Proof shows something is always true
What you will learn
Know the rule, then use it
These are the short notes. Read each one, then check you can use it in the worked example below.
Method
In algebraic proof: represent general cases using algebra (2n for even, 2n+1 for odd, 3n for multiples of 3)
Represent two consecutive even numbers
Let them be 2n and 2n + 2, where n is any integer
Multiply them out
2n(2n + 2) = 4n(n + 1)
Use the fact that one of n or n+1 must be even
Either n or n + 1 is even, so n(n + 1) is always even
Watch out
Students use specific numbers as proof (for example '2 × 4 = 8')
In algebraic proof: represent general cases using algebra (2n for even, 2n+1 for odd, 3n for multiples of 3)
Students use specific numbers as proof (for example '2 × 4 = 8')
Prove that the product of any two consecutive even numbers is divisible by 8.
Represent two consecutive even numbers: Let them be 2n and 2n + 2, where n is any integer.
Multiply them out: 2n(2n + 2) = 4n(n + 1).
Use the fact that one of n or n+1 must be even: Either n or n + 1 is even, so n(n + 1) is always even.Write n(n + 1) = 2k.
Conclude divisibility by 8: Product = 4 × 2k = 8k, which is a multiple of 8.∴ The product of any two consecutive even numbers is divisible by 8.
Product = 8k for some integer k, so divisible by 8
Build up to the hardest questions
Do them in order. If you miss a step, read the solution, then redo the question without looking.
WorkedreasoningProve that the product of any two consecutive even numbers is divisible by 8.
4 marks4 minsalgebraic-proof-workedShow solution
Prove that the product of any two consecutive even numbers is divisible by 8.
- 1.Represent two consecutive even numbers: Let them be 2n and 2n + 2, where n is any integer.
- 2.Multiply them out: 2n(2n + 2) = 4n(n + 1).
- 3.Use the fact that one of n or n+1 must be even: Either n or n + 1 is even, so n(n + 1) is always even.Write n(n + 1) = 2k.
- 4.Conclude divisibility by 8: Product = 4 × 2k = 8k, which is a multiple of 8.∴ The product of any two consecutive even numbers is divisible by 8.
Product = 8k for some integer k, so divisible by 8
- M1: represent two consecutive even numbers
- M1: multiply them out
- M1: use the fact that one of n or n+1 must be even
- M1: conclude divisibility by 8
- A1: Product = 8k for some integer k, so divisible by 8
g. '2 × 4 = 8'). This only shows one case, not all cases. Every step must use the general algebraic form.The conclusion must quote the original claim.
DiagnosticrecallProve that the sum of any three consecutive integers is divisible by 3.
1 mark2 minsalgebraic-proof-q1Show solution
Prove that the sum of any three consecutive integers is divisible by 3.
- 1.Spot the skill: In algebraic proof: represent general cases using algebra (2n for even, 2n+1 for odd, 3n for multiples of 3).
- 2.Use the represent two consecutive even numbers stage first, then multiply them out.
- 3.Keep the final answer visible: n + (n+1) + (n+2) = 3n + 3 = 3(n+1), divisible by 3.
n + (n+1) + (n+2) = 3n + 3 = 3(n+1), divisible by 3
- M1: use the correct in algebraic proof: represent general cases using algebra (2n for even, 2n+1 for odd, 3n for multiples of 3).expand and simplify, then draw a clear conclusion. never use specific numbers — that is verification, not proof.
- A1: n + (n+1) + (n+2) = 3n + 3 = 3(n+1), divisible by 3
g. '2 × 4 = 8'). This only shows one case, not all cases. Every step must use the general algebraic form.The conclusion must quote the original claim.
EasyprocedureProve that the sum of a number and its square is always even.
2 marks3 minsalgebraic-proof-q2Show solution
Prove that the sum of a number and its square is always even.
- 1.Spot the skill: In algebraic proof: represent general cases using algebra (2n for even, 2n+1 for odd, 3n for multiples of 3).
- 2.Use the multiply them out stage first, then use the fact that one of n or n+1 must be even.
- 3.Keep the final answer visible: n + n2 = n(n + 1); one of n or n+1 is even, so product is even.
n + n2 = n(n + 1); one of n or n+1 is even, so product is even
- M1: use the correct in algebraic proof: represent general cases using algebra (2n for even, 2n+1 for odd, 3n for multiples of 3).expand and simplify, then draw a clear conclusion. never use specific numbers — that is verification, not proof.
- A1: n + n2 = n(n + 1); one of n or n+1 is even, so product is even
g. '2 × 4 = 8'). This only shows one case, not all cases. Every step must use the general algebraic form.The conclusion must quote the original claim.
MediumreasoningDisprove: 'All prime numbers are odd.'
