Disproving Universal Statements

From a Universal Claim to a Disproof

A universal statement makes a demand at every input in its stated domain. To disprove it, we do not need to understand every input; we need to identify one permitted input at which the demand fails. The previous tutorial, “Constructing Counterexamples,” explained how to verify such an input. Here we focus on organizing the disproof: how the domain can narrow the search, how a necessary consequence can expose a failure, and how to present the reasoning so the claim’s precise failure is clear.

Begin by separating the statement into its domain and its property. For example, “\(x^4\geq x^2\) for every real \(x\)” has domain \(\mathbb{R}\) and property \(x^4\geq x^2\). A disproof must use a real number and show that this inequality does not hold there. An input outside the domain, or an input at which the inequality happens to hold, cannot do the job.

Theorem (Negation of Quantified Statements), established earlier in this course, gives the logical form behind a disproof: denying a universal statement means finding an allowed input at which its property is false. The Counterexample Criterion for a Conditional Claim adds the corresponding requirement for implications: the hypothesis must hold and the conclusion must fail. These results tell us what a disproof must establish; the choice of input and the verification still depend on the particular claim.

Use a Smaller Domain to Find the Failure

A universal statement over a large domain also makes a claim on every subset of that domain. This observation lets us focus a search. If the statement concerns all real numbers, for instance, we may look only at a convenient interval. A failure within that interval is automatically a failure among the real numbers. There is no need to test the rest of the domain once one such failure is verified.

Proof. Suppose \(P(x)\) holds for every \(x\in A\). Take any \(x\in B\). Since \(B\subseteq A\), this \(x\) also belongs to \(A\). The assumption therefore gives \(P(x)\). As \(x\) was arbitrary in \(B\), \(P\) holds for every element of \(B\).

Equivalently, if \(P\) fails at some \(b\in B\), it cannot be true that \(P\) holds for every \(x\in A\): that universal claim would imply \(P(b)\), since \(b\in B\subseteq A\). This proves both assertions. \(\square\)

A useful search often begins with a subset chosen for a reason. For a claim about real numbers, the values \(0\), \(1\), and \(-1\) can simplify expressions; intervals such as \((0,1)\) can reveal how powers compare; and a claim about integers may be easier to test on even or odd integers. Choosing a smaller domain is a way to make the search manageable, not a substitute for checking that the chosen input is allowed and actually makes the claim false.

Look for a Necessary Consequence

Sometimes the stated property is complicated, but anything satisfying it would also have to satisfy a simpler condition. Such a condition is called a necessary consequence: if the original property holds at an input, then the consequence must hold there too. If we find an allowed input where that consequence fails, the original property cannot hold at that input.

Proof. Let \(a\in A\) satisfy the stated condition that \(R(a)\) is false. Suppose, for contradiction, that \(P(x)\) holds for every \(x\in A\). Then \(P(a)\) holds because \(a\in A\). By the assumed implication from \(P(x)\) to \(R(x)\), it follows that \(R(a)\) holds. This contradicts the choice of \(a\), for which \(R(a)\) is false. Hence \(P\) cannot hold for every element of \(A\). \(\square\)

When using this method, the implication from the original property to the proposed consequence must itself be justified. It is not enough that the consequence seems related to the claim. A valid disproof needs both links: truth of the original property would force the consequence, and the chosen input makes that consequence false.

Verify Every Condition in a Conditional Claim

A universal conditional statement has an additional feature: it makes a demand only at inputs satisfying its hypothesis. To disprove such a claim, first check that the chosen input makes the hypothesis true. Then check that it makes the conclusion false. An input where the hypothesis fails does not contradict the conditional.

The same care applies when a statement includes more than one condition. If it claims that every object satisfying \(P(x)\) also satisfies \(Q(x)\), an input for which \(P(x)\) is false is irrelevant, even if \(Q(x)\) is false there too. The Counterexample Criterion for a Conditional Claim records this distinction. Use it as a checklist, rather than treating every failure of the conclusion as a disproof.

From Testing to a Complete Disproof

Testing examples can suggest where a universal claim fails, but the final argument must verify a specific failure. A small table of values, a graph, or a pattern in early cases may help locate a candidate; none alone shows that the claim is false unless it exhibits an allowed input and checks the property there. Conversely, once one such input is verified, there is no need to check more cases.

A common pitfall is to stop at a promising candidate. For instance, a number that makes an expression smaller does not disprove a claimed inequality unless the calculation shows that the inequality is reversed or otherwise false. Another is to overlook the domain: a non-integer cannot refute a statement restricted to integers. For conditional statements, a third pitfall is to verify only that the conclusion fails while leaving the hypothesis unchecked.

A disproof is often short, but it should be complete. State the chosen input, establish that it is allowed, and show the exact calculation or logical consequence that contradicts the claim. The Restriction to a Subset theorem helps narrow where to look; the Necessary-Consequences Test helps simplify what to check. Both techniques serve the same purpose: turning a broad universal assertion into a specific, verifiable failure.