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How do you prove a mathematical statement is always true, and how do you disprove one that is not?

Methods of proof: proof by deduction, proof by exhaustion, disproof by counter-example, and proof by contradiction, including the irrationality of root 2 and the infinitude of primes.

A focused answer to the OCR A-Level Mathematics A proof content, covering proof by deduction, proof by exhaustion, disproof by counter-example and proof by contradiction, with the standard results that root 2 is irrational and that there are infinitely many primes.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

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

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What this dot point is asking
The answer
Examples in context
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What this dot point is asking

OCR wants you to understand and produce mathematical proofs using four methods: proof by deduction, proof by exhaustion, disproof by counter-example, and proof by contradiction. You must know the standard contradiction proofs that $\sqrt{2}$ is irrational and that there are infinitely many prime numbers, and you must lay out logical reasoning clearly with correct notation, because this is overarching theme OT1 and is rewarded across every paper.

The answer

Proof by deduction

A deductive proof starts from known facts or definitions and reasons step by step to the result. Each line must follow logically from the previous one. You must prove the statement for the general case, not just check examples.

A common error is to "prove" a statement by checking one or two numerical cases. That is not a proof; it only shows the statement holds for those cases.

Proof by exhaustion

Proof by exhaustion splits the problem into a finite number of cases and checks each one. It works only when the cases genuinely cover every possibility.

Disproof by counter-example

To disprove a "for all" statement you need only one case where it fails. You do not have to explain why it fails in general.

For example, the statement " $n^2 > n$ for all real $n$ " is false: take $n = \tfrac{1}{2}$ , then $n^2 = \tfrac{1}{4} < \tfrac{1}{2}$ .

Proof by contradiction

Assume the statement is false, then deduce something impossible. The contradiction shows the assumption was wrong, so the original statement is true. The two named results you must know are below.

Examples in context

Constructing the infinitude-of-primes proof

Try this

Q1. Prove by deduction that the product of two even numbers is divisible by $4$ . [3 marks]

Cue. Write them as $2a$ and $2b$ ; the product is $4ab$ .

Q2. Disprove: "for all integers $n$ , $n^2 + n + 1$ is odd." [2 marks]

Cue. This one is actually always odd; instead disprove " $2^n - 1$ is prime for all $n$ " using $n = 4$ , giving $15$ .

Exam-style practice questions

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

OCR 20194 marksProve by contradiction that

\sqrt{2}

is irrational. You may assume that if

n^2

is even then

n

is even.

Show worked answer →

Assume the opposite: $\sqrt{2}$ is rational, so $\sqrt{2} = \dfrac{a}{b}$ where $a$ and $b$ are integers with no common factor and $b \neq 0$ (M1, set up the contradiction with a fraction in lowest terms).

Square: $2 = \dfrac{a^2}{b^2}$ , so $a^2 = 2b^2$ . Hence $a^2$ is even, so $a$ is even, say $a = 2k$ (M1).

Then $(2k)^2 = 2b^2$ gives $4k^2 = 2b^2$ , so $b^2 = 2k^2$ . Hence $b^2$ is even, so $b$ is even (A1).

But then $a$ and $b$ share the factor $2$ , contradicting "no common factor". The assumption is false, so $\sqrt{2}$ is irrational (A1, the explicit contradiction and conclusion).

Markers reward the lowest-terms set-up, deducing $a$ even, deducing $b$ even, and stating the contradiction with a conclusion.

OCR 20213 marksShow, by means of a counter-example, that the statement 'for all integers

n

n^2 - n + 11

is prime' is false.

Show worked answer →

A counter-example needs a single integer for which the expression is not prime (M1).

Try $n = 11$ : $n^2 - n + 11 = 121 - 11 + 11 = 121 = 11^2$ (M1).

Since $121 = 11 \times 11$ is not prime, the statement is false (A1).

Markers reward choosing a value that breaks the claim, the correct evaluation, and the explicit statement that the result is composite so the claim fails. Choosing $n = 11$ (or $n = 0$ , giving $11$ , which is prime, would not work) is the key insight.

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Sources & how we know this

OCR A Level Mathematics A (H240) specification — OCR (2017)