## Archive for December, 2010

### more than 4K characters for k. nowak

(i couldn’t put this in a comment
for this “log laws” post
at f(t); too long.)

it’s the old principles-versus-procedures problem.
students hate general principles.

i’ve had tutees pay me hundreds
to supervise their homework
who very clearly “tune out”
whenever i try to explain
what’s going in in a general way.

it’s worst when it’s about a topic
they’ve been exposed to before…
the famous “math phobia”. panic
sets in when you realize that
you’re going to expose your ignorance
yet again… about fractions, variables, or
vocabulary (“equations” and “expressions”,
for example)… to name some common
weaknesses for *beginning* algebra.

the tendency is to believe that whatever
this “general principle” thing is that
teachers keep wanting to come back to
that, if we’d only present it in plain english,
then they’d know how to work the problems
on the test (which had darn well better
be *just like the ones we practiced*).

so we give in and show ’em the “procedures”:
FOIL instead of “repeated distributive law”,
for example… and they *pass* the tests
but get *further behind* in the grand scheme.

“i’ve done OK this far by disregarding
all the math-brains-only *theory*…
and now i’m failing. but i’ll *never*
be a math-brain… good heavens,
i’d have to go back and understand
an *awful lot* of stuff i’ve been
given a pass on all these years!…
it’s the pedagogy! it’s my learning
style! it’s my teacher! math sucks!”

the phenomenon persists at all levels
(that i’ve passed through in learning
math up to my published thesis…
when, sure enough, i still felt myself
an imposter compared to the “real
mathematicians” i’d been working

“logs” are a particularly interesting case:
otherwise-well-prepared calculus students
are often very weak on logs. hell, every
tenth student has decided they’ll just
go ahead and pretend “ln(x)” is to be
replaced with “1/x” somewhere along
the line no matter how many times
it’s marked wrong.

the abstract, proof-heavy style
of math-rightly-so-called?
the typical student will never
*feel* ready for *any* new idea.
(this should be considered in
any discussion of “developmental
barriers”).

do-the-same-thing-on-both-sides.
logs-are-inverse-exponentials.

here’s a common-log equation:

log(4) + log(25) = log(x).

and here’s its solution.

10^[log(4) + log(25)] = 10^[log(x)]
10^[log(4)] * 10^[log(25)] = x
4*25 = x
x=100.

now, kate’s worksheets are great
and i’m not suggesting that anyone
make my approach into their
exhibit A for day-two-of-logs.
so don’t get me wrong.

the point i set out to make when i started…
if i can remember that far back… is that
*no* approach can overcome the simple
fact that our courses are designed to
go *much too quickly* for median-level
students. the good news is that we
can spot talent pretty easily this way:
anybody who can keep up typically
hasn’t “hit the wall” where math
gets *hard*; give ’em an A and
pretend we’ve done something
to be proud of.

my “discrete mathematics” students
last quarter were all calc 2 vets:
future programmers and engineers
and whatnot. and *almost* all of
’em learned quite a bit about
writing proofs. the *easy* thing
here is (of course!) the logic.
(i mean the symbol manipulation;
what all the p’s and q’s have to do
with the “logical structure” of
a given proof is much more obscure
than it appears to non-teachers.)

if *anybody* i’ve ever worked with was
“show using the definition of logs
that log(a) + log(b) = log(ab)”
as an exercise, it’ll’ve been these
talented hardworking students.

but the resistance was palpable and i folded.
there was… as always… much too much
material (i also had to skip stuff like
“manipulations with sigmas” and
“complete induction”… the class
emerged ready to take on maybe
a third of the exercises in the sections
covered [mostly the easiest ones]).

there *is* no developmental appropriateness
in “real world problem solving”: one needs
a technique that *works* wherever it’s found.

though, is pretty much “preparing students
for yet another class”… and one technique
that “works”, alas, is *hide your weaknesses”.

that idea of “apply some inverse-thingy
on both sides of the equation”? one of
a handful of big ideas in algebra
(along with “variables”, “graphs”,
and “roots of polynomials” and…
well, not much else…)

