Terminating and Recurring Decimals
Every fraction becomes either a terminating decimal, which stops, or a recurring decimal, which repeats a block of digits forever. This NSW Year 9 Advanced maths subtopic revises the Stage 4 work (NESA outcome MA4‑FRC‑C‑01) and explains why each kind happens.
You will write repeating decimals in dot notation, convert fractions such as \(\dfrac{5}{12}\) and \(\dfrac{4}{7}\) to recurring decimals, and decide from the prime factors of the denominator, without dividing, whether a fraction terminates. You will order recurring decimals, find the \(100\)th digit of \(\dfrac{3}{7}\), and explain why sevenths repeat in blocks of at most six digits. It is part of Year 9 Advanced Working with Numbers.
Every question with a fully worked solution.
- Terminating and Recurring Decimals - Video - Terminating and Repeating Decimals Watch
Theory
A fraction becomes either a terminating decimal, such as \(\dfrac{9}{50} = 0.18\), or a recurring decimal, such as \(\dfrac{11}{12} = 0.91\dot{6}\). This Year 9 Advanced maths page revises the skills from earlier years and extends them: dot notation, predicting from the denominator whether a decimal terminates, counting decimal places, finding digits far along a recurring decimal and ordering recurring decimals.
A terminating decimal stops, such as \(0.046875\). A recurring decimal repeats a digit or block of digits forever, such as \(0.153846153846\ldots\). Every fraction gives one or the other, so both kinds are rational numbers.
Dot notation shows which digits repeat. \(0.91\dot{6}\) means \(0.91666\ldots\): only the \(6\) repeats. \(0.\dot{1}5384\dot{6}\) means the whole block \(153846\) repeats; the dots sit on its first and last digits.
Whether a fraction terminates depends only on its denominator, once the fraction is in simplest form. A decimal place is a power of \(10 = 2 \times 5\), so the fraction terminates exactly when the denominator has no prime factors other than \(2\) and \(5\).
A calculator shows only about ten digits and rounds the last one. \(0.8333333333\) is not exact; the exact value is \(0.8\dot{3}\).
Dot notation marks the repeating digits. A single dot means one digit repeats; two dots mark the first and last digits of a repeating block:
Some books draw a bar (vinculum) over the whole block instead: \(0.\overline{246}\) means the same as \(0.\dot{2}4\dot{6}\).
For a fraction \(\dfrac{a}{b}\) in simplest form:
How to work with terminating and recurring decimals
- Simplify the fraction, then factorise the denominator. Only 2s and 5s: it terminates. Any other prime: it recurs.
- To convert, divide the numerator by the denominator, bringing down zeros. List each remainder.
- Stop when a remainder repeats. The digits produced since that remainder first appeared form the repeating block.
- Write dot notation: digits before the block stay plain; dot one digit, or the first and last digits of a longer block.
- To compare, write each number out to several places and compare digit by digit from the left.
(i) \(\dfrac{11}{12}\)
(ii) \(\dfrac{2}{13}\)
Divide, bringing down a zero each time, and watch the remainders. When a remainder repeats, the digits repeat.
(i)
| \(110 \div 12\) | \(=\) | \(9\ \text{r}\ 2\) |
| \(20 \div 12\) | \(=\) | \(1\ \text{r}\ 8\) |
| \(80 \div 12\) | \(=\) | \(6\ \text{r}\ 8\) |
| \(80 \div 12\) | \(=\) | \(6\ \text{r}\ 8\) |
Remainder \(8\) repeats straight away, so only the \(6\) recurs: \(\dfrac{11}{12} = \mathbf{0.91\dot{6}}\).
(ii)
| \(20 \div 13\) | \(=\) | \(1\ \text{r}\ 7\) |
| \(70 \div 13\) | \(=\) | \(5\ \text{r}\ 5\) |
| \(50 \div 13\) | \(=\) | \(3\ \text{r}\ 11\) |
| \(110 \div 13\) | \(=\) | \(8\ \text{r}\ 6\) |
| \(60 \div 13\) | \(=\) | \(4\ \text{r}\ 8\) |
| \(80 \div 13\) | \(=\) | \(6\ \text{r}\ 2\) |
Remainder \(2\) is the starting number again, so the block \(153846\) repeats. Dot its first and last digits: \(\dfrac{2}{13} = \mathbf{0.\dot{1}5384\dot{6}}\).
