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3/3/2003 | An ever-shrinking class accompanied me as I finished the
elementary "technology" of limits. So here's an outline:
- Fundamental facts
- Limits are unique
- The set of numbers in a convergent sequence is bounded*
- Only tails matter
- Limits and arithmetic
Theorems about sums and products and reciprocals and quotients - Limits and order
- If (x
_{n}) converges to x then: x>0 implies x_{n}eventually>0.
- If (x
_{n}) converges to x then: x_{n}>=0 for all large enough n implies x>=0. - The Squeeze Theorem**
- If (x
- Examples
Polynomials, rational functions, and more coming today!***
We showed that if (x _{n}) converges to x and if S={x_{n} : n in N} then
the set S is bounded. Is the converse true? That is, if the set of
elements of a sequence are bounded, need the sequence converge?
Ms. Greenbaum suggested that we look at the sequence whose elements
were defined by the formula x_{n}=(-1)^{n}. Then (x_{n}) "alternates" between
+1 and -1, and it seems hard to believe that the sequence could
converge. To show that (x_{n}) does not converge to x, however,
we would need to verify the following: there is some epsilon>0 so
that for every N in N, there is n>N so that |x_{n}-x|>=epsilon.
Note that the choice of epsilon might well depend on x. In this case,
there is a clever way to see that convergence can't occur. It
depends on the fact that the distance from -1 to 1 is 2. If we choose
epsilon to be 1, then for large n, the distance from x_{n} to x should be
less than 1. But the values of x_{n} are +/-1. So go from 1 to -1 "by
way of" x, and get a contradiction. More precisely, we will use the
triangle inequality. Here is how:
If (x
Now let's try **the famous
Here is a somewhat poetic (?) view of the theorem. n increases as
pictures of the real line descend. I am trying to illustrate how
b
I remarked that a proof similar to this but with more complicated
details can be used to verify the following theorem:
Now we can do a more complicated version of the Math 152
workshop:
The I'll go on to 3.3 next time. I encouraged people to send in answers to review questions, and I will post these answers. Due a week from today (Monday, March 10) are these textbook problems: 3.2: 4, 9, 11, 13a, 16. | ||||||||||||||||||||||||||||||

2/27/2003 | We continued to trek through the rough landscape of
limits. My online dictionary defines the verb "trek" as 1. travel or make one's way arduously ("trekking through the forest"). 2. esp. [hist.] migrate or journey with one's belongings by ox-wagon. 3. (of an ox) draw a vehicle or pull a load. The only ox in sight is the instructor.
We proved the reciprocal theorem. A reward was given to all students present when this proof was completed.
Whew! Now I remarked that we can do a few more examples, officially.
These examples are still rather baby calculus, but we can
I then tried to "perturb" this theorem a bit. I asked what happened if we changed the inequalities.
What about the converse? So we could look at some assertion similar
to:
In fact x does
A simple repertoire of examples is very useful to check
ideas! My online dictionary gives this as one definition of
"repertiore":
Again, I tried to examine what would happen if we changed the inequalities a bit.
We are quite close to moderately sophisticated mathematics. Here is an
almost realistic scenario. Suppose we want to solve some sort of
equation, f(x)=0. We might have a root-finding technique (or even
several of them) which creates a sequence of better and better
approximations to a root. Well we might have the following setup:
(x
We decided, in fact, that in this case the limits would be equal. That
is, x=y. Why would this occur? Well, we need all the information, but
the key is the estimate with 1/n. We do know:
Now we look at z I handed out review problems for the first exam, which will be given in a week. | ||||||||||||||||||||||||||||||

2/26/2003 | I continued to try to assemble a technological base to study
sequences. This is rather intricate, and some parts of it have details
which are not fun.
The final fundamental result that we need is this:
We see what happens with addition, multiplication, etc. Most of the results here are very familiar to any long-suffering calculus student, but the interest in this course is building the detailed proof structure. Addition is fairly easy.
That one isn't too bad. Slightly more difficult is multiplication.
