#### Chapter 1 Elementary Arithmetic

Section 1.2 Fractional Arithmetic

# 1.2.1 Calculating with Fractions

A fraction is a rational number written in the form $\frac{\text{numerator}}{\text{denominator}}$, where numerator and denominator are integers, and the denominator is $\ne 0$. Examples are:

$\frac{1}{2} , \frac{5}{-10} , \frac{-17}{12} , \frac{1}{23} , \frac{4}{6} , \frac{-2}{3} , \dots .$

It can be seen very quickly that a single number can have an arbitrary number of equivalent representations. For example:

$\frac{12}{36}=\frac{1}{3}=\frac{24}{72}=\frac{-12}{-36}=\frac{3}{9}=\frac{2}{6}=\frac{120}{360}=\dots .$

The different representations transform into each other by reducing and expanding, respectively.
##### Info 1.2.1

Fractions are reduced by dividing numerator and denominator by the same non-zero integer.
Fractions are expanded by multiplying numerator and denominator by the same non-zero integer.

##### Example 1.2.2
Three friends like to share a pizza. Tom eats $\frac{1}{4}$ of the pizza, Tim eats $\frac{1}{3}$ of the pizza. How much of the pizza is left for their friend Sven, who always has the biggest appetite?
The solution is found by means of fractional arithmetic: First, we have to add two fractions, to decide how much of the pizza Tim and Tom already ate:

$\frac{1}{4}+\frac{1}{3}=\frac{1·3}{4·3}+\frac{1·4}{3·4}=\frac{3}{12}+\frac{4}{12}=\frac{7}{12} .$

Here, we can already identify the two most important steps: First we have to expand the two fractions to the so-called least common denominator, or, as one also says, we have to create like fractions. Then, if the fractions have the same denominator, we can add them by adding their numerators and maintaining the same denominator. From the result that Tim and Tom ate $\frac{7}{12}$ of the pizza, we can calculate how much of the pizza is left for Sven by subtracting $\frac{7}{12}$ from 1:

$1-\frac{7}{12}=\frac{12}{12}-\frac{7}{12}=\frac{5}{12} .$

Again, we must expand the fractions to the least common denominator. Then we have to subtract the two numerators. Indeed, the two friends have left much of the pizza for the always hungry Sven.soweit

The reducing of fractions can be practised in the training exercises below.
##### Exercise 1.2.3
Kürzen Sie soweit möglich: $\frac{6270}{1235}$$=$
.

It becomes more difficult if indeterminate expressions (e.g. variables) occur in numerator and denominator. These can be reduced or cancelled just like numbers (but not with numbers), for example, we get

$\frac{4{x}^{2}{y}^{3}+3{y}^{2}}{10{y}^{2}}\mathrm{ }=\mathrm{ }\frac{4{x}^{2}y+3}{10}$

by cancelling the term ${y}^{2}$ from numerator and denominator. The following training exercise has been extended to fractions including indeterminate expressions.
##### Exercise 1.2.303
Kürzen Sie soweit möglich: $\frac{36{y}^{2}}{81{y}^{2}-81{y}^{2}{z}^{3}}$$=$
.

##### Info 1.2.603

The least common denominator of two fractions is the least common multiple (lcm) of the two denominators.
The least common multiple (lcm) of two numbers is the smallest number that is divisible by both numbers.
The greatest common divisor (gcd) of two numbers is the largest number that divides both numbers without remainder.

If the determination of the lcm is too complicated, the simple product of the denominators can be used instead of the lcm in the following calculation rule:
##### Info 1.2.604

Fractions are added/subtracted by finding a common denominator and then adding/substracting the numerators, i.e.

$\frac{a}{b}±\frac{c}{d}=\frac{ad±bc}{bd} , bd\ne 0 .$

Usually fractions are expanded to the least common denominator.

For example, the least common multiple of $6=2·3$ and $15=3·5$ is the number $2·3·5=30$. However, the product is $6·15=90$. Thus, you can calculate

$\frac{1}{6}+\frac{1}{15}\mathrm{ }=\mathrm{ }\frac{5}{30}+\frac{2}{30}\mathrm{ }=\mathrm{ }\frac{7}{30}$

but also

$\frac{1}{6}+\frac{1}{15}\mathrm{ }=\mathrm{ }\frac{15}{90}+\frac{6}{90}\mathrm{ }=\mathrm{ }\frac{21}{90}$

and finally reduce the last fraction to $\frac{7}{30}$.
##### Example 1.2.605
The least common multiple for the least common denominator is the smallest number that can be divided by all denominators involved. If these denominators have no factors in common, the least common multiple is simply the product of all denominators:

