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Concept 02 / 04Mathematical Reasoning

Conditional and Biconditional Statements: Converse, Inverse and Contrapositive

The implication p -> q and its truth table, why a false hypothesis makes the implication true, the many English forms (if, only if, necessary, sufficient), the converse, inverse and contrapositive, the biconditional, and negating an implication.

Figure 1
Conditional (implication). p→qp \rightarrow q ('if pp then qq', 'pp implies qq') is false in exactly one case: pp true and qq false. In all other cases it is true, as the figure shows. pp is the hypothesis (antecedent) and qq the conclusion (consequent). Why is p→qp \rightarrow q true when pp is false? Think of it as a promise: 'If you score above 9090, you will get a prize.' The promise is broken only if you score above 9090 and get no prize. If you score below 9090, the promise is not broken, whatever happens. This 'vacuous truth' is also why 'ϕ⊆A\phi \subseteq A' is true: 'if x∈ϕx \in \phi then x∈Ax \in A' can never have a true hypothesis. Equivalent English forms of p→qp \rightarrow q: 'If pp, then qq'; 'qq if pp'; 'pp only if qq'; 'pp is sufficient for qq'; 'qq is necessary for pp'; 'qq whenever pp'. The phrase 'only if' is the trickiest: 'pp only if qq' means p→qp \rightarrow q (pp cannot happen without qq), not q→pq \rightarrow p. For example, 'A number is divisible by 44 only if it is even' means 'divisible by 44 →\rightarrow even'. Related implications. Converse: q→pq \rightarrow p. Inverse: ∼p→∼q\sim p \rightarrow \sim q. Contrapositive: ∼q→∼p\sim q \rightarrow \sim p. The contrapositive is logically equivalent to the original: both are false only when pp is true and qq false. The converse and inverse are equivalent to each other but not to the original. For 'If a quadrilateral is a square, then it is a rectangle' (true): the converse 'If it is a rectangle, it is a square' is false, while the contrapositive 'If it is not a rectangle, it is not a square' is true. Implication in terms of other connectives: p→q≡∼p∨qp \rightarrow q \equiv \sim p \vee q. Consequently, its negation is ∼(p→q)≡p∧∼q\sim(p \rightarrow q) \equiv p \wedge \sim q. The negation of an implication is not another implication: the negation of 'If it rains, the match is cancelled' is 'It rains and the match is not cancelled'. Biconditional. p↔qp \leftrightarrow q ('pp if and only if qq', 'pp is necessary and sufficient for qq', often written 'iff') is true when pp and qq have the same truth value. It is equivalent to (p→q)∧(q→p)(p \rightarrow q) \wedge (q \rightarrow p). Its negation is p↔∼qp \leftrightarrow \sim q, equivalently the exclusive or. Definitions in mathematics are biconditionals even when written with 'if'. Key Takeaways: - p→qp \rightarrow q is false only for pp true, qq false. - 'pp only if qq', 'pp sufficient for qq', 'qq necessary for pp' all mean p→qp \rightarrow q. - Contrapositive ∼q→∼p\sim q \rightarrow \sim p is equivalent to p→qp \rightarrow q; converse and inverse are not. - p→q≡∼p∨qp \rightarrow q \equiv \sim p \vee q; ∼(p→q)≡p∧∼q\sim(p \rightarrow q) \equiv p \wedge \sim q. - p↔q≡(p→q)∧(q→p)p \leftrightarrow q \equiv (p \rightarrow q) \wedge (q \rightarrow p): true when both have the same value. - ∼(p↔q)≡p↔∼q\sim(p \leftrightarrow q) \equiv p \leftrightarrow \sim q. - Pitfall: reading 'pp only if qq' as q→pq \rightarrow p. Brain-Teaser 1: Write the converse, inverse and contrapositive of 'If x2=4x^{2} = 4, then x=2x = 2', and decide which are true (for real xx). Solution: The statement is false (x=−2x = -2 is a counterexample). Converse: 'If x=2x = 2, then x2=4x^{2} = 4': true. Inverse: 'If x2≠4x^{2} \neq 4, then x≠2x \neq 2': true (it is the contrapositive of the converse). Contrapositive: 'If x≠2x \neq 2, then x2≠4x^{2} \neq 4': false, with the same counterexample x=−2x = -2. The pattern confirms that a statement and its contrapositive always share a truth value, as do the converse and inverse. Brain-Teaser 2: Negate 'If a triangle is equilateral, then it is isosceles'. Solution: ∼(p→q)≡p∧∼q\sim(p \rightarrow q) \equiv p \wedge \sim q: 'The triangle is equilateral and it is not isosceles.' The original is true (for every triangle), so this negation is false. Writing the negation as 'If a triangle is not equilateral, then it is not isosceles' is the common error; that is the inverse, a different statement altogether. Brain-Teaser 3: Express 'You can access the internet from campus only if you are a computer science major or you are not a freshman' symbolically, with aa: access, cc: CS major, ff: freshman. Solution: 'aa only if (cc or not ff)' is a→(c∨∼f)a \rightarrow (c \vee \sim f). Its contrapositive, (∼c∧f)→∼a(\sim c \wedge f) \rightarrow \sim a, reads 'If you are a freshman who is not a CS major, you cannot access the internet from campus', which makes the meaning clearer. Translating 'only if' correctly and checking with the contrapositive is a reliable habit.

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Conditional and Biconditional Statements: Converse, Inverse and Contrapositive — DPP

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