AQA A-level Mathematics 7357, section A (Proof), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.
Prerequisites: algebraic manipulation from GCSE. Each lesson develops its own method before independent transfer.
These are original teaching examples, not official AQA material.
1.2
Proof 证明
What would explain this observation?
The formula n² + n + 41 gives prime numbers for n = 1, 2, 3, all the way to 39. A student concludes it is always prime. How can one number destroy forty confirmations?
Start with a prediction. State what kind of argument could settle a claim about EVERY case, and what could settle one about A SINGLE case.
Build the model
A deductive proof 演绎证明 starts from stated definitions and follows logically: odd integers 2a+1 and 2b+1 sum to 2(a+b+1), which is even — proved for every pair.
A counterexample 反例 refutes an all-values claim with one valid failing case: at n = 41, the value is 41 × 43 = 1763, not prime. A proof by contradiction 反证法 assumes the negation and derives an impossibility: if √2 = p/q in lowest terms, then p² = 2q² forces p even, and writing p = 2r forces q even too — contradicting lowest terms, so √2 is irrational.
deduction: Proving a claim by logical steps from accepted facts; counterexample: A single valid case that disproves a universal claim; contradiction: Assuming the negation and deriving an impossibility.
Choose evidence that can test it
Checking n = 0, 1, 2, …, 39 is strong evidence but not a proof: the claim concerns EVERY positive integer. A proof must cover all cases at once — by general algebra, by exhaustion of a complete finite list, or by contradiction.
Before accepting a proof, check its structure: does it start from assumptions rather than from the conclusion? Does the contradiction attack the stated assumption? Checking examples verifies arithmetic but cannot replace the general argument.
Work from known quantities
State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
Known: the candidate counterexample n = 41. Substitution: n² + n + 41 = 41² + 41 + 41 = 41(41 + 1 + 1) = 41 × 43 = 1763. The value has a factor of 41, so it is not prime — the universal claim fails at exactly one case.
Example:
Test whether the sum of two even numbers is always even using deduction: write the two numbers as 2a and 2b and show the structure. (Answer: 2a + 2b = 2(a + b), which is a multiple of 2, hence even for every pair — no examples needed.)
Check the conclusion and its limits
Each method has a required shape: deduction needs general variables, exhaustion needs a genuinely complete case list, and contradiction needs the impossibility to follow from the negation itself. A proof about integers does not automatically cover real numbers, and an identity proved for all values differs from an equation true only at particular solutions.
Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Warn:
Checking a claim for the first hundred integers proves it for every positive integer. This claim is false: a finite sample is evidence, never a proof — n² + n + 41 survives 40 checks and fails at the 41st, so only a general argument or a counterexample settles a universal claim.
Key:
Proof: deduction works from definitions with general variables; a counterexample kills an all-values claim with one case; contradiction assumes the negation and derives an impossibility. Examples verify and disprove — they never prove a universal.
