From Mathematics to Art to Mathematics
A mini case study on the process of doing mathematics
Stephen Schiffman and Gale Rhodes, October 28, 2021
The typical organization of an article in a mathematics journal takes this form: theorem followed by its proof. Usually, however, that is not the way the work was conceived. Over the course of one weekend in the Spring of 2021, Gale and Steve did some mathematics and came up with a theorem (to be revealed at the end of this paper).
Sequence of Events Leading to the Theorem
1) Playing: Gale saw a photo of his grandson in front of an artwork inspired by mathematics. In turn, the artwork inspired him to make up a math problem for his grandson.
2) Inspiring Gale to formulate a problem we might pose to an OLLI class that we teach in Fall 2021.
3) Solving Gale’s problem.
4) Generalizing Gale’s problem.
5) Finding a partial solution to the general problem.
6) Discovering a serendipitous connection with another piece of mathematics that solves the general problem.
7) Capturing that discovery in the form of a theorem.
Notes for Each Event
1) This is the photo of Gale’s grandson in front of a work by Sol LeWitt. The artwork itself is inspired by the idea of superimposing mathematical curves upon each other. Look at the photo and you will see that LeWitt superimposes pairs of curves to form his drawing (shown at the bottom of the photo). For example, the top left corner of the drawing is curve 2 superimposed on curve 1, notated in the exhibit by the symbol 1/2. Gale then posed to his grandson the problem, “draw 5/6/7/8."
2. The question Gale proposed for our class is: “Draw 1/2/3; how many different [x/y/z] are there?”
3. Here is a “brute force” solution that enumerates all 56 possible [x/y/z]
4. The generalized problem is: given 2 integers n and N with 1<=n<=N, how many n-tuples (x1, x2, ... xn) are there where 1<=x1<x2<...<xn<=N ? In Gale’s original problem, n=3 and N=8.
5. Steve: “Instead of trying immediately for a general solution (i.e. any n and N), I stuck to the case where n=3 which I had solved by brute force using the table in note 3. I restricted myself to the somewhat more general question: how many monotonically increasing triples of numbers are there in the set of numbers 1 through N, where N is any integer greater than or equal 3. Doing some algebra - involving counting sums, and sums of sums - I arrived at the formula N(N-1)(N-2)/6. When N = 8, this gives 8*7*6/6 = 56, agreeing with the answer to Gale’s original question. This gave me confidence that the formula N(N-1)(N-2)/6 was correct. But I wasn’t looking forward to working out a formula for the general case of any n because the algebra looked messy, ugly and boring”.
6. Steve: “A most enjoyable aspect of doing mathematics is the sudden feeling of mathematical epiphany - when something that seemed mysterious suddenly reveals itself to become crystal clear. I had the niggling feeling that I’d encountered the formula N(N-1)(N-2)/6 before, with its multiplication of a series of numbers in decreasing order, but the 6 in the denominator made me a bit quizzical: why 6?
And so, falling asleep it hit me: Gale’s original question is the same as the question: “how many hands of 3 cards are possible in a deck of 8 playing cards?”
You may recall from studying basic combinations and permutations that the way you answer this is to note that there are 8 ways to choose the first card, 7 ways to choose the second card and 6 ways to choose the third card. Thus there are 8*7*6 ways to choose three cards, but any hand of three cards can be rearranged in 3! ways; therefore there are 8*7*6/3! possible hands. But 3!=6, so 8*7*6/3! = 8*7*6/6 =56. I now realized that my 6 was really a 3! in hiding. Same formula”.
The question of hands of cards is exactly the same question as Gale posed, but translated into a different setting: if you look at the “brute force” list (note 3) you will see that each entry has a different set of 3 numbers. You can interpret each entry as a hand of three playing cards chosen from a deck of playing cards numbered 1 through 8. Each entry corresponds to a hand because if you take the three numbers in the entry and “shuffle them around” (rearrange their order) there will be 3! ways to rearrange them but there is only one arrangement that puts them in increasing order of size (which is what you see as the entry). Thus each entry corresponds to a unique hand of 3 cards. Finally, any possible “hand of 3 cards” shows up as one of the 56 entries because any subset of 3 different numbers chosen from the set of numbers 1 through 8 shows up as an entry.
Thus, by thinking of hands of cards, the answer to the general problem (i.e. any n and N) is given by
N(N-1)(N-2)...(N-n+1)/n!
Theorem
7. Here is a statement of the theorem (don’t be frightened by the notation!) we arrived at:
Why is this true? Because the left- and right-hand sides are BOTH solutions to the problem: Given 2 integers n and N with 1<=n<=N, how many n-tuples (x1, x2, ... xn) are there where
1<=x1<x2<...<xn<=N ?
The left-hand side (the iterated summands) is the way one notates how to compute the answer by using the “brute force” method shown in note 3.
The right-hand side is the answer one gets by recognizing that the problem is the same problem as “how many hands of n cards are possible in a deck of N cards”.
You get two different formulas that are answers to the same problem; therefore the formulas must be equal.
Final Note
Steve: “If I had recognized immediately that Gale’s question was the same as the question of counting the number of hands of 3 cards in a deck of size 8, I would have arrived at the answer 56 = 8*7*6/3! immediately, and arrived at the formula for the answer to the generalized problem immediately. But in that case -- since I would not have had to construct a table and count entries by brute force - I would have missed out on arriving at our theorem”.
