $\frac{7}{15}$

11
Worked Example
Convert $0.\overline{63}$ into a fraction in simplest form.
Solution
  1. Step 1: Let $x = 0.636363\ldots$ The repeating block has 2 digits, so multiply by 100.
  2. Step 2: $100x = 63.636363\ldots$
  3. Step 3: Subtract: $100x - x = 63$, so $99x = 63$.
  4. Step 4: $x = \frac{63}{99} = \frac{7}{11}$.

Answer: $\frac{7}{11}$

12
Worked Example
Express $\frac{1}{7}$ as a decimal and explain why it repeats.
Solution
  1. Step 1: $7$ is a prime other than 2 and 5, so $\frac{1}{7}$ cannot terminate.
  2. Step 2: Long division gives $1 \div 7 = 0.142857142857\ldots$
  3. Step 3: The block 142857 repeats forever.

Answer: $0.\overline{142857}$ (repeating)

Key Points

  • Rational numbers $\frac{p}{q}$ ($q \ne 0$) give either terminating or repeating decimals — never non-repeating
  • Reduce to lowest terms first; only the denominator decides the type
  • Terminating iff the reduced denominator is $2^{m}\times 5^{n}$; a terminating decimal ends after $\max(m, n)$ places
  • Any other prime factor (3, 7, 11, …) forces a repeating decimal
  • Repeating decimals use a bar over the block: $0.\overline{3}$, $0.\overline{142857}$
  • Every repeating decimal converts back to a fraction by aligning the blocks and subtracting
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.A rational $\tfrac{p}{q}$ (in lowest terms) has a terminating decimal iff $q$ is of the form:
Explanation: Only factors $2$ and $5$ allow termination.
Q2.$\tfrac{3}{8}$ is:
Explanation: $8=2^3$, so it terminates.
Q3.$\tfrac{1}{3}$ is:
Explanation: $0.\overline3$.
Q4.$\tfrac{13}{3125}$ is:
Explanation: $3125=5^5$, so it terminates.
Practice more MCQs → 📝 Assignment · 35 marks