\problemdata
{S00 ps1}
{circular reasoning}
{false proof}

\begin{problem} Consider the following theorem: Suppose $x$ is a real number
and $\frac{2x-5}{x-4}=3$ then $x=7$.
\begin{problemparts}
\problempart What is wrong with the following proof? Suppose $x=7$.  Then
\[
\frac{2x-5}{x-4}=\frac{2(7)-5}{7-4}=\frac{9}{3}=3.
\]
Thus, if $\frac{2x-5}{x-4}=3$ then $x=7$.

\solution{ This proof is a typical example of circular
reasoning.  We suppose $x=7$ and then conclude that $x=7$.  So, all we
have done is not contradict our supposition.  Unlike a proof by
contradiction, which shows our assumption is false, {\em not} arriving
a contradiction does {\em not} show our assumption is true.

In particular, the proof is of the form:
\begin{center}
\begin{tabular}{l}
        $P(x)$ \\
        $P(x) \implies Q(x)$ \\ \hline
        $\therefore Q(x) \implies P(x)$
\end{tabular}
\quad where 
\begin{tabular}{l}
        $P(x) \;\; \equiv\;\; x=7$ \\
        $Q(x) \;\; \equiv\;\; \frac{2x-5}{x-4}=3$ \\
\end{tabular}
\end{center}
The given ``proof'' does demonstrate that the second premise is true.  
However, the conclusion does not follow.
}

\problempart Give a correct proof of this theorem.

\solution{
There are many ways to properly construct this proof.  We demonstrate a direct
proof using \emph{modus ponens}.

\begin{center}
\begin{tabular}{l}
        $Q(x)$ \\
        $Q(x) \implies P(x)$ \\ \hline
        $\therefore P(x)$
\end{tabular}
\quad where 
\begin{tabular}{l}
        $P(x) \;\; \equiv \;\; x=7$ \\
        $Q(x) \;\; \equiv \;\; \frac{2x-5}{x-4}=3$ \\
\end{tabular}
\end{center}
Proposition $Q(x)$ is given.  To complete the proof, we must now show that the
second premise (the implication) is a tautology.  
We do this using basic algebra and arithmetic:
\begin{align*}
Q(x) \equiv \frac{2x-5}{x-4} = 3 \;\;
        & \:\Longrightarrow \;\; 2x-5 = 3(x-4) \\ 
        & \Longleftrightarrow \;\; 2x-5 = 3x-12 \\
        & \Longleftrightarrow \;\; -5+12 = 3x-2x \\
        & \Longleftrightarrow \;\; 7 = x                &\equiv P(x) 
\end{align*}
}
\end{problemparts}
\end{problem}
