Counterexamples in Sets, Conditionals, and Everyday Statements

A proper subset of a set is any subset that contains fewer elements than the original set. The universal statement claims that every subset is proper, which would mean no subset could ever equal the original set. But every set is a subset of itself by definition of “subset.” When a subset is exactly equal to the original set, it is not proper. Therefore, the set itself provides a direct contradiction because it is a subset that fails the stated condition of being proper, proving the universal claim false.

The statement asserts that divisibility by 2 implies divisibility by 4. To disprove it, we need a number that is divisible by 2 but not divisible by 4. The number 6 is divisible by 2 since 6 ÷ 2 = 3, but 6 ÷ 4 does not yield an integer. Because the hypothesis is satisfied but the conclusion fails, 6 provides the exact structure needed to refute a conditional statement, making it the correct counterexample.

A counterexample is simply a single instance that contradicts a universal statement. There is no requirement that only one such instance exists. In fact, many universal statements are contradicted by a large collection of counterexamples. Uniqueness is not a condition; any one example is sufficient to show the claim is false, but there can be infinitely many such examples. Thus, counterexamples do not need to be unique.

To disprove the universal claim that every limit exists, we need an example of a limit that fails to exist. The expression lim x→0 1/x does not approach a single finite value; it diverges to positive or negative infinity depending on the direction of approach. Because this limit fails to exist in the usual real-number sense, it contradicts the universal statement and serves as an appropriate counterexample.

The statement claims that every integer is even, meaning each integer must be divisible by 2. The number 3 is an integer, but it is not divisible by 2, because 3 ÷ 2 does not yield an integer. Since an integer exists that does not satisfy the property of being even, the universal statement is proven false. Therefore, 3 is a valid counterexample.

A counterexample directly contradicts a universal statement by showing a specific failing case. Indirect proofs, such as proofs by contradiction, involve assuming the opposite of what you want to prove and deriving a contradiction. Counterexamples do not follow that structure; they simply provide a direct disproof. Thus, counterexamples are not considered indirect proofs and instead provide a straightforward falsification.

A convergent sequence approaches a single finite limit. The universal statement claims that all sequences converge, so a counterexample must be a sequence that does not converge. An oscillating sequence, such as (−1)ⁿ, continually switches between values and fails to approach any single limit. Because such a sequence does not converge, it contradicts the universal statement and provides the needed counterexample.

The statement asserts that every cat is black. To contradict it, we need an example of an animal that is both a cat and not black. A white cat satisfies the definition of a cat while lacking the stated property of being black. Therefore, a white cat provides a direct contradiction to the universal statement and serves as a valid counterexample.

In a proof by contradiction, one begins by assuming that a universal statement is true. To derive a contradiction, it is often effective to exhibit a specific case that violates the statement. This violating case functions as a counterexample within the framework of the contradiction argument. Therefore, counterexamples can indeed be used as part of proofs by contradiction when demonstrating that the assumption leads to inconsistency.

Prime numbers must have exactly two positive divisors, while composite numbers must have more than two positive divisors. The number 1 has exactly one positive divisor: itself. It meets neither definition, so it is neither prime nor composite. Because the statement claims that every number must be in one of these two categories, the existence of 1 provides a counterexample and disproves the universal claim.

The statement asserts that the sum of any two irrational numbers must always be irrational. However, if we add √2 and −√2, both of which are irrational, their sum is 0, which is rational. This example contradicts the universal claim by providing a pair of irrational numbers whose sum is not irrational. Therefore, √2 + (−√2) is a valid counterexample.

A rectangle is defined as a quadrilateral with four right angles. A square is a special type of rectangle with the additional requirement that all four sides are equal. To disprove the conditional statement, we need a rectangle that is not a square. A rectangle with width 2 and length 4 satisfies the definition of a rectangle but lacks equal side lengths, contradicting the conclusion. Hence, it is a valid counterexample.

The statement claims that if n is prime, then n + 1 cannot be prime. If we choose n = 2, then n is prime, and n + 1 = 3, which is also prime. This means the hypothesis is true while the conclusion is false, which is precisely the structure of a counterexample to an implication. Therefore, n = 2 disproves the statement.

The claim asserts that multiplying a number by another number always results in something larger. However, multiplying by 0.5 cuts the number in half. For example, 10 multiplied by 0.5 gives 5, which is smaller than 10. Because multiplication by 0.5 produces a number that is not larger than the original, it contradicts the universal statement and serves as a valid counterexample.

Both 3 and 5 are odd integers. Their sum is 8, which is an even number, not an odd one. Since the hypothesis is satisfied while the conclusion fails, this pair provides the necessary contradiction to disprove the conditional statement. Therefore, x = 3 and y = 5 constitute a correct counterexample.