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# 1.2 Rational exponents and surds

## 1.2 Rational exponents and surds (EMBF5)

The laws of exponents can also be extended to include the rational numbers. A rational number is any number that can be written as a fraction with an integer in the numerator and in the denominator. We also have the following definitions for working with rational exponents.

• If $$r^n = a$$, then $$r = \sqrt[n]{a} \quad (n \geq 2)$$
• $$a^{\frac{1}{n}} = \sqrt[n]{a}$$
• $$a^{-\frac{1}{n}} = (a^{-1})^{\frac{1}{n}} = \sqrt[n]{\frac{1}{a}}$$
• $$a^{\frac{m}{n}} = (a^{m})^{\frac{1}{n}} = \sqrt[n]{a^m}$$
where $$a > 0$$, $$r > 0$$ and $$m,n \in \mathbb{Z}$$, $$n \ne 0$$.

For $$\sqrt{25} = 5$$, we say that $$\text{5}$$ is the square root of $$\text{25}$$ and for $$\sqrt[3]{8} = 2$$, we say that $$\text{2}$$ is the cube root of $$\text{8}$$. For $$\sqrt[5]{32} = 2$$, we say that $$\text{2}$$ is the fifth root of $$\text{32}$$.

When dealing with exponents, a root refers to a number that is repeatedly multiplied by itself a certain number of times to get another number. A radical refers to a number written as shown below.

The radical symbol and degree show which root is being determined. The radicand is the number under the radical symbol.

• If $$n$$ is an even natural number, then the radicand must be positive, otherwise the roots are not real. For example, $$\sqrt[4]{16} = 2$$ since $$2 \times 2 \times 2 \times 2 = 16$$, but the roots of $$\sqrt[4]{-16}$$ are not real since $$(-2) \times (-2) \times (-2) \times (-2) \ne -16$$.

• If $$n$$ is an odd natural number, then the radicand can be positive or negative. For example, $$\sqrt[3]{27} = 3$$ since $$3 \times 3 \times 3 = 27$$ and we can also determine $$\sqrt[3]{-27} = -3$$ since $$(-3) \times (-3) \times (-3) = -27$$.

A surd is a radical which results in an irrational number. Irrational numbers are numbers that cannot be written as a fraction with the numerator and the denominator as integers. For example, $$\sqrt{12}$$, $$\sqrt[3]{\text{100}}$$, $$\sqrt[5]{25}$$ are surds.

## Worked example 3: Rational exponents

Write each of the following as a radical and simplify where possible:

1. $$18^\frac{1}{2}$$
2. $$(-\text{125})^{-\frac{1}{3}}$$
3. $$4^\frac{3}{2}$$
4. $$(-81)^\frac{1}{2}$$
5. $$(\text{0,008})^\frac{1}{3}$$
1. $$18^\frac{1}{2} = \sqrt{18}$$
2. $$(-\text{125})^{-\frac{1}{3}} = \sqrt[3]{(-\text{125})^{-1}} = \sqrt[3]{\dfrac{1}{-\text{125}}} = \sqrt[3]{\dfrac{1}{(-5)^3}} = -\dfrac{1}{5}$$
3. $$4^\frac{3}{2} = (4^3)^{\frac{1}{2}} = \sqrt{4^3} = \sqrt{64} = 8$$
4. $$(-81)^\frac{1}{2} = \sqrt{-81} =$$ not real
5. $$(\text{0,008})^\frac{1}{3} = \sqrt[3]{\dfrac{8}{\text{1 000}}} = \sqrt[3]{\dfrac{2^3}{10^3}} = \frac{2}{10} = \dfrac{1}{5}$$
temp text

## Worked example 4: Rational exponents

Simplify without using a calculator:

