Disproof by Counterexample

Disproof by counterexample is one of the simplest yet most important techniques in mathematical reasoning. To disprove a universal statement — one that claims something is true for all values — you need only find one single example where the statement fails. This single example is called a counterexample.

The AQA A-Level Mathematics specification (7357) requires students to understand disproof by counterexample as one of the four key proof techniques. Questions on this topic appear regularly in exam papers and are typically accessible marks if you understand the technique.

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The Logic Behind Counterexamples

A universal statement has the form "For all x in S, property P(x) holds." In mathematical notation: ∀x ∈ S, P(x).

To disprove this, you need to find at least one x₀ ∈ S such that P(x₀) is false. In notation: ∃x₀ ∈ S such that ¬P(x₀).

One counterexample is sufficient to disprove a universal statement. It does not matter if the statement holds for millions of other values — a single failure is enough.

Conversely, no number of confirming examples constitutes a proof of a universal statement (unless you check every case, which is proof by exhaustion).

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Finding Counterexamples: Strategy

When looking for a counterexample to disprove a statement, try the following systematic approach:

1. Try small values first: n = 0, 1, 2, 3, ...
2. Try boundary cases: n = 0, n = 1, negative numbers.
3. Try special values: prime numbers, perfect squares, powers of 2.
4. Consider edge cases: What if n = 0? What if n is negative? What about n = 1?
5. Think about what could go wrong with the statement.

Exam Tip: Most A-Level counterexample questions can be answered with small, simple values. Do not overthink it — try n = 1, 2, 3, or common values like 0 and −1 first.

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Worked Examples

Example 1: Disprove the statement "n² > 2n for all positive integers n."

Try n = 1: n² = 1 and 2n = 2. Since 1 < 2, we have n² < 2n, not n² > 2n.

Counterexample: n = 1. When n = 1, n² = 1 < 2 = 2n, so the statement is false. ∎

Also try n = 2: n² = 4 and 2n = 4. Since 4 = 4, we have n² = 2n, not n² > 2n.

So n = 2 is also a counterexample (equality, not strict inequality).

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Example 2: Disprove the statement "2n + 1 is prime for all non-negative integers n."

Try values:

• n = 0: 2⁰ + 1 = 2 — prime ✓
• n = 1: 2¹ + 1 = 3 — prime ✓
• n = 2: 2² + 1 = 5 — prime ✓
• n = 3: 2³ + 1 = 9 = 3 × 3 — not prime ✗

Counterexample: n = 3. When n = 3, 2³ + 1 = 9 = 3², which is not prime. ∎

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Example 3: Disprove the statement "n² − n + 41 is prime for all positive integers n."

This is a famous example. Let us try a few values first:

• n = 1: 1 − 1 + 41 = 41 — prime ✓
• n = 2: 4 − 2 + 41 = 43 — prime ✓
• n = 3: 9 − 3 + 41 = 47 — prime ✓

In fact, this expression gives primes for n = 1 through n = 40! But:

• n = 41: 41² − 41 + 41 = 41² = 1681 — not prime (it equals 41 × 41) ✗

Counterexample: n = 41. When n = 41, n² − n + 41 = 41², which is not prime. ∎

Key Lesson: Even if a statement holds for the first 40 values, it can still be false. This is why proof (not just checking examples) is essential in mathematics.

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Example 4: Disprove the statement "the sum of two irrational numbers is always irrational."

Consider √2 and −√2. Both are irrational.

Their sum is √2 + (−√2) = 0, which is rational.

Counterexample: √2 and −√2 are both irrational, but their sum is 0, which is rational. ∎

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Example 5: Disprove the statement "if n is a positive integer and n divides (a × b), then n divides a or n divides b."

Try n = 6, a = 4, b = 3: then a × b = 12, and 6 | 12. But 6 ∤ 4 and 6 ∤ 3.

Counterexample: 6 divides 4 × 3 = 12, but 6 does not divide 4 and 6 does not divide 3. ∎

(Note: this statement is true when n is prime — this is a key property of primes.)

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Example 6: Disprove the statement "for all real numbers x, if x² > 9 then x > 3."

Try x = −4: x² = 16 > 9, but x = −4 < 3.

Counterexample: x = −4 satisfies x² = 16 > 9, but x = −4 is not greater than 3. ∎

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Example 7: Disprove the statement "the square root of a sum equals the sum of the square roots: √(a + b) = √a + √b for all positive reals a, b."

Try a = 1, b = 1: √(1 + 1) = √2 ≈ 1.414, but √1 + √1 = 1 + 1 = 2.

