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How do you prove a mathematical statement is always true, and how do you disprove one?

Methods of proof including proof by deduction, proof by exhaustion, disproof by counter-example and proof by contradiction, applied to statements about numbers and inequalities.

A focused answer to the AQA A-Level Mathematics proof content, covering proof by deduction, proof by exhaustion, disproof by counter-example and proof by contradiction, with the irrationality of root 2 and the infinitude of primes.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Proof by deduction
  3. Proof by exhaustion
  4. Disproof by counter-example
  5. Proof by contradiction

What this dot point is asking

AQA wants you to understand and use the structure of mathematical proof, including proof by deduction, proof by exhaustion, disproof by counter-example, and proof by contradiction (including the classic proofs that 2\sqrt{2} is irrational and that there are infinitely many primes). The marks here are for rigour and clear logical layout, not just reaching the answer: each step must follow from the last, and you must state a clear conclusion.

Proof by deduction

You start from known facts or definitions and use algebra and logic to reach the required conclusion. The crucial point is generality: represent the objects with letters so the argument covers all cases at once. An even number is 2n2n, an odd number is 2m+12m + 1, and consecutive integers are nn and n+1n + 1. Testing one numerical example proves nothing.

Proof by exhaustion

You split the statement into a finite number of cases and verify each one. This is valid only when the cases are finite and you check them all. For example, to prove that no square number ends in 77, note that any integer ends in one of the digits 00 to 99; squaring each gives final digits 0,1,4,9,6,5,6,9,4,10, 1, 4, 9, 6, 5, 6, 9, 4, 1, none of which is 77. Having checked every case, the claim is proved.

Disproof by counter-example

To disprove a general ("for all") statement, a single example where it fails is enough, and is all that is needed. For instance, the claim that n2n+41n^2 - n + 41 is prime for all positive integers fails at n=41n = 41, where it equals 41241^2, which is not prime. One counter-example settles it; you should not list several.

Proof by contradiction

You assume the opposite (the negation) of what you want to prove, then show this assumption forces an impossible result, so the original statement must be true. The layout matters: state the assumption, reason carefully, and name the contradiction explicitly.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20196 marksPaper 1, Section A. (a) Prove that the sum of the squares of any two consecutive integers is odd. (b) Disprove, by means of a counter-example, the statement that n2+n+11n^2 + n + 11 is prime for all positive integers nn.
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For (a), let the consecutive integers be nn and n+1n + 1. Then n2+(n+1)2=n2+n2+2n+1=2n2+2n+1=2(n2+n)+1n^2 + (n+1)^2 = n^2 + n^2 + 2n + 1 = 2n^2 + 2n + 1 = 2(n^2 + n) + 1. Since n2+nn^2 + n is an integer, the sum has the form 2k+12k + 1 and is therefore odd. For (b), test values: it gives primes for small nn, but at n=10n = 10, n2+n+11=100+10+11=121=112n^2 + n + 11 = 100 + 10 + 11 = 121 = 11^2, which is not prime, so the statement is disproved by this single counter-example. Markers reward general algebra (not a single example) for the proof, a clear concluding sentence, and exactly one valid counter-example for the disproof.

AQA 20215 marksPaper 1, Section A. Prove by contradiction that there is no greatest even integer. Set out your assumption, the logical steps, and the contradiction clearly.
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Assume the opposite: suppose there is a greatest even integer, and call it NN. Then N+2N + 2 is also an integer, and since NN is even, N=2kN = 2k for some integer kk, so N+2=2k+2=2(k+1)N + 2 = 2k + 2 = 2(k + 1), which is even. But N+2>NN + 2 > N, so N+2N + 2 is an even integer greater than the supposed greatest even integer NN. This contradicts the assumption that NN was the greatest. Hence the assumption is false, and there is no greatest even integer. Markers reward stating the assumption explicitly, deriving a larger even integer, and naming the contradiction before concluding.

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