“look: to undo an *addition*, we do
a *subtraction*… to undo a multiplication,
we do a *division*… and these ideas
were *hard* when we tackled ’em
back in some earlier course.
a little later, we decided that
we wanted to ‘undo’ squares,
and introduced ‘square roots’;
the beginning-algebra folks
get lost around the bend in
here somewhere) we’ll need
to undo powers-of-the-variable
like x^K, and introduce Kth roots.
well, then, doggone it, why shouldn’t
the same undo-the-operation
strategy work for K^x?”

i’ve ranted out this mini-lecture
not just to get it out of my system
(for today), but to demonstrate
that one need *not* have any
*formal* understanding of
“inverse function” to use the
never-stop-harping-on-general-principles
method that i’m more or less advocating here.

(indeed, inverses-treated-formally
is a classical crux. “never mind
*why* i solve x = f(y) when i want
a formula for f^{-1}(x)… just show
me *how*!”… one of the most
glaring examples of the
i’ll-never-understand-the-reasons
phenomenon known to me.)

okay. this was fun.
haven’t i got a *job* or something?

### what then are we to do?

here’s a long thread (at KTM) about barry garelick’s “EducationNews.org” post it isn’t the culture, stupid (with throbbing googol ad).

my take? “we” can’t even *consider*
getting teachers competent in math
for *everyone*… and yet are committed
to create (some emperor’s-clothes version of)
the illusion of “equal opportunity”. so teachers
committed to the “math has the authority”
principle have to be careful not to let it be
known very far outside their classroom walls
(or work outside the public-funding arena).

if this isn’t “culture”, make the least of it.

*********************************
no, wait. more thoughts from a different coffeehouse.
that’s not my take; just my life story.
my *take*: “it’s the corporate hegemony (obviously)”.

how come perfectly well-heeled middle-class parents
can’t get local governments to provide instruction
in perfectly basic skills… no matter how much
they’re taxed in the process? well, at least in part,
because none of the actual players has any interest
in doing-it-on-the-cheap. cf: the ever-worsening
“healthcare” disaster (one can’t call it a “crisis” or
even an “emergency”… crises are mere moments,
and the situation emerged a long time ago…).

like i said in the KTM thread:
don’t you know there’s a *war* on?

(i didn’t add “stupid”… plenty of *very*
intelligent and even articulate people
are content to ignore gorillas at any weight.)

### Examine Other Beauties

5. Let $Y = \{0,3,6\}$ and $Z = \{0,5\}$ again;
compute a. $Y\times Z$ and b. $Z \times Y$.

Obviously $Y\times Z =$
{
(0,0), (0,5),
(3,0), (3, 5),
(6,0), (6, 5)
}

and

$Z\times Y$=
{
(0,0), (0,3), (0,6),
(5,0), (5,3), (5,6)
};

what’s more, almost all my students
knew it on the day (typically in the form
$Y\times Z =\{ (0,0), (0,5), (3,0), (3, 5), (6,0), (6, 5)\}$
and
$Z\times Y =\{(0,0),(0,3),(0,6),(5,0),(5,3),(5,6)\}$
). it bothers me
not at all that they don’t seem much
conventions; the main thing is that
they know… and are willing to *admit*
that they know (in writing) how a
“cross-product” (what our text,
alas, insists on calling a “cartesian”
product) can be written out in
“set-of-ordered-pairs” notation.

6. Let $U = \{0, 1, 2, 3, 4, 5, 6\}$ again.
Write out the Relation $R$ defined on $U$ by $xRy \equiv 3 | (x - y)$.

Don’t forget that “a-b” is meaningful even when $b \ge a$; otherwise there are no systematic errors classwide. OK. I’m a liar.
There’s this. Remember that the answer is (i) a set (in our context, written as a list of items, between set brackets, and separated by commas), of ordered pairs (each of which, in our context, is written between ordinary parentheses… what’s weird about all this is that it’s perfectly easy to get students to care about keystroke-perfect code when a robot grades their work but they’ll mortgage their god-damn grandmothers to get you to give ’em some special deal as soon as some hint of “humanity” comes into the picture.).

