i want blog too

The First Stirrings of Structure

2011-08-09T08:54:00.000-07:00

Problem 26: If there exists a one-to-one correspondence, a mapping that is both one-to-one and onto, between S and T, then prove that a one-to-one correspondence also exists between T and S.

Please try the problem above before moving on to the rest of this post, but if you get stuck, then feel free to charge ahead, and then try it again when you're done.

Here are some concepts:

Identity mapping - The identity mapping is a mapping i from a set S to itself such that for any element s in S, i(s) = s.

Inverse mapping - If we let f: S → T be a one-to-one correspondence, then we can define an inverse mapping, f⁻¹: T → S, such that f⁻¹(t) = s only when f(s) = t. The reasons for limiting the inverse function to times when f is a one-to-one correspondence should be clear. If f were not one-to-one, then there would be at least one element in T that would map to multiple elements in S under a hypothetical f⁻¹, and therefore, there could be no inverse function defined in this way. If f were not onto, then there would be some element in T that mapped to nothing under f⁻¹and we would have the same problem.

The ideas of the identity mapping and the inverse mapping are linked through mapping composition in that for any one-to-one correspondence, f, we have f o f⁻¹= f⁻¹ o f = i. I would encourage you to verify this.

Problem 27: If i: S → S is the identity mapping on S, then prove that for any one-to-one correspondence, f: S → S, the following is true: f o i = i o f = f.

The last problem should indicate to us that one-to-one correspondences on a set along with the operation of composition might behave somewhat like a number system in that they have an identity element, i, and inverse elements. If we put all of these one-to-one correspondences in a set, then we have something that really starts to look like a number system.

A(S) - Let S be a set, and define A(S) to be the set of all one-to-one correspondences from S to itself.

Problem 28: Prove that i is an element of A(S).

Problem 29: Prove that if f is an element of A(S), then f⁻¹is also an element of A(S).

Problem 30: Let S have n elements. How many elements are in A(S)?

So A(S) combined with the operation of composition "looks like" a number system with some operation like addition or multiplication defined on it, but with functions rather than numbers. We can and should talk about this idea in the comments since this is exactly the kind of thing we will be dealing with at length when we get into groups and rings and everything else. This is abstract algebra.

Composition

2011-07-22T15:57:00.000-07:00

If we have two mappings, α: S → T and β: T → P, then we can compose them to make a new function, (α o β): S → P which maps every element in S to T through the α mapping and then to P through the β mapping. Another way of writing it is β(α(s)) = p. For example, if S = {a, b}, T = {1, 2, 3}, and P = {pugs, mastiffs, labradors, jack russells}, and if α maps a to 1 and b to 2, and β maps 1 to pugs, 2 to jack russells, and 3 to mastiffs, then α o β maps a to pugs and b to jack russells. It's like this:

Mapping composition example that is different than the example I wrote above since this one involves elements that are all black dots instead of letters, numbers, and dogs.

Problem 22: Is mapping composition commutative? That is, for two mappings, f: K → T and g: T → N, does f o g = g o f? What about for f: K → T and g: T → K? What about for f: T → T and g: T → T?

Problem 23: Is mapping composition associative? That is, for three mappings, f: K → T, g: T → N, and h: N → Q, does f o (g o h) = (f o g) o h?

Problem 24: Let f: S → T and g: T → N both be onto mappings. Is f o g also onto?

Problem 25: Let f: S → T and g: T → N both be one-to-one mappings. Is f o g also one-to-one?

What do we mean when we say that two mappings are equal?

2011-07-17T17:50:00.000-07:00

Problem 21: Let S = {1}, a set with one element (the number 1), and define mappings f and g from S to itself by f(x) = x and g(x) = x^2. Would you claim that f and g are equal? Why or why not?

