Thanks - Ashley

We can thus consider the set of all sets and subsets of an infinite set as having a cardinality equal to the cardinality of the base set (in this case, the Natural numbers set N). But, what about that 1-1 correspondence between the set of all Real numbers to the set of all Natural numbers? If we can divide the set of all Natural numbers up into an infinite number of subsets whos cadinality is equal to the parent set then how can we fail to find a corresponding element in the set of Natural numbers to the set of Real numbers? I have shown that I can divide the Natural numbers into an infinite set of infinite sets. Further more, each subset of infinite cardinality can contain completely unique elements from any other division of the parent set. So, before I have even made a dent in exhausting the elements of the Natural numbers, I can divide off a subset that could itself be divided off into a collection of infinite sets just to cover the elements between 0 and 1 in the Real number set. If there aren't enough, I can divide that sub sub sub set again and still have an infinily divisible set to cover the correspondance with. I would never lack for a unique element from the Natural numbers set to match up with an element to the Real numbers set.

I know that the numbers in the Natural number set would become absolutly enormous by the time we covered the interval from 0 to 1 in the Reals but, so what, I still have plenty to move to the next interval. In fact I still have the same number of elements in the Natural numbers that I had before I used up an infinitely small fraction of them to cover all previous intervals. I am not going to run out of Natural numbers by placing them in a 1 to 1 correspondence with any other set!

2e^x -2e^(-x)/4

I set it equal to zero, but neither parts of the fraction can ever equal zero, right?

If this is the case, how do I tell when it is increasing or decreasing?

Theorem 2 requires two halfs. For each half, you can do whatever proof technique (direct, contradiction, or contrapositive) makes sense to you. You can do each half differently. This one is probably the hardest of the four theorems (in my opinion) despite looking easy on the surface.

Try weak induction for Theorem 3. This problem is one of my favorites. To get some intution, take the picture that I drew for 5 points and ask yourself how many additional lines you would need to add if you added a 6th point. And then how many new lines if you added a 7th? How many new lines would you need to add to a circle with $k$ points and all the lines if you added one more point? Once you set this one up correctly, it is very short.

For Theorem 4, use strong induction and mimick the proofs we've done involving the Fibonacci sequence (except this isn't the Fibonacci sequence). If you know how to phrase the first sentence of the inductive step correctly, there aren't really any tricks. Don't get bogged down in the indices. The recurrence relation says: each term is equal to 5 times the previous term minus 6 times the terms before that. For example, $a_{11}=5a_{10}-6a_{9}$. Despite appearances, this one is quite friendly.

Thank you,
Saddened Exams Students (Ian & Tabby)

9) If f''(x) doesn't ever equal 0, what steps should we take to describe the concavity?

EDIT: I figured this out too. -_-

y= 6x+sin(3x)—-> I understand how to find the first derative using the chain rule and get

y'=6+ 3cos(3x)—> this is where I am forgetting how to use the rules to get the second derative the 6 is a constant so it goes away, but how do I deal with the 3(cos)3x I tried using the product rule then the chain rule but I am confusing myself.

Once I do get that derative how do I deal with that -sinx that I will eventually get

14) The equation for the least material used would be $A=h(2\pi r)+8\pi r^2$, or (height)(diameter of the top / bottom) + (surface area of the top + bottom). Since the volume = 1, I made the secondary equation $1=h \pi r^2$ which can be solved for h: $h=\frac{1}{\pi r^2}$. Plugging that back into the original surface area equation, I got: $A=\frac{2\pi r}{\pi r^2}+8r^2$ or just $A=8r^2+\frac{2}{r}$.

Taking the derivative of that, I got $A'=16r-\frac{2}{r^2}$ or $\frac{16r^3-2}{r^2}$.

Set that equal to 0: $0=16r^3-2$ (the $r^2$ in the denom. is multiplied out).

Do some algebra and $r=\frac{1}{2}$.

