Proof

What you'll learn

How to use a letter to represent any possible case.
How to show a result is always odd, even, or divisible by a chosen number.
How to deal with brackets and squared terms without losing the proof.
How to finish with a clear conclusion.

1. Start with a general number

In proof, you are not just checking examples. You are showing that something works for every possible case.

Instead of choosing a particular number, choose a letter such as $n$ . That letter stands for a general number.

Definition

Core language

An algebraic proof uses letters and algebra to prove a statement is always true.
An integer is a number with no fractional or decimal part, such as -3, 0, 7 or 12.
A positive integer is an integer greater than zero.
Consecutive integers are integers that follow one after another, such as $n$ , $n + 1$ , $n + 2$ .

A number line shows consecutive integers as equally spaced steps of 1.

An even integer can be written as $2 k$ for some integer $k$ .

Even and odd integers can be represented as pairs, with an odd integer having one extra unit.

An odd integer can be written as $2 k + 1$ for some integer $k$ .

Key Idea

The proof idea

If your final expression has the right algebraic form, you have proved the result for every allowed value, not just for examples you tried.

Example

Two neighbouring integers

Show that adding two integers next to each other always gives an odd answer.

Two neighbouring integers can be written as n and n+1 on a number line.

Let the smaller integer be $n$ . The next integer is $n + 1$ .
Add them and simplify:
$n + (n + 1) = 2 n + 1$
The expression $2 n + 1$ is in the form “even plus 1”, so it is odd.
Therefore, the sum of two consecutive integers is always odd.

2. Proving something is a multiple

A common GCSE proof target is to show that an answer is divisible by a number.

Definition

Multiple

A multiple of 6 is any number that can be written as $6 k$ , where $k$ is an integer. More generally, a multiple of $m$ can be written as $mk$ .

For example, to prove something is a multiple of 6, your aim is to rearrange it into:

\times \text{an integer}

Consecutive even numbers

Consecutive even integers go up by 2, not by 1. A safe way to write three of them is:

2n,2n+2,2n+42n,\quad 2n+2,\quad 2n+4

Example

Three even integers in a row

Show that the sum of three consecutive even integers is always divisible by 6.

Three consecutive even integers are spaced by 2 and can be written as 2n, 2n+2 and 2n+4.

Let the three even integers be $2 n$ , $2 n + 2$ and $2 n + 4$ .
Add them:
$2 n + (2 n + 2) + (2 n + 4) = 6 n + 6$
Factorise by taking out 6:
$6 n + 6 = 6 (n + 1)$
Since $n + 1$ is an integer, $6 (n + 1)$ is a multiple of 6.
Therefore, the sum is always divisible by 6.

Common Mistake

Using $n$ , $n + 2$ , $n + 4$ too loosely

Writing $n$ , $n + 2$ , $n + 4$ only guarantees numbers spaced by 2. It does not guarantee they are even unless you also say that $n$ is even. Starting with $2 n$ avoids this problem.

Numbers spaced by 2 can all be odd if the starting number is odd, so using 2n is safer for even numbers.

3. Expand, simplify, then factorise

Some proof questions give you an expression with brackets and ask you to prove it is always a multiple of something.

Definition

Expanding and factorising

To expand means to remove brackets by multiplying out.
To factorise means to rewrite an expression using common factors, usually putting brackets back in.
An expression is a piece of algebra, such as $4n+3)^2-(4n-3)^2$ .

The usual proof flow is:

Expand carefully.
Collect like terms.
Factorise to reveal the required multiple.

Example

A difference of two squared brackets

Show that $4n+3)^2-(4n-3)^2$ is always divisible by 12 for positive integer values of $n$ .

The two squared brackets differ only by adding or subtracting 3 from the same central term 4n.

Expand the first square:
$4n+3)^2=16n^2+24n+9$
Expand the second square:
$4n-3)^2=16n^2-24n+9$
Subtract the second result from the first:
$(16n2+24n+9)−(16n2−24n+9)=16n2+24n+9−16n2+24n−9=48n\begin{aligned} (16n^2+24n+9)-(16n^2-24n+9) &=16n^2+24n+9-16n^2+24n-9\\ &=48n \end{aligned}$
Factor out 12:
$48 n = 12 (4 n)$
Since $4 n$ is an integer, the expression is always a multiple of 12.