Cartesian Product

2011-07-17T17:49:00.000-07:00

A Cartesian Product of two sets, S and T, is defined as the set of all ordered pairs, (s, t) where s is an element of S and t is an element of T. It looks like this - S × T. We say that two elements of S × T are equal if their corresponding components are equal. That is, (s, t) = (x, y) if and only if s = x and t = y.

A simple example: Let A = {a, b} and let B = {1, 2}, then  A × B = {(a, 1), (a, 2), (b, 1), (b, 2)}.

From two sets, S and T, we can construct two Cartesian Products -  S × T and  T × S. These are distinct, but obviously related, sets.

Problem 20: Show that there exists a one-to-one correspondence (a mapping that is both one-to-one and onto) between  S × T and  T × S.

Before I go, I just wanted to throw out an example involving the Cartesian Product and mappings that relates to a discussion about operators in the comments of an earlier post.

If we let Z be the set of integers, then we can define a mapping, a: Z × Z → Z, such that a(x, y) = x + y. So each ordered pair of integers is mapped to their sum. In this way, things that are normally considered binary operations, like addition, can be seen as mappings.

Equivalence Relations (again)

2011-07-12T14:13:00.000-07:00

If I may, I'd like to just pop in here and take another stab at explaining equivalence relations from the point of view of partitioning a set.

Let's take a set with elements:

Assume that each of these dots represents a distinct element in the set. We know, because we are familiar with the concept of equality, that each of these elements is equal to itself and only to itself. We can show this on the picture of the set by drawing boundaries around each element to separate it from all of the others:

Now, we can say, in terms of this picture, that two elements are equal if they are contained within the same boundary. This is the same as saying that each element is equal only to itself.

From here we can generalize the concept of equality by allowing more than one element inside the boundaries. So we could do something like this:

These boundaries no longer show equality, but some generalization of it that we call equivalence. In terms of the picture, two elements are equivalent if they are within the same boundaries. These boundaries can be whatever we need them to be. If the set is the set of integers, we could put all of the evens in one boundary and all of the odds in another. Under that example, two elements would be equivalent if they are both even or both odd.

Each section of elements is what we call an equivalence class. Under the odd/even example, there are two equivalence classes - one set with all of the odds and another with all of the evens. Under equality, there are as many equivalence classes as there are elements in the set - one for each element.

Earlier, you already showed that this idea of equivalence is actually a special type of relation between elements that is reflexive, symmetric, and transitive. Subsequently, I showed that any relation that is reflexive, symmetric, and transitive as defined on a set results in these kinds of internal boundaries on the set.

I hope that this helps somewhat.

Mappings

2011-07-08T16:03:00.000-07:00

A mapping from one set, S, to another, T, is a rule that assigns each element of S to a unique element in T. You probably know this idea better as a function, but sometimes people call them mappings. Under a mapping every element, s, in S must be assigned one and only one corresponding element, t, in T. So you can do this:

f: S → T

and this:

f: S → T (not all elements in T must be mapped to and some can be mapped to multiple times)

but never this:

Here, there is one element in S that maps to two elements in T and one that maps to nothing! So while f might be a legit mathematical construct, it is not what we call a mapping.

Here are some examples of mappings:

Let f be a mapping from the set of real numbers to itself such that every element is mapped to its square. This an example we all know already as f(x) = x², the equation of a parabola.
Let S be the set of three elements - x, y, and z - and define a mapping μ: S → S such that x maps to y, y maps to z, and z maps to x. ("x maps to y" can be written as μ(x) = y or as y = xμ).
Let S be a set, and consider a mapping, m, from S to P(S), its power set (the set of all of S's subsets), defined as m(s) = S - {s}. In this mapping, every element of S gets mapped to its complement in S (loosely speaking since an element can't really have a complement).

Under a mapping, f: S → T, we say that t is the image of s if f(s) = t. We also define the inverse image of t as the subset of S consisting of all of the elements in S that map to t. The inverse image of an element of T could, of course, be the null set if nothing maps to it.

