Introduction to Algebraic Proof Revision Notes | GCSE AQA Higher Further Maths
An algebraic proof uses logical algebraic steps, with 2n for even and 2n + 1 for odd numbers, to show a statement is always true. 12 questions answered.
Updated 14 Mar 2026 · Revise Right Team
Introduction to Algebraic Proof is part of Algebraic proof in GCSE AQA Higher Further Maths.
3 min
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What this lesson covers
An algebraic proof uses logical algebraic steps, with 2n for even and 2n + 1 for odd numbers, to show a statement is always true. 12 questions answered.
Define what constitutes a valid algebraic proof and distinguish it from verification by example
Use algebraic expressions to represent general integers, consecutive integers, odd numbers, and even numbers
Construct a complete algebraic proof to show that a statement about numbers is always true
Identify and correct common errors in attempted algebraic proofs
Key ideas to keep in view
Use this lesson to stay inside Algebraic proof while connecting the detail back to the wider Algebra topic.
Algebra
Algebraic proof
GCSE AQA Higher Further Maths
Revision notes
Read the core explanations from Introduction to Algebraic Proof before testing yourself from memory.
1. Why Checking Examples Is Never Enough
Here is a claim: "The sum of any two odd numbers is always even." You could try $3 + 5 = 8$, $7 + 11 = 18$, $1 + 99 = 100$ — all even. But does that prove it? No. You have only checked three cases out of infinitely many.
Algebraic proof closes that gap. Instead of picking numbers, you work with expressions that stand for any integer. When the algebra works out, the result holds for every possible case simultaneously — that is the power of proof.
The exam command word here is prove: show through logical algebraic steps that a statement is true for all integers. If the question says "show that", the target result is given to you — your job is to reach it from the starting expressions without working backwards from the answer.
2. The Algebraic Toolkit: Representing Integers
Before writing any proof, you need the right building blocks. The key is choosing expressions that capture the structure of the numbers you are describing — even, odd, consecutive, and so on.
Notice that two different odd numbers in the same proof must use the same letter $n$ only if they are related (like consecutive odds). If you need two independent odd numbers, use $2m+1$ and $2n+1$ with different letters. Mixing this up is one of the most common errors in GCSE proofs.
3. Writing a Complete Algebraic Proof
A proof has a clear structure: state your expressions, expand and simplify the algebra, then interpret the result. Every line must follow logically from the one before it.
The final line is where you land the conclusion. You must explicitly state what the simplified expression shows — for example, "this is $2 \times (\text{integer})$, so the result is always even." Without that sentence, you have not completed the proof.
Worked example: Prove that the sum of two consecutive integers is always odd
Prove that the sum of any two consecutive integers is always odd.
Let the integers be $n$ and $n + 1$, where $n$ is any integer. — Starting with a general expression covers every possible pair of consecutive integers at once. [M1 — correct algebraic representation]
$n + (n + 1)$ — Set up the expression for the sum of the two integers.
$n + n + 1 = 2n + 1$ — Collect like terms. [M1 — correct simplification]
$2n + 1$ is of the form $2 \times (\text{integer}) + 1$, which is always odd. — This is the conclusion step — you must state explicitly why the expression proves the claim. [A1 — complete conclusion]
Answer: The sum of any two consecutive integers equals $2n + 1$, which is always odd. Proved.
Worked example: Prove that the sum of two odd numbers is always even
Prove that the sum of any two odd numbers is always even.
Let the two odd integers be $2m + 1$ and $2n + 1$, where $m$ and $n$ are any integers. — Using different letters $m$ and $n$ allows the two odd numbers to be completely independent of each other. [M1 — correct representation with two different variables]
$(2m + 1) + (2n + 1)$ — Set up the expression for the sum.
$2m + 2n + 2 = 2(m + n + 1)$ — Factor out 2. [M1 — correct factorisation]
$2(m + n + 1)$ is $2 \times (\text{integer})$, so it is always even. — Since $m$ and $n$ are integers, $m + n + 1$ is also an integer. Multiplying any integer by 2 gives an even number. [A1 — complete conclusion]
Answer: The sum of any two odd numbers equals $2(m + n + 1)$, which is always even. Proved.
4. Proving Results About Squares and Products
Proofs often involve squaring expressions or multiplying them together. The technique is the same — define general expressions, expand carefully, then interpret. The extra step here is expanding brackets correctly before simplifying.
Worked example: Prove that the product of two consecutive even integers is divisible by 8
Prove that the product of any two consecutive even integers is always divisible by 8.
