A *fraction* is an expression of the form $\dfrac{p}{q}$ where $p$ and $q$ are integers or algebraic expressions.

We call $p$ the *numerator* and $q$ the *denominator*.

Fractions of the form $\dfrac{a}{b}$, where $a$ and $b \neq 0$ are integers make up the set of *rational numbers* $\mathbb{Q}$. Such fractions are also called *arithmetic fractions*.

Arithmetic fractions can be evaluated as a real number on dividing $a$ by $b$. For example, evaluating $\frac{1}{2}$ gives $0.5$.

**Examples**: $\dfrac{3}{4},\;\;\;\dfrac{7}{225},\;\;\;\dfrac{1}{100000},\;\;\;-\dfrac{500}{2300}$

If the numerator and/or the denominator of a fraction is an algebraic expression, then the fraction is called an *algebraic fraction*.

**Examples**: $\dfrac{x+1}{x+2},\;\;\;\dfrac{y^2+xy-2}{x^2+xy-1},\;\;\;\dfrac{3x+3}{6},\;\;\;\dfrac{x^2+2x+1}{x^2+3x+2}$

A fraction remains the same if we multiply both the numerator and denominator by the same non-zero expression.

**Examples**:

\begin{align} \dfrac{1}{4}&= \dfrac{1 \times 2}{4\times 2} = \dfrac{2}{8}\\ \dfrac{x}{x+1}&=\dfrac{x \times (x-1)}{(x+1)\times (x-1)}=\frac{x^2-x}{x^2-1} \end{align} In the last example we require that $x \neq \pm 1$ as otherwise the fraction is **undefined**.

We can **simplify** a fraction by cancelling all common factors of the numerator and the denominator. A fraction which has been simplified it is said to be written in its **simplest form** or **lowest form**.

An arithmetic fraction is in its simplest form if the numerator and denominator do not have any common integer factors.

**Note**: When simplifying algebraic fractions we must always be aware of the possibility of dividing by zero.

Simplify: $\dfrac{24}{36}$

$24$ and $36$ have a factor of $4$ in common so we can cancel $4$ from the top and bottom: \[\dfrac{24 \div 4}{36\div 4} = \dfrac{6}{9}\] Since $6$ and $9$ have a factor of $3$ in common, we can cancel $3$ from the top and bottom: \[\dfrac{6\div3}{9\div3}=\dfrac{2}{3}\] Since $2$ and $3$ do not have any common factors, this fraction is now in its simplest form.

**This is the only way that you can simplify fractions.**

Simplify $\dfrac{x^2+3x+2}{x^2+4x+3}$.

\begin{align} \dfrac{x^2+3x+2}{x^2+4x+3}&=\dfrac{(x+2)(x+1)}{(x+3)(x+1)}\\ &=\dfrac{x+2}{x+3},\;\;x \neq -1 \end{align} On cancelling the common factor $x+1$. Note that we are careful to point out the value of $x$ at which the expression is not defined.

If two or more fractions have the same denominator, they are said to have a **common denominator**.

Adding and subtracting fractions is straightforward if they all have a common denominator: we just add or subtract the numerators.

**Example**: $\dfrac{1}{5} + \dfrac{3}{5} = \dfrac{4}{5}$

**Example**: \begin{align} \frac{y+z-xy}{y+z+1}+\frac{1+xy}{y+z+1}&=\dfrac{(y+z-xy)+(1+xy)}{y+z+1}\\ &=\dfrac{y+z+1}{y+z+1}\\ &=1,\;\;y+z+1 \neq 0 \end{align}

