Wow, I can't believe I've ignored doing this for almost a year! I suppose the biggest news is that I got married four weeks ago yesterday. My wife's name is Erin and she is a first year graduate student at U of L in the Women's and Gender Studies Masters Program. I finished my first year as a Math grad student and got to teach my very own course this summer (it was a lot of work, but a great experience). Erin and I both start class tomorrow--we are excited and nervous.
Those are the biggest pieces of news...now onto some math: I wanted to include at least a little math, since this is a math blog, but since I've been running around getting married and moving to a new apartment (its very nice), there hasn't been much math of note. So, I'll share this interesting occurrence that happens to involve numbers.
Yesterday, Erin and I went to Bed Bath and Beyond for two reasons: we had to return some pots and pans that we got duplicates of from our wedding registry, and we had to pick up a bridal shower gift for Erin's friend whose wedding is in October. The cashier told us to leave the pots at the desk and do our shopping, and then she'd do the return and the sale at the same time.
We picked out the items we wanted to give Erin's friend, and we estimated it to be about $43-$45 (about how much we wanted to spend.) I didn't know how much the pots and pans were, but Erin wanted to get a Yankee Candle (awesome things!) for ourselves as well. We picked out a nice fragrance, then took all of our stuff up to the counter and the cashier rang us up and announced the total we owed: $0.98! Maybe not the most amazing thing in the world, but I thought it was pretty cool that we had just happened to buy a candle that brought our total to almost exactly one dollar more that what we returned.
A funny postscript to the rather mediocre story: The cashier wrapped our glass Yankee Candle in paper, then dropped it in a bag for us. We were all surprised to hear a crash from the floor--yes, the candle had broken, but it was because the bag had a giant slash through both sides of it. The cashier almost let a string of curses fly, telling us that whoever opened the box of plastic bags was going to get punched in the mouth. Apparently someone was a little overexcited with the Exacto knife and had sliced through quite a few bags when opening the box. We got a new candle, a new bag, and left the store only $0.98 poorer. Yay!
Well, I'm off to get a good night sleep before the first day of school tomorrow!
Happy mathing!
Sunday, August 21, 2011
Wednesday, September 29, 2010
A Homework Sampler
This time around, I decided to give all those readers out there an idea of the kind of problems I get for homework. What follows is one of three problems that I will turn in tomorrow in my Real Analysis class. However, I think that the question (at least) and most of the solution is pretty accessible to the masses. I'd love to hear feedback about how well you think the solution was written--keeping in mind that I could have included much more detail. Mentioning things like the fact that the area of the big square is one because its sides are of length one is unnecessary in the solutions I will hand in. I'd also love to hear if you have any questions about the problem, the solution, the class, the subject, or anything! Ok, now to figure out how to include a pdf file...
It seems like that worked. We'll see! If not, I'll post again and possibly include a link you can go to. If it worked, you can download the file by right-clicking it if it is too small to read in your browser.
A bit more information about this problem: it is approximately the average difficulty (probably a bit easier than normal). The second problem in this set was twice as long and twice as dense and complex. However, the third problem was about the same difficulty as this one, but took about half as much space to solve.
I have had analysis before, and I have had algebra before. However, my statistic background is sorely lacking. Therefore, I am finding statistics to easily be my hardest class. We took a test last Monday, and I did not do quite as well as I hoped. However, I've heard this teacher's first test is always hard. I am doing well on the homework, though, so overall I feel good about the whole situation.
Hopefully soon I will be able to post an update on the research stuff--and you never know when something funny (like Sting) will come up. Happy mathing!
It seems like that worked. We'll see! If not, I'll post again and possibly include a link you can go to. If it worked, you can download the file by right-clicking it if it is too small to read in your browser.
A bit more information about this problem: it is approximately the average difficulty (probably a bit easier than normal). The second problem in this set was twice as long and twice as dense and complex. However, the third problem was about the same difficulty as this one, but took about half as much space to solve.
I have had analysis before, and I have had algebra before. However, my statistic background is sorely lacking. Therefore, I am finding statistics to easily be my hardest class. We took a test last Monday, and I did not do quite as well as I hoped. However, I've heard this teacher's first test is always hard. I am doing well on the homework, though, so overall I feel good about the whole situation.
Hopefully soon I will be able to post an update on the research stuff--and you never know when something funny (like Sting) will come up. Happy mathing!
Wednesday, September 15, 2010
Sting is NOT a Mathematician
…as evidenced by his misuse of numbers in quite a few songs by The Police.
