Recognise the four proof types the spec names - direct deduction, cases, contradiction, disproof by counterexample - and be able to reorder a scrambled proof, spot a circular argument, and justify a conjecture made from small cases with a general argument.
Proof types (Prf1)
- Direct: start from what is given and deduce step by step ('\(n(n + 1)\) is a product of consecutive integers, so it is even').
- Cases: split by parity, by remainder mod \(3\), by sign; the cases must cover everything.
- Contradiction: assume the statement is false (the hypothesis true and the conclusion false), derive an impossibility. Classic: \(\sqrt2\) irrational, infinitely many primes.
- Counterexample: one instance with the hypothesis true and the conclusion false disproves a 'for all' claim; it says nothing about an 'exists' claim.
- Checking small cases (\(n^2 - n + 41\) is prime for \(n = 1, \ldots, 40\) but not \(41\)) is evidence, not proof.
Working with proofs (Prf2-Prf4)
- Ordering steps: the assumption comes first, the contradiction last; each line must use only earlier lines.
- Deducing from given implications: use the contrapositive ('\(f(x) > 0 \Rightarrow g(x) > 0\) and \(g(2) = -1\)' gives \(f(2) \leq 0\)).
- Backwards working (start from the result, reach a truth) is valid only if every step is reversible and the proof is then presented forwards.
- Conjecture from a pattern, then justify algebraically: \(T_n + T_{n+1} = (n + 1)^2\) by expanding.
- A proof for \(\sqrt2\) copied for \(\sqrt4\) breaks where the parity argument no longer applies - identify which step fails.
Must-know (no formula booklet)
- counterexample: hypothesis true, conclusion false
- contradiction: assume not-conclusion
- cases must be exhaustive
- small cases \(\neq\) proof
Classic traps
- 'If \(x^2 > 9\) then \(x > 3\)' is false: \(x = -4\). Taking square roots loses the negative branch.
- Assuming the claim to prove the claim is circular even if every step 'works'.
- An 'exists' statement cannot be disproved by a counterexample; it needs a proof that no example exists.
Worked example
Q. Prove: if \(n^2\) is even then \(n\) is even.
A. Contrapositive: if \(n\) is odd, \(n = 2k + 1\), then \(n^2 = 4k^2 + 4k + 1\) is odd.
How it has been examined
Practice papers · Paper 2Question 5counting counterexamples to a primality claimquestion papersit online
Specimen papers · Paper 2Question 8sum of four consecutive integers is a multiple of 6question papersit online