Algebraic Proof
Algebraic proof is a way of showing a mathematical statement is true for every possible case, using general algebraic expressions instead of testing individual numbers. Integers are written as n, consecutive integers as n and n+1, even numbers as 2n and odd numbers as 2n+1. It is a core AQA Level 2 Further Maths topic, tested through show-that and prove questions worth up to 6 marks.
Before you start
Make sure you're comfortable with these topics first:
Method
- Represent the general case algebraically: let n be an integer, write consecutive integers as n and n+1, even numbers as 2n, and odd numbers as 2n+1.
- Expand and simplify the expression fully, collecting like terms carefully.
- Factorise the result to reveal the structure you need, such as a factor of 2 for even numbers or a factor of k for a multiple of k.
- Link the factorised form explicitly to the definition being proved, stating why it is even, odd, or a multiple of k.
- For proof by contradiction, assume the opposite of what you want to prove is true, then use algebra to reach a logical contradiction.
- Finish with a clear concluding sentence that restates the original claim as proved.
Worked example
Prove that the sum of three consecutive even numbers is always a multiple of 6.
- Let the three consecutive even numbers be 2n, 2n + 2 and 2n + 4, where n is an integer.
- Add them together: 2n + (2n + 2) + (2n + 4) = 6n + 6.
- Factorise: 6n + 6 = 6(n + 1).
- Since n is an integer, n + 1 is also an integer.
- So 6(n + 1) is 6 multiplied by an integer, which means it is always a multiple of 6.
Practice questions
Type your answer and press Check to be marked straight away, or reveal the answer and mark yourself.
Q1Show that the sum of two consecutive odd numbers is always a multiple of 4.Show answer
Answer: (2n+1) + (2n+3) = 4n + 4 = 4(n+1), which is a multiple of 4 for any integer n
Q2Show that (n + 3)^2 - n^2 simplifies to 6n + 9.Show answer
Answer: 6n + 9 (expand (n+3)^2 = n^2+6n+9, then subtract n^2)
Q3Show that the sum of four consecutive integers is always even.Show answer
Answer: n+(n+1)+(n+2)+(n+3) = 4n+6 = 2(2n+3), which is even
Q4Prove that the square of an odd number is always odd. Let the odd number be 2n+1.Show answer
Answer: (2n+1)^2 = 4n^2+4n+1 = 2(2n^2+2n) + 1, which is of the form 2k+1, so always odd
Q5Prove that n^2 + 3n is always even for any integer n.Show answer
Answer: n^2+3n = n(n+3); if n is even the product is even, and if n is odd then n+3 is even (odd+odd), so the product is always even
Q6Prove by contradiction that sqrt(3) is irrational.Show answer
Answer: Assume sqrt(3) = p/q in lowest terms; squaring and rearranging shows 3 must divide both p and q, contradicting lowest terms, so sqrt(3) is irrational
Exam-style questions
Written in the style of a GCSE Further Maths exam paper, with a full mark scheme.
Prove that the sum of five consecutive integers is always a multiple of 5.
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Tick each line you got. Your score builds from the marks on the scheme.
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Prove that (n + 4)^2 - (n - 4)^2 is always a multiple of 16, for any integer n.
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Nothing ticked yet - 4 available
Prove by contradiction that if n^2 is even, then n must be even, for any integer n.
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Nothing ticked yet - 5 available
See real GCSE Further Maths past-paper questions, with official mark schemes →
Free printable worksheet
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