## Monday, March 7, 2011

### Pseudocontext Versus Real: Using the Pythagorean Theorem

[Going through my old drafts, and deleting or finishing up... This one's just a week old.]

My textbook has a section called "Solving Equations by Factoring, and Problem Solving". In this section we get a somewhat artificial problem where an object is thrown upward (at a speed which is a multiple of 16 feet per second) from a tall building (whose height is a multiple of 16), for "an action shot" in a movie, providing the equation  h(t) = -16t2 + 80t + 576, which conveniently factors. That's ok with me, I love talking about gravity.

In fact, after we did the book's problem, I tried to make up another gravity problem, this time on the moon. We were still using feet, which I said would never happen on the moon, where sensible scientists would always use metric measurements. My problem didn't factor though (I got to remind students about using the quadratic formula), one more bit of evidence of how careful the problem creator had to be to make up a problem that would factor. Still, gravity is a great way to start thinking about quadratics.

The rest of the 'problems' in this section of the text (which are really exercises in using the Pythagorean Theorem) are complete hooey. Please don't ask me, "If one leg is 7 more than the other and the hypotenuse is one more than the longer leg", to find the sides. How contrived can you get?!  If you want students to practice their algebra skills in relation to the Pythagorean Theorem, how about getting them to think about the 4000-year-old puzzle of generating Pythagorean Triples? After seeing just a few examples, starting with 32+42=52 and 52+122=132, you might wonder ... If we set the hypotenuse to be one more than the longer leg, will that tell us anything?

a2 + b2 = c2 becomes
a2 + b2 = (b+1)2.
Now we get a2 + b2 = b2+2b+1,
or  a2 = 2b+1, which we can write as b = (a2-1)/2.
This seems to be a condition on the legs: a must be odd, so that its square minus 1 will be divisible by 2, and b can then be figured from a. Hmm, will this work for any odd number?

Isn't that a better way to practice your algebra skills?

1. So what seems 'real' to some can be pseudocontext, where math itself can actually be real context?! :)

2. If we hunt a difference of 2 instead, ie c = b + 2, then we arrive at
b = (a^2 - 4)/4 or b = (a/2)^2 - 1
Not only must a be even, but it must also be a multiple of 4 to generate a primitive triple. (why?)
8,15,17
12,35,37
16,63,65
20,99,101

Difference of 3? b = (a^2 - 9)/6, etc

Just yesterday a kid put up a 9,40,41 triangle, (cutesy Law of Cosines problem), and guessed that it was right, and I told the class that I was going to show them (another day, not enough time) how to generate.... (and then you posted!)

Jonathan

3. I love love love this Pythagorean Triple question! Fun for me to think about and hopefully use next year. 