Fractions

A division is often represented as a fraction

\boxed{a:b = \frac{a}{b}}

where the naming is as follows

\frac{\text{numerator}}{\text{denominator}}=\text{quotient}

Examples:

$4:2=\frac{4}{2}$
$1:3 = \frac{1}{3}$
$(2x+y):(z+3)=\frac{2x+y}{z+3}$
$(ab):(abc)=\frac{ab}{abc}$
$\frac{2}{3}:4 = \frac{\frac{2}{3}}{4}$

Fractions are terms. Thus we want to understand again the possible operations we can apply to terms involving fractions without changing them (that is, the original and modified term are equal). This is discussed in the following section.

Losing the fraction line

Since $a:1=a$ and $a:a=1$ , it follows

\boxed{\frac{a}{1}=a} \quad \text{and}\quad \boxed{\frac{a}{a}=1}

Examples:

$\frac{3}{1}=3$
$\frac{-10}{-10}=1$
$\frac{3x-3+2y+b}{1}=3x-3+2y+b$
$\frac{2x+1}{2x+1}=1$

Multiplication rule for fractions

The multiplication of two fractions can be written as one fraction in the following way: multiply the two numerators and multiply the two denominators:

\boxed{\frac{a}{b}\cdot \frac{u}{v} = \frac{a\cdot u}{b\cdot v} }

or without the multiplication dots:

\frac{a}{b}\frac{u}{v} = \frac{a u}{b v}

Examples:

$\frac{2}{3} \cdot \frac{5}{7} = \frac{2\cdot 5}{3\cdot 7} = \frac{10}{21}$
$\frac{3}{5} \cdot \frac{x}{4} = \frac{3x}{5\cdot 4}=\frac{3x}{20}$
$\frac{x}{y} \cdot \frac{(x+1)}{z} = \frac{x(x+1)}{yz}$
$\frac{1}{a}\frac{1}{b}=\frac{1\cdot 1}{ab}=\frac{1}{ab}$
$2\cdot \frac{3}{4}= \frac{2}{1}\cdot \frac{3}{4}=\frac{2\cdot 3}{1\cdot 4}=\frac{6}{4}$

A special case of the rule above is this one:

\boxed{\frac{a}{b}=a\cdot \frac{1}{b}=a\frac{1}{b}}

Click on the right for an explanation of this last statement.

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We have

\frac{a}{b}=\frac{a\cdot 1}{1\cdot b}=\frac{a}{1} \cdot \frac{1}{b} = a\cdot \frac{1}{b}

Examples

$3\cdot \frac{1}{2}=\frac{3}{2}$
$a\cdot \frac{1}{a}=\frac{a\cdot 1}{a}=\frac{a}{a}=1$
$\frac{(x-1)(x+1)}{(x-1)}=(x+1)\frac{x-1}{x-1}=(x+1)\cdot 1=x+1$

Exercise 1

Write as a single fraction:

$x \frac{d}{p}$
$\frac{u}{k} y$
$3y\frac{9z}{8}$
$\frac{u}{i}:x$
$\frac{t}{k}\frac{a}{b}$
$\frac{2ax}{3p}\frac{5c}{7z}$
$(x+y)\frac{a-b}{2k}c$
$\frac{2p+x}{c-d}(u+f)$
$\frac{e+f}{m+n}\frac{x-2y}{x+y}$

Solution

$x \frac{d}{p}=\frac{x}{1}\frac{d}{p}=\frac{xd}{p}$
$\frac{u}{k} y=\frac{u}{k}\frac{y}{1}=\frac{uy}{k}$
$3y\frac{9z}{8}=\frac{3y}{1}\frac{9z}{8}=\frac{27yz}{8}$
$\frac{u}{i}:x=\frac{\frac{u}{i}}{x}=\frac{u}{ix}$
$\frac{t}{k}\frac{a}{b}=\frac{at}{bk}$
$\frac{2ax}{3p}\frac{5c}{7z}=\frac{10acx}{21pz}$
$(x+y)\frac{a-b}{2k}c = \frac{c(a-b)(x+y)}{2k}$
$\frac{2p+x}{c-d}(u+f)=\frac{(2p+x)(u+f)}{c-d}$
$\frac{e+f}{m+n}\frac{x-2y}{x+y}=\frac{(e+f)(x-2y)}{(m+n)(x+y)}$

