1MA1 · A6AlgebraHigherGrades 6–9

Algebraic Proof

Checking that $3^{2} - 1$ , $5^{2} - 1$ and $7^{2} - 1$ are all multiples of 8 is convincing — but it is not a proof, because there are infinitely many odd numbers you have not checked. An $\textbf{algebraic proof}$ uses a letter to represent any odd number at once, so one argument settles every case.

45 min Algebraic Expressions

The big picture

The key move is choosing the right representation: $2n$ for any even number, $2n + 1$ for any odd number, $n, n + 1, n + 2$ for consecutive integers. Once the statement is written in algebra, expanding and factorising reveal why it is true.

Edexcel proof questions are usually 3 or 4 marks, and the final mark goes to a clear concluding statement that links the algebra back to the claim. Showing examples, however many, earns nothing for "prove".

What you'll learn

Represent even, odd, consecutive and multiple numbers algebraically
Prove statements about integers using algebra
Prove that an identity holds by showing both sides are equal
Disprove a statement with a counterexample

Representing numbers

If $n$ is any integer: $2n$ is any even number; $2n + 1$ (or $2n - 1$ ) is any odd number; $n$ , $n + 1$ , $n + 2$ are consecutive integers.

Consecutive even numbers: $2n$ , $2n + 2$ . Consecutive odd numbers: $2n + 1$ , $2n + 3$ .

A multiple of 5 can be written $5n$ ; a number one more than a multiple of 3 is $3n + 1$ .

To show an expression is a multiple of $k$ , rewrite it as $k \times (\text{an integer})$ . To show it is odd, write it as $2 \times (\text{an integer}) + 1$ .

Your turnSame method, new numbers — try it before moving on.

The first of two consecutive even numbers is $2n$ . Write a simplified expression for their sum.

Your answer

Show worked solution

Solution

1The next even number is $2n + 2$ .
2Sum: $2n + 2n + 2 = 4n + 2$ .

Answer: $4n + 2$

Tip — Use different letters for numbers that are not related. Two unrelated odd numbers are $2m + 1$ and $2n + 1$ , not both $2n + 1$ — otherwise you have assumed they are equal.

Proofs about integers

Write the numbers in the claim algebraically, carry out the operation (add, multiply, square), then factorise to expose the required factor or form.

End with a sentence: " $2(n + 1)$ is 2 multiplied by an integer, so the sum is even."

Example 1Prove that the sum of three consecutive integers is always a multiple of 3.

1Let the integers be

n

n + 1

and

n + 2

2Sum:

3n + 3

3Factorise:

3(n + 1)

Answer

3(n + 1)

is 3 times an integer, so the sum is always a multiple of 3.

Example 2Prove that the square of any odd number is one more than a multiple of 8.

1Let the odd number be

2n + 1

. Its square is

4n^{2} + 4n + 1 = 4n(n + 1) + 1

n

and

n + 1

are consecutive, so one of them is even; hence

n(n + 1) = 2k

for some integer

k

3So the square is

4(2k) + 1 = 8k + 1

AnswerThe square of an odd number is

8k + 1

: one more than a multiple of 8.

Why this works

The second proof needed one extra fact — that a product of consecutive integers is even. Proofs often hinge on a small observation like this, which is why reasoning in words alongside the algebra matters.

Proving identities

To prove an identity, start from one side (usually the more complicated) and manipulate it until it matches the other. Do not move terms between sides as if solving an equation.

Alternatively, simplify both sides separately to the same expression.

Example 1Prove that

(n + 3)^{2} - (n - 3)^{2} \equiv 12n

1Expand:

(n^{2} + 6n + 9) - (n^{2} - 6n + 9)

2Remove brackets:

n^{2} + 6n + 9 - n^{2} + 6n - 9

3Simplify:

12n

AnswerLeft side

= 12n =

right side, so the identity holds.

Counterexamples

To show a statement is $\textbf{false}$ , one example where it fails is enough — a $\textbf{counterexample}$ .

