Converting Repeating Decimals into Fractions

Rational numbers, when written as decimals, are either terminating or non-terminating repeating decimals.

Some examples of terminating decimals are \(0.343, 43.5, 2.556753,\) etc. These can easily be converted to fraction by multiplying and dividing it by 10 sufficient times.

For example, \(2.556753 = \frac{2556753}{1000000}.\)

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 \(a=45.\) \( _\square\)