$\left(\frac{5}{4^{-1}-9^{-1}}\right)^{\frac{1}{2}}$

### Write the fraction with positive exponents in the denominator

$\left( \dfrac{5}{\frac{1}{4} - \frac{1}{9}} \right)^{\frac{1}{2}}$

### Simplify the denominator

\begin{align*} &= \left( \dfrac{5}{\frac{9-4}{36}} \right)^{\frac{1}{2}} \\ &= \left( \dfrac{5}{\frac{5}{36}} \right)^{\frac{1}{2}} \\ &= \left( 5 \div \frac{5}{36} \right)^{\frac{1}{2}} \\ &= \left( 5 \times \frac{36}{5} \right)^{\frac{1}{2}} \\ &= (36)^{\frac{1}{2}} \end{align*}

### Take the square root

\begin{align*} &= \sqrt{36}\\ &= 6 \end{align*}

## Rational exponents and surds

Textbook Exercise 1.3

Simplify the following and write answers with positive exponents:

$$\sqrt{49}$$

$$\sqrt{49} = 7$$

$$\sqrt{36^{-1}}$$

\begin{align*} 36^{-\frac{1}{2}} &= \frac{1}{36^{\frac{1}{2}}} \\ &= \frac{1}{\sqrt{36}} \\ &= \frac{1}{6} \end{align*}

$$\sqrt[3]{6^{-2}}$$

\begin{align*} 6^{-\dfrac{2}{3}} &= \frac{1}{6^{\frac{2}{3}}}\\ &= \frac{1}{\sqrt[3]{6^2}} \\ &= \frac{1}{\sqrt[3]{36}} \end{align*}

$$\sqrt[3]{-\dfrac{64}{27}}$$

\begin{align*} \sqrt[3]{-\dfrac{64}{27}} &= \sqrt[3]{-\frac{4^3}{3^3}} \\ &= \sqrt[3]{ \left( -\frac{4}{3} \right)^3} \\ &= -\frac{4}{3} \end{align*}

$$\sqrt[4]{(16x^4)^3}$$

\begin{align*} \sqrt[4]{(16x^4)^3} &= \left( (16x^4)^3 \right)^{\frac{1}{4}} \\ &= \left( 2^4x^4 \right)^{\frac{3}{4}} \\ &= 2^3x^3 \\ &= 8x^3 \end{align*}

Simplify:

$${s}^{\frac{1}{2}}÷{s}^{\frac{1}{3}}$$

\begin{align*} s^{\frac{1}{2}} \div s^{\frac{1}{3}} &= s^{\frac{1}{2}} \times \frac{1}{s^{\frac{1}{3}}} \\ &= s^{\frac{1}{2} - \frac{1}{3}} \\ &= s^{\frac{3}{6} - \frac{2}{6}} \\ &= s^{\frac{1}{6}} \end{align*}

$${\left(64{m}^{6}\right)}^{\frac{2}{3}}$$

\begin{align*} \left( 64m^6 \right)^{\frac{2}{3}} &= \left( 2^6 \right)^{\frac{2}{3}} \left( m^6 \right)^{\frac{2}{3}} \\ &= \left( 2^4 \right) \left( m^4 \right) \\ &= 16m^4 \end{align*}
$$\dfrac{12m^{ \frac{7}{9}}}{8m^{- \frac{11}{9}}}$$
\begin{align*} \dfrac{12m^{\frac{7}{9}}}{8m^{-\frac{11}{9}}} &= \frac{3}{2} m^{ \frac{7}{9} - \left( - \frac{11}{9} \right)} \\ &= \frac{3}{2} m^{\frac{18}{9}} \\ &= \frac{3}{2}m^{2} \end{align*}

$$\left(5x\right)^0 + 5x^0 - \left(\text{0,25}\right)^{-\text{0,5}} + 8^{\frac{2}{3}}$$

\begin{align*} \left(5x\right)^0 + 5x^0 - \left(\text{0,25}\right)^{-\text{0,5}} + 8^{\frac{2}{3}} &= \left(1 \right) + 5(1)- \left( \frac{1}{4} \right)^{-\frac{1}{2}} + \left(2^3 \right)^{\frac{2}{3}} \\ &= 6 - 4^{\frac{1}{2}} + 4 \\ &= 10 - 2 \\ &= 8 \end{align*}