I think the feasible domain would just be $[0,\infty]$, and the 1st derivative test shows that there's a minimum at $r=\frac{1}{2}$.

The area of the top or bottom of this cylinder would be $\pi\frac{1}{2}^2$ or $\frac{\pi}{4}$. The height would be $\frac{1}{\pi\frac{1}{2}^2}$ or $\frac{4}{\pi}$.

The ratio of height to radius would be $\frac{\frac{4}{\pi}}{\frac{1}{2}}$ or $\frac{8}{\pi}$ (which is the only part that I got right, implying that I did the calculations right for the wrong formula? or something).

19) With x being the length of the side of the square you're cutting out, the equation for the volume of the box would be $V=x(1-2x)(\frac{1}{2}-2x)$ or $\frac{8x^3-6x^2+x}{2}$.

The derivative of this would be $\frac{(24x^2-12x+1)(2)}{4}$.

Equal to 0: $0=24x^2-12x+1$.

$x=.3943, x=.1057$

Feasible domain: $[0,\frac{1}{4}]$, get rid of the first solution for x.

1st der. test, x is a local max.

$h=.1057$
$w=.2886$
$l=.7886$

I set up the equation for the area of the rectangle. A=xy. I also have that x is equal to 20, because the problem states that one of the sides of the rectangle is along the diameter. I'm not sure where to go from here. I know I need another equation but I don't know if I need the circumfrence of the semi circle or the area. I also don't know what to do with this second equation.

I was doing 1. (C) and ran into the problem of one of my critical numbers being a negative root. What do I do with that? I thought you mentioned something about it in class, but I couldn't remember.

Also on 1. (D) How do I solve for the critical points of:

So when x = 0, $y'$ wouldn't exist, right? And when $y'$ DNE at x, then x is a critical value?

Using 0 as a critical value, I'm getting that x < 0 as negative and x > 0 as positive, implying that there is a local min at x = 0, but the answer in the book doesn't support this.

EDIT: Looking over my notes again, I understand where my confusion is coming from, but I may still need some clarification. The picture you used (example a, a crazy looking function) has an asymptote where there is intuitively no max or min. But then soon after in the notes I have written down that x = c is a critical value when $f'(c)=0$ or DNE.

Proof:
Let $H=[a_{H},b_{H}]$ and $K=[a_{K},b_{K}]$ where $H$ and $K$ are mutually exclusive.
Since $H$ and $K$ are both closed point sets they each have a left most point and a right most point (by the definition of a closed point set).
If there is a point $p \in H$ between any two points $m$ and $n$ of $K$, then by axiom 1.6, there is a point $q$ between both ($m$ and $q$) and ($q$ and $n$). Thus $H \cup K$ is not a closed interval.
If $H<K$ then by axiom 1.6 there exists some point $p_x$between the right most point of $H$
and the left most point of $K$ where $p_x \notin H \cup K$ and again, no colosed interval exists.
By symetrical arguement for the case where $K<H$ there is a point between the the left most point and right most point of $K \cup H$ that is not a member.
$\therefore$ the union of two mutually exclusive closed point sets in not a closed interval.

Proof:

Let $f$ be a function with domain $D$.
Let $S = [a_{S},b_{S}] \in D$.
Let there be a point $p \in S$.
Let there be a point set $Q = \{m|m = (x,f(x)) = (x,f(p))\}$ such that $Q \in S$.
The set $Q$ is bounded by $S$.
If $Q$ is a closed point set then $Q$ has a left most point and a right most point.
Since $Q$ is bounded by $S$, then the left most point of $Q$ can not be less then the left most point of $S$.
Thus the left most point of $Q$, $a_{Q} \geq a_{S}$.
By symetrical arguement the same can be said for the right most point of $Q$.
This shows that if $a_{Q} = a_{S}$ and $b_{Q} = b_{S}$ then $Q$ is closed.
For the case of $a_{Q} > a_{S}$ or $b_{Q} < b_{S}$, then since $f$ is continuous there are no
points $f(x) < a_{Q}$ where $f(x) \in Q$ thus $a_{Q}$ is the left most point of $Q$.
By symetrical arguement, $b_{Q}$ has no points greater within $Q$.
$\therefore Q$ is a closed point set.

