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


 * $\mathbf V := \tuple {\map {V_x} {\mathbf x}, \map {V_y} {\mathbf x}, \map {V_z} {\mathbf x} }$

Then:
 * $\map {\operatorname {div} } {\curl \mathbf V} = 0$

where:
 * $\curl$ denotes the curl operator
 * $\operatorname {div}$ denotes the divergence operator.

Proof
From Curl Operator on Vector Space is Cross Product of Del Operator and Divergence Operator on Vector Space is Dot Product of Del Operator:

where $\nabla$ denotes the del operator.

Hence we are to demonstrate that:
 * $\nabla \cdot \paren {\nabla \times \mathbf V} = 0$

Let $\mathbf V$ be expressed as a vector-valued function on $\mathbf V$:


 * $\mathbf V := \tuple {\map {V_x} {\mathbf r}, \map {V_y} {\mathbf r}, \map {V_z} {\mathbf r} }$

where $\mathbf r = \tuple {x, y, z}$ is the position vector of an arbitrary point in $R$.

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

Hence:

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

and the same mutatis mutandis for the other partial derivatives.

The result follows.