## Archive for November, 2010

### 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.

### photos belonging here

in my other blog (you may have to scroll a little).

### i am awesome. somebody buy me a drink.

“midterm” report.

actually we’re more like two-thirds
of the way through (my one-shot
Discrete Math class at Big State U;
try to keep up)… but the custom
on this campus is to refer to
all-exams-other-than-the-final
as “midterms” (which strikes me
weird since where i came up we
exams in the middle of the term
[and called ’em “midterm” exams]).
maybe it’s a semesters-versus-quarters
thing. (but i’ll bet you money that
this usage persists long after the
upcoming switchover to semesters…)

i’ll go ahead and remark since it’s
on my mind that i still haven’t been
compensated financially for any of
this work. we walk by faith not sight.
there’s, what else, this third-party
computer interface whereby one is
required to divulge information
supposedly certain events then
occur over the internet… “automatic
deposits” i’ve heard ’em called…
and one’s later transactions are
“covered”. well, i’ve tried to
co-operate. twice now; we’ll see
with what success. anyhow, by some
miracle my phone (don’t get me started)
eventually… days late… delivered a
“voicemail” message from somebody in
(the tellingly nay damningly named)
“human resources” office… somehow
the e-check was refused at my bank and so
(according to this guy, on the phone, later)
they’ve sent an actual 3-D hardcopy check.
to my official address where i go
only about once a week so i’ll see
a third of the quarter’s pay and
much the most money i’ve earned
in over a year so i’m pretty jazzed.

meanwhile, there’s a pile of grading:
the second “midterm”. on writing proofs.
and i’ll have ’em marked up by tuesday
evening when class meets but for now
i’m gonna let ’em age a bit.

the first exam was about logic and went
quite well… this is a *very* well-prepared
group by my standards (even in the pros
[from ’92 to ’96] i sort of specialized
in the survey-for-nonmajors stuff; once
i was sent down to the minors, it was
mostly “remedial” stuff) and the course
is pretty well-designed for students
at their level of the game.

my best move was finding the online course notes
for the guy who co-ordinated the course a few
quarters back (and taught at least one section
of his own): there is, more or less of course,
*much more material* covered by the sections
of the text we’re given to go over this quarter
than one can find time in class to discuss,
so i’ve followed this much-more-experienced
instructor’s choices of topics-to-stress
(as revealed in his exam-prep problem sets)
pretty closely and, so far knock wood,
with pretty good results.

meanwhile, the whole thing is overall
much mathier than the only *other*
Discrete Math class i ever taught
(back in the pros at Churchy Small
Liberal Arts College [now styling
itself a “University” but i digress]).
stuff like algorithms and data structures.
they’re quite a bit alike in requiring
students to be very careful with
definitions and stuff like that
i suppose… and i haven’t looked
at my notes from that long-ago
stuff for quite a while so they
might be even more alike than i think…
but some readers might naively believe
there’s some actual area of study
called “discrete math” and i’ll just
go ahead and strongly hint here that
they think it in error.

you get enough stuff in one of these
monster textbooks to keep even the
brightest of a class of college kids
busy for *years*. and for much the same
reason that to get a decent phone,
you’ve gotta sign on for years at
a time (even if you have no very
realistic idea of being able to
*pay* for it for that long let
alone of getting any useful service
without working like a wageslave
for untold amounts of very frustrating
time and no certain reward whatever):
“more waste faster” rules the roost.
but like i said. don’t get me started.

anybody worthy of the name “former math prof”
could bang out a text that would cost a small
fraction of what ours does and serve their
class just as well. i’ve done something
very like this myself for one of those
survey-for-nonmajors courses. there’s
even some overlap between that long-ago
work and the course at hand (as i’ve already,
in some sense, remarked).

and anybody worthy of name
“fit to survive in the present
economic climate” could probably
find a way to make it *pay*.

but concerning that of which
we cannot speak, it’s best
to STFU.

• ## (Partial) Contents Page

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