Wrong Implications - Forms of Argument

Mathematical Writing - Vivaldi Franco 2014

Wrong Implications
Forms of Argument

Inappropriate handling of implications results in common mistakes. Instead of proving an implication, we may end up proving its converse; or we may assume the statement we are meant to prove, deduce from it a true statement, and believe we’ve completed the proof. These faulty deductions—of which we now show an example—are sometimes called circular arguments, or non sequiturs.6

Prove that $$\sqrt{2}+\sqrt{6}<\sqrt{15}$$

WRONG PROOF.

$$\begin{aligned} \sqrt{2}+\sqrt{6}<\sqrt{15}&\Rightarrow (\sqrt{2}+\sqrt{6})^2<{15}\\&\Rightarrow 8+2\sqrt{12}<15\\&\Rightarrow 2\sqrt{12}<7\\&\Rightarrow 48<49. \qquad \square \end{aligned}$$

We were supposed to prove $$P$$, where $$P=(\sqrt{2}+\sqrt{6}<\sqrt{15})$$. Instead we have assumed $$P$$, and correctly deduced from it the true statement $$48<49$$. However, the deduction $$P\Rightarrow $$ TRUE (unlike the deduction $$P\Rightarrow \textsc {False}$$) gives us no information about $$P$$, from Table (4.9). Indeed, had we started from the false statement $$\sqrt{2}+\sqrt{6}<-\sqrt{15}$$, we would have reached exactly the same conclusion.

There are two methods for fixing this problem.

First method: retracing the steps. We regard the chain of deductions displayed above as ’rough work’; then we start from the end and prove the chain of converse implications.

PROOF.

$$\begin{aligned} 48<49&\Rightarrow \sqrt{48}<\sqrt{49}\\&\Rightarrow 2\sqrt{12}<7\\&\Rightarrow 8+2\sqrt{12}<15\\&\Rightarrow (\sqrt{2}+\sqrt{6})^2<{15}\\&\Rightarrow \sqrt{2}+\sqrt{6}<\sqrt{15} \end{aligned}$$

where in the first and the last implications we have taken the positive square root of each side. We have proved the implication $$\textsc {True}\Rightarrow P$$, from which we deduce that $$P$$ is TRUE. $$\square $$

Clearly, we could not have come up with such a proof without having done the ’rough work’ first; the next method does not require this.

Second method: contradiction.

PROOF. Let us assume $$\lnot P$$:

$$\begin{aligned} \sqrt{2}+\sqrt{6}\geqslant \sqrt{15}&\Rightarrow (\sqrt{2}+\sqrt{6})^2\geqslant {15}\\&\Rightarrow 8+2\sqrt{12}\geqslant 15\\&\Rightarrow 2\sqrt{12}\geqslant 7\\&\Rightarrow 48\geqslant 49. \end{aligned}$$

We have proved that $$\lnot P\Rightarrow \textsc {False}$$, which implies that $$\lnot P$$ is $$\textsc {False}$$, that is, $$P$$ is TRUE. $$\square $$