P2.1 - Proof
- Syllabus
- 2019
- Topic
- P2.1
- Level
- AS
A mathematical proof starts from stated assumptions and uses valid logical steps to show that the conclusion must follow for every object in the stated domain.
| Stage | What it must establish |
|---|---|
| assumptions | domain, definitions and given conditions |
| representation | a general form such as 2k for an even integer or 2k+1 for an odd integer |
| logical steps | equalities or implications justified without assuming the result |
| conclusion | the exact claim, with its domain or condition retained |
To prove 2x2+8≥8x for every real x, move everything to one side: 2x2−8x+8=2(x−2)2. Because a real square is non-negative, 2(x−2)2≥0, so the stated inequality follows for every real x.
When multiplying, dividing, taking roots or cancelling factors, state the condition that makes the step valid. For example, dividing an inequality by a positive quantity preserves its direction; dividing by a quantity of unknown sign does not.
Checking several values can suggest a universal statement but cannot prove it. Avoid circular reasoning: the required conclusion may appear only after it has been derived from the assumptions.
Proof by exhaustion works when the allowed possibilities can be divided into a finite, complete set of cases. Each case must be checked, and the cases must leave no allowed value uncovered.
| Situation | Exhaustive cases |
|---|---|
| parity of an integer | n=2k or n=2k+1 |
| remainder after division by 3 | n=3k, 3k+1 or 3k+2 |
| single-digit primes | 2,3,5,7 |
To prove n2+n is even for every integer n, exhaust the two parity cases. If n=2k, then n2+n=2k(2k+1), which has a factor 2. If n=2k+1, then n2+n=(2k+1)(2k+2)=2(2k+1)(k+1), also even. Every integer is even or odd, so the proof is complete.
For a literal finite set, the single-digit primes are exactly 2,3,5,7. The values of p2+p are respectively 6,12,30,56, all even. Naming the complete allowed set is what turns these checks into a proof.
Trying a few convenient examples is not exhaustion. State why the cases are mutually sufficient, evaluate every case under the same claim, and finish with a conclusion covering the original domain.
A universal claim says a conclusion holds for every allowed input. To disprove it, one counterexample is enough: choose an input that satisfies the claim's conditions but makes its conclusion false.
| Counterexample requirement | Evidence to show |
|---|---|
| allowed input | the chosen value belongs to the stated domain |
| failed conclusion | substitution gives a result that contradicts the claim |
| explicit verdict | identify the value as a counterexample and state that the universal claim is false |
Consider the claim ‘n2−n+1 is prime for every positive integer n’. Take n=5, which is in the stated domain. Then 52−5+1=21=3×7, so the result is composite. Therefore n=5 is a counterexample and the universal claim is false.
A useful search targets values likely to expose the weakness, but the final disproof must show the calculation and the failed property. A valid counterexample need not be the smallest one.
One successful example cannot prove a statement about all inputs, while one genuine failure can disprove it. An input outside the stated domain or a value that still satisfies the conclusion is not a counterexample.