Anyway, math is probably one of my favorite topics. Why? Because I beat my older sister at it:-). Basically, here's the story (I've told it a million times, so you can skip it if you've heard it before). Anyway, one day when I was 4 years old (note: I just turned 4, because we moved when I was 4 in May and my birthday is in February, but I was still in the old house, known as the "green house"). Anyway, I was 4 years old and my sister, Katherine, age 6, was in public school at that time. Katherine walked in the door and I was busy working on a math problem. The problem (I remember exactly) was 8888+8888. Those of you who are not 100% relient on calculators can see that that problem requires a mathematical technique known as "carrying" (in fact, you have to carry multiple times!). As far as I can tell, carrying is a topic taught in 2nd grade (if you disagree, please tell me. I'm just going with what I saw on the Internet in teacher curriculum). Anyway, Katherine did not know how to do the problem. Katherine was bigger, stronger, and louder than I and she was bossy and I was somewhat submissive, but I finally had something that Katherine didn't, so I decided to like math.
To give you an idea of how geeky I was at math, when we'd go on vacation, a favorite pass time of mine was multiplying fractions. I'd get my mother to write a bunch of fractions to multiply with each other (maybe 10 or so fractions all multiplied with each other). I would then spend my time reducing the fractions. Once I had reduced all the common factors, I would multiply. It was great fun.
Other than my parents, I can count my other math teachers on one hand (as of this writing). They are Dr. Kelly and Dr. McGowan of CCSU, and Dr. Madych, Dr. Kyle Kneisl, and Dr. You of UConn (Dr. Kneisl left UConn and teaching, which is too bad). I'd like to thank Dr. Kelly for teaching me mathematical precision and finesse. My mother had tried for years to get me to write out my steps in a clear manner, but I would always do all the steps in my head (I would be able to look at most problems and solve them in my head).
Anyway, at the moment I've plowed through the 3 step calculus sequence, differential equations, and a proofs writing class which also introduced some more abstract higher math topics. In the fall I should be taking combinatorics (counting), linear algebra, and a math history class.
Note to people who are interested in math and have a good grasp of some college math, I highly recommend the book "Journey Through Genius" by William Dunham. Dr. Kneisl recommended this book to me when I was in his MATH 213Q class, and while UConn was out drinking on Spring Weekend in X Lot, I was in the physics building reading that book. Some people say that abstract math is just chasing symbols across a page, but unlike other sciences, once the chasing is done, the result is always the same (and will remain the same forever). That book contains some of the most profound symbol chasing of all time. Nothing complex, but that's the beauty of it.
If there is any topic I would consider a higher education in, it would be math. Right now I am contemplating the possibility of applying to math graduate school to get a PhD in math. I already have some ideas for my dissertation, but the problem is that many geniuses of tried their hands at those problems and have gotten stuck. The problem is that if I got a PhD in math, I wouldn't get a job involving math (see the "Michael the businessman" section for an explanation).
So, for those of you who have a high enough math education, I have a problem for you. This was one of the bonus problems on Dr. Kneisl's 213Q final for Spring 2003. He didn't actually ask for a proof, but he just asked us to "collect evidence" that a statement was true. I wasn't able to furnish a full proof on the exam, but I did get one later. He claimed a graduate school education was needed to get the answer right, but I think everything needed to solve that problem was covered in MATH 213Q. I will try to explain this problem in full detail. Suppose we are dealing with the first 'n' numbers (the first 5 numbers would be 0, 1, 2, 3, and 4, but NOT 5). For our purposes no other numbers exist. Now we are dealing with multiplication. It's just like normal multiplication, except that if we get up to or above 'n', we have to adjust somehow, as for our purpose no number larger than or equal to 'n' exists. So, what we'll do is we'll subtract off 'n' until we get a number that does exist for our purposes. For example, with the first 5 numbers (n=5) let's say we want to multiply 3*4, that would give us 12, but 12 isn't a number in our small world. So, let's subtract off 'n', so 12-n=12-5=7, but that's still out of our world. So, let's do it again, 7-n=7-5=2. Ah, 2 is a number in our world. So, when n=5, 3*4=2 (pretty cool, huh?). Now, let's say n=6. I want to play around with squares, so I'm going to square every number in our new universe. 0*0=0, 1*1=1, 2*2=4, 3*3=9, but that's too big, so 3*3=9-6=3, 4*4=16, but that's too big, so 4*4=16-6=10-6=4, and 5*5=25, but that's too big again, so 5*5=25-6-6-6-6=1. Now, we notice something interest, some elements are their own squares! When we're dealing with the real numbers, the only numbers which have this property are 0 and 1, but here we see 3*3=3 and 4*4=4. So, a geek might ask himself, in a general case, how many numbers will there be that have this property? Suppose 'n' has 'k' unique prime factors, the answer, as it turns out, is that there are 2^k elements with this property (2^k means 2 to the power of 'k'). So, 8=2*2*2*2, which only has 1 unique prime factor (2), so there is 2^1=2 elements that have that property. 60=2*2*3*5, which has 3 unique prime factors (2, 3, and 5), so there will be 2^3=8 elements that are their own square. The question is, how do you prove this?
I'll give a couple hints for the above problem. The sum of 2 choose 0 + 2 choose 1 + 2 choose 2=2^2=4 (a similar sum for 3 would equal 2^3=8). That can be proven for the general case using induction (or you can take my word for it). Also if one number, 'c' has a certain set of prime factors, the next number up, c+1, will share NONE of c's prime factors.
That's all I have for math at the moment...
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