Converting recurring decimals - Higher
A recurring decimal exists when decimal numbers repeat forever. For example, \(0. \dot{3}\) means 0.333333... - the decimal never ends.
Dot notation is used with recurring decimals. The dot above the number shows which numbers recur, for example \(0.5 \dot{7}\) is equal to 0.5777777... and \(0. \dot{2} \dot{7}\) is equal to 0.27272727...
If two dots are used, they show the beginning and end of the recurring group of numbers: \(0. \dot{3} 1 \dot{2}\) is equal to 0.312312312...
Example
How is the number 0.57575757... written using dot notation?
In this case, the recurring numbers are the 5 and the 7, so the answer is \(0. \dot{5} \dot{7}\).
Example
Convert \(\frac{5}{6}\) to a recurring decimal.
Divide 5 by 6. 5 divided by 6 is 0, remainder 5, so carry the 5 to the tenths column.
50 divided by 6 is 8, remainder 2.
20 divided by 6 is 3 remainder 2.
Because the remainder is 2 again, the digit 3 is going to recur:
\(\frac{5}{6}\) = 0.8333鈥= \(0.8 \dot{3}\)
Algebra can be used to convert recurring decimals into fractions.
Example
Convert \(0. \dot{1}\) to a fraction.
\(0. \dot{1}\) has 1 digit recurring.
Firstly, write out \(0. \dot{1}\) as a number, using a few iterations (repeats) of the decimal.
0.111111111鈥
Call this number \(x\). Written as an equation \(x = 0.1111111 \dotsc\)
If this number is multiplied by 10 it will give a different number with the same digit recurring.
So if:
\(x = 0.11111111 \dotsc\) then:
\(10x = 1.11111111 \dotsc\)
Notice that after the decimal points the recurring digits match up. So subtracting these equations gives:
\(10x - x = 1.11111 \dotsc - 0.11111 \dotsc\)
So: \(9x = 1\)
Dividing both sides by 9 gives:
\(x = \frac{1}{9}\)
So, \(0. \dot{1}\) = \(\frac{1}{9}\).
Question
Show that \(0. \dot{1} \dot{8}\) is equal to \(\frac{2}{11}\).
\(0. \dot{1} \dot{8}\) has 2 digits recurring.
First, write the recurring decimal as a long number. Use a few iterations (it doesn't matter exactly how many are used).
\(0. \dot{1} \dot{8} = 0.181818 \dotsc\)
Call this number \(x\).
Shown as an equation \(x = 0.181818 \dotsc\)
If this number is multiplied by 100 it will give a different number with the same digits recurring. So if:
\(x = 0.181818 \dotsc\) then:
\(100x = 18.181818 \dotsc\)
Notice that after the decimal points the recurring digits match up. So subtracting these equations gives:
\(100x - x = 18.181818 \dotsc - 0.181818 \dotsc\)
So, \(99x = 18\)
Dividing both sides by 99 gives:
\(x = \frac{18}{99}\)
9 is a common factor of 18 and 99.
So, \(x = \frac{18}{99}\) simplifies to \(\frac{2}{11}\).
So, \(0. \dot{1} \dot{8}\) converts to \(\frac{2}{11}\).
Question
Show that \(0.2 \dot{8}\) is equal to \(\frac{13}{45}\).
First, write the recurring decimal as a long number. Use a few iterations (it doesn鈥檛 matter exactly how many are used).
\(0.2 \dot{8} = 0.288888 \dotsc\)
Give this number a name (\(x\)):
\(x = 0.288888 \dotsc\)
1 digit recurs, so multiply by 10.
So, \(x = 0.288888 \dotsc\)
\(10x = 2.88888 \dotsc\)
Subtracting these equations gives:
\(10x - x = 2.88888 \dotsc - 0.28888 \dotsc\)
So, \(9x = 2.6\)
Multiplying both sides by 10 gives:
\(90x = 26\)
Dividing both sides by 90, (to get the value of \(x\)):
\(x = \frac{26}{90}\)
2 is a common factor of 26 and 90, so:
\(x = \frac{26}{90} = \frac{13}{45}\)
\(0.2 \dot{8}\) as a fraction is \(\frac{13}{45}\).
Terminating, recurring and irrational decimals
Some decimals terminate which means the decimals do not recur, they just stop. For example, 0.75.
To find out whether a fraction will have a terminating or recurring decimal, look at the prime factors of the The bottom part of a fraction. For 鈪, the denominator is 8, which represents 'eighths'. when the fraction is in its most simple form. If they are made up of 2s and/or 5s, the decimal will terminate. If the prime factors of the denominator contain any other numbers, the decimal will recur.
Some decimals are irrational, which means that the decimals go on forever but not in a pattern (they are not recurring). An example of this would be \(\pi\) or \(\sqrt{2}\).