Is
$\vec \nabla \cdot \vec A $
$\mbox{where } \vec\nabla = \frac{\partial}{\partial x}\hat i + \frac{\partial}{\partial y}\hat j + \frac{\partial}{\partial z}\hat k$
$\mbox{where } \vec A = \mbox{constant vector}$
$ = 0\mbox{?}$
Is there a proof somewhere that shows this?