Here's the last major algebraic result about limits (we'll use it
to get an additional statement about quotients, of
course!):
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2/24/2003 | Several announcements: Because the material of Workshop #4 is both important and complicated, I will meet with students in 6:30 PM on Thursday, February 27, in Hill 425, and I hope to write out complete solutions to all seven of these problems. As a spectator sport alone this should be fun (for you, maybe, and less so for me!). If someone takes good notes, I can scan them and put them on the web. We will have an exam a week from Thursday (in class on March 3). I will give out a review sheet this Thursday instead of more workshop problems. The review sheet will contain full information about the exam (the format, sections to be covered, etc.). It will also have sample problems, mostly problems I will get from other instructors' past exams. The reading for the week is sections 3.1 and 3.2 and 3.3.
I again wrote the definition of the
I tried an example: suppose x
There are several comments to make about this formal proof. It seems really irritating, somehow redoing things we've already done. And it seems long and clumsy. In its defense I remark, as I did in class, that we right now have essentially no "technology" to do proofs about limits: all we have is the definition of limit, and that's the only way we can verify that a sequence has a limit.
I changed the problem a bit, from x
I remarked that the balance of the week would be spent in developing "technology" to handle limits, so we won't have to verify every limit statement by direct checking of the definition. That would indeed be rather tedious. The agenda will be: - General properties of limits ("simple" things, but in 311
everything must be
*proved*!) - Limits and algebra (such as limits of sums, quotients, etc.)
- Limits and order (when the terms of one sequence are less than another then etc.)
- Systematic Examples (so that we don't need to "reinvent the wheel" whenever we see a new limit).
General properties.
Then I went on to ask how sup's and inf's interact with limits. More
specifically, suppose (x
The last general principle I wanted to state was illustrated by the
following example: suppose (x | ||||||||||||||||||||||||||||||

2/20/2003 | I first discussed yesterday's "Question of the Day". Here the
set S was the numbers, x, which could be written as (M-N)/(M+N) where
M and N were elements of N, the positive integers. I asked if S
were bounded above, and if it was, what sup(S) was.
We discussed this question. A suggestion that 1 was an upper bound of
S was made. The assertion (M-N)/(M+N)<=1 would need to be
verified. But this is not too hard, because if we begin with M<M,
certainly a true statement, and then subtract a positive number (N)
from the left-hand side, and add that positive number to the
right-hand side, then M-N<=M+N. Since M+N>0, we can divide by
M+N to get (M-N)/(M+N)<=1. Therefore 1 Then the suggestion was made that 1 is actually sup(S). It isn't totally clear to me how such things are guessed: look at lots of examples, try to see what happens when things get "big", etc. It is hard to explain human invention, and this is a modest example of such invention.
How can one verify that sup(S)=1? I suggested looking at the
difference between 1 and a typical element of S. So consider
1-(M-N)/(M+N)=[(M+N)-(M-N)]/(M+N)=2N/(M+N). I would like this to be
"small" to establish that 1 is the Similarly one can verify that this S is bounded below, and that inf(S)=-1. I remarked that it would be easy to get more examples of sets following this description. I could replace the "formula" x=(M-N)/(M+N) by something like - x=(7M-3N)/(4M+5N). Then S would be bounded, and I think sup(S) would be 7/4 and inf(S) would be 3/5.
- x=(3M
^{2}-4N)/(2M^{2}+N). Then S would be bounded, and I think sup(S) would be 3/2 and inf(S) would be -4. - x=(3M
^{2}-2N)/(5M^{3}+7N). Then I think S would be bounded, and sup(S) would be 3/5 and inf(S) would be -2/7. - x=(5M-7N
^{3})/(3M+2N^{2}). Then I think this S is bounded above, with sup(S)=5/3. But S is not bounded below.
Now we come to the definition of
I decided to do something weird, to try to write the negation of the
statement above. So:
Notice that if we want to prove that a sequence does I asked students to begin reading chapter 3 (at least 3.1 and 3.2), and to hand in textbook problems 2.4: 3, 4a and 3.1: 8, 10 in a week. Another set of workshop problems was also handed out. Also, I announced that an exam would probably be given in about two weeks. | ||||||||||||||||||||||||||||||

2/19/2003 | I discussed the material following the last diary entry's
"Pedagogical Error" statement.
