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.
What a learner can do afterwards
- 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
1 · Read
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.
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.
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.
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.
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.
Four readings, six practice questions and six study cards written for this dataset. OpenStax logic records informed facts and exercise design; their passages and questions are not reproduced.
More to read
State the domain before searching
The domain is part of a mathematical claim. Consider the statement that x squared is at least x. For integer inputs, subtracting x gives x(x-1). If the integer is at least 1, both factors are nonnegative. If it is at most 0, both factors are nonpositive. Their product is nonnegative in either case.
The same statement fails on the real numbers. Input 1/2 gives square 1/4, which is less than 1/2. That input refutes the real-domain claim but cannot refute the integer-domain claim.
Quantifiers determine the search too. A universal claim is refuted by one allowed exception. An existence claim is proved by one allowed witness. Finding that a particular candidate does not work leaves open the possibility that another candidate does.
Before calculating, identify the domain, every hypothesis and the exact conclusion. A calculation cannot settle a claim about objects it excludes.
Check the direction of a claim
To refute 'if P then Q', find a case in which P holds and Q does not. A case with false P cannot refute that conditional. The hypothesis must be checked as carefully as the conclusion.
Reversing a conditional creates a different claim. Every integer divisible by 6 is divisible by 3: writing the integer as 6k also writes it as 3 times 2k. The reverse claim fails. The integer 9 is divisible by 3 but not by 6. It refutes the converse while leaving the original statement intact.
An argument may contain several premises. To refute the argument's claim of valid inference, find a case where every premise is true and the conclusion is false. One true premise is not enough. If a premise is itself a conditional, evaluate that conditional rather than assuming its hypothesis must hold.
Recording each premise separately helps prevent a counterexample to one statement from being mistaken for a counterexample to another.
A repair still needs a proof
Suppose someone claims that integers with equal squares must be equal. Take a=4 and b=-4. Their squares are both 16, but the integers differ. This pair satisfies the hypothesis and violates the conclusion.
Factoring explains the failure. Equal squares give (a-b)(a+b)=0. For real numbers, a zero product has at least one zero factor. Thus a=b or a=-b. This proves a corrected conclusion containing both possibilities.
A different repair restricts both integers to be nonnegative. If a=-b under that restriction, both must be zero, so they are equal. The restriction therefore gives a provable statement.
Removing one counterexample is only a starting point. A proposed restriction may exclude that pair while admitting another exception. Search again, then give an argument covering all remaining allowed objects. The calculation that exposed the failure can help identify what the proof must address.
Know what your tests cover
Small allowed inputs are inexpensive tests. The formula may also suggest a particular target. For n squared plus n plus 7, choose n=6. The value is 36+6+7=49, which equals 7 times 7. This one calculation refutes the claim that the expression is prime for every whole number n.
A successful calculation has a different reach. If a claim concerns only the list 2, 4 and 6, checking those three entries covers the whole domain. A universal claim about that list can be proved by checking them all. If the domain is all whole numbers, the same three tests leave infinitely many inputs unchecked.
Keep track of the scope of a test result. It may refute a universal claim, establish an existence claim, or confirm only a particular case. A search that finds no exception may suggest an argument to try, but it does not itself supply a proof on an infinite domain.
2 · Watch
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