#### Chapter 1 Elementary Arithmetic

Section 1.2 Fractional Arithmetic

# 1.2.2 Converting Fractions

Dividing the denominator into the numerator gives a decimal fraction or a decimal number, respectively, for example,

$\frac{1}{2}=0.5 , \frac{1}{3}=0.33333\dots =0.\stackrel{‾}{3} , \frac{1}{7}=0.\stackrel{‾}{142857} , \frac{1}{8}=0.125 .$

By means of these examples, it can already be seen that the division is either finite, leading to a proper decimal fraction, or the digits of the decimal number repeat in a certain way, leading to an infinite repeating decimal fraction.
Converting decimal numbers to fractions is done using the base-10 positional notation. Each decimal number has the form

 1 2 3 4 5 . 6 7 8 9 TTH TH H T O . t h th tth

with the abbreviations TTH ... ten thousands, TH ... thousands, H ... hundreds, T ... tens, O ... ones, t ... tenths, h ... hundredths, th ... thousandths, tth ...  ten thousandths etc.
Then, the conversion is done as follows:

$\begin{array}{ccc}\multicolumn{1}{c}{4.375}& =\hfill & 4+\frac{3}{10}+\frac{7}{100}+\frac{5}{1000} \hfill \\ \multicolumn{1}{c}{}& =\hfill & 4+\frac{300+70+5}{1000} \hfill \\ \multicolumn{1}{c}{}& =\hfill & 4+\frac{375}{1000} \hfill \\ \multicolumn{1}{c}{}& =\hfill & 4+\frac{75}{200} \hfill \\ \multicolumn{1}{c}{}& =\hfill & 4+\frac{15}{40}=\frac{35}{8} .\hfill \end{array}$

But what about converting an infinite repeating decimal number? It seems that we would have to add an infinite number of fractions, which is in practise of less use, of course. Therefore, in converting infinite repeating decimal numbers to fractions we use a trick:
##### Info 1.2.612

Converting infinite repeating decimal numbers to fractions is done by multiplying the decimal number with a power of ten such that the repeating digits are shifted to the left of the decimal point. This leads to an equation of the form ${10}^{k}·x=x+n$ for the decimal number $x$, that can be solved for $x$: $x=\frac{n}{{10}^{k}-1}$ (which is a simple fraction).

##### Example 1.2.613
The number $0.\stackrel{‾}{6}$ is to be converted to a fraction. For this, you multiply the number by $10$ and subtract from the result the initial number to eliminate the infinite repeating decimal:
 $10$ $·$ $0.\stackrel{‾}{6}$ = $6.\stackrel{‾}{6}$ $-$ $1$ $·$ $0.\stackrel{‾}{6}$ = $0.\stackrel{‾}{6}$ $⇒$ $9$ $·$ $0.\stackrel{‾}{6}$ = $6.0$
From the last relation, it immediately follows after division by $9$:     $0.\stackrel{‾}{6}=\frac{6}{9}=\frac{2}{3}.$

This method also works if not all digits after the decimal point repeat periodically:
##### Example 1.2.614
The decimal number $0.8\stackrel{‾}{3}=0.83333\dots$ is to be converted to a fraction:
 $100$ $·$ $0.8\stackrel{‾}{3}$ = $83.\stackrel{‾}{3}$ $-$ $10$ $·$ $0.8\stackrel{‾}{3}$ = $8.\stackrel{‾}{3}$ $⇒$ $90$ $·$ $0.8\stackrel{‾}{3}$ = $75.0$
Division by $90$ gives the result: $0.8\stackrel{‾}{3}=\frac{75}{90}=\frac{5}{6}$.

Thus, the method is always the same: by multiplying by powers of ten and subsequent subtraction, the infinite repeating decimal is removed.
##### Exercise 1.2.615
Using the above method, find a simple and fully reduced fraction that represents the value $0.45555\dots$.
Answer: $0.4\stackrel{‾}{5}$$=$
.
Enter the fraction as numerator/denominator fully reduced and with positive denominator.

However, in back-of-the-envelope calculations (i.e. if you only want to roughly estimate a magnitude or the ratio of one number to the other without knowing the correct values of the decimal fractions) it is useful to multiply the numbers by the least common denominator instead of converting them to decimals:
##### Example 1.2.616
The fractions $\frac{2}{3}$, $\frac{32}{12}$, and $\frac{12}{15}$ are to be arranged in order of size. For this, the fractions are multiplied by the least common denominator ($60$, in this case). The denominators are cancelled and the fractions are converted to the integers

$\frac{2}{3}·60\mathrm{ }=\mathrm{ }2·20\mathrm{ }=\mathrm{ }40\mathrm{ }\mathrm{ },\mathrm{ }\mathrm{ }\frac{32}{12}·60\mathrm{ }=\mathrm{ }32·5\mathrm{ }=\mathrm{ }160\mathrm{ }\mathrm{ },\mathrm{ }\mathrm{ }\frac{12}{15}·60\mathrm{ }=\mathrm{ }12·4\mathrm{ }=\mathrm{ }48 .$

Arrangement in order of size gives $40<48<160$. With this, we have $\frac{2}{3}<\frac{12}{15}<\frac{32}{12}$, since the multiplication of the fractions by the same integer $60$ does not change the arrangement of the fractions (see section 3.1, which deals with inequalities and their transformation).

##### Exercise 1.2.617
Arrange the fractions $\frac{16}{15}$, $\frac{1}{2}$, $\frac{2}{3}$, $\frac{2}{-3}$, $\frac{60}{90}$, and $\frac{4}{3}$ in order of size:

$<$
$<$
$=$
$<$
$<$
.