We went over some of the workshop problems from workshop #4. I tried
to emphasize that finding examples and understanding the statements
were important. I repeated my comments written in the last diary entry
about the general structure of many of the proofs at this point in the
course (contradiction, inequalities, Archimedean property). In almost
an hour of discussion, we were only able to cover some aspects of
problems 2 and 3. So investigating these problems is very difficult,
and very painstaking.
Somewhat more detail Another example? We considered A=(0,1). Here A is nonempty, and A consists of positive numbers, and I asked why (III) was satisfied. We thought about this, and if a number is between 0 and 1, two-thirds of it would also be, since 2/3 is positive and 2/3 is less than 1. After some inquiry, we were given (0,1) union {2} as another example. This certainly "works".
And another example: we considered S whose elements were 1/n where n
was in
Now I made the game
The last requirements seem to mean that S has lots of holes. The
suggestion was made that we consider an S which is the collection of
For problem #3, the examples we considered seem far more
"routine". For example, we looked at intervals "drifting" to the left:
one example was A Then I tried to verify orally the equality suggested in the proof. I don't know how successful I was. What was the strategy I used? This: after we established that the two sup's mentioned in the problem exist, I also remarked that to prove A=B, we may prove both A<=B and A>=B. This strategy will certainly work for the equality requested in problem 3.
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2/13/2003 | Today's the last day I will try to "cover" material in
chapter 2. I will attempt on Monday to discuss with students
some of the workshop problems.The results today contain further nearly paradoxical consequences of completeness. The discussion will also continue using various proof techniques which are common to the subject:
try to see the structure of these proofs in spite
of all of the details. All of the results below are discussed and
proved in detail in the text (section 2.4).
More special case: 0<x<y & y-x>1. Here the length of the
interval is greater than 1. I'll
What if v-1 is So we have found an integer in (x,y) if y-x>1. Let's go back now to a less specific case: what if we only know that y-x>0? At the suggestion of a student, we will "stretch" the interval (x,y).
I claim there is m in Sigh. So we have "handled" 0<x<y. What about 0=x<y? Well we can use the fact that 0<y/2<y and use the previous case: we can find a rational number between y/2 and y, so there must be one between y and 0. What about the case x<0<y? Well, we can use the previous case and find a rational number between 0 and y, and that will work here. Sigh. Now move "down": we have two further cases: x<y=0 and x<y<0. Both of these can be "transformed" into an earlier case by multiplying the inequalities by -1. Then positive rationals are found in the transformed cases, and changed back to the current cases by multiplying those rationals by -1: so if x<y<0, for example, then 0<-x<-y so that there is n/m with 0<-x<n/m<-y and therefore 0>x>-n/m>y. So there is a rational number in every interval of positive length. Indeed, if you keep chopping up intervals, you can actually find an infinite number of rational numbers inside every interval of positive length!
Now the task is to analyze the irrationals. Well, here is a very
irritating proof of a statement:
So A has a least upper bound. Let's call
The case
The case
We know Therefore we have eliminated the two "undesirable" alternatives, and the Completenenss Axiom has guaranteed us a positive real number whose square is 2.
In calculus we could "create" a square root of 2 by considering the
graph of y=x Now we know one specific irrational number (but in fact, by the counting argument there are lots and lots and lots of them!). The enthusiastic and masochistic student could try the same argument to get a square root of 3 or a cube root of 4, etc. We now can "create" many irrational numbers: if P and Q are rational numbers, and Q is not 0, then I claim that P+Q*sqrt(2) is also irrational. Because if it were rational, say a rational number M, then sqrt(2)=(M-P)/Q would be rational which is false. Using this idea (multiples of sqrt(2) properly scaled) we could prove:
Perhaps a Pedagogical Error!
I may have made a serious error in exposition here. For the purposes of the course, the results I would like students to "internalize" from this lecture are the following: - Any interval of positive length in the reals must contain rational numbers.
- Any interval of positive length in the reals must contain irrational numbers.
infinitely many both rationals and irrationals.)
