Disproving statements
Learn how to disprove statements in Year 11 Specialist Mathematics in Queensland (QCAA). When a claim holds for every case, a single counterexample — one case that fails — proves the whole claim false.
You will learn to find counterexamples by testing small and special values, check they really fail the claim, and see why one supporting example never proves a universal statement — a key idea for proof.
Theory
Disproving statements is a core skill in proof for Year 11 Specialist Mathematics (QCAA, Queensland). A statement that claims something is true for all cases is disproved by a single counterexample — one case that fails. This page shows how to find counterexamples, and why one supporting example never proves a universal claim.
A universal statement claims that a property holds for all members of some set, for example "every prime number is odd" or "for all integers \(n\), \(n^2>n\)". These are the statements the quantifier "for all" introduces.
A counterexample is a single case in which the statement fails: the hypothesis is satisfied but the conclusion is not. To disprove a "for all" statement you need only one counterexample — that one failing case makes the whole claim false.
The reverse does not hold. Checking the statement for one case, ten cases, or a thousand cases does not prove a universal claim, because it may still fail for a case you have not tried. Supporting examples build confidence, but they are not a proof.
So testing cases can only ever disprove a "for all" statement, never prove it. Proving a universal claim requires a general argument that covers every case at once.
A universal statement has the form "for all \(x\), \(P(x)\)". Its negation is an existence statement — there is some \(x\) for which \(P(x)\) fails:
So disproving the claim means producing just one such \(x\), called a counterexample \(a\):
How to disprove a "for all" statement
- Read the claim carefully and confirm it is a "for all" (universal) statement, noting the hypothesis and the conclusion.
- Search for a failing case, testing small, boundary and special values first — \(0\), \(1\), negatives, and the prime \(2\) are common culprits.
- Check the candidate really is a counterexample: it must satisfy the hypothesis but make the conclusion false.
- Conclude that the single counterexample disproves the whole statement — no further cases are needed.
Find a prime that is not odd; test the smallest primes:
| \(3\) | \(=\) | \(\text{odd prime — obeys the claim}\) |
| \(5\) | \(=\) | \(\text{odd prime — obeys the claim}\) |
| \(2\) | \(=\) | \(\text{prime}\) |
| \(2\) | \(=\) | \(\text{even, so not odd}\) |
| \(\Rightarrow\) | \(2 \text{ is a counterexample}\) |
\(2\) is a counterexample: it is prime but even, so the claim is false.
Test values, watching the small ones where squaring does not increase \(n\):
| \(n=2\) | \(:\) | \(4>2 \;\text{true}\) |
| \(n=3\) | \(:\) | \(9>3 \;\text{true}\) |
| \(n=0\) | \(:\) | \(0^2=0\) |
| \(=\) | \(0>0 \;\text{is false}\) | |
| \(\Rightarrow\) | \(n=0 \text{ is a counterexample}\) |
\(n=0\) is a counterexample (\(n=1\) also works).
Look for two irrationals whose sum is rational:
| \(\sqrt{2}+\sqrt{3}\) | \(=\) | \(\text{irrational}\) |
| \(\sqrt{2}+\bigl(-\sqrt{2}\bigr)\) | \(=\) | \(0\) |
| \(0\) | \(=\) | \(\text{rational}\) |
| \(\Rightarrow\) | \(\sqrt{2},\ -\sqrt{2}\ \text{is a counterexample}\) |
\(\sqrt{2}\) and \(-\sqrt{2}\) are irrational but sum to \(0\), which is rational.
Test the odd numbers in turn until one is not prime:
| \(3,\ 5,\ 7\) | \(=\) | \(\text{odd and prime — obey the claim}\) |
| \(9\) | \(=\) | \(3\times 3\) |
| \(9\) | \(=\) | \(\text{odd, but not prime}\) |
| \(\Rightarrow\) | \(9 \text{ is a counterexample}\) |
\(9\) is a counterexample: it is odd and greater than \(1\), but \(9=3\times3\).
Common pitfalls
Frequently asked questions
How many counterexamples do you need to disprove a statement?
Just one. A single case where a "for all" statement fails is enough to make the whole statement false.
Does one example prove a "for all" statement?
No. One supporting example, or even many, does not prove a universal claim, because the statement might still fail for a case you have not tried.
What is a counterexample?
A counterexample is a single case that satisfies the statement’s hypothesis but makes its conclusion false, so it shows the statement is not true for all cases.
How do you find a counterexample?
Test small, boundary and special values first — \(0\), \(1\), negatives and the prime \(2\) are common places where a claim breaks down.
Is 2 a counterexample to "all primes are odd"?
Yes. \(2\) is a prime number and it is even, so it is a prime that is not odd, which disproves the claim.
Why isn’t checking lots of cases a proof?
A "for all" statement covers infinitely many cases, so no finite list of checks can guarantee it. Only a general argument that covers every case proves it.