Terminating and Recurring Decimals
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Theory
Every fraction \(\dfrac{p}{q}\), with \(p\) and \(q\) integers, is a decimal that either terminates (stops) or recurs (repeats a block of digits forever). Simplify the fraction: if the denominator’s only prime factors are \(2\) and \(5\), it terminates. This Year 8 maths page shows how to write, test and convert both kinds.
A terminating decimal stops: \(\dfrac{3}{8} = 0.375\). A recurring decimal repeats the same block of digits forever: \(\dfrac{5}{11} = 0.4545\ldots\). When you divide, a remainder of \(0\) means it terminates; a remainder that comes back means it recurs.
The test. Simplify first, then write the denominator as a product of primes. If the only primes are \(2\) and \(5\), the decimal terminates. Any other prime factor makes it recur.
Terminating, without dividing. Build a power of \(10\): \(\dfrac{9}{40} = \dfrac{9 \times 25}{40 \times 25} = \dfrac{225}{1000} = 0.225\).
Dot notation. Put a dot over the first and last digit of the repeating block: \(0.4545\ldots = 0.\dot{4}\dot{5}\) and \(0.4818181\ldots = 0.4\dot{8}\dot{1}\). The block length is the number of digits in the block. For a mixed numeral, keep the whole number and divide out the fraction: \(2\dfrac{1}{6} = 2.1\dot{6}\).
Back to a fraction. A one-digit block sits over \(9\), a two-digit block over \(99\), a three-digit block over \(999\). Then simplify: \(0.\dot{2}\dot{1} = \dfrac{21}{99} = \dfrac{7}{33}\). Percentages: multiply by \(100\): \(\dfrac{2}{9} = 0.\dot{2} = 22.\dot{2}\%\).
Ordering. Write each number out to the same number of places, then compare digit by digit: \(0.\dot{1} = 0.1111\ldots\) and \(0.\dot{1}\dot{2} = 0.1212\ldots\), so \(0.\dot{1} < 0.12 < 0.\dot{1}\dot{2}\).
A digit far along. Divide its position by the block length; the remainder tells you which digit of the block it is (a remainder of \(0\) means the last digit).
A pattern for elevenths. \(\dfrac{1}{11} = 0.\dot{0}\dot{9}\), \(\dfrac{2}{11} = 0.\dot{1}\dot{8}\), \(\dfrac{3}{11} = 0.\dot{2}\dot{7}\): the two digits of the block add to \(9\).
Test the denominator (in simplest form):
| Fraction | Denominator in primes | Decimal |
|---|---|---|
| \(\dfrac{3}{16}\) | \(2^4\) | \(0.1875\) (terminates) |
| \(\dfrac{7}{20}\) | \(2^2 \times 5\) | \(0.35\) (terminates) |
| \(\dfrac{5}{6}\) | \(2 \times 3\) | \(0.8\dot{3}\) (recurs) |
| \(\dfrac{4}{33}\) | \(3 \times 11\) | \(0.\dot{1}\dot{2}\) (recurs) |
How to write a fraction as a decimal
- Simplify the fraction and test the denominator.
- Divide the numerator by the denominator, adding zeros after the point.
- Stop at a remainder of \(0\), or when a remainder repeats; then dot the repeating block.
How to write a recurring decimal as a fraction
- Put the block over \(9\), \(99\) or \(999\) (one \(9\) for each digit).
- Simplify the fraction.
(i) \(\dfrac{3}{8}\)
(ii) \(\dfrac{5}{11}\)
(i) \(3.000 \div 8\): \(30 \div 8\) is \(3\) remainder \(6\), \(60 \div 8\) is \(7\) remainder \(4\), \(40 \div 8\) is \(5\) remainder \(0\). The remainder is \(0\), so it stops.
(ii) \(5.00 \div 11\): \(50 \div 11\) is \(4\) remainder \(6\), \(60 \div 11\) is \(5\) remainder \(5\). The remainder \(5\) is back, so the digits repeat.