Use the laws to re-write the following expression as a power of $$x$$:

$$x\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}$$
\begin{align*} x \sqrt{x \sqrt{x\sqrt{x\sqrt{x}}}} &= x \times x^{\frac{1}{2}} \times x^{\frac{1}{4}} \times x^{\frac{1}{8}} \times x^{\frac{1}{16}} \\ &= x^{\frac{16}{16}} \times x^{\frac{8}{16}} \times x^{\frac{4}{16}} \times x^{\frac{2}{16}}\times x^{\frac{1}{16}} \\ &= x^{\frac{31}{16}} \end{align*}

### Simplification of surds (EMBF6)

We have seen in previous examples and exercises that rational exponents are closely related to surds. It is often useful to write a surd in exponential notation as it allows us to use the exponential laws.

The additional laws listed below make simplifying surds easier:

• $$\sqrt[n]{a}\sqrt[n]{b} = \sqrt[n]{ab}$$
• $$\sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$$
• $$\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}$$
• $$\sqrt[n]{a^m} = a^{\frac{m}{n}}$$
• $$\left( \sqrt[n]{a} \right)^m = a^{\frac{m}{n}}$$

## Worked example 5: Simplifying surds

Show that:

1. $$\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{ab}$$
2. $$\sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$$
1. \begin{align*} \sqrt[n]{a} \times \sqrt[n]{b} &= a^{\frac{1}{n}} \times b^{\frac{1}{n}} \\ &= (ab)^{\frac{1}{n}} \\ &= \sqrt[n]{ab} \end{align*}
2. \begin{align*} \sqrt[n]{\frac{a}{b}} &= \left( \frac{a}{b} \right)^{\frac{1}{n}} \\ &= \dfrac{a^{\frac{1}{n}}}{b^{\frac{1}{n}}} \\ &= \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} \end{align*}

Examples:

1. $$\sqrt{2} \times \sqrt{32} = \sqrt{2 \times 32} = \sqrt{64} = 8$$

2. $$\dfrac{\sqrt[3]{24}}{\sqrt[3]{3}} = \sqrt[3]{\dfrac{24}{3}} = \sqrt[3]{8} = 2$$

3. $$\sqrt{\sqrt{81}} = \sqrt[4]{81} = \sqrt[4]{3^4} = 3$$

### Like and unlike surds (EMBF7)

Two surds $$\sqrt[m]{a}$$ and $$\sqrt[n]{b}$$ are like surds if $$m=n$$, otherwise they are called unlike surds. For example, $$\sqrt{\frac{1}{3}}$$ and $$-\sqrt{61}$$ are like surds because $$m = n = 2$$. Examples of unlike surds are $$\sqrt[3]{5} \text{ and } \sqrt[5]{7y^3}$$ since $$m \ne n$$.

### Simplest surd form (EMBF8)

We can sometimes simplify surds by writing the radicand as a product of factors that can be further simplified using $$\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}$$.

## Worked example 6: Simplest surd form

Write the following in simplest surd form: $$\sqrt{50}$$

### Write the radicand as a product of prime factors

\begin{align*} \sqrt{50} &= \sqrt{5 \times 5 \times 2} \\ &= \sqrt{5^2 \times 2} \end{align*}

### Simplify using $$\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}$$

\begin{align*} &= \sqrt{5^2} \times \sqrt{2} \\ &= 5 \times \sqrt{2} \\ &= 5\sqrt{2} \end{align*}

Sometimes a surd cannot be simplified. For example, $$\sqrt{6}, \sqrt[3]{30} \text{ and } \sqrt[4]{42}$$ are already in their simplest form.