Fractions and the negative sign

The minus sign can be moved up or down, or taken to the front:

\boxed{\frac{a}{-b}=\frac{-a}{b}=-\frac{a}{b}}

Also, a negative sign in the nominator and denominator can be cancelled

\boxed{\frac{-a}{-b}=\frac{a}{b}}

Examples:

$\frac{1}{-3}=\frac{-1}{3}=-\frac{1}{3}$
$\frac{-(x+2)}{3}={x+2}{-3}=-\frac{x+2}{3}$
$\frac{-(a+3c+1)}{a+3c}=\frac{a+3c+1}{-(a+3c)}=-\frac{a+3c+1}{a+3c}$
$\frac{-(x-1)}{-(x+2)}=\frac{x-1}{x+2}$

Important: In the examples above observe how we use the brackets:

\frac{-(a-4)}{b+5}=\frac{a-4}{-b+5}\quad\text{WRONG!!}

\frac{-(a-4)}{b+5}=\frac{a-4}{-(b+5)}\quad\text{CORRECT!!}

Why are these rules correct? Click on the right for an explanation.

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It follows from the multiplication rule for fractions by noting that $1=\frac{-1}{-1}$ . Thus we have, for example

\frac{-2}{3}=1\cdot \frac{-2}{3}=\frac{-1}{-1}\cdot\frac{-2}{3} = \frac{-1\cdot-2}{-1\cdot 3}=\frac{2}{-3}

Observe how the minus sign moves down. To understand why we can take the minus sign to the front of the fraction, we can write

\frac{-2}{3}=\frac{-1 \cdot 2}{3}=-1 \cdot\frac{2}{3}=-\frac{2}{3}

Exercise 2

What is correct, what not?

$\frac{-4}{5}=-\frac{4}{5}$
$\frac{-3x}{y}=\frac{3x}{-y}$
$-\frac{8z}{11k}=\frac{-8z}{-11k}$
$\frac{-4}{7}=\frac{1}{7}\cdot (-4)$
$\frac{0}{25}=0$
$\frac{25}{0}=0$

Solution

$\frac{-4}{5}=-\frac{4}{5}$ correct
$\frac{-3x}{y}=\frac{3x}{-y}$ correct
$-\frac{8z}{11k}=\frac{-8z}{-11k}$ incorrect
$\frac{-4}{7}=\frac{1}{7}\cdot (-4)$ correct
$\frac{0}{25}=0$ correct, because $0:25=0$ (dividing empty by $25$ stays empty)
$\frac{25}{0}=0$ not correct, actually it is infinity!

Simplify fractions

Equal factors(!!) in the numerator and denominator can be deleted (cancelled down):

\boxed{ \frac{a u}{a v} = \frac{u}{v}}

We can generalise this to subterms, we we denote now by capital letters:

\boxed{ \frac{A\cdot U}{ A\cdot V} = \frac{U}{V}}

Click right for an explanation of this law.

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This follows because

\frac{au}{av}=\frac{a}{a} \frac{u}{v}=1\cdot \frac{u}{v}=\frac{u}{v}

Examples:

$\frac{4}{6}=\frac{2\cdot 2}{2\cdot 3}=\frac{2}{3}$
$\frac{2a+4}{8}=\frac{2(a+2)}{ 2\cdot 4} = \frac{a+2}{4}$
$\frac{ax+ay}{2a}=\frac{a(x+y)}{2 a}=\frac{x+y}{2}$
$\frac{(x+1)x}{(x+1)y}=\frac{x}{y}$

It is important to note that we can only cancel down a number, variable, or subterm if it is separated by a multiplication. For example

\frac{a+x}{b\cdot x}= \frac{a}{b} \quad \text{WRONG!!}

\frac{a+x}{b+x}= \frac{a}{b} \quad \text{WRONG!!}

\frac{a\cdot x}{b\cdot x}= \frac{a}{b} \quad \text{CORRECT!!}

Exercise 3

Simplify the fraction as much as possible:

