Overview
A counterexample is not simply an unusual value or a case where some part of a statement fails. It must satisfy the claim’s assumptions while violating its conclusion. This makes counterexample questions a test of logical structure as well as arithmetic. We will examine divisibility, signs, irrational numbers and rectangle measurements, checking the premise and conclusion separately for every promising choice.
Start with the assumption, not the desired contradiction
The claim “every positive integer divisible by 6 is also divisible by 12” has a clear premise and conclusion. The number 6 meets the premise but is not divisible by 12, so it disproves the claim. The numbers 12, 24 and 36 meet both parts and therefore do not provide counterexamples.
This distinction explains why an arbitrary number not divisible by 12 would not necessarily help: if it also failed divisibility by 6, the claim would say nothing about it. A universal statement can be disproved by one qualifying failure, whereas many confirming examples cannot prove it. In practice, write two checks next to a proposed counterexample: “assumptions true” and “conclusion false.” Both are required. This simple structure becomes especially useful when a claim has several assumptions that are easy to overlook.
Do not violate the premise when testing a sign claim
To disprove “ab > 0 implies a > 0 and b > 0,” choose a = −2 and b = −5. Their product is 10, so the positive-product premise is true, but both factors are negative. The proposed conclusion fails.
The pair (−2, 5) also contains a negative factor, but its product is negative, so it does not satisfy the premise. The pair (0, 5) fails the strict positive-product condition as well. The pair (2, 5) satisfies the conclusion and cannot disprove it. Checking all four illustrates why a wrong choice can look appealing when you focus only on the desired failure. The mathematics of signs supplies the actual counterexample: a positive product can arise from two positive factors or from two negative factors.
Use conjugates to satisfy several number conditions
A claim says that two positive irrational numbers with an integer product must have an irrational sum. The pair 2 + √3 and 2 − √3 satisfies all assumptions. Both are positive because √3 < 2; both are irrational because adding or subtracting an irrational number from a rational number cannot produce a rational result. Their product is 4 − 3 = 1, an integer, but their sum is 4, contradicting the conclusion.
The other candidate pairs—√2 with √8, √3 with √12, and √5 with √20—also have integer products, but their sums simplify to 3√2, 3√3 and 3√5, which remain irrational. They support rather than refute the conclusion. The conjugate pair is effective because multiplication cancels the radical through a difference of squares and addition cancels it through opposite signs. Understanding those cancellations is more reliable than guessing that a complicated-looking pair must be unusual enough.
The positivity check is especially easy to miss for the smaller conjugate, 2 − √3. Opposite signs in a written expression do not mean that its value is negative. Comparing √3 with 2 establishes the sign before the pair is used. In a multiple-assumption claim, keep a short checklist of the assumptions; an elegant cancellation is not enough if one member lies outside the permitted domain.
Compare both measurements in a geometric claim
For rectangles with positive integer sides, greater area does not necessarily imply greater perimeter. The 3-by-8 rectangle has area 24 and perimeter 22; the 2-by-10 rectangle has area 20 and perimeter 24. The first has greater area but smaller perimeter, so this pair disproves the claim. The other offered pairs have both greater first area and greater first perimeter and therefore are not counterexamples.
For transfer, translate a claim into “if assumptions, then conclusion.” Check all assumptions explicitly, including positivity, integer restrictions and strict inequalities. Then compute the quantity that decides the conclusion. In practice, identify why each rejected option fails as a counterexample: it may lie outside the claim’s scope or may satisfy the conclusion after all. This classification teaches the logical method and prevents an attractive numerical contradiction from being used where the original claim never applied.
Apply the method to a new case
A student claims: 'If a rectangle's perimeter is greater than another rectangle's perimeter, its area must also be greater.' Test that universal claim with two explicit rectangles. A 1-by-10 rectangle has perimeter 22 and area 10. A 4-by-4 rectangle has perimeter 16 and area 16. The first perimeter is larger but its area is smaller, so these two rectangles disprove the claim. Comparing a 5-by-5 square with a 4-by-4 square would support the claim for one pair without proving it universally. A useful counterexample must satisfy the premise and fail the conclusion at the same time; merely finding two different areas is insufficient. Check all four measurements in the same units, and keep the order of comparison consistent. The lesson is about which implication the data refute, not about declaring perimeter and area unrelated in every setting.