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
Write \(\frac{5}{6}\) as a 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 skills are needed to turn recurring decimals into fractions.
Example
Convert \(0. \dot{1}\) to a fraction.
Firstly, write out \(0. \dot{1}\) as a number, using a few iterations (repeats) of the recurring digit. Give this number a name (\(x\) is usually used).
If \(x = 0. \dot{1}\) is written in long form, it is: \(x = 0.11111 \dotsc\) (the 1s repeat forever). It is now possible to manipulate the number to create another equation which can be used to find the fraction.
As there is a 1 in every decimal place, it will be difficult to eliminate this, unless another number can be found that also has a recurring 1 in every decimal place. If this happens, one decimal can be subtracted from the other, removing all the recurring numbers.
\(x = 0.11111 \dotsc\) contains only one digit that recurs - the 1. To create another number with recurring 1s, multiply \(x = 0.11111\) by 10. This will give \(10x = 1.11111 \dotsc\)
If one digit recurs, multiply by 10. If two digits recur, multiply by 100. If three digits recur, multiply by 1,000, and so on.
So:
\(x = 0.11111 \dotsc\)
\(10x = 1.11111 \dotsc\)
To solve these two equations and write \(0. \dot{1}\) as a fraction, take \(x\) away from \(10x\) to remove all the recurring decimal places:
\(10x - x = 1.11111 \dotsc - 0.11111 \dotsc\)
So: \(9x = 1\)
Next, divide each side by 9, to get the value of \(1x\):
\(9x = 1\)
\(\div 9\)
\(x = \frac{1}{9}\)
\(0. \dot{1}\) as a fraction is \(\frac{1}{9}\).
Question
Show that \(0. \dot{1} \dot{8}\) is equal to \(\frac{2}{11}\).
First, write the recurring decimal as a long number. Use a few iterations (it doesn't matter exactly how many are used).
\(0. \dot{1} \dot{8} = 0.181818 \dotsc\)
Give this number a name (\(x\)):
\(x = 0.181818 \dotsc\)
Examine how many digits recur. In this case, there are two digits that recur, so multiply by 100.
This gives:
\(100x = 18.181818 \dotsc\)
\(x = 0.181818 \dotsc\)
To solve these two equations and write \(0. \dot{1} \dot{8}\) as a fraction, take \(x\) away from \(100x\) to remove all the recurring decimal places:
This gives:
\(100x - x = 18.181818 \dotsc - 0.181818 \dotsc\)
So, \(99x = 18\)
Next, divide each side by 99, to get the value of \(1x\):
\(99x = 18\)
\(\div 99\)
\(x = \frac{18}{99}\)
Simplify this fraction by taking out any common factors. In this case, 9 is a common factor:
\(x = \frac{18}{99} = \frac{2}{11}\)
\(0. \dot{1} \dot{8}\) as a fraction is \(\frac{2}{11}\).
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’t matter exactly how many are used).
\(0.2 \dot{8} = 0.288888 \dotsc\)
Give this number a name (\(x\)):
\(x = 0.288888 \dotsc\)
Here, we need to create two additional equations where only a recurring part of the number is to the right of the decimal point.
This gives:
\(\begin{array}{rcl} 100x & = & 28.88888 \dotsc \\ 10x & = & ~~ 2.88888 \dotsc \end{array}\)
To solve these two equations and write \(0.2 \dot{8}\) as a fraction, take \(10x\) away from \(100x\) to remove all the recurring decimal places:
This gives:
\(100x - 10x = 28.88888 \dotsc - 2.88888 \dotsc\)
So, \(90x = 26\)
Next, divide each side by 90, to get the value of \(1x\):
\(90x = 26\)
\(\div 90\)
\(x = \frac{26}{90}\)
Simplify this fraction by taking out any common factors. In this case, 2 is a common factor:
\(x = \frac{26}{90} = \frac{13}{45}\)
\(0.2 \dot{8}\) as a fraction is \(\frac{13}{45}\).
More guides on this topic
- NEW: Whole numbers
- NEW: Order of operations and negative numbers
- NEW: Decimals
- NEW: Converting between fractions, decimals and percentages
- NEW: Higher – How to convert recurring decimals
- NEW: How to round numbers
- NEW: What is accuracy in maths?
- NEW: What are fractions?
- NEW: How to add, subtract, multiply and divide fractions
- NEW: Multiples and factors
- NEW: Highest Common Factor and Lowest Common Multiple
- NEW: Laws of indices
- NEW: Negative and fractional indices
- NEW: Standard form
- NEW: Calculations using standard form
- Whole numbers - OCR
- Approximation - OCR
- Decimals - OCR
- Multiples, factors, powers and roots - OCR
- Standard form - OCR
- Laws of indices - OCR
- Fractions - OCR
- Surds - Higher - OCR
- Financial mathematics - OCR
