# A problem for helping kids prove zero/negative exponent properties

This comes via Paul Solomon during our Google Plus hangout, where we discussed approaches to teaching exponential expressions. Paul offered this problem:

Define a function recursively in the following way:

$f(a + b) = f(a)*f(b)$
$f(1) = 2$

Now, what’s f(2)?  The key is that $f(2) = f(1+1) = f(1) * f(1) = 2 * 2.$

And so on for f(3), f(4), and f(100).

Once kids get the hang of this, ask them what f(0) is. And then what f(-1) should be. *

(This can also be run the other way, with logarithms, by the way.)

The function notation, which my students don’t learn in Algebra I, isn’t crucial. It can be replaced with symbols such as [1] = 2, and [a + b] = a*b. This works even better along with some of Park Math’s defining new symbols activities.

Some commentary on this problem: it seems that what Paul is up to in this problem is that he’s using function notation to subtly eliminate some of what makes the idea of zero/negative exponents so frustrating for students. For students who are used to $a^b$ as representing $"b"$ copies of \$a\$ being multiplied together, it’s difficult to interpret zero or negative powers. After all, how do you multiply 0 copies of 2 together? “Isn’t that just nothing? Isn’t nothing 0?” **

But Paul has taken the power notation away from students, and kept just the algebraic structure. He’s giving students a chance to focus in on just the important features, and not to let the power notation distract their reasoning.

There’s a generally applicable lesson here: when the obstacle to a proof is a major conceptual hurdle, consider restating the problem in a form that minimizes the conceptual hurdle.

* Some approaches: $f(1) = f(1 + 0) = f(1) * f(0) = 2$. Or $f(0) = f(0 + 0) = f(0) * f(0)$. Once they have f(0), you can get $f(0) = f(1 + -1) = f(1) * f(-1)$
** For a larger discussion of this issue, see this post from Math Mistakes.