### Quiz of November 23, 2010

1.Let A and B denote sets. Use an “element” argument to prove the “Absorbing Law” .

**Proof:**

Suppose first that . It follows from the definition of intersection that then . Since was chosen arbitrarily in , we have shown that .

Now suppose instead that . By the definition of union, we can then say that also . These two statements together (and the definition of intersection) now give us . So ( was arbtrary… we are “generalizing from the generic particular” again) we’ve shown that .

The two inclusions together show that .

*Part of the point here is that “element” style proofs are direct but somewhat tedious. For “higher level” set equations, one begins with a few well-established “laws” (like this “absorbing” law) and uses them to write out a string of equations (rather than “chasing” an element from side to side as we have done here).*

2.Let the Universal Set (for this problem) be the Irrational Numbers (R – Q) and let and .

Compute the given sets.

*Should be a gift at our level of the game; one sees similar problems in Math for Poets classes: this is * very *basic set theory. Still, I anticipate lots of errors… like leaving off the set braces altogether among others. Did I mention that I’m avoiding actually* grading *these darn things?*

3.Let A and B denote sets Show algebraically that . Give a reason for each equation.

**Proof:**

(Set Difference Law)

(De Morgan’s Law)

(Double Complement Law)

(Commutative Law)

*Much more fun than “chasing” an element like in 1.*

4.Let X = {p, q} and compute the sets (the power set) and (the cartesian product of X with itself).

P(X) = {

{ }, {p}, {q}, {p,q}

}

.

*One typically does* not * see the cartesian product… which I call simply the “cross” product… in the “math for poets” (terminal-introductory) classes. This is too bad. Anyhow, this is basic-basic stuff again; with any luck, the class’ll’ve flattened it.*

5.Write out (as a set of ordered pairs) the “less than or equal relation” given by on the set .

R = {

(0,0), (0,1), (0,2),

(1,1), (1,2),

(2,2)

}

*With, ideally, every comma and parenthesis in place. One may of course write out R = {(0,0), (0,1), (0,2), (1,1), (1,2), (2,2)} but it’s harder that way.*

6.Consider the function given by (for all positive integersn). Name the domain & codomain forf. Isfone-to-one? (Prove your answer.) Isfonto? (Prove your answer.)

(*I meant to have had f(n) = n^2. Rats.*)

The domain is the set of positive integers.

The codomain is the set of integers.

The function *f* is one-to-one. **Proof.** Suppose f(a) = f(b) [for some positive integers a and b]. Since f is the *identity* function (Rats!), we have f(a) = a and f(b)=b;

it follows that a = b. But is the defining condition for “one-to-one”.

But *f* is *not* onto. To show this, we must “construct” an integer (element of the codomain) which isn’t in the “image” (or “range”) of *f*. So consider . If f(x) = -1 for some *x* in the domain, we would have x = -1 by the definition of *f*. But -1 is *not* a positive integer. So there is no such *x*.

7.Now consider given by (for all real numbersx). Repeat the questions of the previous problem (forgrather than forf).

*This would have been more interesting if I hadn’t messed up the previous problem… the point is that the answers depend, not only on the “formula” given, but also on the domain and codomain.*

The domain and codomain are of course and respectively (the Real Numbers and the *Nonegative* Reals).

The squaring function is *not* one-to-one over the Reals; for example (-2)^2 = 2^2 [but -2 \not= 2; note that both *are* Reals]. The squaring function *is* onto the non-negative reals. To see this, let . Then and ; done.

*For the record, the squaring function* is *one-to-one when the domain is “zee-plus” (the positive integers) and is * not * onto the set of integers for this domain. Rats.*

November 30, 2010 at 2:28 am

successfully introduced codomain to high school seniors today. How do I know it was successful? They asked tough questions. Codomain properly bothered them and their comfortable idea of range.

Glad you are writing again! ; )>

November 30, 2010 at 2:24 pm

jd on domain & codomain last year.

December 1, 2010 at 1:55 am

Ooh! I’m a good student. I forgot about that discussion. But it’s contents stayed with me, even if I didn’t think about them, and showed up in the lesson.

JD

December 2, 2010 at 6:55 pm

(corrected to eliminate rats).

\centerline{Math 366. Quiz of November 23, 2010}

Name:

\vfil

{\bf 1.} Let A and B denote sets. Use an “element” argument to prove

the “Absorbing Law” $A \cap (A \cup B) = A$.

