Curl-less fields and divergence-less fields I was wondering if anyone could explain some intuition or proof of the two following statements.
$\nabla \times \textbf{F} = 0 \leftrightarrow \textbf{F} = \nabla V$
$\nabla \cdot \textbf{G} = 0 \leftrightarrow \textbf{G} = \nabla \times \textbf{A}$
I understand what they imply when the fundamental theorems of calculus are considered, I am just curious to why one implies the other.
 A: For the first identity, the key point is $\boldsymbol\nabla \times\mathbf{F} = 0 \rightarrow\oint \mathbf{F}\cdot d\boldsymbol\ell = 0$ for any closed path. This means that for any two paths from $\mathbf{a}$ to $\mathbf{x}$,
$$
\int_\mathbf{a}^\mathbf{x}\mathbf{F}\cdot d\boldsymbol \ell_1 - \int^\mathbf{x}_\mathbf{a}\mathbf{F}\cdot d\boldsymbol \ell_2 = \int_\mathbf{a}^\mathbf{x}\mathbf{F}\cdot d\boldsymbol \ell + \int^\mathbf{a}_\mathbf{x}\mathbf{F}\cdot d\boldsymbol \ell = \oint \mathbf{F}\cdot d\boldsymbol \ell = 0,
$$
i.e., the integral does not depend on the path taken, only on $\mathbf{a}$ and $\mathbf{x}$. Thus, $V(\mathbf{x}) = \int_{\mathbf{a}}^\mathbf{x}\mathbf{F}\cdot d\boldsymbol \ell$ is a well-defined scalar function of $\mathbf{x}$, and it can be quickly verified that $\mathbf{F} = \boldsymbol\nabla V$.
I'm currently having trouble coming up with a similar sort of proof for the second case. If you have to, you can just take the curl of the Biot-Savart law,
$$
\mathbf{A}(\mathbf{x}) = \frac{1}{4\pi}\int\!\!\!\int\!\!\!\int \left[\mathbf{G(\mathbf{x'})}\times\frac{\mathbf{x}-\mathbf{x'}}{|\mathbf{x}-\mathbf{x'}|^3}\right]d^3\mathbf{x'} + \boldsymbol\nabla f
$$
and verify that this is the desired $\mathbf{A}$, but that doesn't give much insight as to where the Biot-Savart law comes from (and also requires certain conditions on $\mathbf{A}$ for the integral to converge, I guess).
