This mini-lesson covers the whole of AQA section A — Proof: the language of proof (⇒, ⇐, ⇔, necessary and sufficient), and the four methods A-level asks for — deduction, exhaustion, disproof by counter-example and proof by contradiction.
Work through each screen, answer the questions as you go (a few are conceptual, most are calculations) and collect ⭐ stars. Everything here is A-level standard, drawn from the AQA 7357 subject content. Press Start when you are ready.
A proof is a watertight chain of reasoning from things already known to the thing you want to establish. At A-level you must write it in the correct language.
The classic trap: "x = 2 ⇒ x² = 4" is true, but "x² = 4 ⇒ x = 2" is false (x could be −2). So x = 2 ⇔ x² = 4 is not a valid statement. Never write ⇔ when only one arrow works.
You also need the standard algebraic disguises: any even number can be written 2k, any odd number 2k + 1, and consecutive integers as n and n + 1, for some integer k or n. Choosing the right disguise is usually the whole battle.
Deduction (sometimes called direct proof) means starting from known facts or definitions and reasoning forwards, in general terms, until the result falls out. Because you work with a general n, the proof covers every case at once.
Consecutive odd numbers can be written 2n − 1 and 2n + 1.
(2n + 1)² − (2n − 1)² = (4n² + 4n + 1) − (4n² − 4n + 1) = 8n.
8n is a multiple of 8 for every integer n, so the difference of the squares of consecutive odd numbers is always a multiple of 8. ∎
Marks are lost for stopping at examples. Showing it works for n = 1, 2, 3 is not a proof — it is only evidence. Only the general algebra earns the marks.
Exhaustion works when the claim only has finitely many cases. You split the possibilities into a complete set of cases and check every one. It is a proper proof — provided the cases really are exhaustive.
Prove that n² + n is even for every integer n, by exhaustion on parity.
Case 1, n even: n = 2k, so n² + n = 4k² + 2k = 2(2k² + k) — even.
Case 2, n odd: n = 2k + 1, so n² + n = (4k² + 4k + 1) + (2k + 1) = 4k² + 6k + 2 = 2(2k² + 3k + 1) — even.
Every integer is even or odd, so the two cases are exhaustive. ∎
Warning: exhaustion is only valid if your cases cover everything. Checking n = 1 to 10 for a claim about all integers proves nothing at all.
Tap a step or clue, then tap the method of proof it belongs to.
To disprove a universal claim ("for all n …") you need just one case where it fails. That single case is a counter-example, and it is a complete disproof.
Note the asymmetry: one counter-example destroys a universal claim, but no number of examples can prove one.
Contradiction is new at A-level (it is not in GCSE). The method: assume the opposite of what you want to prove, reason correctly, and reach an absurdity. Since the reasoning was sound, the assumption must have been false — so the original statement is true.
Assume √2 is rational. Then √2 = a/b for integers a, b with no common factor (lowest terms), b ≠ 0.
Squaring: 2 = a²/b², so a² = 2b². Hence a² is even, and therefore a is even. Write a = 2k.
Then 4k² = 2b², so b² = 2k². Hence b² is even, and therefore b is even.
But now a and b are both even — they share a factor 2. That contradicts "no common factor". ∴ √2 is irrational. ∎
The step students skip: "a² is even ⇒ a is even". Justify it — if a were odd, a = 2k + 1, then a² = 4k² + 4k + 1 would be odd.
The second contradiction proof AQA expects you to have met is Euclid's theorem: there are infinitely many primes.
Assume there are finitely many primes. Then we can list them all: p₁, p₂, …, pk.
Form the number N = (p₁ × p₂ × … × pk) + 1.
Divide N by any prime pi on the list: the remainder is 1. So no prime on the list divides N.
But every integer greater than 1 has a prime factor. So N has a prime factor not on the list — contradicting that the list contained them all. ∴ there are infinitely many primes. ∎
Careful: N itself is not always prime. With the list 2, 3, 5, 7, 11, 13 we get N = 30031 = 59 × 509 — composite, but its prime factors 59 and 509 are missing from the list, which is all the argument needs.
Reading the question tells you the method. AQA questions signal it clearly:
Contradiction favourites: √2 (and √3, √5) is irrational · there is no largest prime · if n² is even then n is even · there are no integers a, b with 6a + 8b = 15 (the left side is always even).
Tap an item on the left, then its partner on the right.
Marks in proof questions are for communication as much as for the idea. A full-mark proof:
Final check: have you actually proved the statement you were asked about, and does every case fall inside your argument?
Language: ⇒ implies · ⇐ is implied by · ⇔ if and only if (necessary and sufficient) · ∴ therefore
Deduction: general algebra — even = 2k, odd = 2k + 1, consecutive = n, n + 1
Exhaustion: a complete, finite set of cases, all checked
Counter-example: one failing case disproves a universal claim — n = 41 kills n² − n + 41
Contradiction: assume the negation → absurdity → the claim is true (√2 irrational; infinitely many primes)
Never: offer examples as a proof of a general statement
That is the whole of AQA section A — Proof. Press Finish to see your score.
You have worked through Proof for AQA A-level Mathematics. 🎉
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Next: test yourself in the Evaluate stage Confidence Quiz, then lock it in with Verify.