3 marks4 minsalgebraic-proof-q3Show solution
Disprove: 'All prime numbers are odd.'
- 1.Spot the skill: In algebraic proof: represent general cases using algebra (2n for even, 2n+1 for odd, 3n for multiples of 3).
- 2.Use the use the fact that one of n or n+1 must be even stage first, then conclude divisibility by 8.
- 3.Keep the final answer visible: 2 is prime and even — counterexample disproves the claim.
2 is prime and even — counterexample disproves the claim
- M1: use the correct in algebraic proof: represent general cases using algebra (2n for even, 2n+1 for odd, 3n for multiples of 3).expand and simplify, then draw a clear conclusion. never use specific numbers — that is verification, not proof.
- A1: 2 is prime and even — counterexample disproves the claim
g. '2 × 4 = 8'). This only shows one case, not all cases. Every step must use the general algebraic form.The conclusion must quote the original claim.
Hardproblem solvingProve that (2n + 1)2 − 1 is always a multiple of 8.
3 marks5 minsalgebraic-proof-q4Show solution
Prove that (2n + 1)2 − 1 is always a multiple of 8.
- 1.Spot the skill: In algebraic proof: represent general cases using algebra (2n for even, 2n+1 for odd, 3n for multiples of 3).
- 2.Use the conclude divisibility by 8 stage first, then represent two consecutive even numbers.
- 3.Keep the final answer visible: 4n2 + 4n = 4n(n+1); n(n+1) is even so 4n(n+1) = 8k.
4n2 + 4n = 4n(n+1); n(n+1) is even so 4n(n+1) = 8k
- M1: use the correct in algebraic proof: represent general cases using algebra (2n for even, 2n+1 for odd, 3n for multiples of 3).expand and simplify, then draw a clear conclusion. never use specific numbers — that is verification, not proof.
- A1: 4n2 + 4n = 4n(n+1); n(n+1) is even so 4n(n+1) = 8k
g. '2 × 4 = 8'). This only shows one case, not all cases. Every step must use the general algebraic form.The conclusion must quote the original claim.
Exam-stylemulti-stepProve that the difference of squares of two consecutive integers is always odd.
4 marks6 minsalgebraic-proof-q5Show solution
Prove that the difference of squares of two consecutive integers is always odd.
- 1.Spot the skill: In algebraic proof: represent general cases using algebra (2n for even, 2n+1 for odd, 3n for multiples of 3).
- 2.Use the represent two consecutive even numbers stage first, then multiply them out.
- 3.Keep the final answer visible: (n+1)2 − n2 = 2n + 1, which is odd.
(n+1)2 − n2 = 2n + 1, which is odd
- M1: use the correct in algebraic proof: represent general cases using algebra (2n for even, 2n+1 for odd, 3n for multiples of 3).expand and simplify, then draw a clear conclusion. never use specific numbers — that is verification, not proof.
- A1: (n+1)2 − n2 = 2n + 1, which is odd
g. '2 × 4 = 8'). This only shows one case, not all cases. Every step must use the general algebraic form.The conclusion must quote the original claim.
Grade 9 stretchproblem solvingProve that the difference between the squares of any two consecutive odd integers is a multiple of 8.
4 marks7 minsproof-g9Show solution
Prove that the difference between the squares of any two consecutive odd integers is a multiple of 8.
- 1.Represent consecutive odd integers as 2n + 1 and 2n + 3.
- 2.Subtract the smaller square from the larger and expand.
- 3.Factor the result to show it has a factor of 8.
(2n + 3)2 - (2n + 1)2 = 8n + 8 = 8(n + 1), so the difference is always a multiple of 8
- M1: represent both consecutive odd integers generally
- M1: expand and subtract correctly
- A1: obtain 8(n + 1)
- C1: conclude it is a multiple of 8
Do not rush straight into arithmetic. Select the relevant method and show a complete chain of working.
Switch between skills
Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.
1Algebraic proof - 2 marksProve that the sum of any three consecutive integers is divisible by 3.Mark answer
n + (n+1) + (n+2) = 3n + 3 = 3(n+1), divisible by 3
2Algebraic notation and substitution - 2 marksFind t² + 4t when t = −2Mark answer
−4
3Simplifying expressions - 2 marksSimplify 2xy + 3x²y − xy + x²yMark answer
3x²y + xy
4Expanding and factorising - 3 marksFactorise fully 6x² + 9xMark answer
3x(2x + 3)
- I can explain the method for algebraic proof.
- I can show clear working without skipping key steps.
- g. '2 × 4 = 8'). This only shows one case, not all cases. Every step must use the general algebraic form.The conclusion must quote the original claim.
This guide follows the OCR GCSE Mathematics J560 specification. Practice questions are original Learnova questions shaped around official content and exam skills.