Any time we meet an irrational number in our work, we have the choice of decimalizing the number or leaving it in symbolic form.

$\begin{align*} &\sqrt{5} \hspace{1.4 cm} \text{the number that when squared is equal to 5, that is} \sqrt{5}\times \sqrt{5}=5\\[10 pt]&\pi \hspace{1.7 cm} \text{the number of diameters that fit on the circumference of a circle}\\[10 pt]&\sin(30^{\circ})\hspace{0.5 cm} \text{the length of the side opposite a 30 degree angle in a right} \\[10 pt]&\hspace{2.0 cm}\text{angled triangle that has hypotenuse length 1}\end{align}$

As can be seen from some of these ‘meanings’ the symbolic form captures a lot of information in a small amount of space on paper. The symbolic form is the exact value.

When we decimalize, we get an idea of the magnitude of the number. Eg, $\sqrt{60} =7.75$ (2 d.p) This can be very useful especially if plotting the value on a graph. However the downside is that we lose accuracy. Rounded decimals are not ‘exact answers’.

Definition

A surd is the square root of a non-square number, or the cube root of a non-cube number etc. In other words, a radical whose value is not an integer.

The numbers $\sqrt{25}$ and $\sqrt[3]{8}$ are not surds because $\sqrt{25}=5$ and $\sqrt[3]{8}=2$ , which are both integers.

On the other hand the numbers $\sqrt{20}$ and $\sqrt[3]{5}$ are surds as $20$ is not a perfect square and $5$ is not a perfect cube.

Simplify a Surd (Square Root)

A surd (square root) can only be simplified if a perfect square is a factor of the radicand (the number under the square root).

The first 12 square numbers are:

$\boxed{1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}$

Example: Simplify $\sqrt{72}$

Let’s list the factors of 72.

The factors of 72 are: $1, 2, 3, \textcircled{4}, 6, 8, \textcircled{9}, 12, 18, 24, \textcircled{36}, 72$

The factors that are also perfect squares are: 4, 9, 36 (also 1 but 1 is not useful here).

The highest is 36.

So, $\sqrt{72}=\sqrt{36\times 2}=\sqrt{36}\times \sqrt{2}=6\sqrt{2}$

Suppose I selected 9 or 4 because I didn’t spot that 36 was a factor. In this case, I end up with another surd that can be simplified, so I just simplify again.

e.g. $\sqrt{72}=\sqrt{9\times 8}=\sqrt{9}\times \sqrt{8}=3\sqrt{8}$

and $\sqrt{8}=\sqrt{4 \times 2}=\sqrt{4} \times \sqrt{2}=2\sqrt{2}$

Therefore, $\sqrt{72}=3\sqrt{8}=3\times2 \sqrt{2}=6\sqrt{2}$

Simplify the square roots by writing as a product of a square number with a non square number:

The Exponent Laws for Surds

As explained on the page Rational Exponents, $\sqrt{a}=a^{\frac{1}{2}}$ .

The procedure to simplify surds works out because it follows the laws of exponents.

$(a\times b)^m=a^m \times b^m$

In the case of square roots,

$\sqrt{a\times b}=(a \times b)^{\frac{1}{2}}=a^{\frac{1}{2}} \times b^{\frac{1}{2}}=\sqrt{a} \times \sqrt{b}$

Rewriting the example $\sqrt{72}$ , we have

$\sqrt{72}=72^{\frac{1}{2}}=(36 \times 2)^{\frac{1}{2}}=36^{\frac{1}{2}}\times 2^{\frac{1}{2}}=\sqrt{36}\times \sqrt{2}=6\sqrt{2}$

Rationalise a Denominator

A fraction that has a rational denominator is considered more simple than a fraction with an irrational denominator.

To rationalise a denominator, we multiply by the number $1$ which retains the value of the fraction, however we choose the form of $1$ to change the appearance of the fraction.

We use the fact that $\frac{a}{a}=1$ for $a\in \mathbb{R}$ with the exception of $a=0$ .

We also use the fact that $\sqrt{b}\times \sqrt{b}=b$ for all $b \in \mathbb{R^+}$

Example 1:

Rationalise the denominator of the fraction $\frac{5}{\sqrt{3}}$ .

We use the facts: $\frac{\sqrt{3}}{\sqrt{3}}=1$ , and $\sqrt{3}\times \sqrt{3}=3$ .

$\begin{align*}&\frac{5}{\sqrt{3}}\\[10 pt]=&\frac{5}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}\\[10 pt]=&\frac{5\sqrt{3}}{3}\end{align}$

The denominator, or the divisor, is now the integer $3$ which is a rational number.

In this following example we use the difference of two squares to eliminate the surds on the denominator. Using FOIL, we have

$(a+b)(a-b)=a^2-ab+ab-b^2=a^2-b^2$

Example 2

Rationalise the denominator of the fraction $\frac{5}{2-\sqrt{3}}$ .

$\begin{align*}&\frac{5}{2-\sqrt{3}}\\[10 pt]=&\frac{5}{2-\sqrt{3}}\cdot \frac{2+\sqrt{3}}{2+\sqrt{3}}\\[10 pt]=&\frac{10+5\sqrt{3}}{4+2\sqrt{3}-2\sqrt{3}-3}\\[10 pt]=&\frac{10+5\sqrt{3}}{1}\\[10 pt]=&10+5\sqrt{3}\end{align}$

As this work is entirely numerical, we can check our answers on the calculator by evaluating the initial and final values to determine if they are indeed equal:

$\frac{5}{2-\sqrt{3}}=5 \div (2-\sqrt{3})=18.66025\dots$

$10+5\sqrt{3}=18.66025\dots\quad \checkmark$

Practice

Try the Corbett Maths textbook questions on surds. Answers here. Worked solutions to select questions here: Solutions Corbett surds.

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