First: From the song “Next To You”: “I’ve had a thousand girls or maybe more…”
Analysis: Let’s assume Sting “has a girl” five days out of every week. That means he has been “having girls” for the past 1000/(52*5) = 3.846 years. This does not seem so outlandish until one realizes the line after the one mentioned is “But I’ve never felt like this before.” This is either testament to how amazing this girl is (despite Sting being probably the most diseased person on the planet), or to how little Sting pays attention to all the other girls he has “had.” (Even if we assume Sting “has a girl” six out of seven days a week, and on the seventh “has” two girls, he has been at it for the past 1000/(52*8) = 2.40 years!)
Second: From the song “Every Little Thing She Does is Magic” : “…a thousand rainy days since we first met…”
Analysis: Let’s assume (generously) that it rains one-third of the days in a year. That means that Sting has known this magic girl for 1000/(365/3) = 8.2192 years. Considering another line in the song mentions calling her up to propose to her, it shows that Sting has no courage whatsoever: he’s been trying to propose to this girl for over 8 years?! (A more conservative and realistic estimate would be that it rains one-fifth of the days in a year, in which case Sting has been wallowing in indecision for 1000/(365/5) = 13.70 years!!)
Finally: From the song “Message in a Bottle”: “A hundred billion bottles washed up on the shore”
Analysis: Currently estimates of the worlds population are around 7 billion—that is, 7,000,000,000 people. So, even if everyone in the world contributes to the cause, each person would have to throw 100,000,000,000/7,000,000,000 = 14.26 bottles into the ocean.
Since this is not inconceivable, let’s consider the actual physicality of 100 billion bottles on a shore. Let’s assume (hopefully reasonably) that one bottle takes up half of one square foot. Let’s also assume that the beach of Sting’s island is 50 feet wide. 100,000,000,000 bottles would then cover 100,000,000,000/(50*2) = 100,000,000 feet = 100,000,000/5280 miles = 18,939.4 miles. This is equal to 1.7 times the length of the coastline of Great Britain--11,073 miles (courtesy of the Ordnance Survey via Wikipedia). Sting would need a pretty big island!
So, it is clear that Sting is not a mathematician. However, I am a tiny bit closer to being one. With about three weeks of grad school under my belt, things are going pretty well. My classes in Analysis and Algebra are fairly easy (right now), and my Statistics course is more challenging than I’d like, but I am holding my own—we’ll see how the test on Monday goes! The recitations that I teach are also going well considering they are full of people who don’t really want to be there. On the whole, life is pretty good. I am excited for many things in my future, but the present is nice too!
Wednesday, August 18, 2010
57 Divided by 3 is 19, Alice and Bob Play a Game, and Other News
I finally was able to get the license plates for my car the other day. They say EWV-5719. I will remember this (at least the digits) by remembering that 57/3=19. That is your mathematical fact of the day. 57/3=19. Commit it to memory now.
In other news, I moved to Louisville yesterday to start my work as a graduate student and a graduate teaching assistant there. I was assigned a class to TA for, but it conflicts with my own class schedule. So, that will have to get fixed. It seems the mathematics secretary is a position in transition. The old secretary is becoming the business manager, and the new secretary is flustered and overwhelmed. So, it may be a bit until everything gets sorted out. It seems that most TAs run recitation sessions, give and grade quizzes, and hold office hours. None of that seems to strenuous and the professors seems laid back and likable. More on that story as it develops.
Now I might try a little math: This is the introduction to the problem I worked on during my REU (Research Experience for Undergraduates) in McMinnville, OR.
Alice and Bob are friends who are going to play a game. The game board is a piece of paper with a graph drawn on it. The graph consists of some open dots, called vertices. The vertices are connected by lines called edges. The game is played by coloring the vertices with crayons. Alice and Bob only have a limited number of crayons, though. Let's say they have r crayons, each a different color. A color (i.e. red) is legal for a vertex as long as none of that vertex's neighbors are red. So, if two vertices are connected by an edge, those two vertices cannot be colored the same color. Alice always goes first. She wins if eventually all the vertices are colored. Bob tries to thwart Alice. He wins if at any point, a vertex cannot be colored. This means that each of the r colors appear among that vertex's neighbors.
Now, each and every graph has a number associated with it in relation to this game. It is called the game chromatic number. "Game" because the regular chromatic number of a graph is the least number of colors needed for Alice to color the graph, without Bob taking his turns. "Chromatic" refers to colors. So, the game chromatic number of a graph is the least number of colors Alice needs to have a winning strategy against Bob. In other words, its not enough for Alice to win a game with, say, six colors. She needs to be able to win every game on that graph with six colors to make six that graph's game chromatic number (she cannot rely on Bob being stupid).
