Divergence of Curl is Zero

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

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

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


 * $\mathbf f := \tuple {\map {f_x} {\mathbf x}, \map {f_y} {\mathbf x}, \map {f_z} {\mathbf x} }$

Then:
 * $\nabla \cdot \paren {\nabla \times \mathbf f} = 0$

where:
 * $\nabla \times f$ denotes the curl of $\mathbf f$
 * $\nabla \cdot \paren {\nabla \times f}$ 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.