Contrapositive Proof Quiz: Rewrite and Prove Logical Implications

A contrapositive reverses the implication and negates both parts: (x∈A∩B)→(x∈A) becomes (x∉A)→(x∉A∩B). The idea is that if the conclusion fails, then the hypothesis must also fail. If x is not in A, it cannot possibly belong to A∩B, since membership in the intersection requires membership in both sets. This correctly mirrors the original implication’s logic.

The contrapositive starts by negating the conclusion: f(x)>0 fails when some x satisfies f(x)≤0. To prove an implication about positivity, it is often easier to show that if positivity does not hold, then the original hypothesis cannot persist. The inequality f(x)≤0 captures exactly the failure of f(x)>0.

Negate the conclusion (¬∃xQ(x) becomes ∀x¬Q(x)) and the hypothesis (¬∀xP(x) becomes ∃x¬P(x)). Switching quantifiers is essential in contraposition involving quantified statements. Universal becomes existential and vice versa, while the internal predicate also gets negated to reflect the logical reversal.

Negate the conclusion: assume a does not divide c, then show the hypothesis cannot hold. If the conclusion “a divides c” turns out false, contraposition requires proving that the hypothesis “a divides b and b divides c” cannot both be true. This approach uses the definition of divisibility to reach a contradiction.

Negating n≥4 gives n2ⁿ gives n!≤2ⁿ. Saying n≤3 is an equivalent rephrasing. Contrapositive proofs rely on correct negation of inequalities. Recognizing that values less than 4 are exactly 0,1,2,3 simplifies working with specific numerical cases rather than general bounds.

Start by negating the conclusion: A⊄C. Then show the hypothesis fails. If there exists an element in A not contained in C, then the implication “A⊆C” is false. A contrapositive argument would show this situation contradicts the original assumption, revealing that the implication must hold.

Negating ∀ε∃δP yields ∃ε∀δ¬P using quantifier‑switching rules. This is a standard manipulation in analysis: the negation of a universal-existential statement flips both quantifiers while also negating the predicate. This structure appears frequently in the definition of discontinuity.

Showing that discontinuity forces non‑differentiability avoids epsilon‑delta complications. The contrapositive approach turns a difficult forward implication about differentiability into a simpler statement: failure of continuity eliminates any possibility of differentiability, which logically must imply continuity.

Negate the conclusion (sum limit differs) and conclude that one hypothesis limit must fail. If the limit of a sum does not equal the sum of the limits, then at least one of the original limiting behaviors must break down. This reversal captures the dependency of the conclusion on the correctness of both hypotheses.

P→Q has the same truth value as ¬Q→¬P, so proving either one proves both. Because the contrapositive retains identical truth conditions, we can freely choose whichever direction is easier to demonstrate. Many proofs use this strategy to circumvent difficult forward arguments.

Negation of 'x=0 and y=0' is 'x≠0 or y≠0', by De Morgan’s law. Negating a conjunction produces a disjunction, reflecting that failure of the conclusion occurs whenever at least one component is false, not necessarily both.

Negate the conclusion: start by assuming √x is rational, then derive that x must be rational. To show the forward implication, we instead assume the opposite of the conclusion and work backwards. If √x were rational, squaring it would force x itself to be rational, contradicting the hypothesis.

If the structure of P→Q is already straightforward, direct proof is easier. Sometimes the forward direction naturally follows from definitions or standard algebraic manipulations. In such cases, contraposition may add unnecessary steps.

To use contrapositive, you assume ¬Q (unbounded), not the hypothesis P (convergent). For sequences, it is often easier to show that if a sequence is unbounded, it cannot be convergent. This is a classic example of a contrapositive being simpler than proving the forward implication.

Contrapositive proof always begins by assuming the negation of the conclusion. The entire method relies on replacing “If P then Q” with “If not Q then not P,” shifting the burden to analyzing failure of the conclusion rather than the original hypothesis.

Negate continuity and differentiability, then reverse their order. Statements about differentiability imply continuity. The contrapositive approach shows that if continuity fails, differentiability must also fail, applying negation before reversing logical direction.

Negating 'n²≥100' yields 'n²

You must show that the hypothesis cannot hold when the conclusion is false. The entire idea is to assume the conclusion fails and deduce that the hypothesis is incompatible with this assumption. Reaching this impossibility confirms the original implication.

Negate 'a and b are odd' to 'a or b is even', then show the product can't remain odd. Using De Morgan’s law, the negation replaces a conjunction with a disjunction. From there, reasoning about parity shows that the product would lose its oddness, leading to a contradiction.

Negate a conjunction via De Morgan: ¬(x>5 ∧ y>5) becomes (x≤5 ∨ y≤5). To form the contrapositive, the conclusion is negated. Turning an “and” into an “or” clarifies the alternative cases under which the original implication’s conclusion fails.