My class decisions (and an introduction to each)

I believe that I have finally settled on which classes I would like to take during my time in Budapest (which actually is pronounced with a 't' sound I have found- in conflict with the online program I was studying Hungarian from). I have been taking classes for the past week without knowing for sure which class I would like to take. You may wonder how this is possible. Well, my program has an introductory three weeks where you can take any three courses you would like in order to better decide which classes you would like to take, as well as to learn more about the different topics in the field of mathematics. I took many, many classes this week, but I found that many were above my level or covered material that I really wasn't interested in studying. I finally decided on four classes: Set theory, Graph theory, Mathematical Problem Solving, and Probability theory. To many of you, this may not mean much, so I'll introduce you to each subject.

Set theory is the study of sets or organizations of elements into distinct units. For example I could organize the numbers 1-10 in one set, 11-20 into another, 21-30 into another and so one. Most of the sets I work with will consist of arbitrary points, and I will work to make conclusions about the properties of these sets. One introductory topic that many people learn about at some point in their lower level education in intersections and unions. An intersection of two sets is the set of elements that is present in both sets (so the sets {1,2,3,4,5} and {2,4,6} would have an intersection set of {2,4}). The union of two sets is the set of elements that is present in either sets (so the union of the two sets listed above would be {1,2,3,4,5,6}). There is no repetition allowed in sets. The empty set is the set consisting of no element. This is just a basic introduction, but throughout this blog I will go into more details on multiple topics in set theory. This is probably my favorite mathematical subject, and I am very excited for this class.

Graph theory is the study of graphs (but not the graphs you're probably thinking of). A graph is a diagram consisting of vertices and edges (or possible no edges). The easiest way that I can think to describe graphs of this type to you without having a picture is to describe how I will apply it to anthropology as well as how I have already applied it. Image a group of people. Each person knows a particular amount of other people in the group. Put a dot on a piece of paper for each person, and then draws lines connecting who knows who. The image you would get is the type of graph that I will be studying. There are multiple types of graphs and many of them are not broken off into particular categories. One category, however, is called a tree. A tree is a graph in which there are no cycles (connections between vertices that form a path that will at some point return to the original starting point). This course will probably be very challenging, but I expect it to be enjoyable and very useful for my future work in anthropology.

Probability theory is much less interesting, in my opinion. It involves ...well probabilities. What is the probability of the king's single sibling being a male? (surprisingly, it is not 50%, but I will get to that in a later post) for example is a common problem type. For the first class, we began with some combinatorics and worked with questions like “how many ways can you pick 3 girls and 5 boys from a group of 5 girls and 8 boys?” and “How many possible ways can you choose twelve cakes from four different types?”. I do not expect this course to be nearly as challenging at graph and set, but I also do not expect it to be as rewarding.

Mathematical Problem Solving, or MPS from this point on, is a class where we as a class work through multiple problems from many different subjects of mathematics. We do this using very interesting methods that simplify the problem in ways that may not be originally clear. We will work on problems primarily from four different divisions: number theory, algebra, geometry, and combinatorics. The problems from this class remind me a lot of the problem types that I would work on during math team in high school, and I have really enjoyed this class so far. Because of how much I have enjoyed it, I would like to work through a problem with you. The problem is as follows:

Let N=2^(20)x3^(15)x5^(10). 1) Find the number of divisors d s.t. (such that) d divides N. 2) Find the number of divisors s.t. d divides N and 10 divides d. 3) find the number of divisors s.t. D divides N and d is a perfect square (the square root of d is an integer (…,-2,-1,0,1,2...)).

1. Since d divides N, we know that d must be of the form 2^x x 3^y x 5^z. Because of the exponents of N, we know that x cannot be larger than 20, y cannot be larger than 15, and z cannot be larger than 15. Thus 0≤x≤20, 0≤y≤15, 0≤z≤10. Thus there are 21 options for x, 16 for y, and 11 for z. Thus, we can conclude that there are 21x16x11 possible values for d.
2. We can use the information give from number one to help solve number two. We still know that d is of the format 2^x x 3^y x 5^z, but now we know more about d. For d to be divisible by 10, we need to have at lease one 2 and one 5. Thus x and z must both be bigger than 1. So we now only have 20 options to x, 16 for y, and 10 for z. This makes the total possible numbers for d equal to 20x16x10
3. This part was one of my favorites. Once again the rule for d applies that it must be of the form 2^x x 3^y x 5^z and the range for each variable is the same as in part 1 ( 0≤x≤20, 0≤y≤15, 0≤z≤10). Now, though, we have the challenge of figuring out how to have d be a perfect square. If you think back to reducing square roots, you may recall that if you have a root with a number who has more than two of the same factors that this factor can be pulled out of the root once (i.e. say we have (24)^.5 [something the the one half power is the same as it being square rooted. This is simply easier to type]. 24 can be factored into 2x4x3 with is then 2³x3. This means that we can pull out two of these 2's and have the value 2x(2x3)^.5). This implies that for us to get a perfect square that we need to have the values for x,y, and z all be even (so that we can then pull out all of the doubles under the square root sign and get an integer). Thus our options for x are 0,2,4,6,8,10,12,14,16,18, and 20, for y are 0,2,4,6,8,10,12,14, and for z are 0,2,4,6,8,10. This counts up to 11 options for x, 8 options for y and 6 options for z. This implies that there are 11x8x6 options for d.

I hope that you enjoyed this problem as much as I did. I will continue to post questions on here that I found to be enjoyable, and that are fairly easy to explain to those of you with a limited background (so, sadly I will probably not be posting very many problems from my graph theory course).

Notice: From now on I will try to post two blogs a week. One on my mathematical studies here in Budapest and one on my life outside of the classroom.