\problemdata
{S00 tut1}
{intro}
{false proof}

\begin{problem} Identify \emph{precisely} what the bugs are in each of
 the following false proofs.

\begin{problemparts}
\problempart\label{log10}
\begin{align*}
    3 &> 2 \\
    3 \log_{10} (1/2) &> 2 \log_{10}(1/2) \\
    \log_{10} (1/2)^3 &> \log_{10} (1/2)^2 \\
    (1/2)^3 &> (1/2)^2 \\
    \mbox{Therefore, } 1/8 &> 1/4!
\end{align*}

\solution{$\log x < 0$, for $0<x<1$, so since both sides of the inequality
``$3 > 2$'' are being multiplied by the negative quantity
$\log_{10}(1/2)$, the inequality in the second line should have been the
reverse of the first line, that is, $<$ instead of $>$.}


\problempart\label{cents}
You are richer than you think:
\[1 \cent = \$0.01 = (\$0.1)^2 = (10\cent)^2 = \$1.\]

\solution{$\$0.01 = \$(0.1)^2 \neq (\$0.1)^2$ because the units $\$^2$ and
$\$$ don't match (just as in physics the difference between $sec^2$ and
$sec$ indicates the difference between acceleration and velocity).}

\problempart\label{millions} Consider the following scheme to make millions.
Obtain a slab of gold $10'' \times 11'' \times 1''$.  Then cut out two
little one-inch triangles at the corners.  These triangles are the
profit.  Finally, slide the two slabs together again and note that the
rectangle is now $11'' \times 10'' \times 1''$---the same size as it
was before we removed the small triangles.  By repeating the process,
we can produce an unlimited supply of gold!


\solution{The picture makes the little triangles look like right
  triangles, but they aren't.  So when the two large pieces left after
  the little triangles are removed are slid together, they do not form
  a rectangle, which means the resulting shape is not the same as the
  original figure. (This illustrates why ``proof by diagram'' is
  mathematically suspicious.)  }


\newpage 
\problempart
\textbf{Claim}: Every polynomial $p(x) = ax^2 + bx +c$ over $\complex$
(the complex numbers) has two roots.

Sounds good, but the claim is false.  Find the counterexamples and then
restate the claim so it is correct.


\begin{proof}
Using the usual formulas for the roots of a quadratic, let
\begin{eqnarray*}
r_1 & ::= & \frac{-b + \sqrt{D}}{2a}, \mbox{ and} \\
r_2 & ::= & \frac{-b - \sqrt{D}}{2a}, \mbox{ where}\\
D & ::= & b^2-4ac.
\end{eqnarray*}

To verify that $r_1$ is a root, note that
\begin{eqnarray*}
a{r_1}^2 & = &  a({\frac{-b +\sqrt{d}}{2a}})^2\\
& = & a\frac{b^2 -2b\sqrt{D} + D}{4a^2}\\
& = & \frac{b^2 -2b\sqrt{D} + b^2 -4ac}{4a}\\
& = & \frac{2b^2 -2b\sqrt{D} -4ac}{4a}\\
& = & \frac{b^2 -b\sqrt{D} -2ac}{2a}
\end{eqnarray*}
and
\begin{eqnarray*}
br_1 & = & b\frac{-b+\sqrt{D}}{2a}\\
 & = & \frac{-b^2+b\sqrt{D}}{2a}
\end{eqnarray*}
so
\begin{eqnarray*}
p(r_1) &=&{r_1}^2 + br_1 + c\\
& = & \frac{b^2 -b\sqrt{d} -2ac}{2a} + \frac{-b^2+b\sqrt{d}}{2a} +c\\
 & = & \frac{-2ac}{2a} + c = - c + c = 0,
\end{eqnarray*}
as claimed.  A similar calculation confirms that $p(r_2)=0$.
\end{proof}


\solution{
The polynomial $0x^2 + 0x + 1$ has no roots.  The proof failed for this
case since it divides by $a$, which is unsound  when $a=0$.

The polynomial $x^2 + 2x + 1$ has only one root.  For $a\neq 0$, the
algebra is ok, but the proof assumes that $r_1 \neq r_2$, which will be
false when $D = 0$.

The proof is correct for polynomials with $a\neq 0 \neq D$ (even
though when $D < 0$, the meaning of $\sqrt{D}$ is ambiguous as we saw
in lecture). So these two conditions are required to make the claim
correct.

}

\end{problemparts}
\end{problem}