Since √2 ≠ 2, the statement is false.

Counterexample: a = 1, b = 1. √2 ≠ 2. ∎

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Common False Conjectures (Exam Favourites)

The following are frequently used in AQA exam questions:

False Conjecture	Counterexample
All prime numbers are odd	2 is prime and even
n² > n for all integers n	n = 0: 0² = 0 = n, not > n; or n = 1: 1² = 1
n² ≥ n for all integers n	n = −1: (−1)² = 1 but need to check... actually 1 ≥ −1 is true. Try the original: "n² > 2n for all positive integers" — use n = 1
If a > b, then a² > b²	a = −1, b = −2: a > b but a² = 1 < 4 = b²
2ⁿ > n² for all n ≥ 1	n = 3: 2³ = 8 < 9 = 3²
n² + n + 41 is always prime	n = 41: gives 41²
The product of two irrationals is irrational	√2 × √2 = 2, which is rational

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Presentation in Exams

When writing a disproof by counterexample:

1. State the counterexample clearly: "Let n = ..." or "Consider n = ..."
2. Show the computation that demonstrates the statement fails.
3. State the conclusion: "This is a counterexample, so the statement is false."

A complete answer might look like:

"Consider n = 1. Then n² = 1 and 2n = 2. Since 1 < 2, we have n² < 2n, which contradicts the claim that n² > 2n. Therefore the statement is false. ∎"

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Summary

• To disprove "for all x, P(x)," find one x where P(x) is false.
• One counterexample is sufficient — you do not need to explain why the statement fails in general.
• Try small values, boundary values, and special cases first.
• Checking examples is not a proof — but finding one failure is a disproof.
• Always present your counterexample clearly with a full computation and concluding statement.

Exam Tip: Counterexample questions are usually worth 2–3 marks on AQA papers. The marks are typically: (1) identifying the counterexample, (2) showing the computation, and (3) stating that the statement is therefore false. Do not skip the computation — even if the counterexample is obvious, you must show that it fails.

A-Level Deep Dive: Disproof by Counter-Example

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section A: Proof. The specification requires students to "use methods of proof, including proof by deduction, proof by exhaustion, disproof by counter-example, and proof by contradiction." Disproof by counter-example is the logical complement of universal proof: if a claim has the form "for all $x \in S$x∈S, $P(x)$P(x) holds", a single $x_0 \in S$x0​∈S for which $P(x_0)$P(x0​) fails is sufficient to disprove the claim entirely. The technique is examined explicitly on Paper 1 and synoptically on Paper 2 wherever a generalisation is offered (sequences, functions, inequalities). The AQA formula booklet contains nothing relevant — counter-example questions test logical literacy, not formula recall.

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Worked example with full mark scheme

Question (8 marks):

(a) Disprove the claim: "Every prime number is odd." (2)

(b) A student claims that "for every real number $n$n, $n^2 > n$n2>n." Disprove this claim by giving two distinct counter-examples and explain why a single counter-example would be sufficient. (4)

(c) Decide whether the following claim is true or false, justifying your answer: "There exists a real number $x$x with $x^2 < 0$x2<0." Explain why a counter-example cannot be used to disprove this claim. (2)

Solution with mark scheme:

(a) B1 — identify the counter-example $p = 2$p=2: $2$2 is prime (its only positive divisors are $1$1 and $2$2) and $2$2 is even, not odd.

B1 — state the conclusion: "since $2$2 is a prime that is not odd, the claim that every prime is odd is false."

(b) Step 1 — choose a counter-example. Take $n = 0$n=0. Then $n^2 = 0$n2=0 and $n = 0$n=0, so $n^2 > n$n2>n is the assertion $0 > 0$0>0, which is false.

M1 — substituting a value of $n$n into the inequality.

A1 — correct conclusion that $n = 0$n=0 violates the inequality.

Step 2 — second counter-example. Take $n = \tfrac{1}{2}$n=21​. Then $n^2 = \tfrac{1}{4}$n2=41​ and $n = \tfrac{1}{2}$n=21​, so the claim says $\tfrac{1}{4} > \tfrac{1}{2}$41​>21​, which is false.

A1 — second valid counter-example with working shown.

Step 3 — explain sufficiency of one. "A universal claim asserts $P(n)$P(n) holds for every $n$n; the negation is $\exists n: \neg P(n)$∃n:¬P(n). Producing one such $n$n exhibits the existential negation directly, so a single counter-example is logically sufficient to disprove the universal claim."

B1 — coherent explanation of why one counter-example suffices (logical-structure mark).

