Divergence of Curl is Zero

Definition
Let $\R^3 \left({x, y, z}\right)$ denote the real Cartesian space of $3$ dimensions..

Let $\left({\mathbf i, \mathbf j, \mathbf k}\right)$ be the standard ordered basis on $\R^3$.

Let $\mathbf f: \R^3 \to \R^3$ be a vector-valued function on $\R^3$:


 * $\mathbf f := \left({f_x \left({\mathbf x}\right), f_y \left({\mathbf x}\right), f_z \left({\mathbf x}\right)}\right)$

Then:
 * $\nabla \cdot \left({\nabla \times \mathbf f}\right) = 0$

where:
 * $\nabla \times f$ denotes the curl of $\mathbf f$
 * $\nabla \cdot \left({\nabla \times f}\right)$ denotes the divergence of the curl of $\mathbf f$.

Proof
From Partial Differentiation Operator is Commutative for Continuous Functions:
 * $\dfrac {\partial^2 f_z} {\partial x \partial y} = \dfrac {\partial^2 f_z} {\partial y \partial x}$

and the same mutatis mutandis for the other partial derivatives.

The result follows.