What is $(-1)^\sqrt{2}$? How do you reason that $(-1)^\sqrt{2}$ is a complex number based on the following statements:


*

*$(-1)^{p/q}$ is a complex number if $p$ is odd and $q$ is even integers.

*Rational numbers are dense in the real line.

*Exponential is a continuous function.


Another question arises as what $(-1)^\sqrt{2}$ represents on the real line?
 A: 
Note that in general $z^c$ is defined as $e^{c\log(z)}$, where $\log(z)$ is the multi-valued complex logarithm given by $\log(z)=\log(|z|)+i\arg(z)$. 

Here, we have
$$\begin{align}
(-1)^\sqrt2&=e^{\sqrt2\log(-1)}\\\\
&=e^{\sqrt2 (i(2n+1)\pi)}\\\\
&=e^{i(2n+1)\pi \sqrt2} \\\\
&=\cos((2n+1)\pi \sqrt2)+i\sin((2n+1)\pi\sqrt2)
\end{align}$$
where $n\in \mathbb{N}$.
If we choose the principal branch of the logarithm (i.e., $n=0$), then $-1=e^{i\pi}$ and we have
$$(-1)^\sqrt2=\cos(\pi\sqrt2)+i\sin(\pi\sqrt2)$$

Now, if $p$ is an odd integer and $q$ is an even integer, then we have
$$\begin{align}
(-1)^{p/q}&=e^{(p/q)\log(-1)}\\\\
&=e^{(p/q) (i(2n+1)\pi)}\\\\
&=e^{i(2n+1)p\pi  /q} \\\\
&=\cos((2n+1)p\pi/q )+i\sin((2n+1)p \pi/q )
\end{align}$$
If $p$ is odd, then certainly $p(2n+1)$ is also odd and $p(2n+1)/q$ is not an integer.  Therefore, $\sin((2n+1)p\pi/q)\ne 0$ and $(-1)^{p/q}$ has a non-zero imaginary part.

By continuity of the complex exponential function, for any $\epsilon>0$, there exists a $\delta>0$ such that 
$$\left|e^{i(2n+1)p\pi/q}-e^{i(2n+1)\sqrt 2\,\pi}\right|<\epsilon$$
whenever $|(2n+1)p\pi /q-(2n+1)\sqrt2 \pi|<\delta$.
By the density of the rational numbers, for this $\delta>0$ there exist integers $(2n+1)p$ and $q$ such that $|(2n+1)p\pi /q-(2n+1)\sqrt2 \pi|<\delta$. 
And we are done!
A: As Mark pointed out in the comment, 
$$(-1)^{\sqrt{2}}=(i^2)^{\sqrt{2}}=i^{2\sqrt{2}}\text{.}$$
Now we know that
$$i^{2\sqrt{2}}=(e^{\pi i/2})^{2\sqrt{2}}=e^{\pi i\sqrt{2}}\text{,}$$
and by De Moivre's formula,
$$e^{\pi i\sqrt{2}}=\cos(\pi\sqrt{2})+i\sin(\pi\sqrt{2})$$
which is complex (and transcendental).
Not sure how to use your given fact, though, since $\sqrt{2}$ is irrational.
