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Square numbers and square roots Cube numbers and cube rootsThis topic is relevant for:

Here we will learn about simple **index notation** including how to write an expression using index notation and how to simplify expressions written in index form.

There are also index notation worksheets based on Edexcel, AQA and OCR GCSE exam style questions, along with further guidance on where to go next if you’re still stuck.

**Index notation** is a way of representing numbers *(constants) *and variables (e.g.

Below are all examples of expressions involving index notations:

\[ 3^{4} \quad \quad \quad a^{5} \quad \quad \quad 2 x^{7} \quad \quad \quad \frac{1}{2}^{2 x} \quad \quad \quad \left(4 y^{2} x^{4}\right)^{7} \quad \quad \quad z^{-\frac{5}{2}} \]

We use index notations, or the plural ‘indices’, to simplify expressions or solve equations involving powers.

E.g.

**Simplify**

We can therefore write this as ^{4}

*This is said as “ 6 to the power of 4”*

E.g

**Simplify** ** x × x × x × x**

We can therefore write this as ^{4}

*This is said as “x to the power of *

E.g

**Simplify**

In this question the term * 4* times.

Therefore you can write this as ^{4}

*This is said as “2y to the power of *

**Term**:

A single number (constant) or variable

E.g.

In the expression

**Coefficients**:

The number which the variable is being multiplied by

E.g.

In ^{3}

**Integer**:

A whole number

E.g.

**Index (also called exponent or powers)**:

The index number is the amount of times you multiply a number/variable by itself. It is normally shown about the base number

E.g.

The index number in ^{4}

*Note: the plural of index is indices*

*Note: you will see index number as a superscript*

**Base Number**:

The number/unknown that is being multiplied by itself an amount of times

E.g.

The base number in ^{4}^{3}

**Square Numbers**:

A number/variable that is* ‘squared’* is multiplied by itself

E.g. ^{2 }*“* or

A square number is found when we multiply an integer (whole number) by itself.

E.g.

\begin{align*}
&1 \times 1=1 \quad \quad \quad \quad \quad \quad \text{Therefore 1 is a square number} \\\\
&2 \times 2=4 \quad \quad \quad \quad \quad \quad \text{Therefore 4 is a square number} \\\\
&3 \times 3=9 \quad \quad \quad \quad \quad \quad \text{Therefore 9 is a square number} \\\\
&10 \times 10=100 \quad \quad \quad \quad \text{Therefore 100 is a square number}
\end{align*}

**Cube Number**:

A number/variable that is ‘*cubed’* is multiplied by itself three times.

E.g.^{3 }*“ 4 cubed”*.

A cube number is found when we multiply an integer (whole number) by itself three times.

E.g.

\begin{align*}
&1 \times 1 \times 1=1 \quad \quad \quad \quad \quad \quad \text{Therefore 1 is a cube number} \\\\
&2 \times 2 \times 2=8 \quad \quad \quad \quad \quad \quad \text{Therefore 8 is a cube number} \\\\
&3 \times 3 \times 3=27 \quad \quad \quad \quad \quad \text{ Therefore 27 is a cube number} \\\\
&10 \times 10 \times 10=1000 \quad \quad \quad \text{Therefore 1000 is a cube number}
\end{align*}

**Identify whether the base numbers for each term are the same***In higher tier questions you may need to manipulate the base numbers first***Identify the operation/s being undertaken between the terms****Follow the rules of index notation to simplifying the expression**

Get your free index notation worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free index notation worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE**Index notation** is part of our series of lessons to support revision on **laws of indices**. You may find it helpful to start with the main laws of indices lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

Simplify ^{2} × 3^{3}

**Identify whether the base numbers for each term are the same**

The base number is

2 **Identify the operation/s being undertaken between the terms**

The terms are being multiplied.