If the denominators are different, just adding or subtracting the numerators together doesn't make sense - what should the denominator of the result be? The solution is to restate both fractions over a common denominator before doing the addition (or subtraction). This common denominator is equal to the product of the denominators of the fractions we are adding (or subtracting). For example, suppose we want to calculate: \[\frac{3}{5}+\frac{7}{10}\] The common denominator will be $5\times 10=50$. To restate the first fraction over this denominator, we must multiply both the numerator and denominator by $10$: \[\frac{3\times 10}{5 \times 10}=\frac{30}{50}\] To restate the second fraction over this denominator, we must multiply both the numerator and denominator by $5$: \[\frac{7\times 5}{10 \times 5}=\frac{45}{50}\] Since the denominators of the two fractions we want to add are the same ($50$), we can just add the numerators and write the result over the common denominator so that we now have one fraction: \[\frac{30}{50}+\frac{45}{50}=\frac{75}{50}\] The final step is to write this fraction in its simplest form by cancelling all common factors in the numerator and denominator (if there are any). The highest common factor of $75$ and $50$ is $25$, so: \[\frac{75}{50}=\frac{75 \div 25}{50\div 25}=\frac{3}{2}\] Thus \[\frac{3}{5}+\frac{7}{10}=\frac{3}{2}\]

In **summary**:

1) Multiply the denominators to get a common denominator.

2) Restate each fraction over the common denominator.

3) Add (or subtract) the numerators and write the result over the common denominator.

4) Simplify the fraction if required.

Calculate $\dfrac{3}{5}+\dfrac{1}{4}$.

\begin{align} \frac{3}{5}+\frac{1}{4} &= \frac{3\times4}{5 \times 4} + \frac{1 \times 5}{5\times4}\\ &= \frac{12}{20} + \frac{5}{20} \\ &=\frac{12+5}{20}\\ &=\frac{17}{20} \end{align}

**Note:** here we multiplied 4 and 5 together to produce the common denominator 20 as 4 and 5 have no common factor apart from 1.

Write $\dfrac{x+1}{y}+\dfrac{y+1}{x}$ as a single algebraic fraction.

\begin{align} \dfrac{x+1}{y}+\dfrac{y+1}{x}&=\dfrac{(x+1)\times x+(y+1)\times y}{xy}\\ &=\dfrac{x^2+x+y^2+y}{xy} \end{align}

Calculate $\dfrac{5}{6}-\dfrac{2}{3}$.

\begin{align} \frac{5}{6}-\frac{2}{3}&= \frac{5}{6} - \frac{2\times 2}{3 \times 2} \\ &= \frac{5}{6} - \frac{4}{6} \\ &=\frac{5-4}{6}\\ &=\frac{1}{6} \end{align}

**Note:** as you can see in this example, it is not always necessary to multiply the denominators together to make a common denominator. Here, $3$ divides evenly into $6$, so it was more reasonable to use $6$ as the common denominator, rather than $6 \times 3=18$.

*If we had used $18$ as the common denominator then we would have got the same result, but in one more step:*

\begin{align} \frac{5}{6}-\frac{2}{3}&= \frac{5\times 3}{6 \times 3} - \frac{2\times 6}{6 \times 3}\\ &= \frac{15}{18} - \frac{12}{18} \\ &=\frac{15-12}{18}\\ &=\frac{3}{18}\\ &=\frac{1}{6} \end{align}

Write $\dfrac{x+y}{xy-x}-\dfrac{xy-1}{y-1}$ as a single algebraic fraction.

In this example we find the common denominator $x(y-1)$ which both denominators divide into.

\begin{align} \dfrac{x+y}{xy-x}-\dfrac{xy-1}{y-1}&=\dfrac{x+y}{x(y-1)}-\dfrac{xy-1}{y-1}\\ &=\dfrac{(y-1)(x+y-x(xy-1))}{x(y-1)^2}\\ &=\dfrac{(y-1)(2x+y-x^2y)}{x(y-1)^2}\\ &=\dfrac{2x+y-x^2y}{x(y-1)} \end{align}

The rule for multiplying fractions is multiply the numerators together and multiply the denominators together:

\[\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}\]

Calculate $\dfrac{3}{7} \times \dfrac{5}{3}$.

\begin{align} \frac{3}{7} \times \frac{5}{3} &= \frac{3\times5}{7\times3}\\ &= \frac{15}{21}\\ &= \frac{5}{7} \end{align} **Note:** Here, $15$ and $21$ are both multiples of $3$, so we divide the top and bottom by $3$ to get the fraction in its simplest form.

