Quiz of November 23, 2010

1. Let A and B denote sets. Use an “element” argument to prove the “Absorbing Law” A \cap (A \cup B) = A.

Proof:
Suppose first that x \in A \cap (A \cup B). It follows from the definition of intersection that then x \in A. Since x was chosen arbitrarily in A \cap (A \cup B), we have shown that A \cap (A \cup B) \subset A.

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

The two inclusions together show that A = A \cap (A \cup B). \bullet

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 X = \{\pi, e, \sqrt2\} and Y = \{e, \gamma, \pi \}.

Compute the given sets.
X \cap X^c
X \cup Y
X - Y
X - Y^c

X \cap X^c = \{ \}
X \cup Y = \{\pi, e, \sqrt2, \gamma \}
X - Y = \{ \sqrt2\}
X - Y^c = \{\pi, e\}

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 (A - B)^c = B \cup A^c. Give a reason for each equation.

Proof:
(A - B)^c
= (A \cap B^c)^c (Set Difference Law)
= A^c \cup B^{cc} (De Morgan’s Law)
= A^c \cup B (Double Complement Law)
= B \cup A^c (Commutative Law)

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

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

P(X) = {
{ }, {p}, {q}, {p,q}
}

X \times X= \{ (p,p), (p,q), (q,p), (q,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 xRy \equiv x \le y on the set \{0, 1, 2 \}.

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 f: {\Bbb Z}^+ \rightarrow {\Bbb Z} given by f(n) = n (for all positive integers n). Name the domain & codomain for f. Is f one-to-one? (Prove your answer.) Is f onto? (Prove your answer.)

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

The domain is the set \Bbb Z^+ of positive integers.
The codomain is the set \Bbb Z 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 [f(a) = f(b)]\rightarrow[a=b] 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 -1 \in \Bbb Z. 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 g: {\Bbb R} \rightarrow [0,\infty) given by g(x) = x^2 (for all real numbers x). Repeat the questions of the previous problem (for g rather than for f).

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 \Bbb R and [0,\infty) 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 x\in [0, \infty). Then \sqrt{x} \in {\Bbb R} and g(\sqrt{x}) = x; 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.

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  1. 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! ; )>

  2. 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

  3. blag

    (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

  4. vlorbik

    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…

  5. 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 invisible
    characters
    lurking around somewhere…

  6. \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.




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