Proof

What you'll learn

How to write integers, odd numbers, even numbers and consecutive numbers using algebra.
How to prove something is always odd, even, a multiple, or a square number.
How to use expansion and factorising clearly in proof questions.
How to finish proofs with a convincing final sentence.

What is a proof?

A proof is a logical argument that shows a statement is true for every value allowed in the question, not just for a few examples.

Checking numbers can help you spot a pattern, but it is not a proof. For example, if a result works for 1, 2 and 3, it might still fail for 100. Algebra lets you represent all possible cases at once.

Definition

Key words for algebraic proof

An integer is a whole number, such as -3, 0, 7 or 25.
A positive integer is a whole number greater than zero.
A multiple of a number is the result of multiplying that number by an integer. For example, $6 k$ means a multiple of 6, where $k$ is an integer.
A square number is a number that can be written as something squared, such as $m^2$ .

Key Idea

The proof mindset

In algebraic proof, your job is usually to rewrite the expression into a form that makes the result obvious, such as $2 k + 1$ for an odd number, $6 k$ for a multiple of 6, or $m^2$ for a square number.

Example

Showing a statement is true for all integers

Prove algebraically that the sum of any two consecutive integers is always odd.

Let the first integer be $n$ . The next consecutive integer is $n + 1$ .
Add the two integers:
$n + (n + 1) = 2 n + 1$
Since $2 n$ is even, $2 n + 1$ is odd.
Therefore, the sum of any two consecutive integers is always odd.

Representing odd, even and consecutive numbers

Before you can prove anything, you need to choose a good algebraic form.

Even and odd numbers

Any even integer can be written as $2 n$ , because it is a multiple of 2.

Any odd integer can be written as $2 n + 1$ or $2 n - 1$ , because it is one more or one less than an even number.

Consecutive numbers

Consecutive means “next to each other in order”.

Consecutive integers: $n$ , $n + 1$ , $n + 2$
Consecutive even integers: $2 n$ , $2 n + 2$ , $2 n + 4$
Consecutive odd integers: $2 n + 1$ , $2 n + 3$ , $2 n + 5$

Common Mistake

Forgetting the gap between evens or odds

Consecutive integers go up by 1, but consecutive even or odd integers go up by 2. So $2 n$ , $2 n + 1$ , $2 n + 2$ are not consecutive even numbers.

Example

Three consecutive even numbers

Prove algebraically that the sum of any three consecutive even integers is always a multiple of 6.

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

Proving something is a multiple

To prove an expression is always a multiple of a number, you should try to factorise it so that number is clearly a factor.

For example:

$4 (3 n + 2)$ is clearly a multiple of 4.
$12 n$ is clearly a multiple of 12.
$8 k + 2$ is not a multiple of 8, but it is 2 more than a multiple of 8.

Tip

How to finish a multiple proof

If you reach something like $12 n$ , write a final sentence such as: “Since $n$ is an integer, $12 n$ is a multiple of 12.” This earns the proof mark because it explains why the factorised form proves the result.

Example

Difference of two squared expressions

Prove that $3n+2)^2-(3n-2)^2$ is always a multiple of 12, for positive integer values of $n$ .

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

Proving something is a square number

Sometimes you are asked to show an expression is always a square number. That means your final answer should look like a bracket squared, such as $(n+3)^2`.

Look out for expressions that factorise into two identical brackets.

Example

Showing a sum is always a square

$n$ is an integer. Prove algebraically that the sum of $(n + 3) (n + 2)$ and $n + 3$ is always a square number.

Write the sum as one expression:
$(n + 3) (n + 2) + (n + 3)$
Notice that both terms have a common factor of $n + 3$ .
Factorise $n + 3$ :
$(n+3)(n+2)+(n+3)=(n+3)((n+2)+1)=(n+3)(n+3)\begin{aligned} (n+3)(n+2)+(n+3) &= (n+3)\left((n+2)+1\right) \\ &= (n+3)(n+3) \end{aligned}$
Write the result as a square:
$n+3)(n+3)=(n+3)^2$
Therefore, the sum is always a square number.

Common Mistake

Expanding when factorising is easier

If the question says “show it is a square number”, factorising is often quicker than expanding. You want to end with something like $(n+a)^2`, not a long quadratic expression.