\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil

{\bf 2.} Let the Universal Set (for this problem) be

the Irrational Numbers (${\bf R}-{\bf Q}$) and let

\centerline{$X=\{\pi, e, \sqrt2\}$\hskip .4in and \hskip .4in $Y=\{e, \gamma, \pi\}\,.$}

Compute the given sets.

\line{

{$X\cap X^c$}\hfil

{$X\cup Y$}\hfil

{$X – Y$}\hfil

{$X – Y^c$}

}

\vfil\vfil\vfil\vfil\vfil\vfil

\eject

{\bf 3.} Let $A$ and $B$ denote sets. Show algebraically

that $(A-B)^c = B \cup A^c$. Give a reason for

each equation.

\vfil

{\bf 4.} Let $X = \{p, q\}$ and compute the sets {\bf a.} $\wp(X)$ (the power set) and

{\bf b.} $X\times X$ (the cartesian product of $X$ with itself).

\vfil

{\bf 5.} Write out (as a set of ordered pairs) the “less than or equal relation”

given by $xRy \equiv x \le y$ on the set $\{0,1,2\}$.

\vfil

\eject

{\bf 6.} Consider the function $f: Z^+ \rightarrow Z$

given by $f(n) = n^2$ (for all positive integers $n$).

{\bf a.} Name the {\it domain} and {\it codomain} for $f$.

{\bf b.} Is $f$ one-to-one? (Prove your answer.)

{\bf c.} Is $f$ onto? (Prove your answer.)

\vfil

{\bf 7.} Now consider $g: R \rightarrow [0, \infty)$

given by $g(x) = x^2$ (for all real numbers $X$).

Repeat parts {\bf a}, {\bf b}, and {\bf c} of the previous

problem (for $g$ rather than for $f$).

\vfil

\eject

\bye

December 3, 2010 at 2:31 pm

Math 366 Quiz 2 1.Each of the arguments(a) through(d) is VALID or NOT VALID; determine which.For VALID arguments, name the valid argumentform (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). (a) If that’s a thermometer, then I’m Bugs Bunny. That’snot a thermometer. •I’m not Bugs Bunny. (b) tis rational or tis irrational. tis not irrational. •tis rational.(c) (∀x ∈R)(x∈ Q)→ (x∈ˆ Q) ∼(t ∈ˆ Q) •∼(t ∈Q) (d) If Pigs Fly, then there’s a Sty in the Sky. Thereis a Sty in the Sky. •Pigs Fly. 2.Use BOTH of(a) a Truth Table AND(b) a sequence of Equivalences (from the textbook-and-handout list…naming a justiﬁcation for each “step”) to prove the equi

Math 366 Quiz 2

1.Each of the arguments (a) through (d)

is VALID or NOT VALID; determine which.

For VALID arguments, name the valid argument 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).

(a)

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

That’s not a thermometer.

•I’m not Bugs Bunny.

(b)

t is rational or t is irrational.

t is not irrational.

•t is rational.

(c)

(∀x∈R) (x ∈ Q) → (x ∈ Q^)

∼(t ∈ Q^)

• ∼(t ∈ Q)

(d)

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

There is a Sty in the Sky.

•Pigs Fly.

2.Use BOTH of

(a) a Truth Table AND

(b) a sequence of Equivalences

(from the textbook-and-handout list…

naming a justiﬁcation for each “step”)

to prove the equi

the SCRIBD document breaks off here…anyway on my equipment it does.

part of the point here is that i don’t

seem to have the original code for

this quiz…

December 3, 2010 at 2:41 pm

another part… what i find somewhat weird…

is that *some* of the “is an element of” signs…

∈ …

come through just fine whereas others are mangled.

but then, one can say the same thing

about “spaces” and whatnot… there

are *usually* some

invisiblelurking around somewhere…characters

December 3, 2010 at 2:58 pm

\hskip1.7in Name: \null\nobreak\leaders\hrule\hskip10pt plus1filll\ \par%

\smallskip

\centerline{Math 366}

\centerline{Exam 1: October 14, 2010.}

\vskip .5in

\parindent=0

{\bf 1.} Classify the arguments.

Specifically, each is

is VALID or NOT VALID;

determine which (explicitly; “V” or “NV” is OK).