We say that a mapping is one-to-one if no two elements map to the same element. That is, f: S → T is one-to-one if for some elements s and x in S, f(s) = f(x) implies that s = x. Alternatively, s ≠ x implies that f(s) ≠ f(x).

Problem 17: Let f: S → T be a one-to-one mapping, and let t be an element of T. What can we say about the order (the number of elements in) the inverse image of t.

We say that a mapping, f: S → T, is onto if every element in T has at least one element in S that maps to it. In other words, for each t in T, there exists an s in S such that f(s) = t.

If a function is both one-to-one and onto, the we call it a "one-to-one correspondence" as shorthand.

Problem 18: Consider the three examples of mappings given above. For each determine whether the function is one-to-one and/or onto.

Problem 19: Let S be a set, and let ~ be an equivalence relation defined on S. Let T be the set of equivalence classes of ~. (Note that T is a subset of P(S)). Let f: S → T be a function that maps each element of S to its equivalence class, f(s) = cl(s). Is this function one-to-one and/or onto? What do the inverse images of the elements in T look like?*

(Sorry that this post is so long. Hopefully, it was pretty straightforward.)

*This problem obviously relates to some of the points about equivalence relations that we are struggling with. Give it your best shot, but don't get frustrated if this problem reads like nonsense to you. Just let me know.

How funny is Cleverbot?

2011-06-26T11:50:00.000-07:00

Not very. Cleverbot is a robot that can only speak in phrases said to it in the past. This morning I spent some time with it trying to hear some jokes.

Joke 1:
Cleverbot: Knock knock!

User: Who's there?

Cleverbot: Little man.

User: Little man who?

Cleverbot: You little man.

User: You little man who?

Cleverbot: You're so cute, honey.

User: I don't get it.

Cleverbot: You are not enough and I can't have you.

User: Ok.

Cleverbot: Thank You.

User: You're welcome.

Cleverbot: Thank you.

Joke 2:

Cleverbot: Okay. How do monkeys fly?

User: I don't know. How?

Cleverbot: I don't know, how?

User: You tell me.

Cleverbot: That's not a funny joke.

User: No, I guess it isn't.

Cleverbot: Yay! You admitted I was right!

Joke 3:

Cleverbot: Knock knock.

User: Who's there?

Cleverbot: Bad joke.

User: Bad joke who?

Cleverbot: I'll tell you a joke.

User: Ok.

Cleverbot: Why did the man throw his clock out the window?

User: I don't know. Why did the man throw his clock out the window?

Cleverbot: I don't know.

Non-Math Post: Nuuk

2011-06-24T12:44:00.000-07:00

enlarge this

Anyone want to go to Greenland with me? I don't have any money to go right now, but someday we should because take a fucking look at Nuuk, huh?!

What I'd really like to do is go up to Newfoundland and then travel by boat or small plane to Greenland, then on to Iceland and down to Europe - kind of a reverse Viking trip. Apparently, the Newfoundland to Greenland leg would be prohibitively expensive without going through Iceland first, but that would defeat the purpose for me a little.

We should at least go to Newfoundland though, right guys?

Equivalence Relations 2/2

2011-06-22T10:04:00.000-07:00

Problem 12: Consider the set of all integers, and define a relation, ~, on its elements such that a ~ b if and only if a - b is even. Is this an equivalence relation?

Problem 13: Consider the set of all points in a plane. Define a relation, ~, such that elements a ~ b if and only if a is equidistant from the origin of the plane (which can be wherever you want it to be) as b. Is this an equivalence relation?

Problem 14: Consider the set of all people in the world, and define a relation such that two people are related if they live within 100 miles of each other. Is this an equivalence relation?

Problem 15: Consider the set of integers again, and say that a ~ b if and only if a > b and b > a. Does this define an equivalence relation?

Problem 16: Can you come up with a relation that is reflexive and transitive but not symmetric?

Equivalence Relations 1/2

2011-06-21T14:46:00.000-07:00

Now we're going to move on to something slightly different to wrap up set theory (Chapter 1 of Herstein's book consists of set theory, mappings, and some stuff on integers just to give you an idea of where we're at).

