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# Surds

## Introduction to surds

The roots of most rational numbers cannot be written as fractions. They are irrational numbers. A surd is the root of a rational number that is irrational. A surd is also called a radical.

$\Tred{\sqrt{2}}$, $\Tred{\sqrt{3}}$, $\Tred{\sqrt{1/2}}$, $\Tred{\sqrt{8}}$ are surds.

$\Tblue{\sqrt{4}} = 2$, $\Tblue{\sqrt{1/9}} = 1/3$ and $\Tblue{\sqrt{9/16}} = 3/4$ are not surds because the roots are rational numbers.

In practice, we don't simplify surds and keep the roots $\sqrt{a}$ in the expressions.

We can simplify $\Tblue{\sqrt{4}}$ to $2$, but not $\Tred{\sqrt{2}}$.

$$1-\Tblue{\sqrt{4}} +\Tred{\sqrt{2}} = 1-\Tblue{2}+\Tred{\sqrt{2}} = -1 + \Tred{\sqrt{2}}.$$

To check if a number is a surd, write it in decimal form with a calculator. If you find recurring decimals, it is not a surd. Otherwise it is.

$\Tred{\sqrt{5}}$ is a surd because it has no recurring decimals.

$$\Tred{\sqrt{5}}\simeq 2.2360679774998$$

The sides of a square with area two have length equal to the square root of two. This is a surd.

## Simplifying surds

We generally simplify surds as the product of positive integers times the surd of the product of prime numbers.

$\sqrt{9}$ simplifies to $3$ and $\sqrt{8}$ simplifies to $2\sqrt{2}$.

We illustrate the simplification process with $\sqrt{24}$.

• Find the prime factorisation of the number $$24 = 2^3\times 3$$
• Put aside the even powers of the prime numbers. $$24 = \Tblue{2^2}\times \Tgreen{2}\times \Tgreen{3}$$
• Take the square root of the even powers by dividing the power by 2. Multiply the numbers with multiplicity 1 and leave them as a surd. $$\sqrt{24} = \sqrt{\Tblue{2^2}}\times\sqrt{\Tgreen{2\times 3}} = \Tblue{2}\sqrt{\Tgreen{6}}$$

Here are other simplified surds.

\begin{align*} &\qquad\sqrt{4} = \Tblue{2},\quad \sqrt{75} = \sqrt{5^3} = \sqrt{\Tblue{5^2}\times \Tgreen{5}} = \Tblue{5}\sqrt{\Tgreen{5}},\\ &\sqrt{72} = \sqrt{2^3\times 3^2} = \sqrt{\Tblue{2^2}\times \Tgreen{2}\Tblue{\times 3^2}} = \Tblue{2\times 3}\sqrt{\Tgreen{2}} = \Tblue{6}\sqrt{\Tgreen{2}}. \end{align*}

## Multiplying and dividing surds

It is easy to compute and simplify the product and division of surds.

$$\sqrt{2}\times\sqrt{8} = \sqrt{16} = 4,\quad\sqrt{\frac{18}{8}} = \sqrt{\frac{9}{4}} = \frac{3}{2}.$$

It uses the formula for the multiplication of square roots. $$\sqrt{x}\sqrt{y} = \sqrt{xy},\quad \frac{\sqrt{x}}{\sqrt{y}} = \sqrt{\frac{x}{y}}.$$

Let's illustrate the process to compute $\displaystyle\frac{\sqrt{18}}{\sqrt{10}}$

• Compute the prime factorisation of each number. $$18 = 2\times 3^2,\quad 50 = 2\times 5$$
• Simplify the product or the fraction. $$\frac{18}{10} = \frac{\Tblue{2}\times 3^2}{\Tblue{2}\times 5} = \frac{3^2}{5}$$
• Take the square root and take out numbers with even power. $$\sqrt{\frac{18}{10}} = \frac{\sqrt{\Tred{3^2}}}{\sqrt{5}} = \frac{\Tred{3}}{\sqrt{5}}$$
• Rationalise the denominator. $$\sqrt{\frac{18}{10}} = \frac{3}{\Tgreen{\sqrt{5}}} = \frac{3\times\Torange{\sqrt{5}}}{\Tgreen{\sqrt{5}}\times\Torange{\sqrt{5}}} = \frac{3\Torange{\sqrt{5}}}{5}$$

## Conjugate surds

The conjugate of the sum of two surds $\sqrt{x}\Tred{+} \sqrt{y}$ is $\sqrt{x} \Torange{-} \sqrt{y}$. We just need to change sign.