Tip

Shortcut for this shape

When two squared brackets differ only by a sign, use $a+b)^2-(a-b)^2=4ab$ as a quick check. Still make sure your working is clear enough for the marks.

4. Showing something is a square number

Sometimes the target is not “multiple of something”, but “square number”.

Definition

Square number

A square number is the result of an integer multiplied by itself. Algebraically, it has the form $k^2$ , where $k$ is an integer.

For these proofs, you usually want to end with a single bracket squared, such as $n+5)^2$ .

Example

Turning a product into a square

Show that $(n + 5) (n + 4) + (n + 5)$ is always a square number when $n$ is an integer.

The expression can be seen as a rectangle of height n+5 and width n+4, plus one more strip to make a square.

Notice that both parts contain the factor $n + 5$ .
Factorise $n + 5$ :
$(n+5)(n+4)+(n+5)=(n+5)((n+4)+1)(n+5)(n+4)+(n+5)=(n+5)\left((n+4)+1\right)$
Simplify inside the bracket:
$(n+5)((n+4)+1)=(n+5)(n+5)(n+5)\left((n+4)+1\right)=(n+5)(n+5)$
Write it as a square:
$n+5)(n+5)=(n+5)^2$
Since $n + 5$ is an integer, $n+5)^2$ is a square number.

Common Mistake

Stopping too early

If the question asks for a square number, do not stop at an expanded form like $n^2+10n+25$ . Rewrite it as a square, such as $n+5)^2$ .

5. Harder proofs: “more than a multiple”

A Grade 8/9-style proof may ask you to show that something is “2 more than a multiple of 8” or “4 more than a multiple of 8”.

That means your target form is:

2 more than a multiple of 8: $8 k + 2$
4 more than a multiple of 8: $8 k + 4$

Example

Squares of neighbouring odd integers

Show that the sum of the squares of two odd integers next to each other is always 2 more than a multiple of 8.

Neighbouring odd integers are two apart, so they can be written as 2n+1 and 2n+3 before squaring.

Let the two odd integers be $2 n + 1$ and $2 n + 3$ .
Square both and add:
$(2n+1)2+(2n+3)2=(4n2+4n+1)+(4n2+12n+9)=8n2+16n+10\begin{aligned} (2n+1)^2+(2n+3)^2 &=(4n^2+4n+1)+(4n^2+12n+9)\\ &=8n^2+16n+10 \end{aligned}$
Rewrite the expression in the target form:
$8n^2+16n+10=8(n^2+2n+1)+2$
Since $n^2+2n+1$ is an integer, $8(n^2+2n+1)$ is a multiple of 8.
Therefore, the sum is always 2 more than a multiple of 8.

Tip

Aim for the target form

Before you start expanding, decide what the final line should look like: odd means $2 k + 1$ , a multiple of 6 means $6 k$ , and 2 more than a multiple of 8 means $8 k + 2$ .

Exam technique

In the exam

Define your variables clearly, for example: “Let the first integer be $n$ .”
Use the correct form: even numbers need $2 n$ , odd numbers need $2 n + 1$ , and consecutive numbers must increase correctly.
Finish with a sentence linking your final algebraic form to the claim, such as “Since this is $6$ times an integer, it is a multiple of 6.”

Self review

Check yourself

Can you write three consecutive odd integers using $n$ ?
Can you explain why $12 (3 n + 1)$ is always a multiple of 12?
Can you prove that the squares of two consecutive integers have an odd sum?

Proof

What you'll learn

How to use a letter to represent any possible case.
How to show a result is always odd, even, or divisible by a chosen number.
How to deal with brackets and squared terms without losing the proof.
How to finish with a clear conclusion.

1. Start with a general number

In proof, you are not just checking examples. You are showing that something works for every possible case.

Instead of choosing a particular number, choose a letter such as $n$ . That letter stands for a general number.

