Converting Repeating Decimals into Fractions

Rational numbers, when written as decimals, are either terminating or non-terminating repeating decimals. Convertig terminating decimals into fractions is straightforward: multiplying and dividing by an appropriate power of ten does the trick. For example, \(2.556753 = \frac{2556753}{1000000}.\) However, when the decimals are repeating, things are a little more difficult. Repeating decimals occur very frequently both when doing simple arithmetic and when solving competition problems, so being able to convert them to fractions is a valuable skill.

Some examples of non-terminating repeating decimals are \(0.12121212121212\ldots\) and \(1.2354354354354\ldots.\) We can represent these decimals in short as \(0.\overline{12}\) and \(1.2\overline{354},\) respectively.

To convert these type of decimals to fractions, we can view the decimal as the sum of (infinite) terms in a geometrical progression. This can be easily understood by some examples.

Write \(0.\overline{34}\) as a fraction.

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Proof 1:

We can write \(0.\overline{34}\) as \(0.3434343434 \ldots\). Now let \(x=0.\overline{34},\) then \[\begin{align} x &= 0.34 + 0.0034 + 0.000034 + \cdots\\ &= \frac{34}{100} + \frac{34}{10000} + \frac{34}{1000000} + \cdots\\ &= 34 \times \left( \frac{1}{100^{1}} + \frac{1}{100^{2}} + \frac{1}{100^{3}} + \cdots \right). \end{align}\] Recognize that this is the sum of infinite terms of a GP which has initial term \(a = \frac{1}{100}\) and common ratio \(r = \frac{1}{100}.\) Since the sum of infinite terms is \(\frac{a}{1-r},\) substituting the values of \(a\) and \(r\) gives \[x = 34 \times \dfrac{\frac{1}{100}}{1 - \frac{1}{100}} = 34 \times \frac{1}{99} = \dfrac{34}{99}. \ _\square \]

Proof 2:

Here is an alternative way to solve this problem:
Let \(x = 0.3434343434 \ldots,\) then \(100x = 34.343434 \ldots.\)
On subtracting the first equation from the second, we have \[99 x = 34 \Rightarrow x = \dfrac{34}{99}. \ _\square \]

Write \(0.\overline{1}\) as a fraction.

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We can write \(0.\overline{1}\) as \(0.1111111111 \ldots\). Let \(x=0.\overline{1},\) then \[\begin{align} x &= 0.1 + 0.01 + 0.001 + \cdots\\ &= \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \cdots\\ &= \frac{\frac{1}{10}}{1-\frac{1}{10}}=\frac{1}{9}. \ _\square \end{align}\]

Write \(0.0\overline{23}\) as a fraction.

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We can write \(0.0\overline{23}\) as \(0.02323232323 \ldots\). Let \(x=0.0\overline{23},\) then \[\begin{align} x &= 0.023 + 0.00023 + 0.0000023 + \cdots\\ &= \frac{23}{1000} + \frac{23}{100000} + \frac{23}{10000000} + \cdots\\ &= \frac{23}{1000} \times \left( 1 + \frac{1}{100} + \frac{1}{100^2} + \cdots \right)\\ &= \frac{23}{1000} \times \frac{1}{1-\frac{1}{100}} \\ &= \frac{23}{1000} \times \frac{100}{99}=\frac{23}{990}. \ _\square \end{align}\]

Write \(4.1\overline{454}\) as a fraction.

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We can write \(4.1\overline{454}\) as \(4.1454454454454 \ldots\). Let \(x=4.1\overline{454},\) then \[\begin{align} x &= 4.1+0.0454 +0.0000454 + 0.0000000454 + \cdots\\ &= \frac{41}{10} + \frac{454}{10000} + \frac{454}{10000000} +\frac{454}{10000000000}+ \cdots\\ &= \frac{41}{10} +\frac{454}{10000} \times \left( 1 + \frac{1}{1000} + \frac{1}{1000^2} + \cdots \right)\\ &= \frac{41}{10} +\frac{454}{10000} \times \frac{1}{1-\frac{1}{1000}} \\ &= \frac{41}{10} +\frac{454}{10000} \times \frac{1000}{999}\\ &=\frac{41}{10} +\frac{454}{9990}=\frac{41413}{9990}. \ _\square \end{align}\]

Which of the following is equal to \(0.\overline{5}+0.\overline{7}?\) \[\] \[\begin{array} &(a)~ 1.\overline{2} &&&(b)~ 1.\overline{3} &&&(c)~ 1.2\overline{3} &&&(d)~ 1.3\overline{2} \end{array}\]

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We can write \(0.\overline{5}\) as \(0.55555555 \ldots\). Let \(x=0.\overline{5},\) then \[\begin{align} x &= 0.5+0.05 +0.005 + 0.0005 + \cdots\\ &= \frac{5}{10} + \frac{5}{10^2} + \frac{5}{10^3} +\frac{5}{10^4}+ \cdots\\ &= \frac{5}{10} \times \left( 1 + \frac{1}{10} + \frac{1}{10^2} +\frac{1}{10^3} + \cdots \right)\\ &= \frac{5}{10} \times \frac{1}{1-\frac{1}{10}} = \frac{5}{10} \times \frac{10}{9}=\frac{5}{9}. \end{align}\] Similarly, if we let \(y=0.\overline{7},\) then we can get \(y=\frac{7}{9}. \) Thus, \[0.\overline{5}+0.\overline{7}=x+y=\frac{5}{9}+\frac{7}{9}=\frac{12}{9}=1+\frac{3}{9}=1.\overline{3}.\]

Therefore, the answer is \(1.\overline{3}.\) \(\ _\square\)

The non-terminating repeating decimal \(3.9\overline{1}\) can be written as a fraction \(\displaystyle{\frac{176}{a}}.\) What is \(a?\)

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Observe that \[\begin{align} 100x &= 391.1111111 \ldots \qquad (1)\\ 10x &= 39.1111111 \ldots . \qquad (2) \end{align}\] Taking \((1)-(2)\) gives \[90x=352 \Rightarrow x=\frac{176}{45},\] which implies \(a=45.\) \( _\square\)