Exercise 1.3

Question

\begin{enumerate}[label=(\alph*)]
\item Write the contrapositive and converse of the following statement: ``If $x < 0$, then $x^2 - x > 0$,'' and determine which (if any) of the three statements are true.
\item Do the same for the statement ``If $x > 0$, then $x^2 - x > 0$.''
\end{enumerate}

Dan Whitman · Accepted Answer

(a) First we claim that the original statement is true.

Proof. Since $x < 0$ we clearly have that $x - 1 < x < 0$ as well. Then, since the product of two negative numbers is positive, we have that $x^{2} - x = x (x - 1) > 0$ as desired. □

The contrapositive of this is, “If $x^{2} - x \leq 0$ , then $x \geq 0$ .” This is of course also true by virtue of the fact that the contrapositive is logically equivalent to the original implication.

Lastly, the converse of this statement is, “If $x^{2} - x > 0$ , then $x < 0$ .” We claim that this is not generally true.

Proof. A simple counterexample of $x = 2$ shows this. We have $x^{2} - x = 2^{2} - 2 = 4 - 2 = 2 > 0$ , but also clearly $x = 2 > 0$ as well so that $x < 0$ is clearly false. □

(b) First we claim that this statement is false.

Proof. As a counterexample, let $x = 1 ∕ 2$ . Then clearly $x > 0$ , but we also have $x^{2} - x = {(1 ∕ 2)}^{2} - 1 ∕ 2 = 1 ∕ 4 - 1 ∕ 2 = - 1 ∕ 4 < 0$ so that $x^{2} - x > 0$ is obviously not true. □

The contrapositive is then “If $x^{2} - x \leq 0$ , then $x \leq 0$ ,” which is false since it is logically equivalent to the original statement.

The converse is “If $x^{2} - x > 0$ , then $x > 0$ ,” which we claim is false.

Proof. As a counterexample, consider $x = - 1$ so that $x^{2} - x = {(- 1)}^{2} - (- 1) = 1 + 1 = 2 > 0$ . However, we also clearly have $x = - 1 < 0$ so that $x > 0$ is not true. □

Exercise 1.3

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Exercise 1.3

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