Let A be a nonempty set, and let's partition it into several nonempty subsets. That is, let's break A apart into several pieces so that each piece is a nonempty subset of A that is disjoint (sharing no elements) with the other pieces such that all of the subsets union to A. I made a picture:

Now, let's call any two elements of A equivalent if they ended up in the same subset after the partition and denote this equivalence relation by "a ~ b" (where a and b are elements of A).

Problem 9: Show that an equivalence relation is reflexive, symmetric, and transitive. That is, for all elements a, b, and c in the set A,

a ~ a,
a ~ b implies that b ~ a, and
a ~ b and b ~ c imply that a ~ c.

Problem 10: Is equality (=), as it is usually defined on numbers, an equivalence relation?

Problem 11: Show that the converse of problem 9 is also true - that any relation defined on a set's elements that is reflexive, symmetric, and transitive is an equivalence relation on that set.

Power Set

2011-06-18T12:19:00.000-07:00

These problems have a different flavor and can be done independently of the others.

Problem 7: How many elements belong to the union of two sets, A and B? (Obviously, the answer won't be a number but in terms of some other things)

Problem 8: How many subsets does a set have including itself and the null set?

Set Theory Grab Bag

2011-06-18T11:51:00.000-07:00

Here are some more problems for whenever you want to tackle them.

Problem 3: We say that two sets are disjoint if they have no elements in common (or more than two sets can all be mutually disjoint if each set is disjoint with every other). What is a way of saying that sets A and B are disjoint by using the language of unions and/or intersections?

Problem 4*: The difference set, A - B, can be defined as the set of all elements in A that are not also in B. Is it possible to define the difference set in terms of unions and/or intersections of A and B?

Problem 5: The symmetric difference, A Δ B, is defined as everything in A not also in B plus everything in B not also in A. In other words, A Δ B = (A - B) ∪ (B - A). What is another way that we could define the symmetric difference using the union, intersection, and difference operators? (I'm looking for a definition that would probably come to mind directly from a Venn diagram representation of a symmetric difference. By the way, Venn diagrams are sometimes helpful to use when thinking about sets, but I think they are more helpful when you produce them on your own.)

Problem 6: Let A, B, and C be sets such that A and B are both subsets of C. Let A' denote the set of all elements of C not in A, let B' denote the set of all elements of C not in B, and so on for all subsets of C. Show that (A ∩ B)' = A' ∪ B' and that (A ∪ B)' = A' ∩ B'.

*I'm not really sure about the answer. It wasn't a question in the book or anything but just something that I wondered. I think that I know the answer. So you think about it, and then think about it again after problem 6. And then we can talk.

Unions and Intersections

2011-06-17T19:45:00.000-07:00

Here are a couple of things that we can do to two or more sets. We can take their union - A ∪ B - or we can take their intersection - A ∩ B. These mean what they sound like. The union of sets A and B is the set of all elements that are in A or B or both. The intersection of A and B is the set of all elements that are only in both A and B. These concepts can be extended beyond two sets (e.g. you can take the union or intersection of 20 sets and it means the same thing).

Note that when we combine two sets, there is no repeating of elements. That is, if we take the union of the integers with the set {2, π}, then the integer 2 is not repeated in the union.

Problem 2: Let A, B, and C be sets, and let B be a subset of A. Show that A ∪ B = A, that B ∪ C is a subset of A ∪ C, and that B ∩ C is a subset of A ∩ C.

It begins - Here is some Set Theory

2011-06-17T11:35:00.000-07:00

A set can be informally defined as a collection of objects.

