Worked examples · Sheet 3

Contrapositives and converses of implications

Theory used: Implication Converse and contrapositive Negation of an implication

Exercise 1

Given, for real $x$ , the implication “if $x > 2$ then $x^{2} > 4$ ”, write the contrapositive and the converse and determine the truth value of each.

Contrapositive.

x^{2} \leq 4 ⟹ x \leq 2 true

Converse.

x^{2} > 4 ⟹ x > 2 false: for x = - 3 one has 9 > 4 and - 3 \leq 2

The contrapositive swaps hypothesis and conclusion and negates both: the negation of $x^{2} > 4$ is $x^{2} \leq 4$ and the negation of $x > 2$ is $x \leq 2$ . The original implication is true, because from $x > 2$ , multiplying the inequality by $x > 0$ , it follows that $x^{2} > 2 x > 4$ ; the contrapositive is equivalent to the original implication, therefore it is true as well.

The converse swaps hypothesis and conclusion without negating them. To determine its truth value one looks for a value whose square is greater than $4$ and whose base is less than or equal to $2$ : the value $x = - 3$ satisfies $x^{2} = 9 > 4$ and negates $x > 2$ , therefore the converse is false. The example shows that the truth of an implication leaves the truth of its converse undetermined.

Exercise 2

Given, for natural $n$ , the implication “if $n$ is a multiple of $10$ , then $n$ is a multiple of $5$ ”, write the contrapositive, the converse and the negation, and determine the truth value of each.

Contrapositive. “If $n$ is a natural number that has nonzero remainder in the division by $5$ , then $n$ has nonzero remainder in the division by $10$ ”: true.

Converse. “If $n$ is a multiple of $5$ , then $n$ is a multiple of $10$ ”: false, for $n = 15$ .

Negation.

\exists n \in N : n is a multiple of 10 \land n has nonzero remainder in the division by 5

false.

The original implication is true: if $n$ is a multiple of $10$ , then $n = 10 k$ with $k$ natural, therefore $n = 5 \cdot 2 k$ is a multiple of $5$ . The contrapositive negates the conclusion and the hypothesis and swaps them; since it is equivalent to the original implication, it is true.

The converse states that every multiple of $5$ is a multiple of $10$ . The number $15 = 5 \cdot 3$ is a multiple of $5$ and the division $15 : 10$ has remainder $5$ , therefore $n = 15$ is a counterexample and the converse is false.

The negation of an implication is a conjunction: hypothesis true and conclusion false. Since the implication holds for every $n$ , its negation requires the existence of a multiple of $10$ with nonzero remainder in the division by $5$ . The original implication is true for every $n$ , hence a number with these properties does not exist and the negation is false.

Exercise 3

Write the negation of

\forall x \in R : x > 1 ⟹ x^{2} > x

and determine which of the proposition and its negation is true.

\neg (\forall x \in R : x > 1 \Rightarrow x^{2} > x) \equiv \exists x \in R : x > 1 \land x^{2} \leq x

x > 1 ⟹ x \cdot x > 1 \cdot x ⟹ x^{2} > x

The original proposition is true.

The negation is built in two steps. The universal quantifier becomes existential; the predicate is an implication, and the negation of an implication is the conjunction of the hypothesis with the negation of the conclusion, therefore $x > 1 \land x^{2} \leq x$ .

For the truth value one proves the original implication. From $x > 1$ it follows in particular that $x > 0$ ; multiplying both sides of the inequality $x > 1$ by the positive number $x$ , the direction is preserved and one obtains $x^{2} > x$ . The implication holds for every real $x$ , therefore the original proposition is true and its negation is false.