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.
Do the same for the statement “If \(x > 0\), then \(x^2 - x > 0\) .”
Solution
The original statement is \[ x<0\Longrightarrow x^2-x>0. \] Its contrapositive is \[ x^2-x\le0\Longrightarrow x\ge0, \] and its converse is \[ x^2-x>0\Longrightarrow x<0. \] The original statement is true because if \(x<0\), then both \(x\) and \(x-1\) are negative, so \[ x^2-x=x(x-1)>0. \] Its contrapositive is therefore also true. The converse is false: \(x=2\) satisfies \(x^2-x>0\) but not \(x<0\).
The original statement is \[ x>0\Longrightarrow x^2-x>0. \] Its contrapositive is \[ x^2-x\le0\Longrightarrow x\le0, \] and its converse is \[ x^2-x>0\Longrightarrow x>0. \] All three are false. Taking \(x=\tfrac12\) disproves both the original statement and its contrapositive since \[ x^2-x=-\tfrac14\le0 \] while \(x>0\). Taking \(x=-1\) disproves the converse since \(x^2-x=2>0\) but \(x\not>0\).