For VALID arguments,

name the valid argument

{\it form} (from the given

list or (be careful!) certain

“Universal” generalizations

of arguments from the list).

For arguments that are

NOT VALID, name the

fallacy illustrated

(Converse Error

or Inverse Error,

for example).

\vfil

{\it (a)}

If you don’t see sharp, you will be flat.

You will be flat.

$\bullet$ You don’t see sharp.

\vfil

{\it (b)}

All mathematicians are reasonable.

My cat is not reasonable.

$\bullet$ My cat is not a mathematician.

\vfil

{\it (c)}

I will die or I will pay taxes.

If I pay taxes then Big Brother is watching.

If I die then Big Brother is watching.

$\bullet$ Big Brother is watching.

\vfil

{\it (d)}

If you let me get a word in edgewise, I will go on.

You will not let me get a word in edgewise.

$\bullet$ I will not go on.

\vfil

{\it (e)}

I’m Jack and I’m loud.

$\bullet$ I’m loud.

\vfil

{\it (f)}

If I didn’t know any better, I’d swear that’s a UFO.

I didn’t know any better.

$\bullet$ I’d swear that’s a UFO.

\vfil\eject

{\bf 2.}

{\it (a.)} Write out a Truth Table (a “complete” Truth Table…

one column for each

variable and one for each calculation) for

the statement form $$p \wedge \sim(q \wedge \sim p)\,.$

(You will need six columns.)

{\it (b.)} Mark the appropriate parts of the table to complete

the “truth-table proof” of the equivalence

$$p \wedge \sim(q \wedge \sim p) \equiv p\,.$$

{\it (c.)}

Write out an English sentence explaining

why this diagram proves the equivalence valid.

\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil

{\bf 3.}

Use a sequence of “steps” from the

given list of Logical Equivalences

to show that

$$p \wedge \sim(q \wedge \sim p) \equiv p\,.$$

Explicitly name the equivalence at each step.

\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil

\vfil\eject

{\bf 4.} Will all be well?

Create symbols appropriately

(“ABW” replacing “All will be well”,

for example). Think through the problem

to discover its Answer.

Then {\it justify} your answer with

a sequence of “steps”;

give an explicit reason

for each of your steps

(premises, previous steps,

or “rules of inference” from

the given list).

{\it a.} I will keep turning the crank.

{\it b.} I’ll get paid soon or I’ll know-the-reason-why.

{\it c.} If I’ll know-the-reason-why then all will not be well.

{\it d.} If I get paid soon then all will be well.

{\it e.} If I keep turning the crank then I’ll get paid soon.

\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil

{\bf 5.}

Write appropriate negations,

simplifying appropriately.

(Recall, or example, that

the negation of $x>0$

is $x\le 0$ and that the

negation of an “implication”

is a “conjunction”).

\vfil

$\bullet (\exists x) P(x) \wedge Q(x)$

\vfil

$\bullet\, n0$

is $x\le 0$ and that the

negation of an “implication”

is a “conjunction”).

\vfil

$\bullet (\exists x) P(x) \wedge Q(x)$

\vfil

$\bullet\, n<-5$ or $n\ge 17 $

\vfil

$\bullet$ If you know what's

good for you then you keep

your mouth shut.

\vfil

$\bullet (\forall x \exists y): y+1 = x$

\vfil\eject

{\bf 6.} {\it (a.)} Write down a Boolean Expression

for the given circuit.

\vfil

{\it (b.)} Write down a Boolean Expression for

the given Input-Output table.

\newline

\centerline{

\vbox{\offinterlineskip

%TRUTH TABLE

%AND, OR, IMPLIES

\halign{\strut\vrule\hfil\sc#\hfil\vrule&&\hfil\sc#\hfil\vrule\cr

\noalign{\hrule}

\quad$p$\quad&\quad$q$\quad&

\quad $r$\quad&\quad ? \quad\cr

\noalign{\hrule}

T&T&T&T\cr

T&T&F&T\cr

T&F&T&F\cr

T&F&F&T\cr

F&T&T&T\cr

F&T&F&T\cr

F&F&T&F\cr

F&F&F&T\cr

\noalign{\hrule}

}

}

}

\vfil

\eject

\bye

*************************************************

\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

**************************************************

oh never mind.