Curl and Divergence definitions - Is this definition mathematically correct? I've been using Stewart's Calculus: Early Transcendentals in my Calculus class, and one of the definitions of curl offered by the book was $$\text{curl } \vec F = \nabla\times \vec F$$ where $\,\nabla=\left\langle\dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z}\right\rangle$. My question is, why are we allowed to put operations into a vector and then "multiply" them with components of $\vec F$ in that way?
 A: I agree with @Ian, this notation is used for memorizing formula rather than as a strict and formal definition of mathematical concept.
Let me quote article about curl from Wikipedia:

The notation $\,\nabla \times \overrightarrow{F}\,$ has its origins in the similarities to the $3$ dimensional cross product, and it is useful as a mnemonic in Cartesian coordinates if $\,\nabla\,$ is taken as a vector differential operator del. Such notation involving operators is common in physics and algebra. However, in certain coordinate systems, such as polar-toroidal coordinates (common in plasma physics), using the notation $\,\nabla \times \overrightarrow{F}\,$ will yield an incorrect result.
Expanded in Cartesian coordinates ... , $\,\nabla \times \overrightarrow{F}\,$ is, for $\,\overrightarrow{F}\,$ composed of $\,\left[\begin{array}{3} F_x,& F_y, &F_z\end{array}\right]\,$:
\begin{align}
\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}
\end{align}
where $\,\mathbf{i},\, \mathbf{j},\,$ and $\,\mathbf{k}\,$ are the unit vectors for the $\,x,\, y,\,$ and $\,z\,$ axes, respectively. This expands as follows:
\begin{align}
\left(\frac{\partial F_z}{\partial y}  - \frac{\partial F_y}{\partial z}\right) \mathbf{i} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) \mathbf{j} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) \mathbf{k}
\end{align}