### Never Give Up. Never Surrender.

4. Let $X = \{0,1,2,3\}$ again; compute $\wp(X)$ (the power set of $X$).

OK. First of all, that’s a “Weierstrass pay” not a script-P; typesetting is always harder than it looks. There’s a “caligraphy” font available I think. But why should I have to know or care? One curly-P looks an awful lot like another and for all I can tell nobody is paying any attention literally.

Anyhow. In context, the power set of {0, 1, 2, 3} is
{
{},
{0}, (1}, {2}, (3},
{0,1}, {0,2}, {0.3},
{1,2}, {1,3}, {2,3},
{1,2,3}, {0,2,3}, {0,1,2}, {1,2,3},
{1,2,3,4}
}
.

Nobody in my class took advantage of my type-oriented “hints”… the carriage-return style I’ve used here (as I also typically do in blackboard work)… and no damn wonder. It’d be helpful if textbooks would talk about how things are actually done in practice but then maybe it would be helpful if everybody clapped for tinkerbell while we’re at it.

The main thing to see here is that, for example, every set-of-three (in this context) is the “complement” of a set-of-one… and so, for example, the set {1,2,3}… a “three-set”… must be in the power set precisely because the set {0} (a “one-set”) has already been “counted”. The same “principle of complementation” can be used more generally to show that, in a Universal Set of n elements, one has exactly as many k-element subsets as there are subsets with n-k elements.

Another good trick in this context is “binary notation”. Many of my students were well aware that the initial string of the natural numbers can usefully be written as (( 0000, 0001, 0010, … 1111))… standard “binary” notations for the natural number sequence ((0, 1, 2, … 15)). By a more-or-less-obvious “place value” notation (order the set; “0” means leave-it-out and “1” means put-it-in), we can set up a one-to-one correspondence of subets-of-{0,1,2,3) [or any other 4-element-set] and the binary “strings” 0000 … 1111). It pleases me immensely that (a handful of) students checked their work on this exercise by this technique explicitly.

5. Let $Y = \{0,3,6\}$ and $Z = \{0,5\}$ again;
compute {\bf a.} $Y\times Z$ and {\bf b.} $Z \times Y$.

6. Let $U = \{0, 1, 2, 3, 4, 5, 6\}$ again.
Write out the Relation $R$ defined on $U$ by $xRy \equiv 3 | (x - y)$.

Let $f(x) = log_3(x)$, $g(x) = 5x+1$. Find (“formulas” for) the following functions.
$f^{-1}$
$g^{-1}$
$f\circ g$
$g \circ f$

### A Man’s A Man Who Looks A Man Right Between The Eyes

3. Let $U= \{0, 1, 2, 3, 4, 5, 6\}$ be the Universal Set (for this problem) and let $X=\{0,1,2,3\}$, $Y=\{0, 3, 6\}$, and $Z=\{0, 5\}$.

Compute the given sets. (Remark: we will have shown that set difference is not associative).

$(X-Y)-Z$
$X-(Y-Z)$
$Y^c \cup Z^c$
$X \cap X^c$

Again (sorry for the child’s play). A poor workman blames his tools; this doesn’t make him wrong. Obviously X – Y is (the set) {1, 2}. It follows then that (X-Y) – Z is also {1, 2}.

Whereas, in light of the fact that Y – Z = {3, 6}, one has X – (Y-Z) = {0, 1, 2}.

Anybody (outside a math class) pretending to want an explanation of what (if anything) has gone wrong is, pretty reliably, actually looking for some weakness whereby they can undermine somebody’s authority.

The classic blunder at this point is to refuse to take such a student seriously. And yet, they’re doing what I would’ve done; what I’ll go out on a limb and guess you would have done. Students didn’t just become stupid because you got an advanced degree any more than kids got to be some inferior order-of-being just because you decided to call yourself a growup.

How are they to know that, here in math class, we mean what we say? When, in every other arena (and for all we know, this one), authority rules?

(Yes, yes… they call it “opinion” and blame it on “the people”. Are we bored yet?)