Proving results about squares of numbers

Many Grade 8/9 proof questions involve square numbers, so it helps to know the common expansions.

Definition

Useful square expansions

$(a+b)^2=a^2+2ab+b^2`
$(a-b)^2=a^2-2ab+b^2`
The difference of two squares is $a^2-b^2=(a-b)(a+b)$ .

You can either expand fully or use the difference of two squares identity. The identity is often faster, but expanding is completely acceptable if it is clear.

Example

Squares of two even integers

Prove algebraically that the sum of the squares of any two even positive integers is always a multiple of 4.

Let the two even positive integers be $2 a$ and $2 b ‘, w h er e$ a $an d$ b$ are positive integers.
Square each integer and add:
$2a)^2+(2b)^2$
Simplify the squares:
$2a)^2+(2b)^2=4a^2+4b^2$
Factorise out 4:
$4a^2+4b^2=4(a^2+b^2)$
Since $a^2+b^2$ is an integer, $4(a^2+b^2)$ is a multiple of 4.

Example

Difference between squares of consecutive integers

Prove algebraically that the difference between the squares of two consecutive integers is equal to the sum of the two integers.

Let the two consecutive integers be $n$ and $n + 1$ .
Find the difference between their squares:
$n+1)^2-n^2$
Expand and simplify:
$(n+1)2−n2=(n2+2n+1)−n2=2n+1\begin{aligned} (n+1)^2-n^2 &= (n^2+2n+1)-n^2 \\ &= 2n+1 \end{aligned}$
Now find the sum of the two integers:
$n + (n + 1) = 2 n + 1$
The difference between the squares is $2 n + 1$ , and the sum of the integers is also $2n+1`.
Therefore, the difference between the squares of two consecutive integers is equal to the sum of those integers.

Proving “more than a multiple”

Some questions ask you to prove a result is “2 more than a multiple of 8” or “4 more than a multiple of 8”.

This means your final expression should look like:

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

where $k$ is an integer.

Example

Squares of consecutive odd numbers

Prove that the sum of the squares of two consecutive odd numbers is always 2 more than a multiple of 8.

Let the two consecutive odd numbers be $2 n + 1$ and $2n+3`.
Square both and add:
$2n+1)^2+(2n+3)^2$
Expand each square:
$(2n+1)2+(2n+3)2=(4n2+4n+1)+(4n2+12n+9)=8n2+16n+10\begin{aligned} (2n+1)^2+(2n+3)^2 &= (4n^2+4n+1)+(4n^2+12n+9) \\ &= 8n^2+16n+10 \end{aligned}$
Rewrite the expression to show “2 more than a multiple of 8”:
$8n^2+16n+10=8(n^2+2n+1)+2$
Since $n^2+2n+1$ is an integer, $8(n^2+2n+1)+2$ is 2 more than a multiple of 8.

Writing a clear final proof

A strong proof usually has three parts:

Define the algebraic form of the number or numbers.
Simplify the expression using expansion, collecting like terms and factorising.
Finish with a sentence that links your final form to the statement in the question.

Exam technique

In the exam

Start by choosing the correct form: $n$ , $n + 1$ for consecutive integers; $2 n$ , $2 n + 2$ for consecutive evens; $2 n + 1$ , $2 n + 3$ for consecutive odds.
Aim for the target form: $2 k + 1$ for odd, $mk$ for a multiple of $m$ , $8 k + 2$ for “2 more than a multiple of 8”, and $p^2$ for a square number.
Always write a final sentence explaining why your algebra proves the result for all allowed integer values.

Self review

Check yourself

Can you explain why $2 n + 5$ is always odd when $n$ is an integer?
If a result simplifies to $6 n + 12$ , how would you show it is a multiple of 6?
What algebraic forms would you use for two consecutive even integers?

Proof

What you'll learn

How to write integers, odd numbers, even numbers and consecutive numbers using algebra.
How to prove something is always odd, even, a multiple, or a square number.
How to use expansion and factorising clearly in proof questions.
How to finish proofs with a convincing final sentence.

What is a proof?

A proof is a logical argument that shows a statement is true for every value allowed in the question, not just for a few examples.

