Converting Repeating Decimals Into Fractions - Brilliant

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.

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 \[99 x = 34 \implies x = \dfrac{34}{99}. \ _\square \]

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

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.

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.

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 \(0.\overline{5}+0.\overline{7}?\)

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

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 \({\frac{176}{a}}.\) What is \(a?\)

Observe that

\[\begin{align} 100x &= 391.1111111 &\qquad (1)\\ 10x &= 39.1111111. &\qquad (2) \end{align}\]

Taking \((1)-(2)\) gives

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

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

True False Reveal the answer

True or False?

\[0.9999 \ldots = 1\]

\(\) Note: The "\(\ldots\)" indicates that there are infinitely many 9's.

The correct answer is: True

Intuition: Visualize a number line. If two real numbers \(x\) and \(y\) on the number line are different, then we should see some space between them. In fact, there would be room to place another real number, namely their average \(\frac{x + y}{2}\). Since no number exists between \(0.999\ldots\) and \(1,\) it must be that they are the same.

Proof: This solution converts a repeating decimal into a fraction in order to calculate the 'fraction' that 0.99999... is equivalent to.

\[ \begin{array} {l r l } \text{Let } & A & = 0. 999 \ldots \\ \text{Multiplying by 10, we get } & 10 A & = 9. 999 \ldots \\ \text{Subtracting } A, \text{ we get } & 9 A & = 9 \\ \text{Dividing by 9, we get }& A & = 1 \ _\square \end{array} \]

For a deeper understanding, read the wiki on why .9999... = 1.