So, let's look at an example. Let's consider a binary tree. A binary tree is a graph where one vertex is designated as the root, and it has at most two neighbors (which are not neighbors of each other), called its children. Each other vertex has at least one neighbor and at most three neighbors (all of which are not neighbors of each other), its parent, and up to two children. I'm going to try to include a picture of an example of a binary tree. If it doesn't work, sketch out the description above.
Well, that seems to have worked! I'm still getting the hang of it. Ok, the red arrow indicates the root (yes, it's at the top). The green arrows are the root's children and are known as the 1st generation. The blue arrow indicates the parent of the vertex pointed to by the purple arrow. Notice that the blue arrowed vertex has only one child, and the purple arrowed vertex has no children. This is perfectly legal. However, the root is the only vertex without a parent.
Now let's color the graph (not by playing the game--this is just Alice coloring right now). How can we do it so no two neighbors have the same color? Let's color the root red. Then it's children must not be red, but they can both be blue, since they are not neighbors themselves. Next, all the children of the blue vertices can be red since they are not neighbors of each other, and they are not neighbors of the root. You should see the pattern: alternating colors as you move down the generations will give a proper coloring of this tree (any tree, in fact). A proper coloring is simply one where no two neighbors have the same color. Let's see if I can add a picture of that.
I apologize for the crudity of the diagrams, but you get the idea.
Now, finally, let's revisit the game. We can easily show that Alice will always win if she and Bob have four crayons. We'll also show that she can win with three colors. But, we'll show that two colors is not enough for her to win.
The first part is easy: no vertex has more than 3 neighbors. Therefore, there are at most three distinct colors among the neighbors of any one vertex. This means that Alice (or Bob) always has the fourth color available to use if necessary. Ok, so four is enough, but what about three?
Let's number the vertices for easy reference:
Here's Alice's Strategy: Color 2.
If Bob colors 1 or anything on the right side, color 3. (If Bob colored 12 or 13, use the same color he used)
If Bob colors 3, color 6.
If Bob colors anything on the left side, color 4 or 5, whichever is closest.
The important vertices are 2, 3, 4, 5, and 6, since these are the only vertices with enough neighbors that would allow Bob to force a win. If, at any time, Alice finds she cannot follow her strategy because the vertex it tells her to color is already colored, she should color one of the important vertices. Once those have been colored, Alice cannot lose.
Now, why can't Alice win with just two colors? Suppose Alice colors a vertex (any of them). Bob moves two vertices away and uses the second color. Now, the vertex in between cannot be colored and Bob automatically wins.
We have shown that 3 colors are enough for Alice to win the game on this graph, and two colors are not enough. So, the game chromatic number of this graph is 3.
It is now very late, and I'm headed to bed. There is a high probability that there is some mistake in the above, although I hope not. It is interesting that if I had given the last vertex in the 2nd generation a second child, then I don't think Alice could win with three colors. I haven't checked this, but I'm pretty sure she would need 4.
This was just an example graph/game. Results in this area are proved about entire classes of graphs, not just individual graphs. For instance, it is known that the game chromatic number of any tree (a graph with no cycles) must be less than or equal to 4. In a future post, I'll describe some variants on the game and some results thereof.
Charlie
In other news, I moved to Louisville yesterday to start my work as a graduate student and a graduate teaching assistant there. I was assigned a class to TA for, but it conflicts with my own class schedule. So, that will have to get fixed. It seems the mathematics secretary is a position in transition. The old secretary is becoming the business manager, and the new secretary is flustered and overwhelmed. So, it may be a bit until everything gets sorted out. It seems that most TAs run recitation sessions, give and grade quizzes, and hold office hours. None of that seems to strenuous and the professors seems laid back and likable. More on that story as it develops.
Now I might try a little math: This is the introduction to the problem I worked on during my REU (Research Experience for Undergraduates) in McMinnville, OR.
Alice and Bob are friends who are going to play a game. The game board is a piece of paper with a graph drawn on it. The graph consists of some open dots, called vertices. The vertices are connected by lines called edges. The game is played by coloring the vertices with crayons. Alice and Bob only have a limited number of crayons, though. Let's say they have r crayons, each a different color. A color (i.e. red) is legal for a vertex as long as none of that vertex's neighbors are red. So, if two vertices are connected by an edge, those two vertices cannot be colored the same color. Alice always goes first. She wins if eventually all the vertices are colored. Bob tries to thwart Alice. He wins if at any point, a vertex cannot be colored. This means that each of the r colors appear among that vertex's neighbors.