Checking numbers can help you spot a pattern, but it is not a proof. For example, if a result works for 1, 2 and 3, it might still fail for 100. Algebra lets you represent all possible cases at once.

Definition

Key words for algebraic proof

An integer is a whole number, such as -3, 0, 7 or 25.
A positive integer is a whole number greater than zero.
A multiple of a number is the result of multiplying that number by an integer. For example, $6 k$ means a multiple of 6, where $k$ is an integer.
A square number is a number that can be written as something squared, such as $m^2$ .

Key Idea

The proof mindset

Example

Showing a statement is true for all integers

Prove algebraically that the sum of any two consecutive integers is always odd.

Let the first integer be $n$ . The next consecutive integer is $n + 1$ .
Add the two integers:
$n + (n + 1) = 2 n + 1$
Since $2 n$ is even, $2 n + 1$ is odd.
Therefore, the sum of any two consecutive integers is always odd.

Representing odd, even and consecutive numbers

Before you can prove anything, you need to choose a good algebraic form.

Even and odd numbers

Any even integer can be written as $2 n$ , because it is a multiple of 2.

Any odd integer can be written as $2 n + 1$ or $2 n - 1$ , because it is one more or one less than an even number.

Consecutive numbers

Consecutive means “next to each other in order”.

Consecutive integers: $n$ , $n + 1$ , $n + 2$
Consecutive even integers: $2 n$ , $2 n + 2$ , $2 n + 4$
Consecutive odd integers: $2 n + 1$ , $2 n + 3$ , $2 n + 5$

Common Mistake

Forgetting the gap between evens or odds

Consecutive integers go up by 1, but consecutive even or odd integers go up by 2. So $2 n$ , $2 n + 1$ , $2 n + 2$ are not consecutive even numbers.

Example

Three consecutive even numbers

Prove algebraically that the sum of any three consecutive even integers is always a multiple of 6.

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

Proving something is a multiple

To prove an expression is always a multiple of a number, you should try to factorise it so that number is clearly a factor.

For example:

$4 (3 n + 2)$ is clearly a multiple of 4.
$12 n$ is clearly a multiple of 12.
$8 k + 2$ is not a multiple of 8, but it is 2 more than a multiple of 8.

Tip

How to finish a multiple proof

Example

Difference of two squared expressions

Prove that $3n+2)^2-(3n-2)^2$ is always a multiple of 12, for positive integer values of $n$ .

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

Proving something is a square number

Sometimes you are asked to show an expression is always a square number. That means your final answer should look like a bracket squared, such as $(n+3)^2`.

Look out for expressions that factorise into two identical brackets.

Example

Showing a sum is always a square

$n$ is an integer. Prove algebraically that the sum of $(n + 3) (n + 2)$ and $n + 3$ is always a square number.

Write the sum as one expression:
$(n + 3) (n + 2) + (n + 3)$
Notice that both terms have a common factor of $n + 3$ .
Factorise $n + 3$ :
$(n+3)(n+2)+(n+3)=(n+3)((n+2)+1)=(n+3)(n+3)\begin{aligned} (n+3)(n+2)+(n+3) &= (n+3)\left((n+2)+1\right) \\ &= (n+3)(n+3) \end{aligned}$
Write the result as a square:
$n+3)(n+3)=(n+3)^2$
Therefore, the sum is always a square number.

Common Mistake

Expanding when factorising is easier

If the question says “show it is a square number”, factorising is often quicker than expanding. You want to end with something like $(n+a)^2`, not a long quadratic expression.

Proving results about squares of numbers

Many Grade 8/9 proof questions involve square numbers, so it helps to know the common expansions.

Definition

Useful square expansions

$(a+b)^2=a^2+2ab+b^2`
$(a-b)^2=a^2-2ab+b^2`
The difference of two squares is $a^2-b^2=(a-b)(a+b)$ .

You can either expand fully or use the difference of two squares identity. The identity is often faster, but expanding is completely acceptable if it is clear.

Example

Squares of two even integers

Prove algebraically that the sum of the squares of any two even positive integers is always a multiple of 4.

