Questions: Proof by Contrapositive

Question 1

To prove 'If n² is odd, then n is odd,' a student assumes n is even and shows that n² must be even. Which proof technique is this?

• A. Direct proof — the student proved the statement from its hypothesis
• B. Proof by contrapositive — the student proved ¬Q → ¬P
• C. Proof by contradiction — the student derived a contradiction from P and ¬Q simultaneously
• D. Proof by exhaustion — the student checked all possible cases

Answer: (B)

The original statement is P → Q: 'If n² is odd (P), then n is odd (Q).' The contrapositive is ¬Q → ¬P: 'If n is even (¬Q), then n² is even (¬P).' The student assumed ¬Q and derived ¬P — the definition of proof by contrapositive. This is not contradiction: the student didn't assume P and ¬Q together and find an impossibility. The contrapositive flows cleanly here because 'n is even → n² is even' (n = 2k gives n² = 4k²) is immediate.

Question 2

In a proof by contrapositive of 'If P, then Q,' what do you assume and what must you derive?

• A. Assume P; derive Q
• B. Assume ¬P; derive ¬Q
• C. Assume ¬Q; derive ¬P
• D. Assume P and ¬Q; derive any contradiction

Answer: (C)

The contrapositive of 'P → Q' is '¬Q → ¬P': assume the negation of the conclusion (¬Q) and derive the negation of the hypothesis (¬P). Option A is a direct proof. Option B proves the inverse (¬P → ¬Q), which is NOT logically equivalent to P → Q. Option D is proof by contradiction — you assume P and ¬Q together and reach an impossibility, rather than proving a positive conclusion.

Question 3

A proof by contrapositive of 'If P then Q' constitutes a complete proof of 'If P then Q,' because the contrapositive and the original conditional are logically equivalent.

• T. True
• F. False

Answer: True

P → Q and its contrapositive ¬Q → ¬P have identical truth tables — they are false in exactly the same situation (when P is true and Q is false). Proving one is therefore the same as proving the other. This is not an indirect workaround; it is a direct consequence of propositional logic. After proving ¬Q → ¬P, you can immediately conclude P → Q.

Question 4

Proof by contrapositive and proof by contradiction are essentially the same technique, since both require negating the conclusion.

• T. True
• F. False

Answer: False

Both methods negate Q, but they differ in structure and what they produce. In a contrapositive proof, you assume ¬Q alone and derive ¬P — you reach a positive conclusion. In contradiction, you assume both P and ¬Q simultaneously and derive any falsehood — you reach an impossibility. Contrapositive is cleaner when ¬Q is productive on its own. Contradiction is needed when the only path forward requires combining all hypotheses. The final moves also differ: contrapositive ends with '¬Q → ¬P, therefore P → Q'; contradiction ends with 'this is impossible, so ¬Q was wrong.'

Question 5

When should you choose proof by contrapositive over a direct proof, and what feature of the statement signals the contrapositive will be the more natural approach?

Model answer: Choose the contrapositive when the negation of the conclusion (¬Q) is a more productive assumption than the hypothesis (P) itself. The signal is in the form of the conclusion: if Q is a positive statement ('x is rational,' 'n is even') whose negation opens concrete algebraic handles, the contrapositive is likely easier. For example, 'If n² is even, then n is even' — assuming 'n is odd' immediately gives n = 2k+1 and n² = 4k²+4k+1 = odd, deriving ¬P cleanly. Proving directly from 'n² is even' to 'n is even' requires a more roundabout argument.

The rule of thumb: if the conclusion has the form 'X has property Y,' its negation 'X lacks property Y' often provides a constructive assumption. The contrapositive is especially natural for proofs involving parity, rationality, divisibility, and set membership, where negations are concrete and algebraically tractable.