Now, each and every graph has a number associated with it in relation to this game. It is called the game chromatic number. "Game" because the regular chromatic number of a graph is the least number of colors needed for Alice to color the graph, without Bob taking his turns. "Chromatic" refers to colors. So, the game chromatic number of a graph is the least number of colors Alice needs to have a winning strategy against Bob. In other words, its not enough for Alice to win a game with, say, six colors. She needs to be able to win every game on that graph with six colors to make six that graph's game chromatic number (she cannot rely on Bob being stupid).
So, let's look at an example. Let's consider a binary tree. A binary tree is a graph where one vertex is designated as the root, and it has at most two neighbors (which are not neighbors of each other), called its children. Each other vertex has at least one neighbor and at most three neighbors (all of which are not neighbors of each other), its parent, and up to two children. I'm going to try to include a picture of an example of a binary tree. If it doesn't work, sketch out the description above.
Well, that seems to have worked! I'm still getting the hang of it. Ok, the red arrow indicates the root (yes, it's at the top). The green arrows are the root's children and are known as the 1st generation. The blue arrow indicates the parent of the vertex pointed to by the purple arrow. Notice that the blue arrowed vertex has only one child, and the purple arrowed vertex has no children. This is perfectly legal. However, the root is the only vertex without a parent.
Now let's color the graph (not by playing the game--this is just Alice coloring right now). How can we do it so no two neighbors have the same color? Let's color the root red. Then it's children must not be red, but they can both be blue, since they are not neighbors themselves. Next, all the children of the blue vertices can be red since they are not neighbors of each other, and they are not neighbors of the root. You should see the pattern: alternating colors as you move down the generations will give a proper coloring of this tree (any tree, in fact). A proper coloring is simply one where no two neighbors have the same color. Let's see if I can add a picture of that.
Now, finally, let's revisit the game. We can easily show that Alice will always win if she and Bob have four crayons. We'll also show that she can win with three colors. But, we'll show that two colors is not enough for her to win.
The first part is easy: no vertex has more than 3 neighbors. Therefore, there are at most three distinct colors among the neighbors of any one vertex. This means that Alice (or Bob) always has the fourth color available to use if necessary. Ok, so four is enough, but what about three?
Let's number the vertices for easy reference:
If Bob colors 1 or anything on the right side, color 3. (If Bob colored 12 or 13, use the same color he used)
If Bob colors 3, color 6.
If Bob colors anything on the left side, color 4 or 5, whichever is closest.
The important vertices are 2, 3, 4, 5, and 6, since these are the only vertices with enough neighbors that would allow Bob to force a win. If, at any time, Alice finds she cannot follow her strategy because the vertex it tells her to color is already colored, she should color one of the important vertices. Once those have been colored, Alice cannot lose.
Now, why can't Alice win with just two colors? Suppose Alice colors a vertex (any of them). Bob moves two vertices away and uses the second color. Now, the vertex in between cannot be colored and Bob automatically wins.
We have shown that 3 colors are enough for Alice to win the game on this graph, and two colors are not enough. So, the game chromatic number of this graph is 3.
It is now very late, and I'm headed to bed. There is a high probability that there is some mistake in the above, although I hope not. It is interesting that if I had given the last vertex in the 2nd generation a second child, then I don't think Alice could win with three colors. I haven't checked this, but I'm pretty sure she would need 4.
This was just an example graph/game. Results in this area are proved about entire classes of graphs, not just individual graphs. For instance, it is known that the game chromatic number of any tree (a graph with no cycles) must be less than or equal to 4. In a future post, I'll describe some variants on the game and some results thereof.
Charlie
Monday, August 9, 2010
What's up with Zero?
As a first-time blogger, I'm going to quickly say "hello" to the internet, and then jump right in.
A few weeks ago, my friend (who is not a math student, but who is taking a philosophy of math course) asked me several questions. I decided to post the questions and most of my response. He told me it was enlightening, so I hope it is for you as well.
His questions included: "I know that our number system heavily relies on zero and the concept of a base of ten, but how can zero really be a number? Sure, it can subtract and add like other numbers, but we cannot divide with it, nor does it multiple like other numbers, where if we reverse the multiplication by division, we get the original number. So why does it qualify as a number if it can't do things that all (I'm presuming all) other numbers can? Do mathematicians just ignore these discrepancies or an am I missing something?"
And here was my response:
"Your question opens a GIANT can of worms, some of which could take a very long time to dicuss and explain. I'm up for an extended discussion, so I'll try to describe as much as I can as simply as I can:
To answer your question very succinctly--YES, zero is weird. But it is useful and important because it is weird. One is useful and important as well, because it is weird. Zero and one are the most important numbers because they do not act like the other numbers.
However, your claim that zero should not be counted as a number (that phrasing needs work), is not founded.