3**Follow the rules of index notation to simplify each expression**

**3 ^{2}**

**3 ^{3}**

Therefore:

\[\begin{array}{l}
3^{2} \times 3^{3} \\
=3 \times 3 \times 3 \times 3 \times 3 \\
=3^{5} \\
\end{array}\]

\[3^{2} \times 3^{3}=3^{5}\]

You will notice the index number

This is a topic we will explore further in lessons on laws of indices.

**Step-by-step guide: Multiplying indices**

Simplify ^{2} × 4x^{3}

**Identify whether the base numbers for each term are the same**

The base number is * x* and is the same in each term.

*Note: the 3 and 4 are the coefficients of the terms*.

**Identify the operation/s being undertaken between the terms**

The terms are being multiplied.

**Follow the rules of index notation to simplify the expression **

**3x ^{2}**

**4x ^{3}**

Therefore:

\begin{array}{l}
3 x^{2} \times 4 x^{3} \\
=3 \times x \times x \times 4 \times x \times x \times x
\end{array}

We can write this as:

\[3 \times 4 \times x \times x \times x \times x \times x\]

Therefore the expression can be simplified to:

\begin{array}{l}
12 \times x^{5} \\
=12 x^{5}
\end{array}

\[3 x^{2} \times 4 x^{3}=12 x^{5}\]

Again here you will notice the index number

Simplify ^{5} ÷ 6^{2}

**Identify whether the base numbers for each term are the same**

The base number is

**Identify the operation/s being undertaken between the terms**

The terms are being divided.

**Follow the rules of index notation to simplify the expression**

**6 ^{5}**

**6 ^{2}**

You can write ^{5} ÷ 6^{2}

\[\frac{6^{5}}{6^{2}} \]

Therefore:

\begin{array}{l}
6^{5} \div 6^{2} \\\\
=\frac{6^{5}}{6^{2}} \\\\
=\frac{6 \times 6 \times 6 \times 6 \times 6 \times 6}{6 \times 6} \\\\
=6 \times 6 \times 6 \\\\
=6^{3}\\\\
\end{array}

\[6^{5} \div 6^{2}=6^{3}\]

You will notice the index number

This is a topic we will explore further in lessons on laws of indices.

**Step-by-step guide: Dividing indices**

Simplify ^{5} ÷ 5y^{2}

**The base number is **

The base number is * y* and is the same in each term.

**Identify the operation/s being undertaken between the terms**

The terms are being divided.

**Follow the rules of index notation to simplify the expression**

**10y ^{5}**

**5y ^{2}**

Therefore:

\[ 10 y^{5} \div 5 y^{2}= \]

\[\frac{10 \times y^{5}}{5 \times y^{2}} \]

You can simplify the coefficient as

You can simplify the

\[\frac{y^{5}}{y^{2}} \quad \text{to} \quad y^{2} \quad \quad \]

(see previous example)

Therefore:

\begin{array}{l}
\frac{10 \times y^{5}}{5 \times y^{2}} \\\\
=2 y^{3} \\\\
\end{array}

\[10 y^{5} \div 5 y^{2}=2 y^{3}\]

You will notice the index number

This is a topic we will explore further in lessons on laws of indices.

Simplify

\[\left(3^{2}\right)^{3}\]

**Identify whether the base numbers for each term are the same**

The base number is

**Identify the operation/s being undertaken between the terms**

The term ^{2}

**Follow the rules of index notation to simplify the expression**

\[\left(3^{2}\right)^{3}\]

is equivalent to **3 ^{2} × 3^{2} × 3^{2}**

Therefore:

\begin{array}{l}
\left(3^{2}\right)^{3} \\\\
=3^{2} \times 3^{2} \times 3^{2} \\\\
=3^{6} \quad \quad \quad \quad \quad \quad
\end{array}

(see example 1 for extended explanation)

\[\left(3^{2}\right)^{3}= 3^{6}\]

You will notice the index number

This is a topic we will explore further in lessons on laws of indices.

**Step-by-step guide: **Brackets with indices

Simplify

\[\left(2 z^{3}\right)^{5}\]

**Identify whether the base numbers for each term are the same**

The base number inside the brackets is* z*.