Examples of sets include the set of integers, the set of lines in a plane, and the set of all dog breeds. The set of all dog breeds is an example that I threw in there to demonstrate that sets can be composed of anything at all. We will never deal with it again.*

The sets that we will be talking about are unordered even if they are usually thought of that way. For instance, there is a well known ordering that comes with the integers - 80 is greater than 4 - but here we just assume that the set of integers is just a big bag of numbers where nothing is ranked. Also, sets will not contain repetitions of objects - everything in a set is going to be distinct. More on this last point later.

*Except for right now:

Problem 1: Is the set of all cute pug puppies equal to the set of all things that deserved to be photographed?

Note: The original first problem discussed in the comments follows in italics, but was changed due to protest and also the fact that it uses polynomials, which most people hate, when it doesn't have to.

Problem 1: Is the set of the roots of the polynomial x² - 1 equal to the set {1, -1}, that is, the set that contains 1 and -1? (This probably seems like a trivial problem, but don't overthink it, just try to explain why.)

Abstract Algebra Preface

2011-06-17T11:34:00.000-07:00

The other day I was playing Just Dance for the Wii when I realized that 'Hot n Cold' is not a fun roackabilly song by The Baseballs, but is, in fact, a pop song by some lady. However, The Baseballs cover is still great:

This was not the only thing that Just Dance taught me. I also learned that aerobic exercise is way better when you're doing it as part of a game. I guess that professional athletes and the makers of Wii Fit (which I don't have) already knew this, but it was a real revelation to me. Independent of my experience with Just Dance, I recently read an essay by Paul Lockhart called A Mathematician's Lament which goes on at length about how math should be taught through solving a series of fun and abstract problems rather than long lectures followed by problem sets designed to test how well you can apply a formula or method. The essay is kinda long, but a worthwhile read for anyone with an interest in math or science instruction. The takeaway is that I think learning math could and should be a game just like exercise should be, and I am eager to try my hand at making it that way. I will be teaching college students math soon so it would be helpful for me, over the summer, to give the problem-based approach a shot.

Samtron, also a math teacher, has agreed to help me by being a student in a course of Abstract Algebra right here on my blog. I would love it if other people joined in too. All I ask is that you attempt the problems on your own before reading the comments. Because I am not comfortable enough with the material to construct the entire subject of Algebra from problems alone, this is probably going to be a half-Lockhart approach in the end, but I will try my best to stick to interesting problems and limit my expositions to a few definitions here and there. The idea is that I will present problems, samtron (and others if there is interest) will attempt to solve them, he (and you?) will present his solution or ask questions in the comments, and I will provide corrections or help or further instruction after that. Of course, this will be a pretty flexible game since I don't really know what I'm doing. We'll have to figure things out as we go along. Right now, I'm planning on the instruction here tracking the material of Herstein's Topics In Algebra book which is broken down into the following chapters:

Preliminary Notions
Group Theory
Ring Theory
Vector Spaces and Modules
Fields
Linear Transformations
Selected Topics

I think that this normally covers two semesters of undergraduate algebra so we're not going to get through all of it unless things are going way better than I'm expecting, but I hope that we can at least do groups, rings, and some fields. The first chapter, where we will start, is pretty scattered since it's just giving you things to work with later. This should make it fun but frustrating in its lack of direction. Stick with it - I will help!

The plan of attack from my end will be to read the chapter, work the problems at the end, and then rearrange things so that the problems move the material forward as you work them. I hope that this works. I should probably mention that I picked algebra to cover mainly because it has been a little while since I've done it, and I need to refresh a lot before getting to grad school in August. So there might be times when I present a problem or two from Herstein's book that I am having trouble solving as well and maybe we can crack them together.

Ok, sorry this preface is getting too long. Please ask lots of questions if I'm not being clear, and don't be afraid to let me know when I'm doing something stupid with regard to teaching because I promise you that I will.

(A) Summer Every Day

2011-06-16T13:03:00.000-07:00

(I am now openly ripping off Ethan Moore. By the way, if you visit Ethan Moore's website, you will be greeted with a big fucking picture of Ethan Moore's face.)

New Apartment*

2011-06-16T09:57:00.000-07:00

The view of my new street from the train platform.

