Saturday, August 05, 2017

boxes and fractions

Last week at MathFest, Dusa McDuff gave an excellent series of lectures on symplectic geometry. I enjoyed the second one the most, because in it she described the solution to a concrete problem that had a beautiful and expected answer, and used several tools of varying levels of difficulty. The result was published in the Annals of Mathematics in 2012; you can get the paper here. I won’t be referring to it for the rest of the post, however. Instead I want to highlight one elementary construction she described during this lecture.

About halfway through the talk, McDuff made a comment along the lines of, “Here’s something I must have learned in elementary school, but I’ve been surprised by how many mathematicians don’t know it.” I certainly don’t recall having seen it, at least not in the generality she described, and I do think people of all ages and mathematical abilities could enjoy playing with it.

Start with any fraction—say, \(30/13\)—and make a rectangle whose side lengths are the numerator and the denominator of the fraction.

Now start marking off squares, as large as possible, inside the rectangle. In this example, we can cut off two \(13 \times 13\) squares at first.
When no more large squares fit, start marking off squares from the remaining rectangle along the side.
Continue until you run out of space. In this example, only one more step in the process is needed.
When you’re finished, you will have filled the rectangle with squares of various sizes.
Count how many squares you have of each size and put those numbers in sequence. In this example, we have two large squares, three medium squares, and four small squares, so the sequence is 2, 3, 4. (Yes, I chose the fraction \(30/13\) to get this sequence.) Now write these numbers into a continued fraction, as follows: \[ 2 + \cfrac{1}{3 + \cfrac{1}{4}}. \] To evaluate a continued fraction, we start at the bottom and work through the nested operations: \[ 2 + \cfrac{1}{3 + \cfrac{1}{4}} = 2 + \cfrac{1}{\cfrac{13}{4}} = 2 + \frac{4}{13} = \frac{30}{13}. \] The result is the fraction we started with!

Here’s another example. If we start with the fraction \(25/7\) and follow the same process, we get this picture.

This time there are four different sizes of boxes. Counting the number of boxes of each size gives the sequence 3, 1, 1, 3, and we can check that \[ 3 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{3}}} = 3 + \cfrac{1}{1 + \cfrac{3}{4}} = 3 + \frac{4}{7} = \frac{25}{7}. \]

I had fun figuring out why this works, so in the interest of keeping this post short, I’ll leave that as an exercise for the reader. A hint: the process of cutting off squares is essentially Euclidean division.

Although we started with rectangles whose side lengths are integers, there’s no reason to restrict the above process to that case. In fact, if this process seems familiar, you may have seen it before in the special case of a golden rectangle, in which only one square of each size can be included:

This is related to the fact that the (infinite) continued fraction of the golden ratio has all \(1\)s: \[ \frac{1 + \sqrt{5}}{2} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \ddots}}} \] which I explained in another way in my last post.

One more example worth considering. Here is the start of the process when the side lengths of the rectangle are \(\pi\) and \(1\):

So there are three of the largest boxes, seven of the next largest, then fifteen of the next size (although the resolution of this image doesn’t let you see that). The fact that the first two steps nearly fill the entire rectangle is why \(3 + 1/7 = 22/7\) is such a good approximation for \(\pi\).

P.S. Dusa McDuff is also married to John Milnor, who gave the Hedrick Lectures in 1965. What other husband-and-wife pairs (or other pairs of family members) have both given the same high-profile math lecture series?

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