** Identify the operation/s being undertaken between the terms**

The term ^{3}

**Follow the rules of index notation to simplify the expression**

\[\left(2 z^{3}\right)^{5} \]

is equivalent to **2z ^{3} × 2z^{3} × 2z^{3} × 2z^{3} × 2z^{3}**

Therefore:

\begin{array}{l}
\left(2 z^{3}\right)^{5} \\\\
=2 z^{3} \times 2 z^{3} \times 2 z^{3} \times 2 z^{3} \times 2 z^{3} \\\\
=32 x^{15}
\end{array}

You will notice the index number

This is a topic we will explore further in lessons on laws of indices.

**Step-by-step guide: Laws of indices**

Below are three further examples involving negative indices, fractional indices and the power of zero.

For more information check out our lessons on these topics.

**Simplify a ^{0}**

If we continue the above pattern below we can see that ^{0}

E.g.

Base Number

\begin{array}{l}
2^{4}=2 \times 2 \times 2 \times 2=16 \\
2^{3}=2 \times 2 \times 2=8 \\
2^{2}=2 \times 2=4 \\
2^{1}=2=2 \\
2^{0}=1
\end{array}

E.g.

Base Number

\begin{array}{l}
3^{4}=3 \times 3 \times 3 \times 3=81 \\
3^{3}=3 \times 3 \times 3=27 \\
3^{2}=3 \times 3=9 \\
3^{1}=3=3 \\
3^{0}=1
\end{array}

This pattern is true for any base number therefore ^{0}

^{0} = 1

**Step-by-step guide: Power of 0**

**Write** **a ^{-1}**

We can continue the pattern shown in example ^{-1}

Remember each time we reduce the power by one we are dividing the value by the base number.

E.g.

Base Number

\begin{array}{l}
2^{2}=2 \times 2=4 \\
2^{1}=2=2 \\
2^{0}=1 \\
2^{-1}=\frac{1}{2}
\end{array}

E.g.

Base Number

\begin{array}{l}
3^{2}=3 \times 3=9 \\
3^{1}=3=3 \\
3^{0}=1 \\
3^{-1}=\frac{1}{3}
\end{array}

Because this pattern is true for any base number, ^{-1}

\[a^{-1}=\frac{1}{a}\]

**Step-by-step guide: Negative indices**

Write a^{½} in another form

Look at the example below, see if you can spot the pattern before moving onto the next one.

\begin{array}{ll}
\left(4^{\frac{1}{2}}\right)^{2}=4^{\frac{1}{2}} \times 4^{\frac{1}{2}}=4^{\frac{1}{2}+\frac{1}{2}}=4^{1}=4 & \therefore\left(4^{\frac{1}{2}}\right)^{2}=4 \\
\left(3^{\frac{1}{2}}\right)^{2}=3^{\frac{1}{2}} \times 3^{\frac{1}{2}}=3^{\frac{1}{2}+\frac{1}{2}}=3^{1}=3 & \therefore\left(3^{\frac{1}{2}}\right)^{2}=3 \\
\left(2^{\frac{1}{2}}\right)^{2}=2^{\frac{1}{2}} \times 2^{\frac{1}{2}}=2^{\frac{1}{2}+\frac{1}{2}}=2^{1}=2 & \therefore\left(2^{\frac{1}{2}}\right)^{2}=2
\end{array}

When you multiply a number by itself it is called ‘squaring a number’.

The inverse of this is called square rooting.

Therefore from the above examples:

\begin{array}{l}
\sqrt{4}=4^{\frac{1}{2}} \\
\sqrt{3}=3^{\frac{1}{2}} \\
\sqrt{2}=2^{\frac{1}{2}}
\end{array}

Because this pattern is true for any base number, ^{½}

\[a^{\frac{1}{2}}=\sqrt{a}\]

**Step-by-step guide: Fractional indices**

**Base Numbers**

A common error is to incorrectly simplify expressions using the index rules where they do not share a base number.

