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.
Proof · language
The language of proof
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.
P ⇒ Q — "P implies Q". If P is true then Q must be true. P is a sufficient condition for Q.
P ⇐ Q — "P is implied by Q". P is a necessary condition for Q.
P ⇔ Q — "P if and only if Q". The implication runs both ways: P is necessary and sufficient for Q.
∴ means "therefore"; ∵ means "because".
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.
Proof · deduction
Proof by deduction
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.
Prove: n² + n is even for every integer nn² + n = n(n + 1) — the product of two consecutive integers. One of them must be even, so the product is even. ∎
Worked example — the difference of consecutive odd squares
Consecutive odd numbers can be written 2n − 1 and 2n + 1.
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.
Quick check
What does ⇔ mean?
?In the statement "P ⇔ Q", what is being claimed?
Proof · exhaustion
Proof by exhaustion
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.
Worked example — case-splitting
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.
Sort it
Which method of proof?
Tap a step or clue, then tap the method of proof it belongs to.
🔁 Exhaustion
❌ Counter-example
💥 Contradiction
Proof · counter-example
Disproof by counter-example
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.
"Every prime is odd" — counter-example: 2. It is prime and even.
"x² > x for all real x" — counter-example: x = 0.5, since 0.5² = 0.25, and 0.25 < 0.5.
"If x² > 9 then x > 3" — counter-example: x = −4, since 16 > 9 but −4 is not greater than 3.
The famous one: n² − n + 41 is prime for n = 1, 2, 3, … all the way up to n = 40. It looks like a theorem. But at n = 41 it gives 41² − 41 + 41 = 41² = 1681, which is 41 × 41 — not prime. Forty successes prove nothing.
Note the asymmetry: one counter-example destroys a universal claim, but no number of examples can prove one.
Quick check
Name that method
?A claim concerns only the integers 1 to 5. You check all five and each one works. Which method of proof is this?
Proof · contradiction
Proof by contradiction
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 ¬P → reason → contradiction → ∴ Pthe contradiction is the proof
Worked example — √2 is irrational
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.
Calculate
Your turn — counter-example
1Disprove the claim "every prime number is odd". Give the single prime number that serves as a counter-example.
Hint: Think of the only even prime. Every other even number has 2 as a factor, so it is not prime.
Calculate
Your turn — the 41 trap
2The expression n² − n + 41 gives a prime for n = 1, 2, 3, … for a long while. Find the smallest positive integer n for which it is not prime.
Hint: Look for an n that makes every term share a factor. Try n = 41: you get 41² − 41 + 41 = 41², which is 41 × 41.
Proof · contradiction
Contradiction: infinitely many primes
The second contradiction proof AQA expects you to have met is Euclid's theorem: there are infinitely many primes.
Euclid's argument
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.
Quick check
Starting a contradiction proof
?You are asked to prove by contradiction that √2 is irrational. What is the correct opening line?
Calculate
Your turn — deduction
3The identity (2n + 1)² − (2n − 1)² = 8n proves that the difference of the squares of consecutive odd numbers is always a multiple of 8. Use it (or direct calculation) to evaluate 15² − 13².
Hint: 15 and 13 are the consecutive odd numbers given by n = 7. So the difference is 8n = 8 × 7. Check directly: 225 − 169.
Proof · choosing a method
Which method should I use?
Reading the question tells you the method. AQA questions signal it clearly:
"Prove that … for all integers n" → deduction with a general n (or exhaustion by parity if odd/even behave differently).
"Prove that … for n = 1, 2, 3, 4, 5" or a claim about a short finite list → exhaustion.
"Disprove the statement" or "Show that the conjecture is false" → find a single counter-example and show the numbers.
"Prove by contradiction that …" → the words are in the question. Start "Assume, for contradiction, that …".
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).
Match it
Match the notation to its meaning
Tap an item on the left, then its partner on the right.
Notation
Meaning
Proof · writing it up
Writing a proof that scores
Marks in proof questions are for communication as much as for the idea. A full-mark proof:
States the assumption explicitly ("Let n be any integer", or "Assume, for contradiction, that …").
Defines the algebra ("let n = 2k + 1 where k is an integer") — do not leave the reader to guess.
Works in general terms — no numerical examples standing in for the argument.
Ends with a clear conclusion sentence tying back to the original claim, not just a line of algebra.
2(2k² + k) is even ✗ → "= 2(2k² + k), which is 2 × an integer, so n² + n is even" ✓the words are worth marks
Final check: have you actually proved the statement you were asked about, and does every case fall inside your argument?
Quick check
Spot the counter-example
?Which single value of x disproves the claim "x² > x for every real number x"?
Calculate
Your turn — an odd result
4Since n² + n = n(n + 1) is the product of consecutive integers, it is always even, so n² + n + 1 is always odd. Evaluate n² + n + 1 at n = 10 and confirm it is odd.
Hint: n² + n + 1 = 100 + 10 + 1. The even part n² + n = 110, so adding 1 must give an odd number.
Calculate
Your turn — Euclid's number
5Suppose, for contradiction, that the only primes were 2, 3, 5, 7 and 11. Euclid forms N = (2 × 3 × 5 × 7 × 11) + 1. Calculate N.
Hint: 2 × 3 × 5 × 7 × 11 = 2310, then add 1. Dividing N by any prime on the list leaves remainder 1 — and in fact 2311 is itself prime, so the list was incomplete.
Recap
The big ideas to know
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.
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Mini-lesson complete!
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You have worked through Proof for AQA A-level Mathematics. 🎉