## Worked example 7: Simplest surd form

Write the following in simplest surd form: $$\sqrt[3]{54}$$

### Write the radicand as a product of prime factors

\begin{align*} \sqrt[3]{54} &= \sqrt[3]{3 \times 3 \times 3 \times 2} \\ &= \sqrt[3]{3^3 \times 2} \end{align*}

### Simplify using $$\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}$$

\begin{align*} &= \sqrt[3]{3^3} \times \sqrt[3]{2} \\ &= 3 \times \sqrt[3]{2} \\ &= 3\sqrt[3]{2} \end{align*}

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## Worked example 8: Simplest surd form

Simplify: $$\sqrt{\text{147}}+\sqrt{\text{108}}$$

### Write the radicands as a product of prime factors

\begin{align*} \sqrt{\text{147}} + \sqrt{\text{108}} &= \sqrt{49 \times 3} + \sqrt{36 \times 3} \\ &= \sqrt{7^2 \times 3} + \sqrt{6^2 \times 3} \end{align*}

### Simplify using $$\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}$$

\begin{align*} &= \left( \sqrt{7^2} \times \sqrt{3} \right) + \left( \sqrt{6^2} \times \sqrt{3} \right)\\ &= \left( 7 \times \sqrt{3} \right) + \left( 6 \times \sqrt{3} \right) \\ &= 7\sqrt{3} + 6\sqrt{3} \end{align*}

### Simplify and write the final answer

$13 \sqrt{3}$

## Worked example 9: Simplest surd form

Simplify: $$\left( \sqrt{20} - \sqrt{5} \right)^2$$

### Factorise the radicands were possible

$\left( \sqrt{20} - \sqrt{5} \right)^2 = \left( \sqrt{4 \times 5} - \sqrt{5} \right)^2$

### Simplify using $$\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}$$

\begin{align*} &= \left( \sqrt{4} \times \sqrt{5} - \sqrt{5} \right)^2 \\ &= \left( 2 \times \sqrt{5} - \sqrt{5} \right)^2 \\ &= \left( 2\sqrt{5} - \sqrt{5} \right)^2 \end{align*}

### Simplify and write the final answer

\begin{align*} &= \left( \sqrt{5} \right)^2 \\ &= 5 \end{align*}

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## Worked example 10: Simplest surd form with fractions

Write in simplest surd form: $$\sqrt{75} \times \sqrt[3]{(48)^{-1}}$$

### Factorise the radicands were possible

\begin{align*} \sqrt{75} \times \sqrt[3]{(48)^{-1}} &= \sqrt{25 \times 3} \times \sqrt[3]{\frac{1}{48}} \\ &= \sqrt{25 \times 3} \times \frac{1}{\sqrt[3]{8 \times 6}} \end{align*}

### Simplify using $$\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}$$

\begin{align*} &= \sqrt{25} \times \sqrt{3} \times \frac{1}{\sqrt[3]{8} \times \sqrt[3]{6}} \\ &= 5 \times \sqrt{3} \times \frac{1}{2 \times \sqrt[3]{6}} \end{align*}

### Simplify and write the final answer

\begin{align*} &= 5\sqrt{3} \times \frac{1}{2\sqrt[3]{6}} \\ &= \frac{5\sqrt{3}}{2\sqrt[3]{6}} \end{align*}

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## Simplification of surds

Textbook Exercise 1.4

Simplify the following and write answers with positive exponents:

$$\sqrt[3]{16} \times \sqrt[3]{4}$$

$$\sqrt[3]{16} \times \sqrt[3]{4} = \sqrt[3]{16 \times 4} = \sqrt[3]{64} = 4$$

$$\sqrt{a^2b^3} \times \sqrt{b^5c^4}$$

$$\sqrt{a^2b^3} \times \sqrt{b^5c^4} = \sqrt{a^2b^8c^4} = ab^4c^2$$

$$\dfrac{\sqrt{12}}{\sqrt{3}}$$

$$\dfrac{\sqrt{12}}{\sqrt{3}} = \sqrt{\dfrac{12}{3}} = \sqrt{4} = 2$$

$$\sqrt{x^2y^{13}} \div \sqrt{y^5}$$

$$\sqrt{x^2y^{13}} \div \sqrt{y^5} = \sqrt{\frac{x^2y^{13}}{y^5}} = \sqrt{x^2y^8} = xy^4$$