$\frac{2ax}{22xby}$
$\frac{2a+x}{22xby}$
$\frac{2ax+cx}{22xby}$
$\frac{14cd}{9cx-5x}$
$\frac{3ax}{33xby-6x}$
$\frac{-2pq}{-5ax}$
$\frac{abx-2ax}{3ab+bx}$
$\frac{21(2x+y)(a-c)}{3(a+c)(2x+y)}$
$\frac{3an-2a^2n^2}{5xan-an}$
$\frac{5ax+ax^2-2a}{a^2+5ax}$

Solution

$\frac{2ax}{22xby}=\frac{2x\cdot a}{2x\cdot 11by}=\frac{a}{11by}$
$\frac{2a+x}{22xby}$ not possible to simplify further
$\frac{2ax+cx}{22xby}=\frac{x\cdot (2a+c)}{x\cdot 22by}=\frac{2a+c}{22by}$
$\frac{14cd}{9cx-5x}$ not possible to simplify further
$\frac{3ax}{33xby-6x}=\frac{3x\cdot a}{3x\cdot (11by-2)}=\frac{a}{11by-2}$
$\frac{-2pq}{-5ax}=\frac{2pq}{5ax}$
$\frac{abx-2ax}{3ab+bx}$ not possible to simplify further
$\frac{21(2x+y)(a-c)}{3(a+c)(2x+y)}=\frac{3(2x+y)\cdot 7(a-c)}{3(2x+y)\cdot (a+c)}=\frac{7(a-c)}{a+c}$
$\frac{3an-2a^2n^2}{5xan-an}=\frac{an(3-2an)}{an(5x-1)}=\frac{3-2an}{5x-1}$
$\frac{5ax+ax^2-2a}{a^2+5ax}=\frac{a(5x+x^2-2)}{a(a+5x)}=\frac{x^2+5x-2}{a+5x}$

Expanding fractions

The reverse of simplifying a fraction is expanding a fraction. We expand a fraction by $x$ if we multiply the numerator and the denominator by $x$ :

\boxed{\frac{a}{b}=\frac{ax}{bx}}

Example:

$\frac{2}{3}=\frac{2\cdot 5}{3\cdot 5}=\frac{10}{15}$ (expanded by $5$ )
$\frac{3}{x}=\frac{3(a+1)}{x(a+1)}$ (expanded by $a+1$ )

It is true that the fraction becomes more complicated when it is expanded. But for the addition and the subtraction of fractions the extension is needed (see later).

Exercise 4

Expand the fraction ...

$\frac{x}{y}$ by $3$
$\frac{a+b}{c}$ by $x$
$\frac{2az}{3b}$ by $x+y$
$\frac{2u+2ax-p^2}{7a-22p}$ by $x$
$\frac{s+t}{2s-t}$ by $s+t$

Solution

$\frac{x}{y}=\frac{3x}{3y}$
$\frac{a+b}{c}=\frac{x(a+b)}{xc}$
$\frac{2az}{3b}=\frac{2az(x+y)}{3b(x+y)}$
$\frac{2u+2ax-p^2}{7a-22p}=\frac{x(2u+2ax-p^2)}{x(7a-22p)}$
$\frac{s+t}{2s-t}=\frac{(s+t)(s+t)}{(s+t)(2s-t)}=\frac{(s+t)^2}{(s+t)(2s-t)}$

Division of fractions, or double fractions

If there is a fraction in the numerator as well as in the denominator, we have a double fraction in front of us, which can be transformed as follows:

\boxed{\frac{\frac{a}{b}}{\frac{u}{v}} = \frac{a}{b}\cdot \frac{v}{u}}

So we can multiply the upper fraction by the inverse of the lower fraction. For example

\frac{\frac{2}{3}}{\frac{4}{7}} = \frac{2}{3}\cdot \frac{7}{4} =\frac{14}{12}

Why does this hold? Click right for an explanation.

