Day 12, MAT115

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Announcements:
- The exams are all graded. I did look over the IMath exams, to see what problems you missed, and whether I thought you might deserve partial credit. So your score might have changed there.
  Sometimes it's hard to figure out how to submit an answer, and so here's the sort of thing that merited full credit:
  
  Binary representation of 32: 0100000
  The "0" out front isn't necessary, but it caused the answer to be marked wrong. But it's certainly correct! So for things like that I was able to intervene.
  By the way, if you want to make a six-figure salary, it's easy: just pad with 0s. So if you make \$70,000, that's the same as \$070,000. (So we're all making six-figure salaries now! I'm making seven-figures, actually; heck, why not: I'm making nine-figures....:)
- If you have any questions about the exams, you may ask me. I'll point out a few things that struck me as I was grading the Canvas part.
  1. Most of you have a pretty good idea of Fibonacci decomposition, and that's good because we're going to be continuing with Fibonaccis in the short term.
  2. The problem that made me saddest was #3. I asked you to explain two concepts: one-to-one correspondence, and uniqueness. And the results were pretty bad.
    Both results are really important, and I thought that I'd hit them hard enough -- perhaps not.
    I'll be making up a key out of your classmates work.
  3. The problem that made me feel the best was #4: many of you really have your Mayan and Babylonian numbers down! That's great.
  4. For the most part, #2 went pretty well. But a fair number of you don't know how to use a tree to do primitive counting; and I did steal too much Easter candy from some of you, because you didn't know how to do your Fibonacci decomposition.
  5. Problem #1 (and generally): don't be sparse with your words. Don't go on if you don't know what you're saying (that just annoys the grader); but if you really understand, go on! Share examples, make connections, etc. etc. The more things you say that make sense, the better I feel about assigning you the points.
- Materials for today:
  1. Here are a couple of videos covering today's material:
    1. Pascal's Facebooks.
    2. Egyptian Multiplication.
  2. Fibonaccis lurk within Pascal's triangle
  3. Derivation of the Golden Rectangle
Last time we talked more about spirals -- Fibonacci spirals, that is -- and they became Golden.
- The Fibonacci spiral leads to rectangles that tend toward a common side ratio of about 1.618.... The "Golden Rectangle" as defined by the Greeks has a side ratio of \[ \varphi=\frac{1+\sqrt{5}}{2} (\approx 1.618) \] called the golden ratio.
- We noticed that Fibonaccis were embedded in Pascal's triangle, and wondered if everything is actually embedded in Pascal's triangle. They'll probably find that all the ships and planes that have been lost in the Bermuda triangle are actually somewhere in Pascal's triangle!
- Now let's look at an example of how we can use Pascal's triangle in the study of "graphs". Consider the concept of simple graphs, which I'll present in the context of Facebook. A simple graph is made up of
  1. vertices (or points, which we will think of as people), and
  2. edges (which are also called arcs, connecting two points -- which we will think of as friendships).
  It turns out that Pascal's triangle gives all the ways of having a "Facebook" for any number of people.
  Now, if we don't care about "who's who", then we can see all the different types of Facebooks we can have:
  - Let's look at some simple cases: 2 or 3 individuals
  - All simple graphs with four vertices.
  - All simple graphs with five vertices.
  A simple graph is a way of connecting people -- either two are connected or not. The question is, how many different ways are there of having Facebook for $n$ distinctly different individuals?
  - Now, if we add names to the "vertices", then we have more Facebooks. If everyone's friends with everyone else, then the graphs we can get will look like this:
    
    Now how many different Facebook configurations are there for, say
    1. 2 individuals?
    2. 3 individuals?
    3. 4 individuals?
    How do we use Pascal's triangle to find the answer for a given number of people? We actually use it in two ways: to find the right row (a story of Gauss -- Prince of Mathematicians), and then to read off the number of different ways of having a given number of Facebook friendships....
  - The upshot is that a row of Pascal's triangle tells us how many different ways we can choose a certain number $p$ of objects from amongst $n$. This is why it was so useful to Pascal in his gambling exercises....
  - In the case of Facebook, we're looking at setting up friendships, which are relationships between n people. So we look to the row containing the number of friendships possible between $n$ people, which is the triangular number $\frac{n(n-1)}{2}$. That row tells us how many different Facebooks we can have with any given number of friendships.
Now, on a completely different note, we're going to travel now to Egypt, to do math like the Egyptians! Today it's multiplication.
- Let's start with a look at the history of a western understanding of Egyptian Mathematics.
- At the heart of Egyptian multiplication is the "Fraudini fact", or the "binary factorization":
  
  Every natural number is either a power of two, or can be expressed as a sum of distinct powers of two in a unique way.
  (I underline "distinct" because you cannot repeat powers: otherwise you could write, for example,
  3=1+1+1 rather than 3=2+1 (which is the unique binary factorization).
- Again this is parallel to our two other factorizations:
  1. The prime factorization: Every natural number (other than 1) is either prime, or can be expressed as a product of prime numbers in a unique way.
  2. The Fibonacci factorization: Every natural number is either Fibonacci, or can be expressed as a sum of non-consecutive Fibonacci numbers in a unique way.
- We'll start with a simple example: multiply 23*42.
  
  1 23
  2
  4
  8
  16
  32
  64 Too big!
- Let's show that we can do a multiplication in either order, by checking the product 42*23.
  
  1 42
  2
  4
  8
  16
  32 Too big!
- Let's try a longer multiplication. Consider, for example, 321*112:
  
  1 321
  2
  4
  8
  16
  32
  64
  128 Too big!
  
  Now add up those rows marked with an asterix (*), and you'll get the answer (35952).
- Examples: Try these:
  1. 43*16
  2. 21*79

Website maintained by Andy Long. Comments appreciated.

1	23
2
4
8
16
32
64	Too big!

1	42
2
4
8
16
32	Too big!

1	321
2
4
8
16
32
64
128	Too big!