Test small numbers, zero, negatives and fractions, which are where general claims most often break.

Example 1Show that the statement "

n^{2} + n + 11

is prime for every positive integer

n

" is false.

1Try

n = 10

100 + 10 + 11 = 121

121 = 11 \times 11

, which is not prime.

Answer

n = 10

is a counterexample.

Your turnSame method, new numbers — try it before moving on.

What is the smallest positive integer $n$ for which $n^2 - n + 11$ is not prime?

Your answer

Show hint

Try a value that makes every term a multiple of 11.

Show worked solution

Solution

1For $n = 1$ to $10$ the values are 11, 13, 17, 23, 31, 41, 53, 67, 83, 101 — all prime.
2For $n = 11$ : $121 - 11 + 11 = 121 = 11^2$ , which is not prime.

Answer: $n = 11$

Tip — A counterexample needs the working that shows it fails — state the value, the result, and why the result breaks the claim.

Think like an examiner

Define what $n$ represents at the start.
Show the factorised form that proves the result.
Finish with a concluding sentence referring back to the statement.

Representations

2n,\ 2n + 1

Even and odd.

n,\ n + 1,\ n + 2

Consecutive integers.

k \times (\text{integer})

Showing a multiple of k.

Watch out for these

Proving by testing several numbers.

Examples illustrate but do not prove; use algebra for the general case.

Using

2n + 1

for two different odd numbers.

Use different letters:

2m + 1

and

2n + 1

Stopping at

6n + 12

without a conclusion.

Write

6(n + 2)

and state that it is a multiple of 6.

Treating an identity like an equation and "solving" it.

Transform one side into the other.

Stretch yourself

Prove that the difference between the squares of two consecutive odd numbers is always a multiple of 8.

Hint — Let the numbers be $2n + 1$ and $2n + 3$ . Subtract their squares, or use the difference of two squares.

Reveal solution

SolutionLet the consecutive odd numbers be

2n + 1

and

2n + 3

. Then

(2n + 3)^{2} - (2n + 1)^{2} = \big((2n + 3) + (2n + 1)\big)\big((2n + 3) - (2n + 1)\big) = (4n + 4)(2) = 8(n + 1)

. Since

n + 1

is an integer, the difference is always a multiple of 8. Check:

7^{2} - 5^{2} = 24 = 8 \times 3

✓.

\blacksquare

Questions students ask

Why isn’t checking lots of numbers enough?

There are infinitely many integers. A pattern holding for a thousand cases can still fail for the next — only a general argument covers them all.

Do I need a concluding sentence?

Yes. The final mark is often for explicitly linking the algebra to the claim.

What does

\equiv

mean?

It means "is identical to" — true for every value of the variable, not just particular solutions.

Key takeaways

Represent evens as $2n$ , odds as $2n + 1$ , consecutive integers as $n, n + 1, \dots$
Manipulate algebraically, then factorise to reveal the required form.
Conclude in words, referring to the statement.
Prove identities by transforming one side into the other.
One counterexample disproves a general statement.

Exam-style questions

Practise it the way it is examined

3 original questions · 9 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 63 marks

Prove that the product of two odd numbers is always odd.[3]
Show mark scheme
- M1Uses $2m + 1$ and $2n + 1$
- M1Expands: $4mn + 2m + 2n + 1$
- C1Writes $2(2mn + m + n) + 1$ and concludes it is odd

Question 2Grade 83 marks

Prove that $(2n + 3)^2 - (2n - 1)^2$ is always a multiple of 8, for any integer $n$ .[3]
Show mark scheme
- M1Expands both squares correctly
- A1Simplifies to $16n + 8$
- C1Writes $8(2n + 1)$ and concludes

Question 3Grade 73 marks

Prove that the sum of any three consecutive even numbers is a multiple of 6.[3]
Show mark scheme
- M1Uses $2n$ , $2n + 2$ , $2n + 4$
- A1Sum $= 6n + 6$
- C1Writes $6(n + 1)$ and concludes

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

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