Converting Repeating Decimals into Fractions

Rational numbers, when written as decimals, are either terminating or non-terminating, repeating decimals. Converting 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 types 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 $\begin{aligned} x &= 0.34 + 0.0034 + 0.000034 + \cdots\\ &= \frac{34}{100} + \frac{34}{10000} + \frac{34}{1000000} + \cdots\\ &= 34 \times \left( \frac{1}{100^{1}} + \frac{1}{100^{2}} + \frac{1}{100^{3}} + \cdots \right). \end{aligned}$ 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 $99 x = 34 \implies x = \dfrac{34}{99}. \ _\square$

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

$\begin{aligned} x &= 0.1 + 0.01 + 0.001 + \cdots\\ &= \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \cdots\\ &= \frac{\frac{1}{10}}{1-\frac{1}{10}}\\\\ &=\frac{1}{9}. \ _\square \end{aligned}$

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

$\begin{aligned} x &= 0.023 + 0.00023 + 0.0000023 + \cdots\\ &= \frac{23}{1000} + \frac{23}{100000} + \frac{23}{10000000} + \cdots\\ &= \frac{23}{1000} \times \left( 1 + \frac{1}{100} + \frac{1}{100^2} + \cdots \right)\\ &= \frac{23}{1000} \times \frac{1}{1-\frac{1}{100}} \\ &= \frac{23}{1000} \times \frac{100}{99}\\\\ &=\frac{23}{990}. \ _\square \end{aligned}$

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

$\begin{aligned} x &= 4.1+0.0454 +0.0000454 + 0.0000000454 + \cdots\\ &= \frac{41}{10} + \frac{454}{10000} + \frac{454}{10000000} +\frac{454}{10000000000}+ \cdots\\ &= \frac{41}{10} +\frac{454}{10000} \times \left( 1 + \frac{1}{1000} + \frac{1}{1000^2} + \cdots \right)\\ &= \frac{41}{10} +\frac{454}{10000} \times \frac{1}{1-\frac{1}{1000}} \\ &= \frac{41}{10} +\frac{454}{10000} \times \frac{1000}{999}\\ &=\frac{41}{10} +\frac{454}{9990}\\\\ &=\frac{41413}{9990}. \ _\square \end{aligned}$

Which of the following is equal to $0.\overline{5}+0.\overline{7}?$

$\begin{array}{c}&(a)~ 1.\overline{2} &&&(b)~ 1.\overline{3} &&&(c)~ 1.2\overline{3} &&&(d)~ 1.3\overline{2} \end{array}$

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We can write $0.\overline{5}$ as $0.55555555 \ldots$. Let $x=0.\overline{5},$ then

$\begin{aligned} x &= 0.5+0.05 +0.005 + 0.0005 + \cdots\\ &= \frac{5}{10} + \frac{5}{10^2} + \frac{5}{10^3} +\frac{5}{10^4}+ \cdots\\ &= \frac{5}{10} \times \left( 1 + \frac{1}{10} + \frac{1}{10^2} +\frac{1}{10^3} + \cdots \right)\\ &= \frac{5}{10} \times \frac{1}{1-\frac{1}{10}} \\ &= \frac{5}{10} \times \frac{10}{9}\\\\ &=\frac{5}{9}. \end{aligned}$

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 ${\frac{176}{a}}.$ What is $a?$

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Observe that

$\begin{aligned} 100x &= 391.1111111 &\qquad (1)\\ 10x &= 39.1111111. &\qquad (2) \end{aligned}$

Taking $(1)-(2)$ gives

$90x=352 \implies x=\frac{176}{45},$

which implies $a=45.$ $_\square$