The conjugate of $1\Tred{+}\sqrt{2}$ is $1\Torange{-}\sqrt{2}$.

The conjugate of $\sqrt{3}\Tred{-} \sqrt{5}$ is $\sqrt{3}\Torange{+}\sqrt{5}$.

Conjugate surds are useful because the product is simple

$$(\Tblue{\sqrt{x}+\sqrt{y}})(\Tgreen{\sqrt{x}-\sqrt{y}}) = (\sqrt{x})^2 - (\sqrt{y})^2 = x - y.$$

Division / reciprocal of conjugate surds simplifies as well

$$\frac{1}{\Tblue{\sqrt{x} - \sqrt{y}}} = \frac{\Tgreen{\sqrt{x}+\sqrt{y}}}{(\Tblue{\sqrt{x} - \sqrt{y}})(\Tgreen{\sqrt{x}+\sqrt{y}})} = \frac{\Tgreen{\sqrt{x} +\sqrt{y}}}{x-y}.$$

Here are a few examples

\begin{align*} (\Tblue{\sqrt{2}+\sqrt{3}})(\Tgreen{\sqrt{2}-\sqrt{3}}) &= 2 - 3 = -1\\ (\Tblue{\sqrt{5}-1})(\Tgreen{\sqrt{5}+1}) & = 5-1 = 4\\ \frac{1}{\Tblue{1+\sqrt{2}}} = \frac{\Tgreen{1-\sqrt{2}}}{1-2} &= -1+\sqrt{2}\\ \frac{3}{\Tblue{\sqrt{7}-1}} = \frac{3(\Tgreen{\sqrt{7}+1})}{7-1} &= \frac{\sqrt{7}+1}{2} \end{align*}

## Rationalising denominators with surds

We don't like having surds in the denominator of fractions. Removing the surd from the denominator is called rationalisation of the denominator.

We prefer $\displaystyle\frac{\sqrt{2}}{2}$ to $\displaystyle\frac{1}{\sqrt{2}}$ and $\displaystyle\frac{\sqrt{6}}{3}$ to $\displaystyle\frac{\sqrt{2}}{\sqrt{3}}$.

• When the denominator is a simple surd, we multiply both the denominator and the numerator by the surd $$\frac{2}{\Tblue{\sqrt{10}}} = \frac{2\Tgreen{\sqrt{10}}}{\Tblue{\sqrt{10}}\times \Tgreen{\sqrt{10}}} = \frac{2\sqrt{10}}{10} = \frac{\sqrt{10}}{5}.$$
• When the denominator is the sum of two surds, we multiply by the conjugate surd

The conjugate of $\Tblue{\sqrt{3} + \sqrt{2}}$ is $\Tgreen{\sqrt{3}-\sqrt{2}}$. This helps us to simplify the following fraction. $$\frac{1}{\Tblue{\sqrt{3} + \sqrt{2}}} = \frac{1}{\Tblue{\sqrt{3} + \sqrt{2}}} \times \frac{\Tgreen{\sqrt{3} - \sqrt{2}}}{\Tgreen{\sqrt{3} - \sqrt{2}}} = \sqrt{3} - \sqrt{2}$$

$$\frac{1}{\Tblue{\sqrt{2}}} + \frac{1}{\Tgreen{\sqrt{3}}} = \frac{\Tblue{\sqrt{2}}}{2} + \frac{\Tgreen{\sqrt{3}}}{3} =\frac{3\sqrt{2}+2\sqrt{3}}{6}$$