1dec14

\begin{document}
\maketitle

\section{General}
Since you have a test on the isometries on material up to F13, and I'm
not sure the grader had time to correct this homework over the break,
for the hourly on W14.

Here are some arch-errors, which some of you still make on occasion. Be
careful to avoid these on the homework and especially on the test and the
final.

\subsection{Circular Reasoning: Assuming the Conclusion}
If anything is to be learned in a geometry course, it is logical thinking.
So, by now, your assuming what you are to prove  is not even worthy of
partial credit any more.

\subsection{Syntax Errors: Writing Non-Equations}
A mathematical equation may be true or false, or nonsense, and therefore
neither true nor false. For example, if the LHS is a transformation and the
RHS is a mixture of lines and points, then this is a non-equation. In a
sense, this is a worse error that the previous. However, it can happen to
people who DO know better, but are thinking faster than they are writing.
On a test, maybe, the grader might try to guess what you mean. On the homework,

\subsection{Plagiarism: Uncritical Sharing of Work}
When I see the same mistake repeated more than once I suspect collaboration.
I said collaboration with classmates on the homework is OK, it is NOT OK on
tests in class or as homework. When you share work you must mention the
other people you worked with.  Nobody does, and there are cases of very
uncritically shared homework on this homework.

Regardless of where you get an answer from:
your co-worker, the back of the book, or some trove of collected solutions
some kind hearted (but usually misguided) former student might have collected,
.... regardless, you're responsible to test the answer critically and correct
any errors in it. So when I see four solutions all with the same incorrect
(and, in part nonsensical) solutions, I am tempted to ....

But such negligence is actually a very bad misuse of resources. Maybe this is
how you got through other math courses, but not mine. You're wasting your
time, the grader's time, and aren't getting much credit for it anyway. The
homework is for learning, not for hoodwinking the grader.

Because you do (and should) collaborate on preparing your journals, it is
your collective duty to make sure what you put into your journals is correct.
This is particularly important for tests on which you may use your Journal.
If a third party were to find the same error on several papers the conclusion
might not be very flattering.

The reason I expect you to repeat the question, or, preferably paraphrase
it is for the grader to know just what you think you're proving. And if
you don't, circling the question not answered is useful to you. I don't
do this here for precisely the same pedagogical reason ... I want you to
look at the questions again to see what you didn't read carefully enough.

\subsection{Using methods from elsewhere}
On a comprehensive examination where no preferred methods are indicated,
or you can't use because you forgot,
you might solve the problem any way you wish,
including counting on your fingers. But not in a homework problem. You
not what you learned a long time ago, or what you think you can figure
out by just arguing geometrically about the problem.

\section{Common Errors and Corrections in the Problems}
\subsection{Problem F}
Most people got from the equation
$\alpha \sigma_C = \sigma_C \alpha$ to
$\sigma_C = \alpha \sigma_C \alpha^{-1} =_{CT}= \sigma_{C^\alpha}$ and
from there to $C = C^\alpha$. But without the universal quantifier,
$(\forall C)$ you can't just conclude that $\alpha = \iota$.

By the way, although it is correct here to just quote the fact that
$\sigma_P = \sigma_Q \implies P=Q$, be sure you also know how to prove
this lemma for the test/final.

\subsection{Problem H}
There is a quick way of deriving commutativity from the hypothesis:

$\sigma_m \sigma_M \sigma_m^{-1} =_{CT}= \sigma_{\sigma_m(M)} =_{(mM)}= \sigma_M$

There was a second question. Using $m$ as one mirror, the co-mirror for
$\sigma_M$ is the perpendicular $h$ to $m$ at $M$. Factoring both sides
yields that both sides equal the same thing, namely $\sigma_h$, and are
therefore equal to each other.

\subsection{Problem K}
This is an if and only if problem, and the only fail-safe method is
to split it into two separate problems, even if the converse turns into
the same step in reverse order (and appropriately different reasons).
So here I'll get you started and you finish it on your own
\begin{eqnarray*}
\iota &=& \sigma_m \sigma_Q \sigma_m \sigma_P \\
& = & \sigma_{Q^m} \sigma_P. \\
\mbox{So   } \sigma_{Q^m} &=& \sigma_P \\
\mbox{So   } Q^m &=& P \\
\mbox{i.e.   } \sigma_m(Q) &=& P \\
\end{eqnarray*}

Now, first justify why the last line implies that $m=\mbox{perbis}(PQ)$.
You can do this geometrically, but also algebraically. For instance,
since the fixed points of the isometry lie on the perbis and also on
the mirror, the mirror has to be the perbis.

Then, reverse the argument. (Don't try to be fancy and attempt getting
the iff in one argument until you're much better geometers.) From the
hypothesis you get to the last line and work your way back up. With a
keyboard and mouse this is not nearly the hardship it was doing math
with pencil and paper. So, on homework, laziness is not equivalent to
efficiency.
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