### there’s going to be a certain amount of fiddling around right in here. of course there’s a *better* way but what we’re after is the *easy* way. standards are great that’s why there’s so many *of* ’em.

$\equiv$$\leftrightarrow$$\Leftrightarrow$.

any luck? three different symbols, right?
an equal-sign-that-didn’t-know-when-to-stop;
a “left-and-right-arrow”; and a *double* leftrightarrow.
right?

okay. it appears to work on *this* set-up;
no telling what *your* mileage looks like
(just be sure that it *will* vary).

these symbols, then.

these symbols *all* stand for versions of
“two things have the same interpretation”.

there’s already a great deal of confusion
about (what i take to be) the *best-known*
symbol in this class: the *sign of equality*, “=”.
i’ve remarked on this phenomenon before.

but at the “math 366” level–my current class–
it becomes very important to be very careful to
*distinguish* between various notions of
“means the same thing as”.

but… how?

1. \centerline{Math 366}
\centerline{Quiz 2}
\vskip .5in
\parindent=0

{\bf 1.}
Each of the arguments
{\it (a)} through {\it (d)}
is VALID or NOT VALID;
determine which.
For VALID arguments,
name the valid argument
{\it form} (Modus Ponens or
Transitivity,
for example, or [be careful!]
their Universal counterparts).
For arguments that are
NOT VALID, name the
fallacy illustrated
(Converse Error
or Inverse Error,
for example).

\vfil
{\it (a)}

If that’s a thermometer, then I’m Bugs Bunny.

That’s {\it not} a thermometer.

$\bullet$ I’m not Bugs Bunny.

\vfil
{\it (b)}

$t$ is rational or $t$ is irrational.

$t$ is not irrational.

$\bullet t$ is rational.

\vfil
{\it (c)}

$(\forall x \in R) (x \not\in Q}) \rightarrow (x \in \^Q)$

$\sim(t \in \^Q)$

$\bullet \sim(t \not\in Q)$

\vfil
{\it (d)}

If Pigs Fly, then there’s a Sty in the Sky.

There {\it is} a Sty in the Sky.

$\bullet$ Pigs Fly.
\vfil\vfil
{\bf 2.}
Use BOTH of {\it (a)} a Truth Table AND {\it (b)} a sequence
of Equivalences (from the textbook-and-handout list… naming
a justification for each “step”) to prove
the equivalence $p\wedge \sim(q \wedge \sim p) \equiv p$.

\vfil\vfil\vfil\vfil
\vfil\vfil\vfil\vfil\vfil\vfil
\vfil\vfil\vfil\vfil
\vfil\vfil\vfil\vfil\vfil\vfil
\vfil\vfil
\vfil\eject

{\bf 3.}Is Owen in debt?

Create symbols appropriately
(“OM” replacing “Owen is a millionaire”,
for example). Think through the problem
a sequence of “steps”
written in these symbols
(and appropriate “logic” code);
each consisting of a three-line
(major premise, minor premise,
conclusion) “argument”. Each
{\it premise} of a given step
should come from the given
information ({\it (a)}–{\it (e)})
or a previous “step”; each
{\it conclusion} will (of course!)
be an instance of some valid
Rule of Inference.
(Compare “Example 2.3.8″ at
Epp, {\it pp.} 56{\it f}).

{\it a.} Owen is a millionaire or Owen is underinsured.

{\it b.} If Owen is underinsured, then Owen is in debt.

{\it c.} If Owen is a millionaire, then Owen is {\it not} in debt.

{\it d.} If Owen is a millionaire, Owen has quit his job.

{\it e.} Owen has not quit his job.
\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil

{\bf 4.}

Write appropriate negations.
It does {\it not} suffice simply to
affix “$\sim$” (or the phrase
“it is not the case that…”
[for example]); the negation
of $x<17$ isn't $\sim(x<17)$,
for example [or $x\not<17$
for that matter], but $x \ge 17$.
\vfil
$\bullet (\forall x) P(x) \rightarrow Q(x)$
\vfil
$\bullet x<0$ or $x\ge 1$
\vfil
$\bullet$ There ain't no life nowhere.

\vfil\eject

\bye

2. http://pdfcast.org/pdf/366q2-pdf

• ## (Partial) Contents Page

Vlorbik On Math Ed ('07—'09)
(a good place to start!)