The Contradiction Strategy

Assume the Opposite—and Make the Conflict Precise

In “The Contrapositive Strategy,” an implication was proved by establishing its logically equivalent contrapositive. Another useful route starts by assuming that the desired conclusion is false and then derives a contradiction. This is the contradiction strategy, also called proof by contradiction or reductio ad absurdum.

The strategy is especially useful when the negation of the claim gives a concrete assumption, but a direct route from the hypotheses to the conclusion is difficult to see. The proof must do more than produce an unexpected calculation: it must identify a statement that cannot be true, such as an inequality contradicting a hypothesis or an object violating a defining property.

For example, if the assumptions imply both \(r>0\) and \(r\leq0\), they cannot all hold. The contradiction shows that the assumptions made for the proof cannot hold together. In an implication proof, the hypothesis \(P\) is retained, while the conclusion \(Q\) is temporarily replaced by its negation. Once \(P\) and \(\neg Q\) have been shown incompatible, the implication follows.

Proof. Suppose \(P\Rightarrow Q\) is true. If \(P\) and \(\neg Q\) both held, the implication would give \(Q\), while the second assumption gives \(\neg Q\). This is a contradiction. Thus \(P\) and \(\neg Q\) cannot both hold.

Conversely, suppose it is impossible for \(P\) and \(\neg Q\) to hold together. Assume \(P\). If \(Q\) were false, then \(\neg Q\) would hold, contradicting the supposition. Therefore \(Q\) holds whenever \(P\) does, which proves \(P\Rightarrow Q\). \(\square\)

This theorem describes the logical structure, but it does not by itself tell us how to find a contradiction. That depends on the subject matter. In real analysis, a contradiction may arise from incompatible inequalities, from the definition of a limit, or from the properties of real numbers or integers. The useful question is: what specific consequence of the temporary assumption can be shown to violate an established fact?

Choosing and Negating the Assumption

For a statement \(P\Rightarrow Q\), the contradiction setup is to assume \(P\) and \(\neg Q\). It is not to assume \(\neg P\) and \(Q\); those assumptions belong to neither the contradiction form nor the contrapositive. In a contrapositive proof, one assumes \(\neg Q\) and proves \(\neg P\). In a contradiction proof of the implication, one assumes both \(P\) and \(\neg Q\), and shows that this combination is impossible. These approaches are logically related, but the proof’s assumptions and target should be stated accurately.

Quantifiers require particular care. To disprove a statement that every object has a property, it is enough to assume that there is an object without that property. For example, the negation of “every \(x\in A\) satisfies \(R(x)\)” is “there exists \(x\in A\) such that \(R(x)\) fails.” More complicated statements must be negated one quantifier at a time, reversing “for every” and “there exists” and negating the condition that follows.

Worked Examples: Producing a Contradiction

This proof illustrates why it helps to identify a contradiction in advance. The assumption of rationality provides a fraction in lowest terms. The equation then forces its numerator and denominator both to be even, which conflicts directly with that defining choice. The proof does not merely show that one particular representation is inconvenient; it rules out the existence of a reduced representation altogether.

Contradiction and the Order of Limits

Contradiction is also effective when a claim concerns the limits of sequences. A strict inequality between two limits can be used to build a fixed positive gap. Convergence then forces sufficiently late terms to preserve enough of that gap to contradict an assumed ordering of the terms.

Proof. Suppose, for a contradiction, that \(a>b\). Let \(\eta=(a-b)/3\), so \(\eta>0\). Since \(a_n\to a\), there is an integer \(N_1\) such that for \(n\geq N_1\), \(|a_n-a|<\eta\). In particular, \(a_n>a-\eta\). Since \(b_n\to b\), there is an integer \(N_2\) such that for \(n\geq N_2\), \(|b_n-b|<\eta\), and hence \(b_n<b+\eta\).

For every \(n\geq\max\{N_1,N_2\}\), these inequalities imply

Therefore \(a_n>b_n\) for all such \(n\), contradicting the hypothesis \(a_n\leq b_n\) for every \(n\). Our temporary assumption \(a>b\) must be false. Since real numbers are ordered, it follows that \(a\leq b\). \(\square\)

What Makes a Contradiction Proof Convincing?

A complete contradiction proof has a clear chain: state the claim, assume its negation, derive consequences from that assumption, identify an incompatibility, and conclude that the assumption was false. The last two steps matter. If the proof presents a surprising consequence but never explains why it is impossible, the contradiction has not been established.

The incompatibility should also be precise. In the irrationality example, “both integers are even” is not a contradiction by itself; the conflict is with the assumption that the fraction is in lowest terms. In the limit theorem, a strict inequality is not inherently impossible; it conflicts with the hypothesis that \(a_n\leq b_n\) for every index. Naming the conflicting facts makes the logic checkable.

A common pitfall is to confuse proof by contradiction with proof by contrapositive. For an implication \(P\Rightarrow Q\), a contrapositive proof assumes \(\neg Q\) and aims to prove \(\neg P\). A contradiction proof assumes \(P\) and \(\neg Q\) together, then derives an impossibility. Either route can be useful, but choosing a route does not excuse an incorrect negation or an unstated hypothesis.

Contradiction is a strategy, not a substitute for definitions or estimates. In the sequence theorem, convergence supplies the specific estimates \(|a_n-a|<\eta\) and \(|b_n-b|<\eta\); without choosing \(\eta\) in relation to the proposed gap, the conflict would not follow. When using this strategy, ask what the negated claim actually gives and what quantitative or logical fact would make that consequence impossible.