$\begin{array}{ccc}\multicolumn{1}{c}{\frac{1}{6}+\frac{1}{10}}& =\hfill & \frac{5}{30}+\frac{3}{30}\mathrm{ }=\mathrm{ }\frac{8}{30}\mathrm{ }=\mathrm{ }\frac{4}{15} , \hfill \\ \multicolumn{1}{c}{\frac{1}{6}+\frac{1}{10}}& =\hfill & \frac{10}{60}+\frac{6}{60}\mathrm{ }=\mathrm{ }\frac{16}{60}\mathrm{ }=\mathrm{ }\frac{4}{15} \text{(also correct)}, \hfill \\ \multicolumn{1}{c}{\frac{4}{15}-\frac{1}{2}}& =\hfill & \frac{8}{30}-\frac{15}{30}\mathrm{ }=\mathrm{ }\frac{8-15}{30}\mathrm{ }=\mathrm{ }-\frac{7}{30} , \hfill \\ \multicolumn{1}{c}{\frac{1}{3}+\frac{1}{9}}& =\hfill & \frac{3}{9}+\frac{1}{9}\mathrm{ }=\mathrm{ }\frac{4}{9} , \hfill \\ \multicolumn{1}{c}{\frac{1}{{2}^{2}}+\frac{1}{{2}^{4}}}& =\hfill & \frac{{2}^{2}}{{2}^{4}}+\frac{1}{{2}^{4}}\mathrm{ }=\mathrm{ }\frac{5}{16} , \hfill \\ \multicolumn{1}{c}{\frac{1}{2}+\frac{1}{3}+\frac{1}{7}}& =\hfill & \frac{21}{42}+\frac{14}{42}+\frac{6}{42}\mathrm{ }=\mathrm{ }\frac{41}{42} .\hfill \end{array}$

The least common denominator can also be found if the denominators include variables. Since the transformations of the fractions have to be correct for all possible values of these variables, they have to be considered as numbers without any common factors:
##### Example 1.2.606
Let $x$ and $y$ be variables, then

$\begin{array}{ccc}\multicolumn{1}{c}{\frac{1}{3}+\frac{1}{x}}& =\hfill & \frac{x}{3·x}+\frac{3}{3·x}\mathrm{ }=\mathrm{ }\frac{3+x}{3·x} ,\hfill \\ \multicolumn{1}{c}{\frac{1}{x}+\frac{1}{y}}& =\hfill & \frac{y}{x·y}+\frac{x}{x·y}\mathrm{ }=\mathrm{ }\frac{x+y}{x·y} , \hfill \\ \multicolumn{1}{c}{\frac{1}{\left(x+1{\right)}^{2}}+\frac{1}{x+1}}& =\hfill & \frac{1}{\left(x+1{\right)}^{2}}+\frac{x+1}{\left(x+1{\right)}^{2}}\mathrm{ }=\mathrm{ }\frac{x+2}{\left(x+1{\right)}^{2}} .\hfill \end{array}$

##### Exercise 1.2.607
Calculate the following sums by means of the least common denominator (or the product of the denominators).
1. $\frac{1}{2}-\frac{1}{8}$$=$ .
2. $\frac{1}{3}+\frac{1}{5}+\frac{1}{6}$$=$ .
3. $\frac{1}{2x}+\frac{1}{3x}$$=$ .
Please use no other operators than the operators for multiplication * and division / during this exercise.

##### Exercise 1.2.608
In the case of like fractions, you may only add or decompose the numerators, for denominators no such rule exists. To convince yourself, calculate the following numbers by finding the least common denominator and reducing as much as possible:
1. $\frac{1}{2}+\frac{1}{3}$$=$
but $\frac{1}{2+3}$$=$ .
2. $\frac{1+2}{5+6}$$=$
but $\frac{1}{5}+\frac{2}{6}$$=$ .

##### Info 1.2.609

Two fractions are multiplied by multiplying the numerators and multiplying the denominators, i.e.

$\frac{a}{b}·\frac{c}{d}=\frac{a·c}{b·d} , bd\ne 0 .$

The division of two fractions is reduced to their multiplication:
##### Info 1.2.610

Two fractions are divided by multiplying the first fraction with the reciprocal of the second fraction, i.e.

$\frac{a}{b}:\frac{c}{d}=\frac{a}{b}·\frac{d}{c}=\frac{a·d}{b·c} , b,c,d\ne 0 .$

The division of two fractions can be expressed as a compound fraction as well:

$\frac{a}{b}:\frac{c}{d}=\frac{\frac{a}{b}}{\frac{c}{d}} .$

##### Example 1.2.611
Taking possible reducing into account, the multiplication and the division of two fractions, respectively, takes the following form:

$\frac{2}{3}·\frac{4}{5}=\frac{2·4}{3·5}=\frac{8}{15} , \frac{2}{3}:\frac{4}{5}=\frac{2}{3}·\frac{5}{4}=\frac{10}{12}=\frac{5}{6} .$