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First note that we have

\frac{u}{v} \cdot \frac{v}{u}=\frac{uv}{vu}=1

Dividing both sides of the equation by $\frac{u}{v}$ we get

\frac{v}{u}=\frac{1}{\;\frac{u}{v}\; }

Thus,

\frac{\frac{a}{b}}{\frac{u}{v}} = \frac{a}{b}\cdot \frac{1}{\frac{u}{v}} = \frac{a}{b}\cdot \frac{v}{u}

But what if there is a fraction only in the numerator or in the denominator? We simply turn the term into a double fraction:

\boxed{\frac{\frac{a}{b}}{u}=\frac{\frac{a}{b}}{\frac{u}{1}}=\frac{a}{b}\frac{1}{u}=\frac{a}{bu}}

\boxed{\frac{a}{\frac{u}{v}}=\frac{\frac{a}{1}}{\frac{u}{v}}=\frac{a}{1}\frac{v}{u}=\frac{av}{u}}

Examples:

$\frac{2}{\frac{3}{4}}=\frac{\frac{2}{1}}{\frac{3}{4}}=\frac{2}{1}\cdot \frac{4}{3}=\frac{8}{3}$
$\frac{\frac{3}{4}}{2}=\frac{\frac{3}{4}}{\frac{2}{1}}=\frac{3}{4}\cdot \frac{1}{2}=\frac{3}{8}$

Exercise 5

Write as a single fraction and simplify the result as much as possible:

$\frac{1}{\frac{3}{2}}$
$\frac{1}{\frac{2x}{5y}}$
$5:\frac{35}{17}$
$\frac{35}{17}:5$
$\frac{9x}{\frac{8x}{7a}}$
$\frac{5y}{\frac{2x}{7y}}$
$\frac{\frac{13}{25}}{\frac{9}{5}}$
$\frac{2cx}{3a}:\frac{5ac}{6y}$
$\frac{\frac{16ux}{3pz}}{\frac{4uy+x}{4u}}$

Solution

$\frac{1}{\frac{3}{2}}=\frac{\frac{1}{1}}{\frac{3}{2}}=\frac{1}{1}\frac{2}{3}=\frac{2}{3}$
$\frac{1}{\frac{2x}{5y}}=\frac{\frac{1}{1}}{\frac{2x}{5y}}=\frac{1}{1}\frac{5y}{2x}=\frac{5y}{2x}$
$5:\frac{35}{17}=\frac{5}{\frac{35}{17}}=\frac{\frac{5}{1}}{\frac{35}{17}}=\frac{5}{1}\frac{17}{35}=\frac{5\cdot 17}{5\cdot 7}=\frac{17}{7}$
$\frac{35}{17}:5=\frac{\frac{35}{17}}{5}=\frac{\frac{35}{17}}{\frac{5}{1}}=\frac{35}{17}\frac{1}{5}=\frac{5\cdot 7}{5\cdot 17}=\frac{7}{17}$
$\frac{9x}{\frac{8x}{7a}}=\frac{\frac{9x}{1}}{\frac{8x}{7a}}=\frac{9x}{1}\frac{7a}{8x}=\frac{9x7a}{8x}=\frac{63a}{8}$
$\frac{5y}{\frac{2x}{7y}}=\frac{\frac{5y}{1}}{\frac{2x}{7y}}=\frac{5y}{1}\frac{7y}{2x}=\frac{35y^2}{2x}$
$\frac{\frac{13}{25}}{\frac{9}{5}}=\frac{13}{25}\frac{5}{9}=\frac{13\cdot 5}{5\cdot 5\cdot 9}=\frac{13}{45}$
$\frac{2cx}{3a}:\frac{5ac}{6y}=\frac{\frac{2cx}{3a}}{\frac{5ac}{6y}}=\frac{2cx}{3a}\frac{6y}{5ac}=\frac{3c\cdot 4 xy}{3c\cdot 5a^2}=\frac{4 xy}{5a^2}$
$\frac{\frac{16ux}{3pz}}{\frac{4uy+x}{4u}}=\frac{16ux}{3pz} \frac{4u}{4uy+x}=\frac{64u^2x}{3pz(4uy+x)}$

Addition and subtraction rule

To add or subtract two fractions, the fractions must first be expanded so that both fractions have the same denominator. Then the numerators can be added or subtracted (just like equal-sized pieces of cake).

