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:
\(\frac{5}{6} = 0.8333 ... = 0.8\dot{3}\)
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...
Call this number \(x\). We have an equation \(x = 0.1111111\)...
If we multiply this number by 10 it will give a different number with the same digit recurring.
So if:
\(x = 0.11111111\)...then
\(10x = 1.11111111âŠ\)
Notice that after the decimal points the recurring digits match up. So subtracting these equations gives:
\(10x~â~x = 1.111111⊠â 0.111111âŠ\)
so \(9x = 1\)
Dividing both sides by 9 gives:
\(x = \frac{1}{9}\)
so \(~ 0. \dot{1} = \frac{1}{9}\)
When 2 digits recur, multiply by 100 so that the recurring digits after the decimal point keep the same place value. Similarly, when 3 digits recur multiply by 1000 and so on.
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 \(x\).
We have an equation \(x = 0.181818 \dotsc\)
If we multiply this number 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 \(\frac{18}{99}\) 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 doesnât 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 \(x = 0.288888...\)
So \(10x = 2.88888...\)
Subtracting these equations gives:
\(10x-x=2.888888⊠â 0.288888\)
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 \(\frac{13}{45}\).