Can someone prove that:
$$\nabla \cdot \mathbf{B} = 0 \implies \mathbf{B} = \nabla \times \mathbf{A}~?$$
I know that $$\nabla \cdot (\nabla \times \mathbf{A}) = 0$$ identically.
But can one prove that if $\mathbf{B} \neq \nabla \times \mathbf{A}$ for any $\mathbf{A}$ then, $\nabla \cdot \mathbf{B} \neq 0?$
This is called Helmholtz theorem, which states that for any vector field $\vec{F}$ that is twice continuously differentiable in a bounded domain, we can perform the decomposition $$ \vec{F} = - \vec{\nabla} \Phi + \vec{\nabla}\times\vec{A} $$ See http://en.wikipedia.org/wiki/Helmholtz_decomposition for a derivation