Proof:

Let $S=\{x|x \in \mathbb{R}, 0<x<1\}$.
Since $S$ is made up of all $x \in \mathbb{R}$ between 0 and 1, $S$is an infinite set, meaning $S$ is not finite (by the definition of infinite).
$S$has a first point to the right equal to $1$where $1 \notin S$. (This is by the completeness axiom and the definition of an open set).
Let there be a point $p \in S$of the form $p = -{1-10^{n} \over 10^{n}}$where $n \in \mathbb{N}$. (note: It is considered evident that for any number $n$, $p$ then can be written in the form $0.999...9_{n}$ where the decimal expansion is equal in length to $n$).
Let there then be a point $q \in S$ where $q = -{1-10^{n+1} \over 10^{n+1}}$.
It is then demonstrable that $p<q$where $p \in S$ and $q \in S$.
As $S \in \mathbb{R}$ and is an infinite set, by the axiom that between any two points there is always another point in an infinite set, then for any value of $n, p<q<1$.
As $n \to \infty$, the length of $p$'s decimal expantion also $\to \infty$.
Since $q > p$, $q = 0.999...$ as $n \to \infty$.
Since $1$ is the first point to the right of $S$ and since $1 \notin S$ and $q \in S$, $q \neq 1$.
$\therefore 0.999... \neq 1$.

Problem 14: It seems like more information is needed here but I really don't even know where to start. Staring at my picture doesn't help.

And so that this post doesn't come off as just a useless vent, I humbly ask for help with problem #1. Are we supposed to be using the formula for the volume of a cylinder? So far I have that r = 20, dV/dt = -25, and dh/dt is what we need to find.

EDIT: Ugh, of course I figured it out right after posting this.

You need to pick one of our four definitions for continuity and probably stick with it throughout the proof. The proof I'm thinking of uses the definition involving open intervals. Here are some hints to get you started:

Let $S=(a,b)$ be an open interval containing $h(p)$. Since $f(p)=h(p)=g(p)$, $f(p)$ and $g(p)$ are also contained in $S$. We are assuming that $f$ and $g$ are continuous. What does our definition of continuity involving open sets tell us in this case? You should say something about two open intervals on the $x$-axis. Now, using the two intervals that you have associated to $f$ and $g$, you need to construct an open interval $T_h$ on the $x$-axis such that if $t\in T_h$, then $h(t)\in S$. At some point, you'll need to encounter a statement like: $a<f(t)\leq h(t)\leq g(t)<b$.

http://jeromyanglim.blogspot.com/2010/10/getting-started-with-writing.html

http://jeromyanglim.blogspot.com/2010/10/getting-started-with-writing.html

http://jeromyanglim.blogspot.com/2010/10/getting-started-with-writing.html

Secondly, I was going over limits with one of the MAC people, and he was talking about graphs. He kept saying we need a graph in order to complete certain limits, and I don't remember ever using a graph to compute limits. Because of this we both ended up confused haha. What are your thoughts on this?

Thridly, will there be limit problems on the test like in section 2.3? And will each problem be solved by either conjugate multiplication, factoring, simplifying, absolute value, or common denominator?