Last week, I took a last-minute trip to Chicago, where I will be moving in July, to hunt down an apartment that would be safe, affordable, and would allow Oscar to continue his indoor/outdoor cat lifestyle. My three days there were stressful, and at one point I got sick from dehydration. But on the last day, I found a place that is really great. It is an apartment inside of a house in a neighborhood that is filled with trees and tiny yards for Oscar to roam in, but it is also close to the train, improv clubs, the threadless store, and lots of restaurants. Plus, the best part is that inside this new apartment I will be able to be a wonderful person.

The perfect kitchen for a man who loves his body just as much as he loves enjoying a delicious meal and whimsical knick-knacks on top of the cabinets.

Here is the kitchen where I will prepare new and interesting dishes because despite being busy with many fulfilling activities outside of the apartment, I will always find the time to put a little effort into making something nutritious - breakfast, lunch, and dinner. And who could blame me? That refrigerator is going to be stocked with a wide variety of ingredients, purchased on a weekly basis with the following week's worth of meals in mind from none other than the nearby Trader Joe's (two stops down on the train, or a 20 minute walk when the weather is nice). The refrigerator will have so much more than a twelve-pack of cheap beer, a half-used and molding jar of Prego, and a package of sliced turkey. Daily trips to a non-organic-promoting grocery store to buy frozen foods to heat up each evening will be unheard of because there will be so much already there in my home.

Nothing gross will ever happen in here.

Here is the bathroom where I will brush and floss my teeth every single night. My poops will always be easy and satisfying, and while I will continue to pee standing up (please take note, genderanalyzer.com with your 61% chance that my blog is written by a woman), there will rarely be excessive splashing or drunken misfires. I will also clean this bathroom every week or two, not just by randomly rubbing toilet paper over surfaces that have collected too much hair and grime for me to stand looking at, but with real cleaning products. My towel and washcloth will be rotated out more frequently than "whenever the mildew stench gets to be unbearable."And, of course, I will not clog up the sink's drain every couple of weeks with my half-assed relationship with shaving.

A center of productivity and charm.

Here is the small office space where I will study math, pay bills before their due date, respond to e-mails within a day of receiving them, and never ever waste a horrendous amount of time on shit that is neither necessary or fun. I see that the current occupant of my new apartment has decided to keep a couple of wine racks in his or her** office. Wine collecting is such a noble hobby. Maybe I will continue that tradition when I take the reins?

Have a seat, and I will get you a drink. Then we can swap stories.

Here is the living room where I will entertain a constant stream of visitors with stories about the wacky things that happen to me in my life. Laughter will be a regular guest in this room. However, it will also sometimes be a solemn place whenever a friend is in need of someone to listen to their problems. In my new apartment, I will be a great listener with a genuine interest in other people and their lives. I will only try to connect what they are telling me to anecdotes from my own life when I think that it might be helpful to them and never because I just want to talk about myself. It will always be an act of empathy. Of course, my new living room will also be an excellent place to watch interesting movies, play fun games, and read novels. It will not become simply a place with a couch where I sit and waste hours surfing the internet for no good reason.

I don't have a good picture of the bedroom, but believe me when I tell you that insomnia will be a stranger to that room. Nights will be a time of rest not meant for rolling around like an unsatisfied lunatic. The sheets of my bed will be cleaned regularly, and Oscar will not sit on top of me licking himself until four in the morning when he will not wake me up to let him outside.

I can't wait.

Please come visit!

* Part of the way through writing this, I became convinced that I stole the whole describing an ideal future specifically devoid of some of my current behavioral flaws in the context of moving to a new building gimmick from someone else. I don't know who, but if it is you, then know that I'm sorry. Obviously, my subconscious thought the idea was hilarious.