### Discrete Math Final: Questions 3, 4, 5, 6, and 7

Very close… embarrassingly close… to the questions of the recent quiz. The “A” students obviously get pretty bored seeing the same thing over and over. And I’ll freely admit that I’ve been more or less ruined as an instructor for these (reasonably well-prepared) students by (what seems like) a lifetime of “remediating” the losers of the system. If you’re gonna be any good at preparing the well-prepared (or so it seems to me right now), you’re gonna have to embrace the “blame the victim” strategy: “we’re going at this pace and it’s your fault if you don’t (‘want to’) keep up”.

Meanwhile, what every (all-too-human, somewhere-in-the-middle) teacher actually does in practice is compromise.

In an ideal world, we’d do “so-and-so”. Certain smile-all-the-time manipulative assholes will therefore pretend that if “so-and-so” doesn’t happen, you’re to blame (and’ll have to be punished severely). It’ll behoove you, then, if you can tell which sides of the cards have the spots on ’em, to pretend that this system makes sense and file certain paperwork alleging to show that so-and-so has been done. “Be realistic.”

Is a syllabus a “contract”? Not in my world. But then, in my world, contracts rightly-so-called can only be entered into by parties of approximately equal power. To put this as forcefully as I know how… though I don’t expect you to get it… if corporations are “persons” in The Law, then actual living beings are something else… and this is obvious. These entities just want to get us on the record admitting that everything is our fault so they can use it against it at their whim. And this is obvious.

Still. Denying the holy spirit is the unforgivable sin; I’m damned if I’m gonna give up hope. This class… as I’ve hinted… came to me wanting to find out more about how math-heads go about their business and willing to work hard to find that out. This was the closest thing to… well… whatever it was I’ve been working so hard all my life to get ready to do… as I’ve got any reasonable hope ever of attaining. It was a blast.

Naturally, most of ’em were, in some sense, “lazy”. One has a lot of commitments; if certain assignments don’t “count”… or remain unassigned altogether… well, then, time is short. Meanwhile, I am committed to the principle that “no ‘serious’ student fails”. (For some hugely subjective value of “serious”.)

So here’s how it works out. I put a bunch of questions on the quizzes and tests that properly belong on some much lower-level tests according to the “hardliner” this-is-college-kid model. And then I’m very generous with “partial credit”. Anybody who’s still failing is really failing… and I’ll typically have a pretty easy time getting ’em to admit this when we talk about it face-to-face. Meanwhile, a lot of on-somebody-else’s-model “undeserving” students are getting Gentleman’s C’s.

The upside… enormous if I took my “math educator” role seriously… is that I encounter actual data about how students of whatever material is at hand actually encounter the given material. The signal-to-noise ratio on this particular subject is very nearly zero in my experience (libraries and the net), so, anyway for me, there’s a great deal to be learned from (carefully) marking papers written by these starting-to-become-advanced students… particularly when the material is of the usually-taken-for-granted variety treated here.

3. Let $U= \{0, 1, 2, 3, 4, 5, 6\}$ be the Universal Set (for this problem) and let $X=\{0,1,2,3\}$, $Y=\{0, 3, 6\}$, and $Z=\{0, 5\}$.

Compute the given sets. (Remark: we will have shown that set difference is not associative).

$(X-Y)-Z}$
$X-(Y-Z)$
$Y^c \cup Z^c$
$X \cap X^c$

at this point wordpress decided to fuck me with a redhot poker. one gets bored. some other time maybe.

{\bf 4.} Let $X = \{0,1,2,3\}$ again; compute $\wp(X)$ (the power set of $X$).
\vfil
{\bf 5.} Let $Y = \{0,3,6\}$ and $Z = \{0,5\}$ again;
compute {\bf a.} $Y\times Z$ and {\bf b.} $Z \times Y$.

(We will have shown
that “forming the Cartesian product” is not commutative.)
\vfil\eject
{\bf 6.} Let $U = \{0, 1, 2, 3, 4, 5, 6\}$ again.
Write out the Relation $R$ defined on $U$ by $xRy \equiv 3 | (x – y)$.
\vfil

{\bf 7.}
Let $f(x) = log_3(x)$, $g(x) = 5x+1$. Find (“formulas” for) the following functions.

