### 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.

… … .

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?

Advertisements

October 6, 2010 at 1:26 pm

\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

to discover its Answer.

Then {\it justify} your answer with

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

October 7, 2010 at 1:19 pm

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