(i) \(\dfrac{3}{8} = 0.375\)
(ii) \(\dfrac{5}{11} = 0.4545\ldots = 0.\dot{4}\dot{5}\)
(i) \(\dfrac{7}{16}\)
(ii) \(\dfrac{20}{45}\)
(iii) \(\dfrac{15}{75}\)
| \(\dfrac{7}{16}\) | \(=\) | \(\dfrac{7 \times 625}{16 \times 625} = \dfrac{4375}{10\,000}\) |
(i) \(16 = 2^4\): \(0.4375\)
(ii) \(\dfrac{20}{45} = \dfrac{4}{9}\) and \(9 = 3^2\): \(0.\dot{4}\)
(iii) \(\dfrac{15}{75} = \dfrac{1}{5}\): \(0.2\)
(i) \(\dfrac{8}{33}\)
(ii) \(\dfrac{11}{18}\)
(i) \(80 \div 33\) is \(2\) remainder \(14\), \(140 \div 33\) is \(4\) remainder \(8\): the remainder \(8\) is back.
(ii) \(110 \div 18\) is \(6\) remainder \(2\), \(20 \div 18\) is \(1\) remainder \(2\): the remainder \(2\) repeats.
(i) \(0.\dot{2}\dot{4}\), block length \(2\)
(ii) \(0.6\dot{1}\), block length \(1\) (only the \(1\) repeats)
(i) \(0.\dot{5}\dot{4}\)
(ii) \(0.\dot{1}3\dot{5}\)
| \(0.\dot{5}\dot{4}\) | \(=\) | \(\dfrac{54}{99} = \dfrac{6}{11}\) |
| \(0.\dot{1}3\dot{5}\) | \(=\) | \(\dfrac{135}{999} = \dfrac{5}{37}\) |
(i) \(\dfrac{6}{11}\) (divide by \(9\))
(ii) \(\dfrac{5}{37}\) (divide by \(27\))
(ii) Write \(\dfrac{2}{9}\) as a percentage.
(iii) Order \(0.\dot{1}\dot{2}\), \(0.12\), \(0.\dot{1}\) from smallest.
(i) \(1 \div 6 = 0.1666\ldots\), so keep the \(2\).
(iii) Write them out: \(0.1212\ldots\), \(0.1200\), \(0.1111\ldots\)
(i) \(2.1\dot{6}\)
(ii) \(0.\dot{2} \times 100 = 22.\dot{2}\%\)
(iii) \(0.\dot{1}\), \(0.12\), \(0.\dot{1}\dot{2}\)
(i) Write each share exactly, in litres.
(ii) Give each share to the nearest millilitre.
\(11\) is prime and is not \(2\) or \(5\), so expect a recurring decimal. \(30 \div 11\) is \(2\) remainder \(8\), \(80 \div 11\) is \(7\) remainder \(3\): repeats.
(i) \(0.\dot{2}\dot{7}\) L
(ii) \(1\) mL \(= 0.001\) L, so keep \(3\) places: \(0.2727\ldots \approx 0.273\) L (\(273\) mL)
Common pitfalls
Frequently asked questions
What is a recurring decimal?
A decimal in which a block of digits repeats forever, such as \(\dfrac{1}{3} = 0.\dot{3}\).
How can you tell if a fraction gives a terminating decimal?
Simplify it. If the denominator has no prime factors other than 2 and 5, the decimal terminates.
How do you write a recurring decimal with dots?
Dot the first and last digit of the repeating block: \(0.454545\ldots = 0.\dot{4}\dot{5}\).
How do you change a recurring decimal to a fraction?
Put the block over as many 9s as it has digits, then simplify: \(0.\dot{2}\dot{7} = \dfrac{27}{99} = \dfrac{3}{11}\).
Are recurring decimals rational?
Yes. Every terminating or recurring decimal can be written as a fraction of two integers.