Simplify the following:

$$\left( \dfrac{1}{a} - \dfrac{1}{b} \right)^{-1}$$

\begin{align*} \left( \dfrac{1}{a} - \dfrac{1}{b} \right)^{-1} &= \left( \dfrac{b - a}{ab} \right)^{-1} \\ &= \frac{ab}{b - a} \end{align*}

$$\dfrac{b-a}{a^{\frac{1}{2}} - b^{\frac{1}{2}}}$$

\begin{align*} \dfrac{b-a}{a^{\frac{1}{2}} - b^{\frac{1}{2}}} &= - \dfrac{a - b}{a^{\frac{1}{2}} - b^{\frac{1}{2}}} \\ &= - \dfrac{ \left( a^{\frac{1}{2}} \right)^2- \left( b^{\frac{1}{2}} \right)^2}{a^{\frac{1}{2}} - b^{\frac{1}{2}}} \\ &= - \dfrac{ \left( a^{\frac{1}{2}} - b^{\frac{1}{2}} \right)\left( a^{\frac{1}{2}} + b^{\frac{1}{2}} \right) } {a^{\frac{1}{2}} - b^{\frac{1}{2}}} \\ &= - \left( a^{\frac{1}{2}} + b^{\frac{1}{2}} \right) \end{align*}

### Rationalising denominators (EMBF9)

It is often easier to work with fractions that have rational denominators instead of surd denominators. By rationalising the denominator, we convert a fraction with a surd in the denominator to a fraction that has a rational denominator.

## Worked example 11: Rationalising the denominator

Rationalise the denominator: $\frac{5x-16}{\sqrt{x}}$

### Multiply the fraction by $$\frac{\sqrt{x}}{\sqrt{x}}$$

Notice that $$\frac{\sqrt{x}}{\sqrt{x}}=1$$, so the value of the fraction has not been changed.

$\frac{5x - 16}{\sqrt{x}} \times \frac{\sqrt{x}}{\sqrt{x}} = \frac{\sqrt{x}(5x - 16)}{\sqrt{x} \times \sqrt{x}}$

### Simplify the denominator

\begin{align*} &= \frac{\sqrt{x}(5x - 16)}{\left( \sqrt{x}\right)^2} \\ &= \frac{\sqrt{x}(5x - 16)}{x} \end{align*}

The term in the denominator has changed from a surd to a rational number. Expressing the surd in the numerator is the preferred way of writing expressions.

## Worked example 12: Rationalising the denominator

Write the following with a rational denominator: $\frac{y-25}{\sqrt{y}+5}$

### Multiply the fraction by $$\frac{\sqrt{y}-5}{\sqrt{y}-5}$$

To eliminate the surd from the denominator, we must multiply the fraction by an expression that will result in a difference of two squares in the denominator.

$\frac{y-25}{\sqrt{y}+5} \times \frac{\sqrt{y}-5}{\sqrt{y}-5}$

### Simplify the denominator

\begin{align*} &= \frac{(y-25)(\sqrt{y}-5)}{(\sqrt{y}+5)(\sqrt{y}-5)} \\ &= \frac{(y-25)(\sqrt{y}-5)}{(\sqrt{y})^2 -25} \\ &= \frac{(y-25)(\sqrt{y}-5)}{y-25} \\ &= \sqrt{y}-5 \end{align*}

## Rationalising the denominator

Textbook Exercise 1.5

Rationalise the denominator in each of the following:

$$\dfrac{10}{\sqrt{5}}$$

\begin{align*} \dfrac{10}{\sqrt{5}} &=\frac{10}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} \\ &=\frac{10\sqrt{5}}{5} \\ &= 2\sqrt{5} \end{align*}

$$\dfrac{3}{\sqrt{6}}$$

\begin{align*} \dfrac{3}{\sqrt{6}} &=\frac{3}{\sqrt{6}} \times \dfrac{\sqrt{6}}{\sqrt{6}} \\ &=\frac{3\sqrt{6}}{6} \\ &=\frac{\sqrt{6}}{2} \end{align*}