Calculate $\dfrac{3x}{y} \times \dfrac{2xy^2}{z}$.

\begin{align} \frac{3x}{y} \times \frac{2xy^2}{z} &= \frac{3x \times 2xy^2}{y \times z}\\ \\ &=\frac{6x^2y^2}{yz}\\ \\ &=\frac{6x^2y}{z} \qquad \textrm{ (On cancelling the common term }y\textrm{ in the numerator and denominator)} \end{align}

The rule for dividing fractions is: \[\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}.\]

This is easily remembered as 'flip the second fraction, then multiply'.

Sometimes this may be written as a fraction on top of a fraction, but the rule is the same: \[\frac{a/b}{c/d}=\frac{a}{b}\times\frac{d}{c}\text{.}\]

Calculate $\dfrac{1}{3}\div\dfrac{1}{4} $.

\begin{align} \frac{1}{3}\div\frac{1}{4} &=\frac{1}{3}\times\frac{4}{1}\\\\ &=\frac{1\times4}{3\times1}\\\\ &=\frac{4}{3}. \end{align}

Calculate $\dfrac{3}{5} \div \dfrac{2}{3}$.

\begin{align} \frac{3}{5}\div\frac{2}{3} &=\frac{3}{5}\times\frac{3}{2}\\\\ &=\frac{3\times3}{5\times2}\\\\ &=\frac{9}{10}. \end{align}

Simplify the fraction $\dfrac{2/3}{2-5/7}$.

Start by changing the fraction to a division symbol to see both parts side-by-side. \[\frac{2}{3} \div \left(2-\frac{5}{7}\right)\] Subtract $\frac{5}{7}$ from $2$ by putting them over a common denominator. \begin{align} 2 - \frac{5}{7} &= 2 \times \frac{7}{7} - \frac{5}{7} \\ &= \frac{14}{7} - \frac{5}{7} \\ &= \frac{14 - 5}{7} \\ &= \frac{9}{7}. \end{align} Now, we have: \[\frac{2}{3} \div \frac{9}{7}\] Use the rule for dividing fractions. \begin{align} \frac{2}{3} \div \frac{9}{7} &= \frac{2}{3} \times \frac{7}{9} \\ \\ &= \frac{2 \times 7}{3 \times 9} \\ \\ &= \frac{14}{27}. \end{align}

Express $\dfrac{4}{1-2x} - 3$ as a single fraction.

First, we put everything over the same denominator $1-2x$, then expand the brackets in the numerator and simplify.

\begin{align} \frac{4}{1-2x} - 3 & = \frac{4}{1-2x} - \frac{3(1-2x)}{1-2x}\\\\ &=\frac{4-3(1-2x)}{1-2x}\\\\ &=\frac{4-3+6x}{1-2x}\\\\ &=\frac{1+6x}{1-2x} \end{align} $1+6x$ and $1-2x$ have no factors in common so the fraction is in its simplest form.

Prof. Robin Johnson computes $\dfrac{2}{3} + \dfrac{4}{5}$ and $\dfrac{5}{6} + \dfrac{1}{4} - \dfrac{7}{30}$.

Prof. Robin Johnson calculates $\dfrac{7}{60}+\dfrac{2}{5}-\dfrac{5}{12}$ and simplifies $\dfrac{1+2x}{(x+1)(x-2)}+\dfrac{3x-2}{x^2-1}$.

Prof. Robin Johnson calculates $2 \div \dfrac{1}{4}$ and $3 \div \dfrac{2}{5}$.

Prof. Robin Johnson simplifies the fraction $\dfrac{3/4}{2-4/7}$.

Prof. Robin Johnson simplifies the combines the fractions $\dfrac{2}{1+x} - \dfrac{3}{2-x}$, into a single fraction.

Prof. Robin Johnson combines the fractions $\dfrac{x^3y^4}{z^5}\times\dfrac{z^3}{x^2y^7}$ and $\dfrac{x^2-1}{x+2}\times\dfrac{2x^2+3x-2}{x+1}$.

This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.

Test yourself: Numbas test on fractions

Test yourself: Another Numbas test on fractions

Test yourself: Numbas test on combining algebraic fractions

- An Algebra Refresher workbook at
**math**centre. - Simplifying Algebraic Expressions workbook at
**math**centre. - Fractions videos, worksheets and quizzes at BBC Skillswise.
- Fractions in Algebra at Maths is Fun.