**Index power of 1**

If no index power is shown you need to remember it is being multiplied by itself ‘one time’ e.g **=** 2^{1}

s**Operation**

A common error is to attempt to simplify an expression using the incorrect operation between the base numbers.

**Coefficients and brackets**

A common error is to not include the coefficients when expanding brackets.

E.g

\[(2 x)^{3}=2 x^{3}\]

*This should be*

\[(2x)^{3}=2x\times 2x\times 2x=8x^{3}\]

**Fractions and decimals**

We can write fractions and decimals using index notation.

E.g.

\begin{array}{l}
0.5^{3}=0.5 \times 0.5 \times 0.5=0.125=\frac{1}{8} \\\\
(\frac{1}{2})^{3}=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}
\end{array}

1. Write 8 \times 8 \times 8 \times 8 using index notation

8^{4}

32

32^{4}

4096^{4}

We are multiplying 8 by itself 4 times which can be written as 8^{4} .

2. Write 3 a \times 3 a \times 3 a \times 3 a using index notation

3a^{4}

(3a)^{4}

12a

12a^{4}

We are multiplying 3a by itself 4 times which can be written as (3a)^{4} .

3. Simplify 5 x^{5} \times 3 x^{6}

8x^{11}

15x^{30}

15x^{11}

8x^{30}

\begin{aligned}
5x^{5} \times 3x^{6} &= 5 \times x \times x \times x \times x \times x \times 3 \times x \times x \times x \times x \times x \times x\\\\
&=15x^{11}
\end{aligned}

4. Simplify z^{5} \div z^{3}

z^{\frac{5}{3}}

z^{-2}

2

z^{2}

\begin{aligned}
z^{5} \div z^{3} &= \frac{z \times z \times z \times z \times z}{z \times z \times z}\\\\
&=z^{2}
\end{aligned}

5. Simplify (3c^{4})^{2}

3c^{8}

6c^{8}

9c^{8}

6c^{6}

\begin{aligned}
(3c^{4})^{2} &= 3c^{4} \times 3c^{4}\\\\
&=3 \times c \times c \times c \times c \times 3 \times c \times c \times c \times c\\\\
&=9c^{8}
\end{aligned}

6. Write g^{-1} in another form

0

\frac{1}{g}

\sqrt{g}

-g

Raising a term to the power of negative 1 is the same as writing 1 over that term. g^{-1}=\frac{1}{g}

7. Find the positive value of 64^{\frac{1}{2}}

8

32

16

\frac{1}{64}

The power of \frac{1}{2} means the square root

\sqrt{64}=8

1. Work out the value of:

a) 3^{3}

b) 3 \times 2^{3}

**(2 Marks)**

Show answer

a) 3 \times 3 \times 3

** = 27 **

** (1)**

b) 3 \times 8

** = 24 **

** (1)**

2. Simplify fully:

a) x^{3} \times x^{3}

b) (2 x)^{4}

**(2 Marks)**

Show answer

a) x^{6}

** (1)**

b) 16 x^{4}

** (1)**

3. Simplify fully:

a) 3 x^{6} \times 2 x^{4} \times x

b) 12 x^{5} \div 3 x^{2}

c) \left(3 x^{2}\right)^{4}

d) 4^{0}

**(7 Marks)**

Show answer

a) 6x^{11}

Correct coefficient

** (1)**

Correct index number for x

** (1)**

b) 4x^{3}

Correct coefficient

** (1)**

Correct index number for x

** (1)**

c) 81x^{8}

Correct coefficient

** (1)**

Correct index number for x

** (1)**

d) 1

** (1)**

You have now learned how to:

- Calculate with roots, and with integer indices
- Simplifying expressions involving sums, products and powers
- Simplify and manipulate algebraic expressions involving indices
- Simplify and manipulate algebraic expressions involving simple negative and fractional indices

- Proof maths
- Functions in algebra
- Sequences

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