(c) The claim is false: $x^2 \geq 0$x2≥0 for all real $x$x, so no real $x$x satisfies $x^2 < 0$x2<0.

B1 — correct verdict with reason.

B1 — explanation: "the claim is existential ($\exists x$∃x), and an existential claim cannot be disproved by a counter-example. Disproof requires a general argument showing the property fails for every $x$x — here, the argument that $x^2 \geq 0$x2≥0 universally."

Total: 8 marks (B2 + M1 A2 B1 + B2).

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): A student writes:

"For all positive integers $n$n, the value of $n^2 + n + 41$n2+n+41 is prime."

(a) Verify the claim for $n = 1$n=1, $n = 2$n=2 and $n = 3$n=3. (2)

(b) By choosing a suitable value of $n$n, disprove the claim. (3)

(c) Explain why verifying many cases — even thousands — does not constitute proof. (1)

Mark scheme decomposition by AO:

(a)

• B1 (AO1.1a) — $n=1: 1+1+41=43$n=1:1+1+41=43 (prime); $n=2: 4+2+41=47$n=2:4+2+41=47 (prime).
• B1 (AO1.1a) — $n=3: 9+3+41=53$n=3:9+3+41=53 (prime).

(b)

• M1 (AO3.1a) — strategy: try $n = 41$n=41 to force the expression to be divisible by $41$41.
• A1 (AO1.1b) — compute $41^2 + 41 + 41 = 41(41 + 1 + 1) = 41 \cdot 43$412+41+41=41(41+1+1)=41⋅43.
• A1 (AO2.4) — conclude: $41 \cdot 43$41⋅43 is composite (it has $41$41 as a proper factor), so the claim fails at $n = 41$n=41.

(c)

• B1 (AO2.2b) — explanation that finitely many verifications cannot establish a claim about infinitely many integers; the claim's domain is unbounded, so empirical checking is at most evidence, never proof.

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. This is a representative AQA proof question — the AO3 strategic mark for choosing the counter-example is what separates a confident counter-example response from a guesswork response.

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Synoptic links

Connects to:

1. Proof by deduction and exhaustion (Pure A): counter-example is the logical mirror image of universal proof. Where deduction shows $\forall n \, P(n)$∀nP(n), counter-example shows $\exists n \, \neg P(n)$∃n¬P(n). Recognising which technique fits a claim is itself an examined skill.

2. Proof by contradiction (Pure A): contradiction often uses a counter-example internally. To prove $\sqrt{2}$2​ is irrational, one assumes $\sqrt{2} = p/q$2​=p/q in lowest terms and derives that $p$p and $q$q share factor $2$2 — a direct contradiction with the lowest-terms assumption.

3. Sequences and series (Pure E): claims about general terms ("$u_n$un​ is increasing for all $n$n") are routinely disproved by exhibiting a single $n$n where $u_{n+1} \leq u_n$un+1​≤un​. The signature is that students must check every claimed property, not just the first few terms.

4. Functions (Pure B): "every function with $f(0) = 0$f(0)=0 satisfies $f(x+y) = f(x) + f(y)$f(x+y)=f(x)+f(y)" is disproved by $f(x) = x^2$f(x)=x2 — a key Cauchy-equation counter-example bridging A-Level and undergraduate analysis.

5. Differentiation and stationary points (Pure G): "if $f'(a) = 0$f′(a)=0 then $f$f has a maximum or minimum at $a$a" is disproved by $f(x) = x^3$f(x)=x3 at $a = 0$a=0 — a stationary point of inflection. This counter-example is examined repeatedly.

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Mark-scheme literacy

Counter-example questions on AQA 7357 split AO marks distinctively because they test reasoning rather than computation:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	30–40%	Substituting the chosen value into the claim, computing both sides, stating the comparison
AO2 (reasoning / interpretation)	40–50%	Identifying the claim's logical form (universal vs existential), explaining why one counter-example suffices, distinguishing disproof from proof
AO3 (problem-solving)	15–25%	Choosing a counter-example strategically — particularly where the "obvious" small values verify the claim and the disproof requires insight ($n=41$n=41 in the $n^2+n+41$n2+n+41 example)

Examiner-rewarded phrasing: "the claim is universal, so a single counter-example suffices"; "since the claim is existential, disproof requires a general argument, not a counter-example"; "we exhibit $n = \dots$n=…, for which $P(n)$P(n) is false; hence the claim is disproved." Phrases that lose marks: "the claim is wrong because it doesn't always work" (no counter-example given); "for some values it fails" (vague — name the value); "I tested many values and they all worked" (verification is not proof).