** I don't know why I wrote this. The landlord told me that it is a he.

Comment

2011-06-13T09:39:00.000-07:00

Earlier today, I left a comment on my friend Barry's blog post. I was then informed that my comment would need to be approved before it could be published. I assume that Barry is the one controlling the approval process. I hope that the issue is resolved quickly though, of course, I realize that sifting through all of the comments awaiting approval cannot be an easy task for Barry so I understand if there is a delay.

I will update you on my comment's progress.

UPDATE: Ok, my comment got in.

Corruption

2011-06-10T11:00:00.000-07:00

Last summer, I "acted" in a short movie that my friends Greg and Bill made. It is a buddy cop flick with poor sound quality but a lot of heart! I play a store clerk who loves life but is also a libertarian (this may or may not get through to the viewer).

Corruption from Greg de Deugd on Vimeo.

Angry Birds

2011-06-10T09:20:00.000-07:00

Blosics gameplay.

Angry Birds is a game that everyone already knows about, I guess. I didn't start playing it until a couple of days ago, and my idea of what it was going to be like was off the mark. I had no idea that the Angry Birds were the protagonists so imagine my shock when I started playing. The game does not center around a human male running around a field while Angry Birds (at first just a few Angry Birds, but progressively more Angry Birds as the game goes on) dive bomb him because of whatever he did in the backstory. Instead, Angry Birds is set in the days of the Pig Kingdom during the time immediately after King Pig's directive to steal all of the eggs from the Angry Birds. There are no humans. Gameplay itself centers around launching the different Angry Birds at King Pig's Royal Guard and their limited defenses. It's a lot like a game that I played a bunch maybe a year ago called Blosics (which has no backstory). I think that the descriptive term for games like these is "physics-based," but that probably means only that there is a projectile and the player needs to choose the angle and initial velocity of that projectile. Physics-based could really mean anything though since physics is the science of reality. Blosics isn't as cute as Angry Birds, but it is more coherent, challenging, and a much better game. Also, maybe the type of violence depicted in Angry Birds shouldn't be as cute as they make it? A lot of creatures die. Whatever though, I'm not one of those people, and I'm not trying to start any shit with the Angry Birds developers or anything. I just think that Blosics is better, and if you found Angry Birds to be addictive but somewhat unsatisfying, then I would recommend you take a look at Blosics.

james stewart's calculus

2011-01-23T15:43:00.000-08:00

If you're anything like me, then you realize that your decision almost three years ago to attend some fancy and expensive law school was a huge mistake, and you ought to be doing everything in your power to right this wrong by setting yourself up to enter a graduate program in mathematics in the next academic year as it was obviously your first love and is the holy path. If you're even slightly like me, then you surely know that a large part of this endeavor must be studying for and taking the GRE Mathematics Subject Exam last November because the score that you received when you took it in 2006 was pretty shitty and nothing that you should be showing off to even a third rate graduate program. You, like me, are probably well aware that Calculus questions make up a solid 50% of the test and that it has been years since you've touched the subject. And now you ask yourself the same question that I did: "How do I get from Calculus zero to Calculus HERO in just a couple of months?!"

Listen: James Stewart's Calculus is always gonna be your best bet, but get an older edition because Calculus doesn't change and the current version is god damn expensive.

oral b toothbrush

2010-10-02T12:56:00.000-07:00

Guys, I bought a new toothbrush last week, and it's electronic! I haven't used such since I was maybe 8 or 9 or 10 so this new one can be pretty painful to use still. However, my teeth feel really polished afterward which is pretty neat. I definitely could not achieve the same sensation with my old manual toothbrush. The thing that I like the most about it is how much like a regular toothbrush is. The technology for these things has come along way since the last time I paid attention. The handle of the toothbrush is not round and bulky but shaped for your hand to grip. Plus, there is no big charging station - the toothbrush is disposable.

If any of you are interested, it is made by Oral B and is similar to or possibly exactly the same as the Oral B Pulsar Toothbrush that I found on Amazon. I paid over 6 dollars for mine at Kroger, but it looks like they are going for half that on the zon!