The first statement follows fairly directly from the Archimedean
Property, and is discussed above. The second result got a very, very
tortuous exposition. The online dictionary defines "tortuous" as
Let me discuss a different way of getting the result desired. The
"counting argument" already establishes that there
This idea should also appeal to a certain "naive" feeling about the
"real line": the real line is or should be "homogeneous" -- any one
piece of it should "look like" another piece. So [0,1] and [53,54] and
[-109,-108] should "look" the same. And that's what the discussion in
the previous paragraph says. We can go further. Look again at
[0,1]. We now know it has irrationals in it. If we break up [0,1] into
intervals, say, [0,1/3] and [1/3,2/3] and [2/3,1], then these look
alike, also. If one of these intervals has irrationals in it, then
translate by +/-1/3 or +/-2/3 to get irrationals in the others. And if
one of those intervals does Of course students must decide if they like this argument. One defect, certainly, is its "non-constructive" nature: no specific irrational is actually exhibited. It is a "soft" argument. But maybe it is easier to understand, and we now do get the density of the irrationals nicely and easily (at least, "nicely and easily" to me -- with much less "sweat" than the sqrt(2) proof above). | ||||||||||||||||||||||||||||||

2/12/2003 | I vowed to (try to) do stuff only in the book
today. And I will finish chapter 2. So I will go from back to front
(!) in chapter 2.
So, beginning backwards:
I remarked that I would
Continuing backwards, I stated the
If I=[a,b] and J=[c,d], then I is a subset of J if and only if c<=a
and b<=d. Note that almost everything from now on in the course
will have inequalities in it, and this is one example.
Therefore the "nested sequence" I - a
_{n}<=b_{n}for all n in**N**(so I_{n}is*not*empty). - a
_{n}<=a_{n+1}and b_{n+1}<=b_{n}for all n in**N**(so I_{n+1}is a subset of I_{n}).
_{n}'s? We will use the Completeness Axiom much as we did
last time. We will apply the axiom to the set L={x=a_{n} : n in N},
the set of all left-handed endpoints of the intervals. There are two
hypotheses to check:I L is not empty. This is true because a_{1} is in
L.II L has an upper bound. Let me be more precise than I was in
class:
I claim that b
(I) and (II) allow us to apply the Completeness Axiom, and we know
that sup(
Now we know that b
1. If I
2. If I
3. If I
4. If I
5. Now I started fussing with various other hypotheses. What if we
remove "bounded" from the hypotheses? A closed unbounded interval is
one of two types: [A,infty)={x in
6. What if we changed "closed" to "open" in the hypotheses, but kept
all the other words: that is, we look at nested sequences of open
bounded intervals. An interval is open if it is of the form (A,B)={x
in Next time I will continue backwards and finish chapter 2. I will verify the density of the rationals, the existence of sqrt(2), and then the density of the irrationals. | ||||||||||||||||||||||||||||||

2/10/2003 |
Students have made some errors in written work (textbook homework,
workshop problems, class work) which are significant and common. I
hope that by pointing them out I can help people avoid these errors. I
also add some general comments about written work in this course.
- You
*cannot*prove a logical implication of the form "If P then Q" by assuming Q and making deductions from that assumption. This procedure is invalid. - Use of the word "equal" or the notation "=" to mean "logically equivalent" or "implies" (either!) is incorrect. It is bad grammar or syntax or ...
- In what a reader might presume is an induction proof, the assertion "n+1 is true" makes no sense. The adjectives true and false commonly apply to statements or assertions. They do not apply to integers.
- Poorly specified referents. Almost every student in the course has
written statements similar to the quoted typical examples which
follow.
- "if we multiply the right hand [the word "side" is missing!] of the inequality" on a page with more than 10 preceding inequalities, none close to the statement given.
- "assume that it holds for n" with no additional clarification of "it".
- Poor proofreading. This is close to a pun, but I mean both "proofreading" with its standard use: to find and correct errors, and reading a proof to make sure it is logically sound. A first casual writeup will almost surely not be adequate in this course. Arguments in 311 need to be precisely stated, and may be rather intricate. Misspelled words, sentence fragments, and circular arguments will detract from your work and may invalidate it.
Please proofread what you hand in. Ideally, you should read and reread and revise almost any formal communication. Neatness and clarity count, as you darn well know if you've tried to read any complicated document.
