--- title: "Counterexamples and Testing a Conjecture" description: "One object satisfying the hypotheses but violating the conclusion refutes a universal claim. Testing can suggest a repair or proof route. Checking every allowed case in a finite domain can prove the c" canonical: https://lightmysky.com/learn/mathematics/counterexamples-and-testing-a-conjecture-mt_SjA4jqNkaM source: https://lightmysky.com/learn/mathematics/counterexamples-and-testing-a-conjecture-mt_SjA4jqNkaM.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Counterexamples and Testing a Conjecture One object satisfying the hypotheses but violating the conclusion refutes a universal claim. Testing can suggest a repair or proof route. Checking every allowed case in a finite domain can prove the claim; testing that leaves cases unchecked cannot do so alone. Subject: Mathematics · Area: Mathematical Thinking · Ages 22 to 24 Page: https://lightmysky.com/learn/mathematics/counterexamples-and-testing-a-conjecture-mt_SjA4jqNkaM ## Ready when they can - Construct a counterexample to a plausible but false statement - Use a counterexample to identify a possible restriction, then justify a repaired statement - Test small and boundary cases, and distinguish testing from proof ## Lesson: Breaking claims to make them true You have checked how definitions handle domains and edge cases. A counterexample must belong to the claim's domain, satisfy every hypothesis, and violate the conclusion. To refute 'all A are B', seek an A that is not B. To refute 'if P then Q', seek true P and false Q. Read the quantifiers and boundaries before choosing a test. *(drawing: Both checks matter: the hypothesis holds and the conclusion fails.)* A prime is an integer greater than 1 whose only positive divisors are 1 and itself. To test 'all primes are odd', check both primality and parity. A nonprime cannot refute a claim about primes. For a claim about two integers, check both entries. One zero factor makes their product zero without forcing the other factor to be zero. To refute 'any two irrationals have an irrational product', find two irrational factors with a rational product. For positive a and b, root a times root b equals root ab. A positive integer has an irrational square root unless it is a perfect square. For example, root 3 times root 12 equals root 36, which is 6. Check both factors and the product. Small, boundary and degenerate cases can be useful tests when the hypotheses allow them. No class of tests always finds a counterexample fastest. For n squared plus n plus c, where c is an integer greater than 1, try n=c-1. Substitution gives c squared, which is composite. With c=11, n=10 gives 121=11 times 11. Successful tests elsewhere do not undo this failure. *(drawing: Choosing n one less than the constant exposes a square greater than 1.)* An unsuccessful search may suggest a proof route, but it does not prove an infinite universal claim. For a differentiable real function defined near c, write f(c+h)-f(c) as h times its difference quotient for nonzero h. As h tends to 0, the quotient has a finite limit and h tends to 0. Their product tends to 0, proving continuity at c. This argument covers every qualifying function. Let f and g be real-valued functions defined near c. Two discontinuous functions can cancel: g=-f makes f+g=0. A repair is: if f is continuous at c and g is discontinuous there, their sum is discontinuous there. If the sum were continuous, g=(f+g)-f would be continuous too, since differences preserve continuity. This contradicts the hypothesis about g. *(drawing: Subtracting the continuous function f exposes the contradiction.)* A restriction needs justification, not just removal of one counterexample. On (0,2), f(x)=x has no maximum: every x has a larger allowed point (x+2)/2. The extreme value theorem gives a valid repair: every continuous real-valued function on a nonempty closed bounded interval [a,b] attains a maximum and a minimum. To refute an argument, make all its premises true and its conclusion false. From 'if P then Q' and 'Q is true', P need not follow: Q may hold when P is false. Check the conditional premise too. It is true whenever P is false, regardless of Q. For a finite domain, checking every allowed case proves a universal claim on that domain. For an infinite domain, finitely many successful tests leave cases unchecked. A claim saying 'there exists' has a different requirement: one qualifying example proves existence. Rejecting one candidate does not show that no qualifying example exists. **Recap.** A counterexample meets the hypotheses and breaks the conclusion. Tests can guide a repair or proof, but a repaired universal claim still needs justification. *Lesson, questions and flashcards written for this dataset. OpenStax Contemporary Mathematics sections 2.1, 2.4 and 2.7, and Introduction to Philosophy section 5.2 informed the study of logic facts and exercise structures; their passages are not reproduced.* ## Practice 14 questions on this page, each with its working shown. ## Needs first - [Definitions as Design Decisions](https://lightmysky.com/learn/mathematics/definitions-as-design-decisions-mt_b6ip8eP0-d) ## Optional connections - [Continuity and Uniform Continuity](https://lightmysky.com/learn/mathematics/continuity-and-uniform-continuity-mt_plWc9raOwz) ## Opens up - [Open Problems as a Genre](https://lightmysky.com/learn/mathematics/open-problems-as-a-genre-mt_qBWcaUWG9J)