Thanks

Proof:

Let set $S = \{x | x \in \mathbb{R}$ and $0 < x < 1\}$.
Let $S$ converge to $1$ by definition of convergence to a point.
By construction and the definition of an open point set, $1 \notin S$.
By the definition of a limit point, $1$ is a limit point of $S$.
Let there be a point $p \in S$ such that there is also a point $b$ where $p < b < 1$.
(an example of such a set of points is where $p = 0.8$ and $b=0.9$, $0.8 < 0.9 < 1$).
By definition of a limit point, $b \in S$ and $b$ is between $p$ and $1$.
By the axiom, for any two points $p$ and $q$ there is a point between them.
By definition of first point to the right, $1$ is the first point to the right of $S$ and greater then every point in $S$.
By the definition of infinite, $S$ is infinite meaning not finite and there are always $n + 1$ points in $S$ where $n \in \mathbb{N}$ showing that $S$ does not meet the definition of finite.
Let $M$ be an open interval $(a, b)$ containing $1$ and $p$ by the definition of a limit point.
For all values of $p < 1$, $b \in M$.
For any value of $p < 0.999...$, $b > p$.
For some $p \leq 0.999...$, $b > p \in S$.
By the definition of infinite there is a value of $b = 0.999...$ for some $M$ where $b \in S$.
Since $b \in S$ and $1 \notin S$, then $b \neq 1$.
$\therefore 0.999... \neq 1$.

Also, section 4.4 Problem #1, I'm getting $\sin{2x}$ for an answer, which doesn't look like the answer in the back of the book.

i changed tangent to sine/cosine and multiplied by the reciprocal so i have

sine/cosine time 1/x

next i multiplied it out and got
sin(x)/cos(x)(x)

im not sure where to go from here

If $p_1, p_2, p_3, \ldots$ is a non-decreasing sequence and there is a point, $x$, to the right of each point of the sequence, then the sequence converges to some point.

Jeff claimed (please correct me if I am wrong) that the word every should have been used (as it is in the Completeness Axiom) instead of each. I claimed that it didn't matter which word was used and that the meaning was the same. I've just thought about this a little more and figured that I might as well share my thoughts.

The quantifier on $x$ indicates that we are talking about a single $x$. This single $x$ is to right of each point in the sequence. That is, if you pick a random $p_i$ in the sequence, $p_i<x$. This implies that $x$ is to the right of every point in the sequence. The $x$ is a "one size fits all" point. So, I maintain it doesn't matter whether we use each or every.

Contrast the statement of the theorem with the following statement:

For each point in the sequence, there is a point $x$ that is to the right.

In this case, the dependence of the quantifiers is reversed and we no longer have a "one size fits all" $x$. Each individual point $p_i$ in the sequence has a point $x_i$ to the right, where I am using a subscript on $x$ to emphasize that this point depends on $p_i$. The point to the right of $p_i$ may be different than the point to the right of $p_j$ (assuming $i\neq j$). This is a very different situation than that in Theorem 1.60.

First, let's recall the statement that we are trying to prove:

If $x$ and $y$ are odd integers, then $xy$ is odd.

To form the contrapositive, we need to negate the hypothesis and conclusion and reverse the order of the implication. The negation of "$xy$ is odd" is "$xy$ is even." However, the negation of "$x$ and $y$ are odd integers" is a bit trickier. Before negating, notice that the original hypothesis is equivalent to "$x$ is an odd integer and $y$ is an odd integer." The negation of this statement is "$x$ is an even integer or $y$ is an even integer". Notice that the "and" in the original statement becomes an "or". The "or" in this case is an inclusive "or", which means one or the other or both. This is an application of a rule of logic called DeMorgan's Law, which you can read more about here. One can quickly write down a truth table to convince themselves this must be the case. In addition, it matches our use of everyday language. Think of some statements that involve "and" and ask yourself what it would mean if the statement was false. (There are other cases to consider, but this should be enough to convince that our model is accurate.) OK, in summary, the contrapositive is:

If $xy$ is even, then $x$ is an even integer or $y$ is an even integer.

To prove the contrapositive directly, you would start your proof with something like: "Assume that $x$ and $y$ are integers such that $xy$ is even." Your goal is to conclude that $x$ is an even integer or $y$ is an even integer.