Let the two even positive integers be $2 a$ and $2 b ‘, w h er e$ a $an d$ b$ are positive integers.
Square each integer and add:
$2a)^2+(2b)^2$
Simplify the squares:
$2a)^2+(2b)^2=4a^2+4b^2$
Factorise out 4:
$4a^2+4b^2=4(a^2+b^2)$
Since $a^2+b^2$ is an integer, $4(a^2+b^2)$ is a multiple of 4.

Example

Difference between squares of consecutive integers

Prove algebraically that the difference between the squares of two consecutive integers is equal to the sum of the two integers.

Let the two consecutive integers be $n$ and $n + 1$ .
Find the difference between their squares:
$n+1)^2-n^2$
Expand and simplify:
$(n+1)2−n2=(n2+2n+1)−n2=2n+1\begin{aligned} (n+1)^2-n^2 &= (n^2+2n+1)-n^2 \\ &= 2n+1 \end{aligned}$
Now find the sum of the two integers:
$n + (n + 1) = 2 n + 1$
The difference between the squares is $2 n + 1$ , and the sum of the integers is also $2n+1`.
Therefore, the difference between the squares of two consecutive integers is equal to the sum of those integers.

Proving “more than a multiple”

Some questions ask you to prove a result is “2 more than a multiple of 8” or “4 more than a multiple of 8”.

This means your final expression should look like:

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

where $k$ is an integer.

Example

Squares of consecutive odd numbers

Prove that the sum of the squares of two consecutive odd numbers is always 2 more than a multiple of 8.

Let the two consecutive odd numbers be $2 n + 1$ and $2n+3`.
Square both and add:
$2n+1)^2+(2n+3)^2$
Expand each square:
$(2n+1)2+(2n+3)2=(4n2+4n+1)+(4n2+12n+9)=8n2+16n+10\begin{aligned} (2n+1)^2+(2n+3)^2 &= (4n^2+4n+1)+(4n^2+12n+9) \\ &= 8n^2+16n+10 \end{aligned}$
Rewrite the expression to show “2 more than a multiple of 8”:
$8n^2+16n+10=8(n^2+2n+1)+2$
Since $n^2+2n+1$ is an integer, $8(n^2+2n+1)+2$ is 2 more than a multiple of 8.

Writing a clear final proof

A strong proof usually has three parts:

Define the algebraic form of the number or numbers.
Simplify the expression using expansion, collecting like terms and factorising.
Finish with a sentence that links your final form to the statement in the question.

Exam technique

In the exam

Start by choosing the correct form: $n$ , $n + 1$ for consecutive integers; $2 n$ , $2 n + 2$ for consecutive evens; $2 n + 1$ , $2 n + 3$ for consecutive odds.
Aim for the target form: $2 k + 1$ for odd, $mk$ for a multiple of $m$ , $8 k + 2$ for “2 more than a multiple of 8”, and $p^2$ for a square number.
Always write a final sentence explaining why your algebra proves the result for all allowed integer values.

Self review

Check yourself

Can you explain why $2 n + 5$ is always odd when $n$ is an integer?
If a result simplifies to $6 n + 12$ , how would you show it is a multiple of 6?
What algebraic forms would you use for two consecutive even integers?

Proof

What you'll learn

What is a proof?

Representing odd, even and consecutive numbers

Even and odd numbers

Consecutive numbers

Proving something is a multiple

Proving something is a square number

Proving results about squares of numbers

Proving “more than a multiple”

Writing a clear final proof

Recap questions

You've reached the end

Proof

What you'll learn

What is a proof?

Representing odd, even and consecutive numbers

Even and odd numbers

Consecutive numbers

Proving something is a multiple

Proving something is a square number

Proving results about squares of numbers

Proving “more than a multiple”

Writing a clear final proof

Recap questions

You've reached the end

Proof

What you'll learn

What is a proof?

Representing odd, even and consecutive numbers

Even and odd numbers

Consecutive numbers

Proving something is a multiple

Proving something is a square number

Proving results about squares of numbers

Proving “more than a multiple”

Writing a clear final proof

Recap questions

You've reached the end

Proof Revision Guide

Proof

What you'll learn

What is a proof?

Representing odd, even and consecutive numbers

Even and odd numbers

Consecutive numbers

Proving something is a multiple

Proving something is a square number

Proving results about squares of numbers

Proving “more than a multiple”

Writing a clear final proof

Recap questions

You've reached the end

Proof Revision Guide