Exercise 35 Page 687

We want to write the repeating decimal $0.3\overline{21}$ as a fraction. To do so, we will start by writing this number as a sum of decimals and as a sum of fractions.

Number	$0.3\overline{21}$
Sum of Decimals	$0.3+0.021+0.00021+0.0000021+\dots$
Sum of Fractions	$\frac{3}{10}+\frac{21}{1000}+\frac{21}{100000}+\frac{21}{10000000}+\dots$

Consider the sum of fractions above. Note that from the $2^{\text{nd}}$ term on, we can think of it as a geometric series that has a first term of $a_1=\frac{21}{1000}.$ Let's find its common ratio $r.$

The common ratio is $r=\frac{1}{100}.$ Let's substitute this value — together with $a_1=\frac{21}{1000}$ — in the formula for the sum of an infinite geometric series.

$S=\frac{a_1}{1-r}$

SubstituteII

$r=\frac{1}{100}$, $a_1=\frac{21}{1000}$

$S=\frac{\frac{21}{1000}}{1-\frac{1}{100}}$

▼

Simplify right-hand side

OneToFrac

Rewrite $1$ as $\frac{100}{100}$

$S=\frac{\frac{21}{1000}}{\frac{100}{100}-\frac{1}{100}}$

SubFrac

Subtract fractions

$S=\frac{\frac{21}{1000}}{\frac{99}{100}}$

FracToDiv

$\frac{a}{b}=a÷b$

$S=\frac{21}{1000}÷\frac{99}{100}$

DivFracByFracDiv

$\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}\cdot \frac{d}{c}$

$S=\frac{21}{1000}\left(\frac{100}{99}\right)$

MultFrac

Multiply fractions

$S=\frac{2100}{99000}$

ReduceFrac

$\frac{a}{b}=\frac{a/300}{b/300}$

$S=\frac{7}{330}$

We found that the series formed by the sum of the terms, starting from the $2^{\text{nd}},$ is $\frac{7}{330}.$ With this information we can express the given number as a sum of two fractions.

$$\begin{array}{c}0.3\overline{21}=\frac{3}{10}+\frac{7}{330}\end{array}$$

Finally, we will add the fractions to obtain the value of $0.3\overline{21}$ expressed as a single fraction.

$0.3\overline{21}=\frac{3}{10}+\frac{7}{330}$

ExpandFrac

$\frac{a}{b}=\frac{a\cdot 33}{b\cdot 33}$

$0.3\overline{21}=\frac{99}{330}+\frac{7}{330}$

AddFrac

Add fractions

$0.3\overline{21}=\frac{106}{330}$

ReduceFrac

$\frac{a}{b}=\frac{a/2}{b/2}$

$0.3\overline{21}=\frac{53}{165}$