Is $(-2)^{\sqrt{2}}$ a real number? 
Is $(-2)^{\sqrt{2}}$ a real number?

Clarification: Is there some reason why $(-2)^{\sqrt{2}}$ is not a real number because it doesn't make sense why it shouldn't be a real number.
Mathematically we can define the value of $\sqrt{2}$ in terms of a limit of rationals. But the problem is is that some sequences will have values that are not defined for $(-2)^q$ where $q$ is a rational, such as $\dfrac{3}{2}$. Is this the reason why we can't define it or is there some other reason?
 A: By definition, $(-2)^\sqrt{2} = \exp(\sqrt{2} \log(-2))$.  This is multivalued because $\log$ is multivalued: we have $\log(-2) = \log(2) + (1 + 2 n) \pi i$ for arbitrary integer $n$.  Thus 
$$ (-2)^{\sqrt{2}} = \exp(\sqrt{2} (\log(2) + (1+2n) \pi i)) = 2^{\sqrt{2}}
\exp((1+2n) \sqrt{2} \pi i)$$
In order for this to be real, we would need $ (1+2 n) \sqrt{2}$ to be an integer, and that is not the case because $\sqrt{2}$ is irrational.
A: No. There are a countably infinite number of possible values of this expression. None of them are real. They are
$$2^{\sqrt{2}}(\cos (2k+1)\pi\sqrt{2} + i\sin(2k+1)\pi\sqrt{2})
$$
The principal value (taking $k=0$) is $2^{\sqrt{2}}(\cos\pi\sqrt{2} + i\sin\pi\sqrt{2})$.
A: $$\left(-2\right)^{\sqrt{2}}=\left(\left|-2\right|e^{\arg(-2)i}\right)^{\sqrt{2}}=\left(2e^{\pi i}\right)^{\sqrt{2}}=2^{\sqrt{2}}e^{\pi\sqrt{2}i}=2^{\sqrt{2}}e^{\left(2\pi k+\pi\sqrt{2}\right)i}$$
With $k\in\mathbb{Z}$
A: I'll answer the general question of how we make sense of such expressions:
The value of such numbers is determined in one of two ways: by functional equations or by power series. A functional equation is any formula that relates the value of a function at two different points. Rules like $a^{b+c}=a^ba^c$ are examples of basic functional equations which can be rewritten as $f(a,b+c)=f(a,b)f(a,c)$ if that is preferred. An example of a function defined this way is $\Gamma(t):=\int_0^\infty x^{t-1}e^{-x}dx$. After applying integration by parts, we see that $t\Gamma(t)=\Gamma(t+1)$ holds, and we can use this to define $\Gamma(t)$ for any complex number that is not of the form $-n$ for $n\in\mathbb{N}$ (including $0$). This slightly weird way of expressing the undefined points is justified by the fact that at $-n$, $\Gamma(t)$ has a simple pole with residue $\frac{(-1)^n}{n!}$
Such values can also be determined by trying to make a power series make sense when values are plugged into it (for example, we might say $e^{i\pi}=-1$ because of what happens when we plug $i\pi$ into the power series for $e^x$). An example of a function defined this way is the matrix function $f(A)=e^A$. You can prove that the power series expression for $e^x$ around $0$ converges when interpreted as a statement about matrices. This function shares many of the properties of the complex function $f(z)=e^z$.
The wikipedia page on Analytic continuation has more info about extending (complex differentiable) functions.
We like these ways of doing things because they usually ensure that the properties we like the functions to have continue to hold. For example, extending via functional equation guarantees that the extended function also satisfies the functional equation, while power series are useful for preserving analytic properties of the function.
To deal with the question at hand now, your expression turns out to satisfy: $$(-2)^\sqrt{2}=2^{\sqrt{2}}(\cos((2k+1)\pi\sqrt{2}) + i\sin((2k+1)\pi\sqrt{2}))$$
which arises out of rewriting $a^x=\exp(x\log(a))$ and using the power series for $\exp(z)$. This is how we have chosen to define calculating generalized exponents. This is a multivalued expression, but for the principal value of $k=0$ we get $(-2)^\sqrt{2}=2^{\sqrt{2}}(\cos(\pi\sqrt{2}) + i\sin(\pi\sqrt{2}))$ which is not a real number because $\sin(\sqrt{2}\pi)\neq 0$.
A: " Is there some reason why $(-2)^{\sqrt 2}$ is not a real number because it doesn't make sense why it shouldn't be."
Well, not having a sensible interpretation is a pretty good reason for something not to be a real number.
1)  $b^n$; $n \in \mathbb N; n \ge 2$ defined as $\prod_{i=1}^n b$ (or "b multiplied by itself n times") is well defined and "makes sense".
2) $b^n$; $n \le 1$ doesn't make sense because you can't multiple a number by itself 0 or a negative number of times.  You can't even multiply a number by itself 1 time.  But multiplication is invertable and is additive in nature so we can extend the definition to include $b^0 = 1$ and $b^{|n|} = (1/b)^{|n|} = (1/b)^{|n|}$ (As well as $b^1 = b$).  So now it does make sense.
3) $b^r$; $r \in \mathbb Q$ doesn't make sense because you can't multiple or divide a number a fractional number of times.  But as we can prove for any non-negative real number $b$ and integer $m \ne 0$ we can find a unique non-negative real number, c, where $c^m = b$.  We call this $c =\sqrt[m]{b}$ but as $(b^a)^m = b^{am}$.  If we called this $c =\sqrt[m]{b} = b^{\frac 1 m}$ we'd have the consistent $(b^{\frac 1 m})^m = b^1 = b$ and we can define $b^{n/m}$ as $\sqrt[m]{b}^n$ !!!!IF!!!! b is non-negative OR m is odd.  The is no real even root of a negative number.  So $(-2)^{3/2}$ is not a real number!  This definition is consistent and now makes sense.  (We have to verify this is truly consistent which I won't do here.)
4) In high school we tend to wave our hands at real numbers.  They are like rational numbers "but with the holes filled in".  So $b^x$ where x is the limit of some sequence of $\{q_i\} \subset \mathbb Q$ can be defined as the limit of {$b^{q_i}$}.  (Okay, a lot of verifying needs to be done that I will omit.)  But if $b < 0$ this sequence isn't defined for any rationals with even denominators and need not converge at all.
So $(-2)^{\sqrt{2}}$ isn't a real number and it wouldn't make sense that it were.
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Do I want to confuse you further?  In the shower I was wondering that as for all $q \in \mathbb Q$ $q = \sup\{r \in \mathbb Q| r \le q;$ r in reduced form has an odd denominator}, if you could have modified the definition of $b^q$ as the limit of {$b^{r_i}$} where $r_i$ are a sequence of rationals with odd denominators that converge to $q$.
If so than we'd have $(-|b|)^q = -|b|^q = -\sqrt[m]{|b|}^n$. And we could define $b^x$ as the limit of $b^{q_i}$ where the $q_i$ converge to x.
Could we do that?  It would mean that $b^{1/m}$ need not be the same as $\sqrt[m] b$, but could we do that?  We'd lose complex analysis but...
Then I realized we could but then we'd no longer have the $b^x b^y = b^{x+y}$ rule.