The definition of "number" is extremely abstract. I think what you mean is: "should zero be counted as an integer?" In which case, your claim is still unfounded. The true definition of an integer is half a semester's-worth into an introductory set theory class, and I don't remember it off the top of my head (even if I did, it wouldn't make sense at this point). However, I know it has nothing to do with being able to add, subtract, multiply, divide, or do anything else to the item in question. It relies only on a number's set theoretical form, which is complex and which I also don't remember exactly.
In any case, you are correct in that zero belongs in some groups of numbers and does not belong in others. Let's look at a few of these groups (which I should really call sets because groups are something entirely different):
The first set we learn about, and the most natural, is called (oddly enough) the natural numbers. It consists of all the positive integers: (1,2,3,4,....). As you can see, zero does not appear, and does not belong in this set.
This set is represented by a captial boldface N with the boldface part open. (Go to http://www-math.univ-poitiers.
The next set is called the whole numbers. It is the natural numbers and zero (0,1,2,3,4,....). Here, zero does belong. The whole numbers and the natural numbers are used for different things.
This set is represented by the lowercase Greek letter omega (the curly w).
The next set is the integers. It is the natural numbers, their negatives (opposites), and zero (...,-3,-2,-1,0,1,2,3,...).
This set is represented by the open boldface captial Z. (Why "Z"? I have no clue. Maybe because it's a sideways N...)
Ok, that's all very familiar. This probably is too: the next set is the rational numbers. This is all values that can be expressed as fractions (or ratios). It includes all the integers, 1/2, -2/3, 47/1098, -5.7, 3.89485, and many others. However, it does not include numbers like pi, e, or the square root of 2. Those are irrational (they cannot be expressed as fractions). In fact, all non-terminating, non-repeating decimals are irrational.
Rationals are represented by the open boldface capital Q (for quotient) and the irrationals don't have a letter all to themselves.
All of the sets so far are subsets of this next set: the real numbers. This is is simply all the rationals and all the irrationals together. Keep in mind that real is just a word we use to describe these numbers. Try to avoid associating the meaning of the English word "real" with these numbers. As one mathematician put it--"how real can any number be? I mean, when's the last time you stubbed your toe on a seven?"
The symbol for the reals is the capital open boldface R.
Ok now on to the non-familiar. The set of complex numbers is a set that contains all real numbers. It also contains all imaginary numbers. And it contains all combinations of real and imaginary numbers. Recall that i is the symbol for the square root of negative one. Imaginary numbers consist of a real number (besides zero) multiplied by i. So, i, 3i, -9i, 4.5i, pi*i, ei, (5/27)i are all imaginary numbers. Complex numbers consist of a real part, and an imaginary part (which are then added together). In mathematicians' terms, they are of the form a+bi where a and b are real numbers. Either or both of a and b can be zero. Therefore, 0 (0+0i), -3 (-3+0i), 5i (0+5i), 3+5i, pi-ei, and 4.08934759-43.90238098...i are complex numbers. Again, "imaginary" and "complex" are just words. Their English meanings have nothing to do with what they are about--imaginary numbers exist just as much as real numbers and are just as (if not more) useful. Think of it this way: there was a time when 0 was considered nonsense, and a time when negative numbers were considered nonsense. Now they are both vital to our society.
The symbol for the complex numbers is the capital open boldface C.
So why did I tell you all of this? Well, the members of each of these sets act differently when in the context of that set. 3 acts differently as a natural number from the way it acts as a whole number, and from the way it acts as an integer. For example, as a natural number, you cannot subtract 3 from 3. What would you get? 0 is not an option. As a whole number, you cannot subtract 5 from 3. -2 is not an option. As an integer, you cannot divide 3 by 5. 0.6 is not an option. So, when you hear that you cannot divide by 0, that's true, but it's not the only restriction that seems unreasonable. It just depends on your focus (which set you are examining or using). Side note: you still cannot divide by zero in the complex numbers. :( However, you may be able to define a set in which you can divide by zero--I don't know, I've never tried."
In this blog I hope to explore many areas of mathematics and life in general. I hope to post things that cover the very basics (even more basic than the above, possibly), and the very complex (possibly some challenging homework problems, or my research). I may also include stories of life as a graduate student, and information about computer science or LaTeX (if you think this is that stuff balloons are made out of, then you're probably not a mathematician, but you are welcome here anyway!), among other things. I also will aim to keep the atmosphere light, and even humorous where possible, because I think that's often why people are turned off by math: "It's so boring!" Well, I hope to show that math (and mathematicians) can be exciting, enlightening, and yes, even entertaining.
Happy mathing!
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