\line{
{$f^{-1}$}\hfil
{$g^{-1}$}\hfil
{$f\circ g$}\hfil
{$g \circ f$}
}
\vfil

### Discrete Math Final: Question 10

10. Prove by induction that for all positive integers $n$, one has $3 | 4^n - 1$.

Proof: Let P(n) denote the proposition $3 | 4^n -1$.

Base Case For n = 1, we have 3|4^1 – 1; P(1) is true.

Induction Step Assume for some $k \in {\Bbb Z}^+$ that P(k) is true: $3 | 4^k - 1$. Then $4^k -1 = 3a$ for some integer a; rewrite this as $4^k = 3a + 1$. We then have

$4^{k+1} - 1 =$
$4\cdot 4^k - 1 =$
$4\cdot(3a +1) - 1=$
$12a-3=$
$3(4a-1)\,.$

But $4^{k+1} - 1 = 3(4a -1)$ implies $3 | 4^{k+1}$ (i.e., P(k+1)). This completes the induction and the proof.

Remarks
My class saw variants on this problem… 6 | 7^n – 1 (\forall n \in Z^+) for example… in (i) a homework and (ii) a previous test. So they were pretty well-prepared for an induction problem of this type. (Know ye not this parable? and how then will ye know all parables?)

Here’s another, somewhat informal, proof of the same result:
$x^n - 1 = (x-1)\cdot(x^{n-1} + x^{n-2} + \dots + x^0)$
(as one sees by “multiplying out”… all but two terms “cancel”). Put 4 in for x; done. I call this proof “informal” because wherever there are dot-dot-dots, there’s generally an induction hiding somewhere if we want to get technical. Anyhow, I shared this proof with the class.

Also the (equally informal) “handshake problem” proof that $\sum_{i=1}^n i = {{n(n+1)}\over 2}$. But, alas, I didn’t require induction proofs from them involving summations (“sigma” notation $\sum$). Another one of the many topics everybody seems to want their students to’ve already learned about somewhere else (in this case, Calc II most likely). It takes time to get students up to speed with summations. Different amounts of time for different students. I was something of a sigmaphobe myself as an undergrad. (I was even worse with “determinants” [another very useful algebraic gizmo that math departments don’t want to teach much about explicitly…].)

The string of equations in my notes hitherto is a little different; rather than solve 4^k – 1 = 3a to get 4^k = 3a + 1, I’ve consistently used the trick of “adding 0”:
$4^{k+1} - 1 =$
$4\cdot 4^k - 1 =$
$4\cdot 4^k - 4 + 3=$
$4(4^k - 1) + 3=$
$4(3a) + 3=$
$3(4a + 1)\,.$
(We’ve “added -4 + 4” in the third line to produce a “4^k -1” in the code… but this is needlessly tricky and, I expect, threw a few students off the track. I got the “right” way from a student paper.)

### Discrete Math Final: Questions 8 and 9

8. Consider the function $f: {\Bbb Z}^+ \rightarrow {\Bbb Z}^+$
given by $f(n) = n+1$ (for all positive integers $n$).

9. Now consider $g: {\Bbb Z} \rightarrow {\Bbb Z}^+$
given by $g(x) = |x| +1$ (for all integers $x$).

1a. The function $f: {\Bbb Z}^+ \rightarrow {\Bbb Z}^+$ given by f(n) = n + 1 is one-to-one.

Proof: Let a and b be positive integers such that f(a) = f(b).

Then a + 1 = b + 1 (use the definition of f to see this).

But then a = b (by “cancelling” 1’s [i.e., by subtracting 1 from each side of the previous equation]).

This proves the assertion.

[Because a and b were chosen arbitrarily from our domain and because “(\forall x, y) [f(x)=f(y)] \implies [x = y]” defines “f is one-to-one”.]

1b. The function $f: {\Bbb Z}^+ \rightarrow {\Bbb Z}^+$ given by f(n) = n + 1 is not onto.

Proof: Note that 1 is a positive integer, and so is in the codomain of f. Now suppose there exists a positive integer n satisfying “f(n) =1”. Some very simple algebra shows that then n=0. But 0 is not a positive integer (and so is not in the domain of f). So there is no such n. This proves the assertion.