If you wanted to finish this up by using a proof by contradiction, there wouldn't be anything wrong with that. In this case, you would write something like: "For sake of a contradiction, assume that it is not the case that either $x$ is an even integer or $y$ is an even integer." This implies that both $x$ and $y$ are odd. Now, do what we have done in lots of problems. What can you conclude?

Another approach could be to first prove as a lemma that if $p$ is prime and divides $xy$ where $x,y\in\mathbb{Z}$, then $p$ must divide either $x$ or $y$. A special case of this is when $p=2$. By the way, if you remove the requirement that the divisor be prime, it is no longer true. Y'all provided a counterexample to this already on a previous homework assignment.

If people ask, I'll provide more hints on the other problems (which are not nearly as difficult as the one discussed above).

3. sqrt(x) / sqrt(625-x^2)

My solution:

(1/2x^-1/2)(sqrt(625-x^2))-(x^1/2)(-x/sqrt(625+x^2))

Over

(sqrt(625-x^2)^2)

For some reason it wont post syntex for me? What did I do wrong!?

Thanks!

All this can be said without using "infinity," potential or actual, or the pseudoconcept of "the number of natural numbers." One simply identifies a systematic way of representing rational numbers, and the rest follows.

I think Will's difficulty with infinities is (as I said elsewhere) ontological, not mathematical. The Zeno argument that best highlights the difficulty is not Achilles v Tortoise, but the much balder "stade" argument (from the Greek adverb for "standing" or "motionless", with a suggestion of punning on "stadion", the noun for "measure"). Take an arrow shot from a bow. It appears to move smoothly through the air. But at each instant of time, it has an exact (even, theoretically, calculable) position, and an exact measure from end to end. At each instant, then, the arrow simply IS as it is, in a fixed place at a fixed time. But time is simply a sequence of instants. If in no instant does the arrow move, then how can it move over any sequence of instants?
Unlike the Achilles, this does not involve us in summing smaller and smaller fractions of an interval, a mathematical task to which the concept of "limit" seems wholely adequate. Here we are summing genuine zeroes, which always sum to naught.

Where, then, is the motion? Zeno, of course, was arguing precisely that in reality there is none, and the fact that it appears to us to move shows that our perceptual world is not the real world. But there are other ways to approach the paradox. One (not so easy to defend) is to argue that there cannot be an instant of time. But however you pick it up, this paradox is about how things are in the world, not about mathematical entities.

First on number 1 part c, how do we begin the proof when there is nothing said about "divides" in the "P" but there is in the "Q"?

Second, on 1 part d, does the square root effect the outcome somehow?

Third, I am not sure how to fill out the "If Q, then P" and "If not P, then not Q" portions of the truth table.

Finally, for number 3 do you want a mathematical example or just a real life example in which the inverse and converse are not neccessarily true?

Sorry for all the questions.
Kayla

EDIT: Oh also, I understand that the derivative would be $\frac{x}{\sqrt{625-x^2}}$, but only because it was done out the "long" way in section 2.1 of the book. I'm basically trying to understand how it's done with the shortcuts we've been using.

EDIT AGAIN: Wow, so I just noticed at the end of the notes that we got on Monday, Dana mentions exactly this. Oops.

First, recall that you can always use Lurch from a computer on campus by following the instructions that were on the original Circle-Dot handout. In case, you don't have the Circle-Dot handout handy, you can get to it here.

If you want to install Lurch on your own computer, it is easy to do. Follow these instructions:

1. Go to http://lurch.sourceforge.net/
2. Click on "Downloads" in the menu at the top of the page.
3. Click on the big arrow next to "Download LurchLite."
4. The download should automatically begin.
5. Once the download has completed, double-click the file (if it doesn't open for you automatically).
6. If you are using a Mac, drag the application to your Applications folder.
7. If you are using a PC with MS Windows, then do whatever you usually do with applications.
8. Once Lurch (LurchLite technically) is installed, you can just double-click the application to get started.

If you have any trouble, post your questions here.