Definition

Core language

An algebraic proof uses letters and algebra to prove a statement is always true.
An integer is a number with no fractional or decimal part, such as -3, 0, 7 or 12.
A positive integer is an integer greater than zero.
Consecutive integers are integers that follow one after another, such as $n$ , $n + 1$ , $n + 2$ .

A number line shows consecutive integers as equally spaced steps of 1.

An even integer can be written as $2 k$ for some integer $k$ .

Even and odd integers can be represented as pairs, with an odd integer having one extra unit.

An odd integer can be written as $2 k + 1$ for some integer $k$ .

Key Idea

The proof idea

If your final expression has the right algebraic form, you have proved the result for every allowed value, not just for examples you tried.

Example

Two neighbouring integers

Show that adding two integers next to each other always gives an odd answer.

Two neighbouring integers can be written as n and n+1 on a number line.

Let the smaller integer be $n$ . The next integer is $n + 1$ .
Add them and simplify:
$n + (n + 1) = 2 n + 1$
The expression $2 n + 1$ is in the form “even plus 1”, so it is odd.
Therefore, the sum of two consecutive integers is always odd.

2. Proving something is a multiple

A common GCSE proof target is to show that an answer is divisible by a number.

Definition

Multiple

A multiple of 6 is any number that can be written as $6 k$ , where $k$ is an integer. More generally, a multiple of $m$ can be written as $mk$ .

For example, to prove something is a multiple of 6, your aim is to rearrange it into:

\times \text{an integer}

Consecutive even numbers

Consecutive even integers go up by 2, not by 1. A safe way to write three of them is:

2n,2n+2,2n+42n,\quad 2n+2,\quad 2n+4

Example

Three even integers in a row

Show that the sum of three consecutive even integers is always divisible by 6.

Three consecutive even integers are spaced by 2 and can be written as 2n, 2n+2 and 2n+4.

Let the three even integers be $2 n$ , $2 n + 2$ and $2 n + 4$ .
Add them:
$2 n + (2 n + 2) + (2 n + 4) = 6 n + 6$
Factorise by taking out 6:
$6 n + 6 = 6 (n + 1)$
Since $n + 1$ is an integer, $6 (n + 1)$ is a multiple of 6.
Therefore, the sum is always divisible by 6.

Common Mistake

Using $n$ , $n + 2$ , $n + 4$ too loosely

Writing $n$ , $n + 2$ , $n + 4$ only guarantees numbers spaced by 2. It does not guarantee they are even unless you also say that $n$ is even. Starting with $2 n$ avoids this problem.

Numbers spaced by 2 can all be odd if the starting number is odd, so using 2n is safer for even numbers.

3. Expand, simplify, then factorise

Some proof questions give you an expression with brackets and ask you to prove it is always a multiple of something.

Definition

Expanding and factorising

To expand means to remove brackets by multiplying out.
To factorise means to rewrite an expression using common factors, usually putting brackets back in.
An expression is a piece of algebra, such as $4n+3)^2-(4n-3)^2$ .

The usual proof flow is:

Expand carefully.
Collect like terms.
Factorise to reveal the required multiple.

Example

A difference of two squared brackets

Show that $4n+3)^2-(4n-3)^2$ is always divisible by 12 for positive integer values of $n$ .

The two squared brackets differ only by adding or subtracting 3 from the same central term 4n.

Expand the first square:
$4n+3)^2=16n^2+24n+9$
Expand the second square:
$4n-3)^2=16n^2-24n+9$
Subtract the second result from the first:
$(16n2+24n+9)−(16n2−24n+9)=16n2+24n+9−16n2+24n−9=48n\begin{aligned} (16n^2+24n+9)-(16n^2-24n+9) &=16n^2+24n+9-16n^2+24n-9\\ &=48n \end{aligned}$
Factor out 12:
$48 n = 12 (4 n)$
Since $4 n$ is an integer, the expression is always a multiple of 12.

Tip

Shortcut for this shape

When two squared brackets differ only by a sign, use $a+b)^2-(a-b)^2=4ab$ as a quick check. Still make sure your working is clear enough for the marks.

4. Showing something is a square number

Sometimes the target is not “multiple of something”, but “square number”.

Definition

Square number

A square number is the result of an integer multiplied by itself. Algebraically, it has the form $k^2$ , where $k$ is an integer.