[It’s very tempting for some beginners to “Let f(n) = 1” rather than use (the more elaborate) “suppose there exists n such that…” construction. But this is precisely what we can’t do (unless we already know that such an n exists…)]

9a. The function $g: {\Bbb Z} \rightarrow {\Bbb Z}^+$ given by $g(x) = |x| +1$ is not one-to-one.

Proof: Note that both 1 and -1 are integers [and so are “domain values”] and that g(1) = g(-1) = 2. Since 1 \not= -1, this proves our assertion.

9b. The function $g: {\Bbb Z} \rightarrow {\Bbb Z}^+$ given by $g(x) = |x| +1$ is onto.

Proof: Let $z \in {\Bbb Z}^+$. Since $z \ge 1$, we also have $z-1 \ge 0$, which in turn implies that $|z - 1| = z - 1$. Note that moreover $z-1 \in {\Bbb Z}$. We can now say that $g(z-1) = |z - 1| + 1 = z -1 + 1 = z$; g is onto.

[This was probably the hardest question on the exam since it calls for careful use of “absolute value”. The absolute value function is both “easy” to work with (I’ve learned that nothing is easy enough to write “easy” on the board without quote marks) and long familiar to all 366 students… but nevertheless; there it is. Only two students (in a class of 24) satisfied me completely on this one. My favorite of the rest used mathematical induction. A very good idea, oh anonymous one (but some details were lacking; minus-one-point)].

### Discrete Math Final: Questions 1 and 2

1. Let A, B, and C denote subsets of some Universal Set. Use an “element” argument to prove the set equation: $(A-B)-C = (A-C)-B$. (“Let $x \in (A-B)-C$…”).

2. Let A, B, and C denote subsets of some Universal Set. Use an “algebraic” argument to prove the set equation $(A-B)-C = (A-C)-B$.

Do not begin your proof with the equation to be proven! (Rather, begin with (A-B)-C = XXXXX [where XXXXX differs from “(A-B)-C” by one “step”]; then write out another step” [a new set expression and a reason we know it’s equal to the latest one… and so on until (A-C)-B is reached).

The necessary “reasons” are all in the list below.

Associative Law (of intersection) $(X\cap Y)\cap Z = X\cap (Y\cap Z)$
Commutative Law (of intersection) $X \cap Y = Y \cap X$

Set Difference Law $X-Y = X\cap Y^c$

(Lecture without diagrams; solutions included somewhere I hope.)

The given equation is true for any sets and so can be proved “within Set Theory”… by which I mean, without any other math (so we don’t need to know anything about counting or adding or exponentiating or any such number-like stuff, for example). This is why an “algebraic” proof is even possible.

To prove sets S and T equal outside the context of Set Theory, one typically shows that (i.)$S \subset T$ and that (ii)$T \subset S$ (i.e., that S is a subset of T and that T is a subset of S); various calculations are typically carried out (“number-like” calculations are common); typically the calculations for the two directions (“S is a subset of T” is one “direction” here) aren’t very much alike. Quite often, in fact, one direction is “easy” and one is harder.

In Quiz #3 we had $A \cap (A \cup B) = A$, for example, and showing that $A \cap (A \cup B) \subset A$ turns out to be (slightly) easier than $A \subset A \cap (A \cup B)$.

The equation for this problem is a non-example of this one-way-is-harder phenomenon because of its “symmetry”.

(The expressions on the two sides of the equation are very much alike… specifically, one may interchange the “B” with the “C” [wherever they appear] to obtain one side from the other. Graphical representations of suchlike situations display “symmetries” literally; typically one may “rotate” or “reflect” certain pictures back on to themselves. This means that certain “measurements” in certain drawings are “the same”; in the word symmetry “sym-” means “same” and “metr” means “measure” [think of the “metric” system to see that this is so].)

