Contrapositive Fundamentals: Logic, Parity, and Common Pitfalls

Read the original implication carefully: "If ab is odd, then a and b are odd." The contrapositive of P → Q is ¬Q → ¬P. Here P is "ab is odd" and Q is "a and b are odd." The negation of "a and b are odd" is "at least one of a or b is even." The negation of "ab is odd" is "ab is even." By De Morgan's Law, ¬(a is odd ∧ b is odd) ≡ (a is even ∨ b is even). The contrapositive is therefore: “If a is even or b is even, then ab is even”. To check each choice, translate them into these forms. Choice A literally states that if a or b is even then ab is even, which matches the contrapositive. Choice D ("If a is odd, then ab is odd") is a true direct statement but it is not the contrapositive of the original implication. Therefore the correct statement to show by contrapositive is choice A, not D

The proof as written is a direct proof of "If n is even, then n² is even," because it begins by assuming n is even and deduces n² is even. Now note logical relationships: the contrapositive of "If n² is odd, then n is odd" is "If n is even, then n² is even." That is, A is the contrapositive of B and so a valid direct proof of A also establishes B via contraposition. Thus the provided derivation establishes both the direct statement A and, by logical equivalence, the contrapositive B, so choice C ("Both A and B") is correct.

To use contrapositive on "If x is irrational, then √x is irrational" we form ¬Q → ¬P where P is "x is irrational" and Q is "√x is irrational." For the negation,: ¬Q is "√x is rational" and ¬P is "x is rational." So the correct starting assumption for a contrapositive proof is "Assume √x is rational" and then deduce that x is rational. That shows ¬Q → ¬P, which is logically equivalent to the original. Therefore B is the correct contrapositive starting point.

The statement "If a is even or b is even, then ab is even" is logically equivalent to "If not (a and b are both odd), then not (ab is odd)" — that is, it is the contrapositive of "If ab is odd, then a and b are both odd." Alternatively, the direct statement "If ab is even, then a or b is even" is a closely related formulation but not the exact form shown: the given step states the implication in the forward direction from even factor to even product; taking negations shows that it supports the contrapositive in choice A. Thus A is the precise match: the given proof step is the contrapositive supporting the claim that if ab is odd then both a and b must be odd.

The statement is P → Q with P: n ≥ 10 and Q: n² ≥ 100. The contrapositive is ¬Q → ¬P. The negation of Q ("n² ≥ 100") is "n²

Let P be "x ≤ 5" and Q be "x² ≤ 25". The contrapositive is ¬Q → ¬P. ¬Q is "x² > 25" and ¬P is "x > 5". So to prove the contrapositive you assume "x² > 25" and show "x > 5." Choices C and D are stronger numeric versions but not the exact logical negation (D says x² ≥ 26, which is stronger than x² > 25 and may exclude real values between 25 and 26; C says x ≥ 6, again stronger than x > 5). The correct logical assumption used in the contrapositive is x² > 25, choice B. Options C and D are incorrect because they add unnecessary constraints (x ≥ 6, x² ≥ 26) that aren't logical negations.

The original statement "If prime p > 2, then p is odd" has a contrapositive formed by negating both the hypothesis and conclusion and reversing the implication. The hypothesis "p is prime and p > 2" negates to "p is not prime or p ≤ 2" (since the negation of a conjunction is a disjunction of negations). The conclusion "p is odd" negates to "p is even". Thus, the contrapositive of 'If p is prime ∧ p > 2, then p is odd' is 'If p is even, then ¬(p is prime ∧ p > 2)', which simplifies to 'If p is even, then p ≤ 2 or p is not prime' (De Morgan's + negation).

Contrapositive proof structure: to prove P → Q, you prove ¬Q → ¬P. That means you begin by assuming ¬Q (the negation of the conclusion) and then you must logically deduce ¬P (the negation of the hypothesis). Choice D describes proof by contradiction, which is a different technique; while some contrapositive proofs may use contradiction inside them, the direct next-step action after assuming ¬Q is to derive ¬P. Therefore C is the correct description of the contrapositive method's next step.

Start with the three standard related statements for a conditional P → Q: the converse is Q → P, the inverse is ¬P → ¬Q, and the contrapositive is ¬Q → ¬P. If someone assumes ¬P and deduces ¬Q, they have proven ¬P → ¬Q, which is exactly the inverse. The inverse is not logically equivalent to the original conditional P → Q, so this method is an error if the goal was to prove P → Q. Therefore the erroneous method proves the inverse (choice A).

"If Q, then P" is the converse of "If P, then Q." Proving the converse does not generally prove the original implication because P → Q and Q → P are independent statements (they are equivalent only when both are true, i.e., when the biconditional holds). The student's mistake is to assume that proving Q → P automatically gives P → Q. Thus the correct identification of the error is that they proved the converse (choice B).

The statement "a and b are both prime" is the conjunction "a is prime AND b is prime." The negation of a conjunction is the disjunction of the negations: ¬(P ∧ Q) ≡ (¬P) ∨ (¬Q). So the negation is "a is not prime OR b is not prime." Interpreting "not prime" in elementary integer contexts usually means "composite" (or possibly 1, which is neither prime nor composite), but the logical form is a disjunction. Choice A and B incorrectly convert the negation into a conjunction, and D ("Neither a nor b is prime") would mean both are not prime, which is stronger than the correct negation. Therefore C is the right logical negation.

Logical equivalence: P → Q is equivalent to ¬Q → ¬P (the contrapositive). If someone demonstrates ¬P → ¬Q, that is the contrapositive of Q → P (swap P and Q), i.e., ¬(Q) → ¬(P) is equivalent to P → Q? Careful reading: the statement given is "¬P → ¬Q", which is indeed the contrapositive of "Q → P". Thus concluding "Q → P" from "¬P → ¬Q" is valid because they are contrapositives of each other. Therefore the conclusion is correct: "¬P → ¬Q" is logically equivalent to "Q → P" (choice A).

The student proved ¬P → ¬Q. This is the Inverse. The contrapositive is ¬Q →¬P ("If n² is even, then n is even"). While the result is true because parity is biconditional, the method used was Proof by Inverse, not Proof by Contrapositive.

Negation of a conjunction follows De Morgan's law: ¬(P ∧ Q) ≡ ¬P ∨ ¬Q. Here P is "x > 5" and Q is "y > 5". The negation of "x > 5" is "x ≤ 5" and similarly for y. Therefore the negation becomes "x ≤ 5 OR y ≤ 5." Choice A is too strong (it asserts both ≤ 5), and B uses strict . So C is the correct logical negation.

This is the textbook contrapositive equivalence fact: P → Q is logically equivalent to ¬Q → ¬P. So if a student successfully proves ¬Q → ¬P, they have indeed established P → Q. Therefore the student's reasoning is correct, matching choice A.