If a vector field is divergenceless and curless, is that vector field = 0? Just a simple question, from the title of the thread, is a vector field = 0 if the divergence is 0 and the curl is 0? I had trouble finding an answer anywhere online, proof of why or why not would be helpful, thanks!
$$\nabla \cdot \vec{F} = 0$$
$$\nabla \times \vec{F} = 0$$
$$\vec{F} = ?$$
 A: No. Just consider a nonzero constant vector field, i.e. 
$$V=(a,b,c)$$
for some nonzero constants $a, b, c$. 
Then the curl of $V$ is zero vector, and the divergence of $V$ is $0$. 
A: No.  To get a feel for what's up, let $\phi$ be a non-constant harmonic function so $\nabla^2 \phi = 0$ but $\nabla \phi \ne 0$.  Set $\mathbf F = \nabla \phi \ne 0$; then   Then $\nabla \times \mathbf F = \nabla \times \nabla \phi = 0, \,$ since the curl of a gradient always vanishes.  Also, $\nabla \cdot \mathbf F = \nabla \cdot \nabla \phi = \nabla^2 \phi = 0$.  The divergence and curl of $\mathbf F$ both vanish, but not $\mathbf F$!
This line of reasoning can, like tape or film, be re-wound and run "backwards":  if $\mathbf F \ne 0$ and $\nabla \times \mathbf F = 0$, then (locally at least) there is a function $\phi$ with $\mathbf F = \nabla \phi \ne 0$; if now we also have $\nabla \cdot \mathbf F = 0$, then $\nabla^2 \phi = \nabla \cdot \nabla \phi = \nabla \cdot \mathbf F = 0$, and $\phi$ is harmonic.
The classic examples of such a field may be found in the elementary theory of electromagnetism:  in the absence of sources, that is, charges and currents, static (non -time varying) electric fields $\mathbf E$ and magnetic fields $\mathbf B$ have vanishing divergence and curl:  $\nabla \times \mathbf B = \nabla \times \mathbf E = 0$, and $\nabla \cdot \mathbf B = \nabla \cdot \mathbf E = 0$; the electrostatic potential function $\phi$ such that $\mathbf E = -\nabla \phi$ with $\nabla^2 \phi = 0$ exists by virtue of these facts; a similar assertion holds for the $\mathbf B$ field.
Hope this helps.  Cheerio,
and as always (as ol' James Clerk M. has taught us),
Fiat Lux!!!
A: For example $\vec{u} = (x+y, x-y, 0)$ has $\vec{\nabla} \cdot \vec{u}=0$ and $\vec{\nabla} \times \vec{u}=0$.
See picture here: 
A: No. For example, if we consider simple 2 dimension field,
$\vec{V} = a (x,-y) $     ,  for any scalar $a$ then 
$ \vec{\nabla} . \vec{V} = 0 $ and 
$ \vec{\nabla} \times  \vec{V} = 0 $ 
