How does this definition of curl make sense? In our maths textbook, curl is defined as $\text{curl} \textbf{F} = \nabla \times \textbf{F}$.  If the $\nabla$ operator isn't a vector, but an operation on a vector, how can it be an operand of a cross product?
$$\textbf{F} \in \mathbb R^3 \\ (\nabla) \in \mathbb R^3 \rightarrow \mathbb R^3 \\ (\times) \in \mathbb R^3 \rightarrow \mathbb R^3 \rightarrow \mathbb R^3 \\ \nabla \times \textbf{F} \in \mathbb R^3$$
This seems as silly to me as:
$$2 \in \mathbb R \\ \text{sin} \in \mathbb R \rightarrow \mathbb R \\ (+) \in \mathbb R \rightarrow \mathbb R \rightarrow \mathbb R \\ \text{sin} + 2 \in \mathbb R$$
Obviously, you can't treat sine as a number and add it to 2.  How come you can treat functions as numbers in the definition of curl?  Finding this cross product makes less sense, since you have the determinant of a matrix containing 3 vectors, 3 functions, and 3 scalars. 
 A: Think about the cross product.
For two vectors, $\vec{A}$ and $\vec{B}$,
$$\vec{A}\times\vec{B}=\begin{vmatrix}\mathbf{i}&&\mathbf{j}&&\mathbf{k}\\A_x&&A_y&&A_z\\B_x&&B_y&&B_z\end{vmatrix}$$
If we were to replace these instead with $\vec{\nabla}$ and $\vec{F}$, then we have
$$\vec{\nabla}\times\vec{F}=\begin{vmatrix}\mathbf{i}&&\mathbf{j}&&\mathbf{k}\\\frac{\partial}{\partial x}&&\frac{\partial}{\partial y}&&\frac{\partial}{\partial z}\\F_x&&F_y&&F_z\end{vmatrix}$$
If you were to combine the components of $\vec{\nabla}$ with the components of $\vec{F}$ in the same way as you would two vectors in the cross product, you will get the correct formula for the cross product - assuming you are using Cartesian coordinates.
This has simply been carried on as notation, a handy way of remembering the formula - it's not by any means formally correct. The $\times$ doesn't denote a separate operation, like the cross product does.
The same thing goes for the divergence, $\vec{\nabla}\cdot\vec{F}$. The $\cdot$ shouldn't be taken as an extra operation on $\vec{\nabla}$ or $\vec{F}$; the formula for divergence simply happens to be extraordinarily similar to the formula for the dot product.
A: $
\mathrm{\nabla}
$
can be regard as a vector
$
\left({\frac{}{\mathrm{\partial}{x}}\mathrm{,}\frac{}{\mathrm{\partial}{y}}\mathrm{,}\frac{}{\mathrm{\partial}{z}}}\right)
$
to participate in calculation