Anyhow, the point for us right now is that in proving (A-B)-C = (A-C)-B the two “directions” are very much alike. This makes it rather a poor choice for Problem #1 as I now realize. I got a little too cute. I wanted to use the same equation for both contexts and rather liked the “algebraic” proof (for requiring associativity and commutativity; for requiring only one other “reason”; for its take-it-apart-and-put-it-back-together quality [a result of the symmetry of the situation]).

Anyhow, here’s a proof (by a student).

(i) Let A, B, and C be sets and $x \in (A-B)-C$. Therefore, $x \in (A-B)$ and $x \not\in C$ by definition of “-“. Next, $x \in A$ and $x \not\in B$ by def of “-“. This gives $x \in (A-C)$ by “-” and $x \not\in B$. Combining, we get $x \in (A-C)-B$ by “-“.

(ii) Let A, B, and C denote sets and $x \in (A-C)-B$. By def. of “-” $x \in (A-C)$ and $x \not\in B$. Further, by “-“, $x\in A$ and $x\not\in C$. Since $x \in A$ and $x\not\in B$, then $x \in (A-B)$ (by “-“). Also $x \not\in C$ so by def of “-” $x \in (A-B)-C$.

We have shown (i) $(A-B)-C \subset (A-C)-B$ and (ii) $(A-C)-B \subset (A-B)-C$. Therefore we are done.

On the actual marked exam there appear three checkmarks on this work and no other marks by me… but of course there’s much here to discuss. One need not (for example) have redefined A, B, and C; I’ve done that already in the problem statement. It’s likely that some other instructor might frown on the declension “by definition of ‘-‘ “/ “by def of ‘-‘ “/ “by ‘-‘ “… but I myself consider it quite tastefully done. Another sign of this student’s good taste is the wrap-up step where she’s reminded the reader that (x\in S)\implies(x\in T) is the form of a proof that S\subset T (I have [tastefully, hopefully] here slipped into my “easy-type” style; for “\in” read “is an element of” [and mentally substitute $\in$]; other backslashed terms stand for [what I hope are “obvious”] other math symbols).

There were some full-credit versions of this problem omitting the wrap-up altogether. Again, other instructors might’ve required it. Grading on writing style still gives me the heebie-jeebies but I’m getting used to it. On this exercise I had plenty of other fish to fry.

Some students gave, essentially, verbal versions of the algebraic proof. (“Let x \in (A-B)-C. Then x\in A, x\not\in B, and x\not\in C. But then x\in (A-C)-B.” [Then say it again backwards.]) Some even went so far as to mention the algebraic “laws” (commutativity and associativity) that apply in the equation-based proof. I considered this at least partly my fault. Some, of course, blanked out entirely or otherwise missed the point of an “element” style proof. This is not my fault.

Anyhow, here is that algebraic proof. Probably most of the right answers… but not all… used the same steps in (essentially) the same order as I did so I’ll just write mine out again here.

$(A-B)-C$
$= (A-B)\cap C^c$ (Set Difference)
$= (A\cap B^c) \cap C^c$ (Set Difference)
$= A \cap (B^c \cap C^c)$ (Associativity)
$= A \cap (C^c \cap B^c)$ (Commutativity)
$= (A \cap C^c) \cap B^c$ (Associativity)
$= (A\cap C^c) - B$ (Set Difference)
$= (A-C)-B$ (Set Difference)

OK. Breakfast.

### Exam of November 4, 2010

1. Compute the sum: $\sum_{i=1}^3 i^3$.

2. Prove by induction: $(\forall n \in {\Bbb Z}^+)6|(7^n - 1)$.
(“For every integer n greater than or equal to 1, six divides $7^n - 1$“.)

3. Prove that whenever a mod 6 = 3 and b mod 6 = 2, it is also true that ab mod 6 = 0.
(Remark: this shows that the “Zero Product Law” $(a \ne 0 \wedge b \ne 0) \rightarrow (ab \ne 0)$ is false in certain number systems.)

4. Prove that $(\forall x, y \in {\Bbb Q})xy \in {\Bbb Q}$.
(“The product of any two rational numbers is a rational number”.)

5. Prove by contradiction that there is no greatest Real number less than 17.

6. Prove that there is an odd integer k such that k mod 7 = 4.

• ## (Partial) Contents Page

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