## Saturday, October 27, 2012

### Proving the Pythagorean Theorem

In a right triangle, where the lengths of the legs are given by a and b, and the length of the hypotenuse is given by c, a2+b2=c2
We use this so much in math, I have no idea where I first saw it. And it's so simple that I never had trouble remembering it. (The quadratic formula, on the other hand, did not make it into my memory banks until after I had started teaching college. For the first few courses I taught, I had to have it written at the top of my notes.) So I've known and used the Pythagorean Theorem for longer than I can remember.

It comes up in beginning algebra, and for years I showed students how to use it to solve ridiculously artificial algebra problems, never once addressing the issue of proof. This seems terribly wrong to me now. Perhaps about 15 years ago, I realized I'd been 'teaching' this to students for about a decade without even knowing its proof. I tried to come up with a proof on my own and had no idea how to start. Since this was before google became a verb (or even a word), I had to search for a book that would show it. I eventually found it in a high school geometry textbook. Luckily it showed a visually simple proof that stuck with me. (There are hundreds of proofs, many of them hard to follow.)

One of the reasons Pythagoras is held in high esteem by mathematicians is his proof of this idea. It had been used long before Pythagoras and the Greeks, most famously by the Egyptians. Egyptian 'rope-pullers' surveyed the land and helped build the pyramids, using a taut circle of rope with 12 equally-spaced knots to create a 3-4-5 triangle (since 32+42=52 this is a right triangle, which is pretty important for building and surveying). But the first evidence we have that it was proven comes from Pythagoras. Ever since the Greeks, proof has been the basis of all mathematics. To do math without understanding why something is true really makes no sense.

Pam Sorooshian is a homeschooler who trusts kids' natural instinct for learning. So she unschooled her kids (who are now grown and doing very well). That means she never required them to learn something they weren't interested in, and never pushed her own interests on them. She has a story in Playing With Math* that really stuck with me. In a talk to other unschooling parents, she said:
Relax and let them develop conceptual understanding slowly, over time. Don't encourage them to memorize anything - the problem is that once people memorize a technique or a 'fact', they have the feeling that they 'know it' and they stop questioning it or wondering about it. Learning is stunted.
It took me decades to wonder about how we know that a2+b2=c2. Now I feel that one of my main jobs as a math teacher is to get students to wonder. But my own math education left me with lots of 'knowledge' that has nothing to do with true understanding. (I wonder what else I have yet to question...) And beginning algebra students are still using textbooks that 'give' the Pythagorean Theorem with no justification. No wonder my Calc II students last year didn't know the difference between an example and a proof.

Just this morning I came across an even simpler proof of the Pythagorean Theorem than the one I have liked best over the past 10 to 15 years. I was amazed that I hadn't seen it before. (Maybe I did see it, but wasn't ready to appreciate it.)

My old favorite goes like this:
• Draw a square.
• Put a dot on one side (not at the middle).
• Put dots at the same place on each of the other 3 sides.
• Connect them.
• You now have a tilted square inside the bigger square, along with 4 triangles. At this point, you can proceed algebraically or visually.
Algebraic version:
• big square = small tilted square + 4 triangles
• (a+b)2 = c2 + 4*1/2*ab
• a2+2ab+b2 = c2 + 2ab
• a2+b2 = c2
•  www.cut-the-knot.org/pythag
Visual version:

To me, that seemed as simple as it gets. Until I saw this:
 from The Step to Rationality, by R. N. Shepard
This is an even more visual proof, although it might take a few geometric remarks to make it clear. In any right triangle, the two acute (less than 90 degrees) angles add up to 90 degrees. Is that enough to see that the original triangle, triangle A, and triangle B are all similar? (Similar means they have exactly the same shape, though they may be different sizes.) Which makes the 'houses with asymmetrical roofs' also all similar. Since the big 'house' has an 'attic' equal in size to the two other 'attics', its 'room' must also be equal in area to the two other 'rooms'.# Wow!

Added note (6-9-13): I've been asked to clarify why the big house must be equal in size to the two smaller ones added together. Since all three houses are similar (exact same shape, different sizes), the size of the room is some given multiple of the size of the attic. More properly, area(square) = k*area(triangle), where k is the same for all three figures. The square attached to triangle A (whose area we will say is also A) has area kA, similarly for the square attached to triangle B. kA+kB=k(A+B), which is the area of the square attached to the triangle labeled A+B. But kA = a2, and kB = b2. So k(A+B) = a2+b2. And it also equals c2, giving us what we sought, a2+b2 = c2.

