The teacher responded by asking (challenging?) for an example, and my first thought was the Pythagorean theorem since it is so ubiquitous in all areas of mathematics. You remember, a

^{2}+ b

^{2}= c

^{2}... that one. Now there may be more existing proofs of this theorem than any other in existence. So I drew a right triangle, made three copies and formed them into a square as shown in the figure. Then, with the power of modern educational technology, I moved two of the triangles to get the second figure. Q.E.D. as we say; thus it is shown. Any student who knows the geometric idea of the area of a square can see that the white area in both pictures is the same since the same amount has been removed from it (four congruent triangles). In the first the white area is a square whose sides are formed by the hypotenuses (hypotenii?), c, of the four triangles. In the second it is divided into two smaller squares, one with sides of length a, one with sides of length b. The area then can be expressed as a^2 + b^2 or as c^2, and it is the same area. Somehow I envision a good sixth grader understanding this (and the next sixth grader I see is in for a sit down/talk to so I can find out).

Afterwards it struck me that this was not the best of examples. The Pythagorean theorem is essentially a geometric idea, its about triangles after all. What about an idea from math that has nothing obvious to connect it to geometry . Partitions seemed to fit the bill, and I thought of a simple example that I was sure elementary kids could play with and confirm, and I hope be convinced by a simple geometric argument.

Ok, so for the uninitiated, a partition of a number is just writing it out as a sum of smaller numbers. In order to count them we usually write them from largest addend to smallest. For example 6 = 3 + 2 + 1, so that is one partition of six. There are obviously others, 2+2+2, or 3 + 1 + 1 + 1 would be two more. A common, and not trivial, question is what is the total number of partitions of six? Most elementary students can take an organized attack and come up with the answer. If we include 6 itself, there are eleven of them. 6; 5+1; 4+2; 4+1+1; 3+3; 3+2+1; 3+1+1+1; 2+2+2; 2+ 2+ 1+ 1; 2+1+1+1+1; and of course 1+1+1+1+1+1.

But an interesting thing pops out if we look at some special limitations. There are exactly 3 ways to partition six into three terms; 4+1+1; 3+2+1; 2+2+2. There are also exactly three partitions that use three as the largest number; 3+3; 3+2+1; 3+1+1+1 . Coincidence? Not a chance. The same thing would happen if you partitioned 25 into four terms (or five or six or pick your favorite number). There are exactly the same number of partitions of a number into n differen terms as there are partitions with n as the biggest value. If you have never seen this proof, try to make it clear to yourself why it is true before you read on.

OK, it seems true, but how do we relate that to geometry, and a visual proof? To the rescue comes a 19th Century British Mathematician, Norman M Ferrers, who was a fellow at Gonville and Caius (pronounced Keys) College just down the road here at Cambridge. He seems to have been the first to make a simple discovery about partitions using dots. Here is the Ferrer diagram for two of the unique partitions of six into three groups.

Reading the number of dots in each row we see that 6 = 4 + 1+1; but if we read down the columns we see 3+1+1+1. In the second diagram we see rows of 2+2+2; but columns of 3+3. Can you see that this would apply to any diagram? If there are three rows that add up to some number, then the columns will give us the same total with a three as the largest number. The same thing would be as clear with any other number of rows.

See, Geometry makes it easy to understand, and to explain. Geometry is good. Now go do your homework!

## 1 comment:

Another great example is from Euclid's second book. I find it difficult to get students to see that they cannot distribute the exponent over addition, e.g. that (a + b)^2 cannot be simplified as a^2 + b^2 (where the caret ^ means "taken to the power of"). Geometrically, is becomes obvious.

The lesson takes little time. I call for a volunteer and write on his pointer finger the letter 'a' above the top knuckle, and 'b' below it. I then do the same thing on the other hand. Then ask the volunteer to touch the ends of the fingers together at right angles. "If the whole finger length is 'c', what is the area of the imaginary rectangle contained between these fingers?" The class immediately answers, "c squared", or else 'c times c'." Then I write on the board, "c = a + b; c^2 = (a + b)^2", and I ask them if they agree. "Can we simplify (a + b)squared," I ask. Several students offer, "a^2 + b^2," which I write on the board and again ask, "does everyone agree?" Usually everyone does, but an advanced student may not.

"Let's see if this is true." Now I ask my volunteer to take the pointer fingers and cross them at the first knuckle at right angles. Then I ask the class to tell me what is the rectangular area contained by the bottom length of each finger? (The volunteer can complete the square by touching the thumbs together.) They immediately see it is 'b' squared. What is the imaginary rectangle contained by the top of each finger. The class immediately sees it is 'a' squared. And what is the area contained by the top of the right-hand finger and the bottom of the left-hand finger? "a times b," they say. And what is the area contained by the top of the left-hand finger and the bottom of the right-hand finger? "Also 'a' times 'b'," they say. "And so," I ask, "if we add up these four rectangles, what do we get?" This is easy-peasy, they think, as they answer, "a squared plus b squared plus a times b plus a times b." "Okay, lets write the whole problem on the board, using straight lines to represent our fingers." Most of the students see, before the diagram is drawn, that (a + b)^2 has got to equal a^2 + b^2 + 2ab.

The lesson can be repeated using the pointer and the ring finger, with different variables written on the ring finger to solve problems such as (a + b)(c + d).

The interesting thing about this lesson is that I will see students crossing their finger for a few days after as they are doing their classroom exercises, thus reinforcing the reason why FOIL works.

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