(i) Does \(\dfrac{27}{150}\) terminate?
(ii) Does \(\dfrac{11}{48}\) terminate?
(iii) How many decimal places does \(\dfrac{3}{64}\) have?
(i) Simplify first: \(\dfrac{27}{150} = \dfrac{9}{50}\), and \(50 = 2 \times 5^2\). Only 2s and 5s, so it terminates: \(\dfrac{9}{50} = \dfrac{18}{100} = 0.18\).
(ii) \(\dfrac{11}{48}\) is already in simplest form, and \(48 = 2^4 \times 3\). The factor \(3\) means it recurs.
(iii) \(64 = 2^6\). Multiply top and bottom by \(5^6\) to make a power of \(10\):
| \(\dfrac{3}{64}\) | \(=\) | \(\dfrac{3 \times 5^6}{2^6 \times 5^6}\) |
| \(\) | \(=\) | \(\dfrac{46\,875}{1\,000\,000}\) |
| \(\) | \(=\) | \(0.046875\) |
A denominator of \(10^6\) gives \(\mathbf{6}\) decimal places.
(i) Find the \(50\)th digit after the decimal point of \(\dfrac{5}{13} = 0.\dot{3}8461\dot{5}\).
(ii) Write in ascending order: \(0.3\dot{7},\ 0.\dot{3}\dot{7},\ 0.37,\ 0.\dot{3}7\dot{3}\).
(i) The block \(384615\) has \(6\) digits. Divide \(50\) by \(6\).
| \(50\) | \(=\) | \(6 \times 8 + 2\) |
After \(8\) full blocks (\(48\) digits), the \(50\)th digit is the \(2\)nd digit of the block \(3, \mathbf{8}, 4, 6, 1, 5\). It is \(\mathbf{8}\).
(ii) Write each one out to \(6\) places and compare digit by digit.
| \(0.3\dot{7}\) | \(=\) | \(0.377777\ldots\) |
| \(0.\dot{3}\dot{7}\) | \(=\) | \(0.373737\ldots\) |
| \(0.37\) | \(=\) | \(0.370000\) |
| \(0.\dot{3}7\dot{3}\) | \(=\) | \(0.373373\ldots\) |
Ascending: \(\boldsymbol{0.37,\ 0.\dot{3}7\dot{3},\ 0.\dot{3}\dot{7},\ 0.3\dot{7}}\).
(i) Find the smallest positive whole number \(k\) for which \(\dfrac{k}{126}\) terminates.
(ii) A calculator shows \(0.8333333333\) for a fraction with denominator \(12\). Find the fraction and write it with dot notation.
(i) \(126 = 2 \times 3^2 \times 7\). The \(3^2\) and the \(7\) must cancel, so \(k\) must be a multiple of \(9 \times 7 = 63\).
| \(k\) | \(=\) | \(\mathbf{63}\) |
| \(\dfrac{63}{126}\) | \(=\) | \(\dfrac{1}{2} = 0.5\) |
(ii) Multiply by the denominator to find the numerator. The display is rounded, so expect a value very close to a whole number.
| \(0.8333333333 \times 12\) | \(=\) | \(9.9999\ldots \approx 10\) |
| \(\dfrac{10}{12}\) | \(=\) | \(\dfrac{5}{6} = \mathbf{0.8\dot{3}}\) |
The calculator showed only \(10\) digits. Only the \(3\) recurs, so the dot goes over the \(3\) alone.
Common pitfalls
Frequently asked questions
What is the difference between a terminating and a recurring decimal?
A terminating decimal stops, such as \(0.18\). A recurring decimal repeats a digit or block of digits forever, such as \(0.91\dot{6}\).
How do you know if a fraction will terminate?
Write it in simplest form and find the prime factors of the denominator. If the only prime factors are \(2\) and \(5\), the decimal terminates. Any other prime factor makes it recur.
How do you write a recurring decimal with dots?
Put a dot over a single repeating digit. For a repeating block, put dots over the first and last digits of the block. Digits before the block get no dots.
How do you convert a fraction to a recurring decimal?
Divide the numerator by the denominator, bringing down zeros. Write down each remainder. When a remainder appears again, the digits from that point repeat.
Is a recurring decimal a rational number?
Yes. Every recurring decimal can be written as a fraction, so it is rational. Irrational numbers such as \(\pi\) never repeat.