I stumbled on the article in which this appeared (The Step to Rationality, by R. N. Shepard) while searching on 'thought experiment weight times distance must equal to balance'. I'm working on a handout for my Calc II students to explain centroid (since the Briggs textbook leaves this topic out). I was wondering if we need experimental evidence to show that the two sides of a teeter-totter will balance only when the weights times distances from the fulcrum are equal on the two sides. I thought maybe we could come up with a thought experiment that would convince us it must be true. I wasn't having any bright ideas, and turned to google. It hasn't solved my centroid question yet, but I love what I discovered.

I think that, even though this proof is simpler in terms of steps, it's a bit harder to see conceptually. So I may stick with my old favorite when explaining to students. Or maybe there's a way to test out which one is more helpful for a deep understanding of both the notion of proof and this theorem in particular.

What do you think?

____________
*Playing With Math: Stories from Math Circles, Homeschoolers, and Passionate Teachers is a collection of great writing about math education (often outside the classroom), from over 30 authors. I've been working on for 4 years now; it will be available within 2 to 4 months.
#I got this language (of houses, attics, and rooms) from a similar description of this proof which I found on Cut-the-Knot.

1. This isn't really a proof, but a really cool demonstration nonethelesshttp: www.youtube.com/watch?feature=player_embedded&v=CAkMUdeB06o

2. Are you familiar with Perigal's Proof?

Take a right triangle with sides a, b, and c (a <= b < c) and form squares A, B, and C on the three sides.

Draw a line through the center of B parallel to side c, dividing B into two equal pieces. Then draw a second line through the center of B perpendicular to the first line, dividing B into 4 equal pieces.

Each of those 4 pieces has: two sides of length c/2 bounding a right angle, two sides that add to length b bounding another right angle, and two angles which add to 180 degrees.

Cut the 4 pieces out and turn them around, so the 4 right corners in the center are now outside. The two non-right from adjacent pieces will form a line, and you'll get a bigger square with a square hole in it. The bigger square has sides of length 2(c/2)=c, the smaller hole has sides of length a -- which you can verify by dropping square A into it.

The video at http://www.youtube.com/watch?v=LtkAIQcACqY gives a good demonstration.

3. Lara, I remember seeing that (maybe a year ago?) and I agree with you - it's cool to watch.

Blaise,what makes that proof attractive to you?

4. It's like a tangram puzzle. By cutting B into 4 identical pieces, you can rearrange them, plus A, to form a square the size of C.

It doesn't rely on showing a chain of congruancies, or the proportionality of similar figures, or similar ideas. It can be made very tactile, actually drawing right triangles squares on the edges, and cutting them out, moving them around.

There are hundreds, thousands,even, of different proofs of the Pythagorean Theorem, some simpler, some complex. There are a lot of proofs. The very interesting page http://www.cut-the-knot.org/pythagoras/index.shtml gives 98. Your old favorite is #9, your new favorite is a variant on #6, and Perigal's comes in at #14.

5. Minor quibble: In your proof with the square surrounded by triangles, you also need to show that the shape on the inside is actually a square rather than a rhombus. It's easy to prove, but important not to leave it out.

6. Good quibble. I used to show that the angles are 90 degrees (because the two angles surrounding each one add to 90 degrees). The last time I showed this proof, I said that since the triangles are all the same, those angles must be, which makes them 90 degrees.

7. Oh, neat proof, I hadn't seen that before. (Also, I really like "since the triangles are all the same, those angles must be, which makes them 90 degrees" -- symmetry arguments are the best. =)

Your question about centroids reminds me of http://www.amazon.com/The-Mathematical-Mechanic-Physical-Reasoning/dp/0691140200 . I don't know if it actually contains something about centroids in particular, I don't think I have a copy any more so I can't check.

8. Thanks for reminding me of that book, Brent. I had bought it, and found it pretty challenging. I knew I wouldn't read it, so I gave it to my colleague who loves physics. If she has it in her office, I can check on the centroid issue.

9. omg! thank you soooooooooooooooooooo much! i actually understand my homework i hope you know your helping people learn out there <3 thank you again!!!! YOUR THE BEST!!(= xoxox

10. Hi Anon,