I began by rewriting a great deal of what I wrote last time. Then I
asked students to write answers to
The online dictionary tells me this about daphne:
Then I turned to a analysis of the bad consequence (b) of
completeness. This is more elaborate. I began with reminders (from the
lecture of 1/23). I reminded people of the definition of a finite set
(a set that had a bijection with {1,2,3,...,n} for some n in
What can "computer science" do? Let me take a somewhat simplistic view
of what computers can do. They run with programs, and programs act on
inputs. Each program is a finite sequence of instructions. Each input
is a finite sequence of symbols. This vision of computation is
"deterministic", by which I mean that the program + the input
completely specify what will happen. This is a coarse (?)
view of programs, but if you accept it a rather startling conclusion
can be gotten. How many programs are there of length, say, 78? No
matter what the computer language, anyone would agree that the total
number of computer programs of a fixed length is finite. Here I don't
even care how big this number is, just that it is finite (a list of
all strings of symbols of a certain length). How many inputs of length
78 are there? Again, finite (another long list). Therefore, the number
of outputs of computer programs which are length 78 acting on inputs
of length 78 is finite. Note that 78 is not special here. The number
of outputs of computer programs which are length N acting on inputs of
length N is finite for any N in
More or less, the discussion of (a) might convince someone that the
Completeness Axiom proves there aren't Although objections (a) and (b) above can be distracting at times, from now on we will officially "believe" in the Completeness Axiom. As for the students, it is, as I remarked in class, something you should believe (!) if only for the purposes of the course (just as a student who believed in creationism might need to learn enough about evolution to answer some test questions correctly). | ||||||||||||||||||||||||||||||

2/7/2003 | Although I've tried to delay it as long as possible, today we
begin the deep stuff in the course. Our aim is
## NO HOLES!When you are done, please read sections 2.3, 2.4, and 2.5.
We begin with a bunch of definitions>
These definitions are actually designed to be tested -- the terms are
defined by asserting that there are w's for which certain comparitive
statements are
(Counter)example(s):
- v is an upper bound of S: if x is in S, then x<=v.
- If w is any upper bound of S, then v<=w. (v is smaller than any other upper bound of S.)
- v is an lower bound of S: if x is in S, then x>=v.
- If w is any lower bound of S, then v>=w. (v is larger than any other lower bound of S.)
We discussed some examples, which are here displayed in tabular form. The examples were accumulated during the class discussion. In addition to least upper bound and greatest lower bound, I also included maximum and minimum, as described previously in these lectures.
The next piece of business is restating the last axiom we need to do "calculus": Completeness Axiom Suppose S is a non-empty subset of R. If S
is bounded above, then S has a least upper bound.
This axiom carries considerable philosophical "baggage". The axiom
will get rid of the "holes", at least if we identify cuts with
holes. In the Dedekind approach, some cuts in How can we "identify" the least upper bound of a set? First, the text calls the least upper bound of S, sup(S), pronounced "soup of S". The greatest lower bound is inf(S). sup and inf are abbreviations of Latin words, supremum and infimum.
sup(S) is unique. By that I mean if both Bob and Charlie are sup's of
S, then Bob=Charlie. That's because if Bob is a sup, then it is an
upper bound. Since Charlie is a To have a sup, a set must have an upper bound. If a set has a maximum, then the maximum is the sup of the set. But a set can have a sup without having a maximum. (Similar statements are true for inf's and minimums.) How can one try to "find" a sup? Well, if S is a non-empty set and if S is bounded above, then the completeness axiom says that S has a sup, and the discussion above declares that the sup is unique. But what "properties" does sup(S) have (besides the definition)? Here is something to think about.
