How to explain students (which are very new to math) why $(-2)^x$ is not a definable function?

Question

Whenever I teach exponential (real) functions, I encounter the question: "But why $(-2)^x$ is not defined? I mean, $(-2)^{-1} = -\frac{1}{2}, (-2)^3 = -8$,..." etc. I guess the true domain of this function is the complex plain. However, I myself, not an expert in complex functions, and the students I teach, are not familiar with he subject at all. So, my questions are:

1. (For myself) If I view this function only on the real line domain, and let A be the set of real points on which it is not defined, what is A? Is it a dense set in $\mathbb R$ is in an $\aleph_0$ set or is it an $\aleph$?

2. (For my students) How would you explain this someone who is not yet familiar even with the definitions of continuity and differentiability (of real function). i.e. what would be your answer to the question: "But why $(-2)^x$ is not defined? I mean, $(-2)^{-1} = -\frac{1}{2}, (-2)^3 = -8$,..."

Thanks

Answer 1

I would say that the function is defined but only on a particular subset of $\mathbb{R}$.

Observe that $$a^x = e^{\ln{a^x}} = e^{x\ln{a}} \text{ .}$$

Also note that for $a>0$ then

$$ \begin{align} &\ln{(-a)} = z\\ \implies & a = -e^z \\ \implies & a = e^{i(2k+1)\pi}e^z \\ \implies & a = e^{z+i(2k+1)\pi} \\ \implies & z = \ln{a}-i(2k+1)\pi = \ln{(-a)} \text{ .}\\ \end{align}$$

Plugging this in for $(-2)^x$, we see

$$ (-2)^x = e^{x(\ln(2)-i(2k+1)\pi)} = e^{-i(2k+1)\pi x} \: e^{x\ln 2} \text{ .}$$

Now it's much more obvious that there is a real output only if there exists $k\in \mathbb{Z}$ such that $x(2k+1)\in\mathbb{Z}$.

As for your students, if they know complex numbers (or even if they don't) then give them some examples of where the function is defined and where it cannot be defined. For example it should be quite obvious that if $x=\frac{1}{2}$ then the function does not produce a real value.

Answer 2

For (1), the "real" definition of $y^x$ is $\exp(x \log(y))$ where you've fixed some branch of the complex logarithm. You're asking when $\exp(x \log(-2))$ is real. First off, $\log(-2) = \log(2) + (2k+1)\pi i$ for any integer $k$, where $\log(2) = 0.69...$ is the usual real number. You have to fix a particular $k$. In any case, you're asking when $\exp(x (\log(2) + (2k+1)\pi i) \not\in \mathbb{R}$ where $x \in \mathbb{R}$, or equivalently when $x(\log(2) + (2k+1)\pi i) \not\in \mathbb{R} + \pi i\mathbb{Z}$. That's precisely when $(2k+1) x$ is not an integer. People often pick the $k=0$ branch, in which case $A = \mathbb{R} - \mathbb{Z}$, which in any case is always uncountable with the cardinality of the continuum.

For (2), starting from how you've framed it, I'd probably say, "Ok, $(-2)^1 = -2$, $(-2)^2 = 4$, and the signs alternate as we increment the exponent by 1. Then what should the sign of $(-2)^{1.5}$ be? Is it positive or negative?" Both "positive" and "negative" don't end up making sense for any number of reasons, so if it's to be defined it has to be something somehow in-between--ultimately a point in the complex plane.

Answer 3

Although we use the same notation $b^k$ to express the two concepts 1) the operation of multiplying $b$ by itself $k$ times and 2) the exponential function $f(x) = b^x$, mathematicians don't actually consider these to be the same thing at all.

The operation of multiplying $b$ by itself $k$ times only makes sense if $k$ is an integer but it doesn't make any sense if $k$ is not an integer.

But the function $f(x) = b^x$ is a function that works for all real $x$.

However if $x$ happens to be an integer these two concepts coincide. That is why we use the same notation and why we don't usually sweat it.

But as continuous function $f(x) = b^x$ won't work if $b$ is negative. For one thing $(-2)^{\frac 12}$, or $(-2)^x$ where $x$ is a fraction with an even denominator, is not defined. There is no real number it could be. Also if $x$ is rational with an odd denominator and an even numerator then $(-2)^x$ is positive. But if $x'$ is ration with an odd denominator and an odd numerator then $(-2)^{x'}$ is negative, even though $x$ and $x'$ may be arbitrarily close together. So $(-2)^x$ can not be continuous. And thirdly for $x$ irrational... we have no way of defining whate $(-2)^x$ is.

To explain to children I'd point out the $(-2)^{\frac 12}$ can not have any value. And I'd point out that because $\frac 13=\frac 5{15}$ and $\frac 6{15}$ are close together that $(-2)^{\frac 13}$ would be equal to $-\sqrt[3]{2}$ a negative number thatn $(-2)^{\frac {6}{15}}$ would be equal to $\sqrt[15]{(-2)^6} =\sqrt[15]{64}$ is a positive number so the function jumps all over the place and disappears and has holes everywhere.

.... SO it just has too many problems that we just don't bother. If $x$ is a variable possibly any real number we only define it for a positive base. It just doesn't work on a negative base.