For these proofs, you usually want to end with a single bracket squared, such as $n+5)^2$ .

Example

Turning a product into a square

Show that $(n + 5) (n + 4) + (n + 5)$ is always a square number when $n$ is an integer.

The expression can be seen as a rectangle of height n+5 and width n+4, plus one more strip to make a square.

Notice that both parts contain the factor $n + 5$ .
Factorise $n + 5$ :
$(n+5)(n+4)+(n+5)=(n+5)((n+4)+1)(n+5)(n+4)+(n+5)=(n+5)\left((n+4)+1\right)$
Simplify inside the bracket:
$(n+5)((n+4)+1)=(n+5)(n+5)(n+5)\left((n+4)+1\right)=(n+5)(n+5)$
Write it as a square:
$n+5)(n+5)=(n+5)^2$
Since $n + 5$ is an integer, $n+5)^2$ is a square number.

Common Mistake

Stopping too early

If the question asks for a square number, do not stop at an expanded form like $n^2+10n+25$ . Rewrite it as a square, such as $n+5)^2$ .

5. Harder proofs: “more than a multiple”

A Grade 8/9-style proof may ask you to show that something is “2 more than a multiple of 8” or “4 more than a multiple of 8”.

That means your target form is:

2 more than a multiple of 8: $8 k + 2$
4 more than a multiple of 8: $8 k + 4$

Example

Squares of neighbouring odd integers

Show that the sum of the squares of two odd integers next to each other is always 2 more than a multiple of 8.

Neighbouring odd integers are two apart, so they can be written as 2n+1 and 2n+3 before squaring.

Let the two odd integers be $2 n + 1$ and $2 n + 3$ .
Square both and add:
$(2n+1)2+(2n+3)2=(4n2+4n+1)+(4n2+12n+9)=8n2+16n+10\begin{aligned} (2n+1)^2+(2n+3)^2 &=(4n^2+4n+1)+(4n^2+12n+9)\\ &=8n^2+16n+10 \end{aligned}$
Rewrite the expression in the target form:
$8n^2+16n+10=8(n^2+2n+1)+2$
Since $n^2+2n+1$ is an integer, $8(n^2+2n+1)$ is a multiple of 8.
Therefore, the sum is always 2 more than a multiple of 8.

Tip

Aim for the target form

Before you start expanding, decide what the final line should look like: odd means $2 k + 1$ , a multiple of 6 means $6 k$ , and 2 more than a multiple of 8 means $8 k + 2$ .

Exam technique

In the exam

Define your variables clearly, for example: “Let the first integer be $n$ .”
Use the correct form: even numbers need $2 n$ , odd numbers need $2 n + 1$ , and consecutive numbers must increase correctly.
Finish with a sentence linking your final algebraic form to the claim, such as “Since this is $6$ times an integer, it is a multiple of 6.”

Self review

Check yourself

Can you write three consecutive odd integers using $n$ ?
Can you explain why $12 (3 n + 1)$ is always a multiple of 12?
Can you prove that the squares of two consecutive integers have an odd sum?

Proof

What you'll learn

1. Start with a general number

2. Proving something is a multiple

Consecutive even numbers

3. Expand, simplify, then factorise

4. Showing something is a square number

5. Harder proofs: “more than a multiple”

Recap questions

You've reached the end

Proof

What you'll learn

1. Start with a general number

2. Proving something is a multiple

Consecutive even numbers

3. Expand, simplify, then factorise

4. Showing something is a square number

5. Harder proofs: “more than a multiple”

Recap questions

You've reached the end

Proof

What you'll learn

1. Start with a general number

2. Proving something is a multiple

Consecutive even numbers

3. Expand, simplify, then factorise

4. Showing something is a square number

5. Harder proofs: “more than a multiple”

Recap questions

You've reached the end

Proof Revision Guide

Proof

What you'll learn

1. Start with a general number

2. Proving something is a multiple

Consecutive even numbers

3. Expand, simplify, then factorise

4. Showing something is a square number

5. Harder proofs: “more than a multiple”

Recap questions

You've reached the end

Proof Revision Guide