Pure · Proof · Year 1

Disproof by Counter-Example

To destroy a universal claim, you don't need a grand argument — you need a single case where it fails. One counter-example is enough to prove a statement false forever.

8 min read 3 worked examples

There's a deep asymmetry at the heart of mathematics, and this lesson is built on it. A statement that claims something about every case — "for all integers…", "every prime…", "any triangle…" — is called a universal statement. To prove a universal statement true, you have to show it holds in every single case, which usually needs a general argument. But to prove it false, you only need to find one case where it breaks.

That single failing case is a counter-example. The logic is airtight: if someone claims "all swans are white" and you produce one black swan, their claim is finished — it doesn't matter how many white swans exist. The word "all" is a promise about everything, and a single exception breaks that promise completely.

Why one is genuinely enough

Students often feel one example is "not enough" — after all, one example can't prove a claim true. And that's right! But disproving is different. A universal claim says "no exceptions exist." To knock it down, you exhibit an exception. One exception is a complete, logically-final disproof — not a hint, not evidence, but a full proof that the statement is false. This asymmetry (hard to prove, easy to disprove) is one of the most important ideas in all of mathematics.

What makes a valid counter-example

A counter-example must do two things, precisely:

1. Satisfy the conditions of the claim. If the statement is about prime numbers, your counter-example must actually be a prime. You can't disprove a claim about primes using a number that isn't prime — that's not a case the claim ever covered.
2. Make the conclusion false. Within those conditions, it must be a case where the thing claimed simply doesn't hold.

Find a single value satisfying both, show clearly why it breaks the statement, and you're done. There's no need to explain why the claim fails in general — the one example carries the entire argument.

Strategy — how to hunt for one

Counter-examples often hide in the "edge" cases people forget: the number 0, 1, negative numbers, fractions, or the case where two things are equal. When testing a claim, try these first — they're where sloppy generalisations tend to break.

Worked example 1

Disprove a claim about primes

Disprove the statement: "For all prime numbers $p$, the value $p^2 + 2$ is also prime."

DisproofFinding one prime that breaks it

Try $p = 3$, which is prime — so it satisfies the claim's condition.

A valid case the statement claims to cover.

Then $p^2 + 2 = 3^2 + 2 = 11$, which is prime — so $p=3$ does not break it.

This case happens to work; keep hunting.

Try $p = 5$, also prime. Then $p^2 + 2 = 25 + 2 = 27$.

Still a valid case to test.

$27 = 3 \times 9$, so $27$ is not prime.

Here the conclusion fails — this is our counter-example.

$p = 5$ is prime but $p^2+2 = 27$ is not, so the statement is false.

One counter-example is a complete disproof.

▪ disproved

Examiner's eye

Notice step 2 — the first value tried actually worked. That's normal. A claim can be true for many cases and still be false overall. You don't need the first number to fail; you need one to. State clearly that your chosen value is valid (prime) and that it breaks the conclusion.

Worked example 2

The edge-case trap · a claim that fails at a boundary

Disprove: "For all real numbers $x$, we have $x^2 > x$."

DisproofA value between 0 and 1

The claim says squaring always makes a number bigger. Test a fraction: $x = \tfrac{1}{2}$.

A real number, so a valid case.

$x^2 = \left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$.

Just computing the square.

But $\tfrac{1}{4} < \tfrac{1}{2}$, so here $x^2 < x$ — the opposite of the claim.

The conclusion fails for this value.

Therefore the statement "$x^2 > x$ for all real $x$" is false.

$x = \tfrac{1}{2}$ is a counter-example.

▪ disproved

Look deeper — why it fails here

Squaring only enlarges numbers bigger than 1 (or below −1). Between 0 and 1, squaring makes numbers smaller, and at $x=0$ or $x=1$ the two sides are equal. The claim quietly ignored this whole region. Understanding why the counter-example exists — not just that it does — is what builds real intuition. The failure lives exactly in the interval $0 \leq x \leq 1$.

Worked example 3

A plausible-sounding claim · that's still false

Disprove: "The sum of any two irrational numbers is irrational."

This one sounds reasonable — irrationals are "messy," so surely their sum stays messy? The counter-example comes from choosing two irrationals designed to cancel.

DisproofTwo irrationals that sum to an integer

Take $\sqrt{2}$ and $-\sqrt{2}$. Both are irrational.

They satisfy the condition of the claim.

Their sum is $\sqrt{2} + (-\sqrt{2}) = 0$.

The two values cancel exactly.

$0$ is rational (it equals $\tfrac{0}{1}$), not irrational.

The conclusion fails.

So two irrationals can sum to a rational — the statement is false.

A single counter-example settles it.

▪ disproved

The key insight

A counter-example doesn't have to be an "accident" you stumble on — often you construct it deliberately. Here we chose two irrationals specifically so they'd cancel. When a universal claim feels shaky, ask: "can I engineer a case that breaks it?" That constructive mindset is a powerful problem-solving tool well beyond proof questions.

Common mistake

A counter-example must satisfy the claim's conditions. If you're disproving a statement about prime numbers, offering the number 4 proves nothing — 4 isn't prime, so it was never a case the claim covered. Always check your example genuinely qualifies before celebrating.

Check yourself

Disprove: "For all integers $n$, the value $n^2 - n + 11$ is prime."
(Hint: try to make the expression share a factor — what happens near $n = 11$?)

Try $n = 11$, an integer — a valid case.

$n^2 - n + 11 = 121 - 11 + 11 = 121$.

$121 = 11 \times 11$, so it is not prime.

The conclusion fails.

So $n = 11$ is a counter-example, and the statement is false.

▪ disproved

Every term of $n^2 - n + 11$ contains a factor of 11 when $n = 11$, forcing the whole expression to be divisible by 11 — a neat example of constructing a counter-example rather than guessing.

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