$$\dfrac{2}{\sqrt{3}} \div \dfrac{\sqrt{2}}{3}$$

\begin{align*} \dfrac{2}{\sqrt{3}} \div \dfrac{\sqrt{2}}{3} &=\dfrac{2}{\sqrt{3}} \div \dfrac{\sqrt{2}}{3} \\ &=\dfrac{2}{\sqrt{3}} \times \dfrac{3}{\sqrt{2}} \\ &=\dfrac{6}{\sqrt{6}} \times \dfrac{\sqrt{6}}{\sqrt{6}} \\ &=\frac{6\sqrt{6}}{6} \\ &=\sqrt{6} \end{align*}

$$\dfrac{3}{\sqrt{5}-1}$$

\begin{align*} \dfrac{3}{\sqrt{5}-1} &=\dfrac{3}{\sqrt{5}-1} \times \dfrac{\sqrt{5}+1}{\sqrt{5}+1} \\ &=\dfrac{3\sqrt{5}+3}{5-1} \\ &= \dfrac{3\sqrt{5}+3}{4} \end{align*}

$$\dfrac{x}{\sqrt{y}}$$

\begin{align*} \dfrac{x}{\sqrt{y}} &=\dfrac{x}{\sqrt{y}}\times \dfrac{\sqrt{y}}{\sqrt{y}} \\ &=\dfrac{x\sqrt{y}}{y} \end{align*}

$$\dfrac{\sqrt{3} + \sqrt{7}}{\sqrt{2}}$$

\begin{align*} \dfrac{\sqrt{3} + \sqrt{7}}{\sqrt{2}} &=\dfrac{\sqrt{3}+\sqrt{7}}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} \\ &=\dfrac{\sqrt{3}\sqrt{2}+\sqrt{7}\sqrt{2}}{2} \\ &=\dfrac{\sqrt{6}+\sqrt{14}}{2} \end{align*}

$$\dfrac{3\sqrt{p} - 4}{\sqrt{p}}$$

\begin{align*} \dfrac{3\sqrt{p} - 4}{\sqrt{p}} &=\dfrac{3\sqrt{p}-4}{\sqrt{p}} \times \dfrac{\sqrt{p}}{\sqrt{p}} \\ &=\dfrac{3\left ( \sqrt{p} \right )^2-4\left ( \sqrt{p} \right )}{p} \\ &=\dfrac{3p-4\sqrt{p}}{p} \end{align*}

$$\dfrac{t-4}{\sqrt{t} + 2}$$

\begin{align*} \dfrac{t-4}{\sqrt{t} + 2} &=\dfrac{t-4}{\sqrt{t}+2} \times \dfrac{\sqrt{t}-2}{\sqrt{t}-2} \\ &=\dfrac{\left ( t-4 \right )\left ( \sqrt{t}-2 \right )}{t-4} \\ &=\sqrt{t}-2 \end{align*}

$$\left( 1 + \sqrt{m} \right)^{-1}$$

\begin{align*} \left( 1 + \sqrt{m} \right)^{-1} &=\frac{1}{1+\sqrt{m}}\times \frac{1-\sqrt{m}}{1-\sqrt{m}} \\ &=\frac{1-\sqrt{m}}{1-m} \end{align*}

$$a \left( \sqrt{a} \div \sqrt{b} \right)^{-1}$$

\begin{align*} a \left( \sqrt{a} \div \sqrt{b} \right)^{-1} &=a\left ( \sqrt{a} \times\frac{1}{\sqrt{b}} \right )^{-1} \\ &=a\left ( \frac{\sqrt{a}}{\sqrt{b}} \right )^{-1} \\ &=a\frac{\sqrt{b}}{\sqrt{a}} \times \frac{\sqrt{a}}{\sqrt{a}} \\ &=\frac{a\sqrt{ab}}{a} \\ &=\sqrt{ab} \end{align*}