OCR section 1.01 Proof asks you to construct and present a mathematical argument, not just get an answer. You need four tools: deduction, exhaustion, disproof by counter-example and — new at A-level — proof by contradiction.
Every claim below is either proved or destroyed. Work through each screen, answer the questions as you go and collect ⭐ stars. Press Start when you're ready.
A proof is a watertight chain of reasoning from things already known to the thing you claim. Marks are lost for verifying examples and calling it a proof — checking that a statement works for n = 1, 2, 3 proves nothing about n = 4.
Watch out: x² = 9 ⇐ x = 3, but not ⇒, because x could be −3. Direction matters. Only write ⇔ when both directions genuinely hold.
Deduction is the workhorse: start from definitions, do algebra, arrive at the result. Every line must follow from the last.
Let the odd numbers be 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. ∴ proved. ■
Complete the square: x² − 6x + 10 = (x − 3)² + 1.
(x − 3)² ≥ 0 for all real x, so (x − 3)² + 1 ≥ 1 > 0. ∴ proved. ■
Presentation marks: define your letters (let n be an integer), finish with a concluding statement in words. OCR examiners award the final mark for that conclusion.
If a statement covers only finitely many cases, you may prove it by checking every single one. The skill is splitting an infinite problem into a finite number of cases.
The final digit of n² depends only on the final digit of n, so there are just 10 cases — the digits 0 to 9.
0² = 0, 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64, 9² = 81.
Final digits: 0, 1, 4, 9, 6, 5, 6, 9, 4, 1 — the set of possible last digits is {0, 1, 4, 5, 6, 9}. 7 never appears, so no square ends in 7. ■
Key move: exhaustion works only because we reduced infinitely many n to 10 cases. Another classic split is by remainder: every integer is 3k, 3k + 1 or 3k + 2 — three cases, and you are done.
To disprove a universal statement you need exactly one case where it fails. One is enough — and one is all you should give.
This is famously true for n = 1 up to n = 40, which is exactly why testing a few values is dangerous.
Take n = 41: 41² − 41 + 41 = 41² = 1681 = 41 × 41, which is not prime.
So the statement is false. ■
Exam technique: state the counter-example, evaluate it, and say explicitly why it breaks the claim. ‘n = 41 works’ scores nothing; ‘n = 41 gives 1681 = 41², which is not prime, so the statement is false’ scores everything.
To prove statement S: assume S is false, reason correctly, and reach something impossible. Since the reasoning was sound, the only faulty step was the assumption — so S must be true.
Assume √2 is rational. Then √2 = a/b for integers a, b with no common factor (lowest terms).
Square: 2 = a²/b², so a² = 2b². Hence a² is even, so a is even (odd² is odd). Write a = 2k.
Then (2k)² = 2b² ⇒ 4k² = 2b² ⇒ b² = 2k², so b² is even and 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 engine of the proof: if a² is even then a is even. Prove it by contrapositive: if a were odd, a = 2m + 1, then a² = 4m² + 4m + 1 = 2(2m² + 2m) + 1, which is odd.
The second proof OCR names by name. It is Euclid’s, dressed as a contradiction.
Assume there are finitely many primes, and list them all: p₁, p₂, …, pₖ.
Construct N = p₁ × p₂ × … × pₖ + 1.
Dividing N by any prime on the list leaves remainder 1, so no listed prime divides N.
But every integer greater than 1 has at least one prime factor. So N has a prime factor not on the list — contradicting that the list was complete. ∴ there are infinitely many primes. ■
Subtlety worth a mark: N itself need not be 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 both missing from the list. The proof only claims N has a new prime factor.
Tap an argument, then tap the method of proof it uses.
Tap a false statement on the left, then the counter-example that kills it.
Deduction: start from definitions and reason algebraically to the result; finish with a written conclusion
Exhaustion: split into finitely many cases and check every one — e.g. the last digit of n² has only 10 cases
Counter-example: one failing case destroys a universal statement — evaluate it and say why it fails
Contradiction: assume the opposite, reason correctly, hit an impossibility
Know by name: √2 is irrational (the a, b common-factor contradiction) and the primes are infinite (Euclid’s N)
Notation: ⇒, ⇐, ⇔ — only claim ⇔ when both directions really hold
That is the whole of OCR 1.01 — and the reasoning habits every other topic is marked against. Press Finish to see your score.
You've worked through Proof for OCR A-level Mathematics A. 🎉
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Next: test yourself in the Verify stage.