If w=sup(S), consider w-(1/2). Is w-(1/2) an upper bound of S? If it
were an upper bound (subjunctive courtesy of Ms. Greenbaum), then it
would be an upper bound less than w (since 1/2 is positive). But w is
the least upper bound (the second part of the definition of least
upper bound) so this is not possible. Therefore w-(1/2) is
I more or less constantly use this theorem to find sup's of
sets. Here's a picture (?) of how I think (!!):
I think of the upper bounds of a set S as a collection of walls
"marching down" from far to
the right, getting closer and closer to S. They begin to pile up, and
where they "stop" is sup(S). The content of the theorem is that if I
want to move the walls the least little bit to the left, something in
S will end up on the wrong (right) side. Now of course the picture I
have drawn is somewhat simple. The S seems to have 3 pieces, and in
most of the more interesting applications of all this that we will
consider, S will have The ideas I am outlining in this lecture took almost a century to shape. Although the reasoning is all displayed, it is often subtle, and may be difficult to understand. Here's one significant application.
I note that without the completeness axiom there are "models" of
ordered fields where
I will verify this next time. The following textbook problems will be due next Thursday: 2.3: 5 and 6 and 2.4: 2 and 8. | ||||||||||||||||||||||||||||||

2/5/2003 | I returned textbook homework (graded on the basis of 4 points
per problem). I returned the proofs of Bernoulli's inequality. I urged
students who had not done well on the proof of Bernoulli's inequality
and who also had time and energy to do the following: please read the
proof in the book. Please read my comments on the proof that you
handed in. Please then, flip the page of what you handed in, and in at
most 10 minutes, write a proof of Bernoulli's inequality. I will
regrade what you hand in tomorrow. I urge you to do this. I want your
work to improve and I am willing to help. I also gave out the next workshop problems. I strongly urged people to work in a group and to proofread each other's work. This is good practice. Covered with virus particles of the common cold, I persisted in trying to instruct. I proceeded further in the book. I discussed the content of section 2.2, absolute value. In this I totally followed the text, having little energy for anything else. The proofs I used are almost all in the text. The proof of the Triangle Inequality is quite "cute", and when I have taught the Triangle Inequality in more elementary courses, I have verified it using an argument of cases depending upon the "sign" of various terms. The most important results I deduced we will use repeatedly in the course. Here they are: - |AB|=|A| |B|.
- |A+B| is less than or equal to |A|+|B| (the
*Triangle Inequality)* - |A|-|B| is less than or equal to |A+B| (a sort of
*reverse*triangle inequality)
I tried feebly to apply the results on absolute value in a fashion
that will be typical in this course. Consider the function
f(x)=x
An upper bound is clear with the triangle inequality:
|x
An underestimate is a bit trickier. We have |A|-|B|<=|A+B|. To
analyze |x
The "split" should be different. Take A=-15x A "beginning" student in this subject might find this all highly unsatisfactory. You can get some information with a certain sequence of algebraic "manipulations" and not get any with another, very parallel collection of manipulations. How can one tell what to do?
I enlarged upon this by asking for positive over- and under- estimates
of the same |f(x)| for x in the interval from [100,200]. Here the
overestimate proceeds exactly as before with the triangle inequality,
and we know that
|f(x)|<=(200) Note that the decomposition used for |f(x)| depended on the domain. Sigh. How can one tell what to do? I also mentioned that we can "invert" or take the multiplicative inverse of all of these inequalities and obtain positive over- and underestimates of 1/|f(x)| on [1,2] and [100,200]. Frequently one can concatenate (online dictionary: v.tr. link together (a chain of events, things, etc.) estimates of this type to work with rational functions, which are quotients of polynomials.
I did a problem from the textbook: 2.2: 8(a). For which x is
|x-1|>|x+1|. Several strategies were suggested for dealing with
this inequality. We could use Trichotomy and an analysis of
cases. That is, see where x-1>0 and etc.: lots of work. Or we could
use the following idea: A>B>=0, if and only if
A
So we investigate |x-1|>|x+1|, an inequality involving non-negative
numbers, by equivalently considering its square:
(|x-1|) We'll move on next time. | ||||||||||||||||||||||||||||||

2/3/2003 | The instructor asked the students to write a proof of
Bernoulli's inequality.
Then we proved the following: - If a
_{j}<a_{n+1}, then M=a_{n+1}. - If a
_{j}=a_{n+1}, then M=a_{j}. - If a
_{j}>a_{n+1}, then M=a_{j}.
