Converting recurring decimals - Higher
Using dot notation
A recurring decimal exists when A number that uses powers of 10 as place value. In the example of 0.82, the 8 represents tenths and the 2 represents hundredths. 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.
First, convert the A fraction is a part of a whole, for example 1/2. to a decimal by division:
\(\frac{5}{6}~=~0.8333...~=~0.8\dot{3}\)
Algebra skills are needed to turn 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 repeats of the decimal.
0.111111111鈥︹赌
Call this number \(x\). We have an equation \(x~=~0.1111111鈥)
If we multiply this number by 10 it will give a different number with the same digit recurring.
So if:
\(x~=~0.1111111鈥) then
袄(10虫词=词0.1111111鈥)
Notice that after the decimal points the recurring digits match up. So subtracting these equations gives:
袄(10虫词-词虫词=词1.111111...词-词0.111111鈥)
So \(9x~=~1\)
Dividing both sides by 9 gives:
\(x~=~\frac{1}{9}\)
So \(0. \dot{1}~=~\frac{1}{9}\)
Question
Prove 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 repeats (it doesn't matter exactly how many are used).
\(0. \dot{1}\dot{8}~=~0.181818...\)
Call this number \(x\).
We have an equation \(x~=~0.181818...\)
If we multiply this number by 100 it will give a different number with the same digits recurring. So if:
\(x~=~0.181818...\) then
\(100x~=~18.181818...\)
Notice that after the decimal points the recurring digits match up. So subtracting these equations gives:
\(100x~-~x~=~18.181818...~-~0.181818...\)
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
Prove that \(0.2 \dot{8}\) is equal to \(\frac{13}{45}\).
First, write the recurring decimal as a long number. Use a few repeats (it doesn't matter exactly how many are used).
\(0.2 \dot{8}~=~0.28888888...\)
Give this number a name (\(x\)):
\(x~=~0.28888888...\)
1 digit recurs, so multiply by 10.
So \(x~=~0.28888888...\)
\(10x~=~2.8888888...\)
Subtracting these equations gives:
\(10x~-~x~=~2.888888...~-~0.288888...\)
So, \(9x~=~2.6\)
Multiplying both sides by 10 gives:
\(90x~=~26\)
Dividing both sides by 90, (to get the value of \(x\)):
\(x~=~\frac{26}{90}\)
Simplify the fraction \(\frac{26}{90}\) to its simplest form:
\(x~=~\frac{13}{45}\)
\(0.2 \dot{8}\) as a fraction is \(\frac{13}{45}\).
- 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 \(\sqrt{2}\) which is also called a surd.