\boxed{\frac{a}{b} +\frac{u}{v} = \frac{av}{bv}+\frac{bu}{bv}=\frac{av+bu}{bv}}

Examples:

$\frac{2}{3}+\frac{4}{7}=\frac{14}{21}+\frac{12}{21}=\frac{26}{21}$
$\frac{3}{8}-\frac{5}{6}=\frac{9}{24}-\frac{20}{24}=-\frac{11}{24}$
$\frac{a}{b}+\frac{c}{d}=\frac{ad}{bd}+\frac{bc}{bd}=\frac{ad+bc}{bd}$
$\frac{a}{2cx}+\frac{b}{3dx}= \frac{3d\cdot a}{3d\cdot 2cx}+\frac{2c\cdot b}{2c\cdot 3dx}=\frac{3da+2cb}{6cdx}$
$3+\frac{1}{4}=\frac{12}{4}+\frac{1}{4}=\frac{13}{4}$
$a+\frac{2}{b}=\frac{ab}{b}+\frac{2}{b}=\frac{ab+2}{b}$

Exercise 6

Write as a single fraction and simplify the result as much as possible:

$\frac{4}{15}+\frac{7}{18}$
$\frac{5}{12}-\frac{11}{42}$
$\frac{2p}{7}+\frac{9f}{7}$
$\frac{5z}{8s}-\frac{2z}{8s}$
$\frac{9x+y}{7ab}+\frac{6x-2y}{7ab}$
$\frac{2a-3b}{5kt}-\frac{a-2b}{5kt}$
$\frac{x}{z}+\frac{p}{n}$
$\frac{k}{2py}+\frac{n}{3px}$
$\frac{4}{5xz}-\frac{7}{18xy}$
$\frac{a-b}{6xt}-\frac{2a+b}{9ty}$
$\frac{3t}{10acd}+\frac{5s}{6bd}-\frac{7v}{15ab}$
$\frac{4a}{45xz}-\frac{7b}{30xyz}+\frac{2c}{75zy}$

Solution

$\frac{4}{15}+\frac{7}{18}=\frac{24}{90}+\frac{35}{90}=\frac{59}{90}$
$\frac{5}{12}-\frac{11}{42}=\frac{35}{84}-\frac{22}{84}=\frac{13}{84}$
$\frac{2p}{7}+\frac{9f}{7}=\frac{2p+9f}{7}$
$\frac{5z}{8s}-\frac{2z}{8s}=\frac{5z-2z}{8s}=\frac{3z}{8s}$
$\frac{9x+y}{7ab}+\frac{6x-2y}{7ab}=\frac{9x+y+6x-2y}{7ab}=\frac{15x-y}{7ab}$
$\frac{2a-3b}{5kt}-\frac{a-2b}{5kt}=\frac{2a-3b-a+2b}{5kt}=\frac{a-b}{5kt}$
$\frac{x}{z}+\frac{p}{n}=\frac{xn}{zn}+\frac{pz}{nz}=\frac{xn+pz}{nz}$
$\frac{k}{2py}+\frac{n}{3px}=\frac{3kx}{6pxy}+\frac{2ny}{6pxy}=\frac{3kx+2ny}{6pxy}$
$\frac{4}{5xz}-\frac{7}{18xy}=\frac{72y}{90xyz}-\frac{35z}{90xyz}=\frac{72y-35z}{90xyz}$
$\frac{a-b}{6xt}-\frac{2a+b}{9ty}=\frac{3y(a-b)}{18xyt}-\frac{2x(2a+b)}{18xyt}=\frac{3ay-3yb-4ax-2bx}{18xyt}$
$\frac{3t}{10acd}+\frac{5s}{6bd}-\frac{7v}{15ab}=\frac{9bt}{30abcd}+\frac{25acs}{30abcd}-\frac{14cdv}{30abcd}=\frac{9bt+25acs-14cdv}{30abcd}$
$\frac{4a}{45xz}-\frac{7b}{30xyz}+\frac{2c}{75zy}=\frac{40ay}{450xyz}-\frac{105b}{450xyz}+\frac{12cx}{450xyz}=\frac{40ay-105b+12cx}{450xyz}$