_{something}. In case 1, M is greater than or equal to
a_{n+1} and M>a_{j} which is in turn greater than or equal to all the
other a_{n}'s. In case 2 and case 3 are similar (and were discussed in
detail in class!). Therefore P(n+1) is proved.We have completed our math induction proof and the theorem is true. Comment I certainly do not recommend the outline of this
proof as an algorithm to be implemented to find the maximum or to sort
a list of 100,000 numbers. There are much faster ways to do such
things!I commented that the logical outline of this proof (math induction!) is what's needed for most of the workshop problems. Problems 1 and 2 need more effort for the base case, while problem 3 needs more work for the inductive step.
Most students successfully answered the question I asked last time, to
find a set which had a max and no min. One simple answer is the
following: {x in
Consider - Neither
**L**nor**R**are empty (I forgot this in class until Mr. Benson called it to my attention! I thank him again and regret that I was careless!) - The union of
**L**and**R**is all of**F**. (Every number is in one of the pieces!) - The intersection of
**L**and**R**is empty. (No number is in both pieces!) - (Most important) If l is in
**L**and r is in**R**, then l<r.
Examples of cuts:
In the second example,
It is almost "clear" (considering the positive and negative integers)
that any
We discussed this a while. Let's consider such an example. Suppose the
maximum of
But what about the
In fact,
Dedekind's original discussion of cuts is available:
I began discussing absolute value. I defined it as in section 2.2 of
the textbook. I asked if "absolute value" was a function with domain
A function from A to B with domain equal to A is a subset W of
A | ||||||||||||||||||||||||||||||

1/30/2003 |
After an analysis of the moral positions of the instructor and the
students (showing that the students are much superior to the
instructor) I began the class. I first wanted to establish some common
methods for showing that "things" must be 0. In this, and in all that
follows in the course, we will assume that F is an ordered field (it
satisfies the 9 algebraic axioms and the 3 order axioms). I will
reserve using R for the real numbers until we get to the completeness
axiom.
Let's discuss the status of
Then I drew some illegal pictures (officially illegal in this course, that is). I attempted to show the part of the "real line" just (?) to the right of 0 and showed by picture that there was no point to the right of 0 which was closest. I attempted to show the part of the line just to the left of 1 and tried to illustrate that there was no point just to the left of 1 which was closest.
Next time I will prove that any finite nonempty set has a maximum and a minimum. But I will begin the class by asking students to write one of the following (I'll allow about 10 minutes): - Prove: If ab>0, then either a>0 and b>0 or a<0 and b<0.
- Briefly discuss this implication:
If a<b, then (1/b)<(1/a). Is this always true? Are there simple conditions which will guarantee that it is true? Are there examples showing that it is sometimes not true? - Prove Bernoulli's inequality: if x>-1 and n is in
**N**, then (1+x)^{n}is greater than or equal to 1+nx.
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1/29/2003 | I postponed the due date for the textbook homework until
tomorrow. I want to continue working with the algebraic axioms a bit
more.
Then I did some textbook problems. (Note that the "observation" above is 1b in section 2.1.) I think I did the following:
I again wrote the order axioms, and stated that we were now going to investigate their consequences. (O1) and (O2) are called "P is closed under addition and multiplication."
Now I can assert the following: the complex numbers cannot be made
into an
Also the integers mod 5 cannot be made into an ordered field. THis is
because 1 must be in
We can add inequalities: if a<b and c<d, then a+c<b+d.
We showed with an example (a=0,b=1,c=-1,d=0) that a similar
I returned the Entrance Exam and gave out the first workshop problems which must be done by groups of students. | ||||||||||||||||||||||||||||||

1/27/2003 | I wrote the axioms for a field, and for an ordered field, and
for a complete ordered field. Here they are:
All this will take quite a while to digest. Today we will primarily work on the first group of axioms, The algebraic axioms.
Binary operation on a set s is a function whose domain is S
Then students proved:
The students proved:
I then defined some conventional words. "-a" (minus a) is the unique additive inverse of a. "b-a" ("b minus a", subtraction) is just b+(-a). And if a is not equal to 0. 1/a (1 over a) is the unique multiplicative inverse of a. And b/a (b divided by a) is exactly bx(1/a). And usually I'll abbreviate axb by a·b or even just ab.
