When does thinking $(-8)^{1/3} = -2$ result in problems for an undergraduates?

Question

In high school we learn that the cube root of $-8$ is $-2$. Much later some of us learn about the single valued natural logarithm of a complex number, and that $w^z = e^{z\cdot Lz(w)}$ when $w$ and $z$ are complex. That gives a different result for the cube root of $-8$. When would what we learn in high school not serve an undergraduate majoring in physics or engineering?

Answer 1

This is a complicated question, and there are a number of articles written on the topic in the math education literature. Here are some of the entries that I would recommend (taken from the bottom of this answer):

• Goel, Sudhir K., and Michael S. Robillard. "The Equation: $-2 = (-8)^\frac{1}{3} = (-8)^\frac{2}{6} = [(-8)^2]^\frac{1}{6} = 2$." Educational Studies in Mathematics 33.3 (1997): 319-320.

• Tirosh, Dina, and Ruhama Even. "To define or not to define: The case of $(-8)^\frac{1}{3}$." Educational Studies in Mathematics 33.3 (1997): 321-330.

• Choi, Younggi, and Jonghoon Do. "Equality Involved in 0.999... and $(-8)^\frac{1}{3}$" For the Learning of Mathematics 25.3 (2005): 13-36.

• Woo, Jeongho, and Jaehoon Yim. "Revisiting 0.999... and $(-8)^\frac{1}{3}$ in School Mathematics from the Perspective of the Algebraic Permanence Principle." For the Learning of Mathematics 28.2 (2008): 11-16.

As the OP alludes to, there are in fact different definitions of the (single-valued) principal root in the real and complex contexts. It's quite common on SE for there to be confusion, cross-talk, and frustration between specialists in the two different domains, each assuming a different context and definition. I suppose that interacting with any of those types of discussions could count as an example for the OP's purpose.

Answer 2

That gives a different result for the cube root of -8.

It doesn't give a different result, it just gives two additional roots that are complex numbers, for a total of three roots.

A physics or engineering student in the US probably first learns about complex numbers in high school, but never sees any interesting applications. Then in college classes they get such applications (possibly not until they get to upper-division classes), and then they do have to do a certain amount of unlearning of facts that hold for the reals but not the complex numbers. (This unlearning should of course have happened in high school when they first saw complex numbers, but it doesn't really.) These applications would include, for example, analyzing linear filters in an electrical engineering class.

In my experience the particular issue you ask about has never been a problem for my physics students. The main problem with their high school background is that they have been drilled endlessly in the cartesian representation, so they have trouble transitioning to the polar representation, which is more natural and convenient for the applications we do. For example, it takes them 10 minutes to compute $1/i$, because they're doing it using some algorithm they've been taught for computing $(a+bi)/(c+di)$, rather than just visualizing $1/i$ on the unit circle.

Answer 3

The equation $(-8)^{1/3}=-2$ in isolation is taught in early algebra. Later, in precalculus, on learns about the Fundamental Theorem of Algebra. At this point, one starts to understand that this equation should be seen more generally in the context of roots of the polynomial $x^3+8=0$. One learns there are three roots, and falls back on the fact that $(-2)^3=-8$ to find one such root. Using the factor theorem and polynomial division with $\frac{x^3+8}{x+2}$, one reduces the problem of finding the other two roots to using the quadratic equation to solve $x^2-2x+4=0$. Typically these two solutions in precalculus are left in the form $x=1\pm i\sqrt{3}$. This is where many students stop developmentally, as described in Ben Crowell's answer.

Students going further in fields such as physics, engineering, and mathematics will develop further mathematically, and become comfortable with the polar representation of complex numbers. At this stage, the student will see the solutions of $z^3+8=0$ best represented by $z=2\exp\left(\frac{(2k+1)i\pi}3\right)$ with $k=0,1,2$.

So as others have pointed out in comments and answers, teaching that $(-8)^{1/3}=-2$ in elementary algebra is not going to be a problem for those who take more mathematics to become physicists, engineers, and mathematicians. In fact, teaching this arithmetic fact is the basic foundation for a future higher level understanding.

Answer 4

When would what we learn in high school not serve an undergraduate majoring in physics or engineering?

It would begin to be a problem when you enter any field where complex numbers appear. Obvious places where complex numbers appear would be in linear ODEs, Fourier transforms, etc. In quantum mechanics, wave functions are complex valued.

In many of these situations, making the cube root single valued would mean that you miss real valued solutions so some equations. Here's an analogue to your example: $$y'''=-8y$$ As it turns out, the solutions are generated by those of the form $y=e^{rx}$ where $r$ is some constant. (I believe physicists call this an ansatz.) When you plug this form into the ODE, you get the requirement that $r^3=-8$. If you take the real valued solution you get $y=e^{-2x}$.

That misses the other two real valued generators: $$ y= e^{x} \cos\left(\sqrt{3}x\right) \quad \text{and} \quad y= e^{x} \sin\left(\sqrt{3}x\right)$$ which are formed from the complex root solutions: $y=e^{2 \omega_3 x}$, $y=e^{2 \omega_3^2 x}$ where $\omega_3 = e^{i\pi/3}=\cos(\pi/3)+i\sin(\pi/3)=\frac{1+i\sqrt{3}}{2}$.

If you only had the real root, you would think that the system rapidly settles down to $0$. However, while this is possible, the generic behavior of a solution is an increasingly wild oscillation.