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

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*}

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. $$$\sqrt{\frac{18}{10}} = \frac{3}{\Tgreen{\sqrt{5}}} = \frac{3\times\Torange{\sqrt{5}}}{\Tgreen{\sqrt{5}}\times\Torange{\sqrt{5}}} = \frac{3\Torange{\sqrt{5}}}{5} $$$

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*}

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. $$$\frac{1}{\Tblue{\sqrt{3} + \sqrt{2}}} = \frac{1}{\Tblue{\sqrt{3} + \sqrt{2}}} \times \frac{\Tgreen{\sqrt{3} - \sqrt{2}}}{\Tgreen{\sqrt{3} - \sqrt{2}}} = \sqrt{3} - \sqrt{2}$$$

Fractions with surds can be added or subtracted as regular fractions. However, the denominator should be rationalised first.

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

Here is another example.

\begin{align*} \frac{\sqrt{2}}{\Tblue{1+\sqrt{3}}} - \frac{5}{\Tgreen{\sqrt{2}}} &=\frac{\sqrt{2}(\Tblue{1-\sqrt{3}})}{1-3} - \frac{5\Tgreen{\sqrt{2}}}{2} \\ &=\frac{\sqrt{2}(\sqrt{3}-1)}{2} - \frac{5\sqrt{2}}{2} \\ &=\frac{\sqrt{2}(\sqrt{3}-1-5)}{2} \\ &=\frac{\sqrt{2}(\sqrt{3}-6)}{2} \\ &=\frac{\sqrt{6}}{2} -3\sqrt{2} \end{align*}