P2.1 - Proof

Syllabus
2019
Topic
P2.1
Level
AS

Learning objectives

A proof links assumptions to a necessary conclusion

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 2k2k for an even integer or 2k+12k+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+88x2x^2+8\ge8x for every real xx, move everything to one side: 2x28x+8=2(x2)2.2x^2-8x+8=2(x-2)^2. Because a real square is non-negative, 2(x2)202(x-2)^2\ge0, so the stated inequality follows for every real xx.

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 covers every possible case

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=2kn=2k or n=2k+1n=2k+1
remainder after division by 3 n=3kn=3k, 3k+13k+1 or 3k+23k+2
single-digit primes 2,3,5,72,3,5,7

To prove n2+nn^2+n is even for every integer nn, exhaust the two parity cases. If n=2kn=2k, then n2+n=2k(2k+1)n^2+n=2k(2k+1), which has a factor 2. If n=2k+1n=2k+1, then n2+n=(2k+1)(2k+2)=2(2k+1)(k+1)n^2+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 2,3,5,72,3,5,7. The values of p2+pp^2+p are respectively 6,12,30,566,12,30,56, 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.

One valid counterexample disproves a universal statement

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 ‘n2n+1n^2-n+1 is prime for every positive integer nn’. Take n=5n=5, which is in the stated domain. Then 525+1=21=3×7,5^2-5+1=21=3\times7, so the result is composite. Therefore n=5n=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.