A similar result is true for multiplicative inverses:
Here is a more subtle result which uses (D).
I remarked that we could already get some reward for this abstraction.
Solving linear equations
Next time, a bit more about algebra and then on to order. | ||||||||||||||||||||||||||||||

1/23/2003 | I briefly discussed possible responses to the fears
(worries?) of the survey. I urged students to question me as often as
they wish, see me outside of class, read the book, work on homework,
etc. Most students are experienced course takers (if that's the
phrase!) and I urge them to work actively here.
In this lecture I will discuss matters which are not vital to the
understanding of the central subject matter of the course, but which
certainly form part of the context that was known to people
originating this "central subject matter". the ideas almost all come
from the work of Georg Cantor in the late 19
If f:A->B is a function, I reviewed what injective or one-one and
surjective or onto and bijective mean. In a loose sense, if f were
injective, then (maybe!) A is "smaller" than B. If f were surjective,
then A might be called "larger" than B. Finally, the existence of a
bijection between A and B should mean, roughly, that A and B are the
same size: we have, with f, a way of pairing up the elements of A and
B. Formally, we say that A and B are the same size (the technical
phrase the
Then I ran through a string of definitions, all in the book. The empty
set (I'll use
A set is
Then I looked at the mapping D:
I want big sets.
By now you should be confused, as confused as many of Cantor's
contemporaries were when they first saw all of this. "Clearly"
In fact,
If S is a set, then the
Therefore Now that you are confused I will stop. What I'd like 311 students to get out of this is that the most naive measure of sizes of sets already leads to complications. I would like you to have some idea of answers to the following questions. - When do two sets have the "same size"?
- What's a finite set?
- What's an infinite set, and what are some of the "peculair" properties of an infinite set?
- What is a countably infinite set, and what are some examples?
- Are there infinite sets which are "bigger" than countably infinite sets?
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1/22/2003 | Introduction of the instructor and the course.
I introduced myself and mentioned the course name. I requested student
information.
I asked for a session of
I was asked what my reasons might be for a student to take the course. I had anticipated this question and replied: - The course is required. (I reported myself as surprised that
only 3 students wrote that, but I was assured that it was so
well-known that it wa
_{n}'t worth including.) - Learning intricate math is a pleasure and informative (combination of the responses about learning and pleasure given above).
- This is the "canon" in math (constructed by dead white European males, mostly in the 19th cent and a little in the 20th). This is what every person in math is expected to know, everywhere in the world now. Even if you don't like it, you should know enough about it to be knowledgeable about what you don't like. (Note: the online dictionary gives this as one meaning of "canon": "a collection or list of sacred books etc. accepted as genuine.").
- Historically, errors were made -- as calculus/analysis began to be applied to more and more situations, there were serious problems with applications and extensions of "clear" results. Serious attention to details, as in this course, will lead to a decrease of errors.
Although we will prove almost everything in the course, I wanted to
give some background and make sure we agreed on notation. So we
started with - 1 is in S
- the successor of any element in the set is in the set.
N.
This defining property is used in proofs by "mathematical
induction". It is equivalent to the following "well-ordering
property": every
The using mathematical induction and the "successor" function,
operations of addition and multiplication can be defined on
The equation a+x=b can be solved only "sometimes" in
The next desire is to solve equations of the form a·x=b. This
leads us to the "larger" (?) set
I mentioned one of the most famous results of mathematics: sqrt(2) is
irrational. Or, more formally, there is no rational number a/b whose
square is 2. This is Theorem 2.1.4 in the text (please look there if
you've never seen it!). This led me to consider the following
function, whose domain is the rationals,
It took a while to understand this function (I should give credit to
T. W. Körner from whose Cambridge University lectures I
"borrowed" this example). I emphasized that this function is
Another example, similar in spirit, is the function
G(x)=x
The major content of the course is the transition or enlargement of I handed out the Entrance Exam. I urged students to begin reading chapter 1, and remarked that 1.1 and 1.2 should be familiar to them. I would try to discuss, more or less informally, the content of 1.3 next time, and would really begin the serious, proof-emphasizing part of the course next week. |

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