Vector Field vs. Gradient Field? 
Suppose we have a gradient field $\vec{F}$. Is there thus a vector field $\vec{G}$ such that the curl of ($\vec{G}$) = $\vec{F}$?

So, I'm trying to find examples/counter-examples. If we take a vector field and div(curl($\vec{F}$)) $\ne$ 0, then we know this isn't the case. This is my thinking. Can anyone guide me through?
 A: The "usual" result is that this is impossible: it's a direct consequence of the Hodge decomposition of vector fields, and can be derived by assuming that
$$F = \nabla \phi = \nabla \times G$$
and noticing that
$$\langle F, F\rangle = \langle \nabla \phi, \nabla \times G\rangle = -\langle \phi, \nabla \cdot (\nabla \times G)\rangle = \langle \phi, 0\rangle = 0,$$
so that $F=0$. Here we have used the fact that the gradient and divergence are adjoint operators, which can be proven using integration by parts:
\begin{align*}
\langle \nabla \phi, v\rangle &= \int_{\mathbb{R}^3} \nabla \phi \cdot v\,dA \\
&= \int_{\mathbb{R}^3} \left[\nabla \cdot (\phi v) - \phi \nabla \cdot v\right]\,dA\\
&= - \langle\phi, \nabla \cdot v \rangle,
\end{align*}
where the first term of the second line vanishes by Stokes's theorem, provided that $\|p\|^2\phi(p) v(p)$ vanishes as $\|p\|\to \infty.$
However since you have not said anything about the behavior of $F$ as $\|p\|\to\infty$, the "usual" result does not necessarily apply. $$F = (0,0,1)$$
is both the gradient of
$$\phi = z$$
and the curl of
$$G = \frac{1}{2}(-y, x, 0).$$
