Universal claims have a weakness
A statement that says something is true for every permitted case carries a heavy burden. To prove it, the reasoning must cover the entire domain. To disprove it, one valid counterexample is enough.
This makes counterexamples one of the most efficient tools in mathematics. They do more than say “wrong”. A well-chosen counterexample reveals which hidden assumption failed and often points toward a better statement.
Example is not counterexample
If someone claims that the square of every integer is non-negative, showing 3²=9 is an example supporting the claim. It does not prove the universal statement. A counterexample would have to be an integer whose square is negative. No such integer exists, but establishing that requires reasoning about the structure of integer multiplication.
If someone claims that x²>x for every real x, x=1/2 is a counterexample because 1/4 is not greater than 1/2. The claim is therefore false, regardless of how many large positive values worked before it.
Attack the quantifier
Words such as all, every, always, never invite counterexample testing. “Every quadrilateral is a rectangle” fails with a non-rectangular parallelogram. “The product of two positive numbers is greater than both numbers” fails with 1/2×1/2=1/4. “A larger perimeter means a larger area” can fail for rectangles.
Worked case: perimeter does not determine area
Claim: rectangles with the same perimeter have the same area. Take a 6 cm by 4 cm rectangle. Its perimeter is 20 cm and its area is 24 cm². Take an 8 cm by 2 cm rectangle. Its perimeter is also 20 cm, but its area is 16 cm². One pair of valid rectangles destroys the claim.
The counterexample also teaches something deeper: perimeter constrains the sum of adjacent sides, while area depends on their product. Equal sums need not produce equal products.
Worked case: cancelling is conditional
Claim: (x²−4)/(x−2)=x+2 for every real x. Factorisation gives (x−2)(x+2)/(x−2), which simplifies to x+2 only when x−2≠0. At x=2 the original expression is undefined. Therefore the unrestricted “for every real x” claim is false.
The repaired statement is: for every real x with x≠2, (x²−4)/(x−2)=x+2. The counterexample did not destroy the useful algebra. It exposed the missing domain condition.
Worked case: averages can conceal variation
Claim: two data sets with the same mean have the same spread. Consider A={5,5,5,5,5} and B={1,3,5,7,9}. Both have mean 5. A has range 0; B has range 8. The claim is false.
The example shows why a measure of centre cannot by itself describe variation. A counterexample can therefore challenge an interpretation as well as an algebraic statement.
Worked case: diagrams can manufacture false confidence
Claim: if two angles look equal in a diagram, they are equal. Draw any deliberately distorted figure with two visibly similar angles whose actual measures differ. Appearance is not a mathematical condition. Equality must follow from given information, construction, theorem or measurement when measurement is the task.
Choose counterexamples intelligently
Random searching can work, but strategic values are better. Try 0, 1 and −1. Try fractions between 0 and 1. Try negative numbers. Try equal quantities and unequal quantities. Try boundary values. For algebraic fractions, try values that make a denominator zero. For geometry, try a special case and a deliberately non-special case.
The best counterexample is usually simple enough that nobody can dispute the calculation. Complexity can hide the reason the claim failed.
Counterexample laboratory
- “Adding the same positive number to numerator and denominator leaves a fraction unchanged.” Find a counterexample.
- “If ab=0, then a=0 and b=0.” Test the claim.
- “Squaring a number makes it larger.” Find at least three different kinds of counterexample.
- “Every number divisible by 6 is divisible by 3.” Can you find a counterexample? If not, explain why.
- “Every number divisible by 3 is divisible by 6.” Find a counterexample.
Solutions
For the fraction claim, 1/2 becomes 2/3 after adding 1 to numerator and denominator, so it changes. For ab=0, a=0 and b=5 gives a product of zero without b=0; the correct conclusion is that at least one factor is zero.
For squaring, 1/2 becomes 1/4, −2 becomes 4 which is numerically greater but changes sign, and 1 remains 1. The precise comparison depends on the original domain and meaning of “larger”.
A number divisible by 6 has the form 6k=3(2k), so it is divisible by 3; no integer counterexample exists. The reverse fails: 3 is divisible by 3 but not by 6.
Counterexamples and examinations
When asked whether a statement is always true, do not assume the expected answer is yes. Search first. If the claim is false, a valid counterexample plus a clear explanation can be decisive. If the claim survives your search, that still does not constitute proof unless the task permits exhaustive checking of a finite set.
Counterexample thinking also improves checking. Instead of asking only “Does my method work on this question?”, ask “Under what circumstances would my method fail?” That question exposes conditions and assumptions.
From destruction to refinement
The mature response to a counterexample is often to repair the theorem. “x²>x for every real x” becomes “x²>x when x<0 or x>1.” “Cancelling x−2 is always allowed” becomes “cancelling x−2 is allowed when x≠2.” Mathematics grows by making its claims more exact.
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