What is incomplete exhaustion?

Proof by exhaustion only works if your cases cover every possibility; missing a case breaks the proof.

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How do we prove a mathematical statement is always true, or show it is false?

Methods of proof, including proof by deduction, proof by exhaustion, disproof by counter-example, and proof by contradiction.

A CCEA A-Level Mathematics answer on the methods of proof - direct proof by deduction, proof by exhaustion, disproof by counter-example and proof by contradiction - with worked examples including the irrationality of root 2.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

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

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The answer
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What this dot point is asking

CCEA wants you to use the standard methods of proof: proof by deduction (a direct chain of logic), proof by exhaustion (checking every case), disproof by counter-example (one case that fails), and proof by contradiction (assume the opposite and derive an impossibility). Rigorous proof is a distinct A2 skill, tested both on its own and within other topics.

The answer

Proof by deduction

For instance, to prove the sum of two odd numbers is even, write them as $2a + 1$ and $2b + 1$ ; their sum is $2a + 2b + 2 = 2(a + b + 1)$ , a multiple of $2$ , hence even.

Proof by exhaustion

For example, to show every integer square ends in $0, 1, 4, 5, 6$ or $9$ , you can check the last digit of $n^2$ for each possible last digit of $n$ from $0$ to $9$ .

Disproof by counter-example

Proof by contradiction

A proof by contradiction assumes the negation of what you want to prove, then derives a logical impossibility. Since the reasoning is valid but the conclusion is absurd, the assumption must be false, so the original statement is true. The classic example is the irrationality of $\sqrt{2}$ .

Worked example: contradiction with primes

Prove there is no largest prime

Prove by contradiction that there is no largest prime number.

step Assume the opposite

Suppose there is a largest prime, and call it $p$ . Then the complete list of primes is $2, 3, 5, \dots, p$ .

step Construct a new number

Let $N = (2 \times 3 \times 5 \times \dots \times p) + 1$ , the product of all the primes plus one.

step Derive the contradiction

Dividing $N$ by any prime on the list leaves remainder $1$ , so $N$ is not divisible by any of them. Hence either $N$ is itself prime, or it has a prime factor not on the list. Either way there is a prime larger than $p$ , contradicting the assumption that $p$ was the largest.

step Conclude

The assumption is false, so there is no largest prime: the primes are infinite.

Examples in context

Example 1. Testing a conjecture. Faced with " $n^2 + n + 41$ is always prime", you might verify it for $n = 1, 2, 3, \dots$ and be tempted to believe it; but $n = 41$ gives a multiple of $41$ . This is why a pattern, however convincing, needs a proof, not just examples.

Example 2. Irrational and rational sums. Proving that the sum of a rational and an irrational number is irrational is a neat contradiction: assume the sum is rational, then the irrational equals a difference of two rationals, which is rational, a contradiction. Contradiction is the natural tool when a direct route is hard.

Try this

Q1. Disprove: "all prime numbers are odd". [1 mark]

Cue. $2$ is prime and even, a counter-example.

Q2. Prove that the product of two even numbers is divisible by $4$ . [3 marks]

Cue. $2a \times 2b = 4ab$ , a multiple of $4$ .

Q3. What is the first step of a proof by contradiction? [1 mark]

Cue. Assume the opposite (the negation) of the statement to be proved.

Exam-style practice questions

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

CCEA 20205 marksProve by contradiction that

\sqrt{2}

is irrational.

Show worked answer →

Assume the opposite: that $\sqrt{2}$ is rational, so $\sqrt{2} = \frac{a}{b}$ where $a$ and $b$ are integers with no common factor (the fraction is in lowest terms).

Then $2 = \frac{a^2}{b^2}$ , so $a^2 = 2b^2$ . Hence $a^2$ is even, which means $a$ is even, so write $a = 2k$ .

Substituting: $(2k)^2 = 2b^2$ , so $4k^2 = 2b^2$ and $b^2 = 2k^2$ . Hence $b^2$ is even, so $b$ is even.

But then $a$ and $b$ are both even, sharing a factor of $2$ , contradicting the assumption that the fraction was in lowest terms.

The assumption must be false, so $\sqrt{2}$ is irrational.

Markers reward the assumption, deriving that both $a$ and $b$ are even, identifying the contradiction with the lowest-terms assumption, and the conclusion.

CCEA 20184 marksDisprove the statement: for all positive integers

n

, the value

n^2 - n + 11

is prime. Then prove that

n^2 + n

is always even.

Show worked answer →

To disprove, find a single counter-example. Try $n = 11$ :

$11^2 - 11 + 11 = 121$ , which is $11^2$ , not prime. So the statement is false.

For the second part, $n^2 + n = n(n + 1)$ , a product of two consecutive integers. One of any two consecutive integers is even, so the product is even for every positive integer $n$ .

Markers reward a valid counter-example for the first claim, and the consecutive-integers argument for the second.

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Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)