Let the integers be $2n$ and $2n + 2$, where $n$ is any integer. — Consecutive even integers differ by 2, so $2n$ and $2n + 2$ is the correct representation. [M1]
$2n \times (2n + 2)$ — Set up the multiplication.
$2n \times 2(n + 1) = 4n(n + 1)$ — Factor 2 out of $(2n + 2)$ first. This keeps the algebra cleaner. [M1 — correct factorisation]
Either $n$ or $n + 1$ must be even, since they are consecutive integers. — One of any two consecutive integers is always even. This is the insight that unlocks the proof.
Since one of $n$ or $n+1$ is even, write it as $2k$. Then $4n(n+1) = 4 \times 2k \times (\text{other factor}) = 8k(\text{other factor})$. — The product contains a factor of $4 \times 2 = 8$, so it is always divisible by 8. [A1 — complete conclusion]
Answer: The product of any two consecutive even integers is $4n(n+1)$. Since $n$ and $n+1$ are consecutive, one is even, introducing another factor of 2. The product is therefore always divisible by 8. Proved.
5. Guided Practice
Now it is your turn to apply the method. Work through each problem using the structure: define expressions, expand and simplify, then state the conclusion explicitly.
6. Exam Technique and Common Mistakes
Algebraic proof questions are mark-rich but frequently dropped. The mistakes below account for the majority of lost marks at GCSE. Read each one carefully — recognising the error pattern is half the battle.
Algebraic proof brings together every algebra skill you have built at GCSE — expanding brackets, factorising, and collecting like terms — and asks you to use them with precision and purpose. Master the structure (define, expand, conclude) and you have a reliable method for any proof question the exam can set.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-14.
An algebraic proof shows through logical algebraic steps that a statement is true for all integers, using expressions that stand for any integer. Checking examples is only verification, not proof: a single counterexample can disprove a statement, but no finite list of examples can prove it.
What is a counterexample?
A counterexample is a single case that shows a statement is false. One counterexample is enough to disprove a statement, whereas proving a statement true for every integer needs algebraic proof, not a list of examples.
How do you write odd and even numbers in algebra?
In algebraic proof, use 2n for an even number and 2n + 1 for an odd number, where n is any integer; consecutive integers are n and n + 1, and consecutive even integers are 2n and 2n + 2. For two independent odd numbers, use different letters: 2m + 1 and 2n + 1.
How do you prove that the sum of two odd numbers is even?
Let the odd numbers be 2m + 1 and 2n + 1, where m and n are any integers. Their sum is 2m + 2n + 2 = 2(m + n + 1). Since m + n + 1 is an integer, this is 2 × (integer), so the sum of any two odd numbers is always even.
How do you prove that the sum of two consecutive integers is odd?
Let the consecutive integers be n and n + 1, where n is any integer. Their sum is n + n + 1 = 2n + 1, which is of the form 2 × (integer) + 1, so it is always odd. The final line must state explicitly why the expression proves the claim.
How do you prove the product of two consecutive even numbers is divisible by 8?
Let the integers be 2n and 2n + 2, so their product is 2n × 2(n + 1) = 4n(n + 1). Since n and n + 1 are consecutive, one of them is even, which introduces another factor of 2, so the product always contains a factor of 4 × 2 = 8.
Why is checking examples not a proof?
Checking examples is not a proof because it covers only a few cases out of infinitely many; showing a statement is true for specific numbers is called verification. A single counterexample can disprove a statement, but no finite list of examples can prove it.
What is the difference between 'prove' and 'show that'?
'Prove' means showing through logical algebraic steps that a statement is true for all integers. 'Show that' gives the target result, so the task is to reach it from the starting expressions without working backwards from the answer.
How do you write consecutive even numbers in algebra?
Two consecutive even integers are 2n and 2n + 2, where n is any integer, because consecutive even integers differ by 2. Their product 2n(2n + 2) = 4n(n + 1) is always divisible by 8, since one of n and n + 1 is even.
Why do you use different letters for two odd numbers in a proof?
Two independent odd numbers in a proof need different letters, such as 2m + 1 and 2n + 1, so they are not forced to be related. Using the same letter n only works for related numbers like consecutive odds, and mixing this up is one of the most common errors.
How do you write the conclusion of an algebraic proof?
The conclusion of an algebraic proof states explicitly what the simplified expression shows, for example 'this is 2 × (integer), so the result is always even'. Without that sentence, the proof is not complete.
What is the structure of an algebraic proof?
An algebraic proof has three stages: define general expressions, expand and simplify, then state the conclusion. For two consecutive integers n and n + 1, the sum n + (n + 1) = 2n + 1, which is 2 × (integer) + 1 and so always odd.