দেওয়া আছে, ভেক্টর ক্ষেত্রটি হলো, $$ \vec{V} = V_x \hat{i} + V_y \hat{j} + V_z \hat{k} $$ আমাদের দেখাতে হবে যে, $\vec{\nabla} \cdot (\vec{\nabla} \times \vec{V}) = 0$। প্রথমে, কার্ল (curl) বা $\vec{\nabla} \times \vec{V}$ নির্ণয় করি: $$ \begin{aligned} \vec{\nabla} \times \vec{V} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ V_x & V_y & V_z \end{vmatrix} \\ &= \hat{i} \left( \frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z} \right) - \hat{j} \left( \frac{\partial V_z}{\partial x} - \frac{\partial V_x}{\partial z} \right) + \hat{k} \left( \frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y} \right) \end{aligned} $$ এখন, ডাইভারজেন্স (divergence) বা $\vec{\nabla} \cdot (\vec{\nabla} \times \vec{V})$ নির্ণয় করি: $$ \begin{aligned} \vec{\nabla} \cdot (\vec{\nabla} \times \vec{V}) &= \left( \hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z} \right) \cdot \left[ \hat{i} \left( \frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z} \right) - \hat{j} \left( \frac{\partial V_z}{\partial x} - \frac{\partial V_x}{\partial z} \right) + \hat{k} \left( \frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y} \right) \right] \\ &= \frac{\partial}{\partial x} \left( \frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z} \right) - \frac{\partial}{\partial y} \left( \frac{\partial V_z}{\partial x} - \frac{\partial V_x}{\partial z} \right) + \frac{\partial}{\partial z} \left( \frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y} \right) \\ &= \frac{\partial^2 V_z}{\partial x \partial y} - \frac{\partial^2 V_y}{\partial x \partial z} - \frac{\partial^2 V_z}{\partial y \partial x} + \frac{\partial^2 V_x}{\partial y \partial z} + \frac{\partial^2 V_y}{\partial z \partial x} - \frac{\partial^2 V_x}{\partial z \partial y} \end{aligned} $$ যেহেতু আংশিক অন্তরীকরণের ক্রম পরিবর্তনযোগ্য (অর্থাৎ $\frac{\partial^2}{\partial x \partial y} = \frac{\partial^2}{\partial y \partial x}$), তাই পদগুলো কাটাকাটি গিয়ে শূন্য হয়: $$ \begin{aligned} \vec{\nabla} \cdot (\vec{\nabla} \times \vec{V}) &= \left( \frac{\partial^2 V_z}{\partial x \partial y} - \frac{\partial^2 V_z}{\partial y \partial x} \right) + \left( \frac{\partial^2 V_y}{\partial z \partial x} - \frac{\partial^2 V_y}{\partial x \partial z} \right) + \left( \frac{\partial^2 V_x}{\partial y \partial z} - \frac{\partial^2 V_x}{\partial z \partial y} \right) \\ &= 0 + 0 + 0 \\ &= 0 \end{aligned} $$ অতএব, $$ \vec{\nabla} \cdot (\vec{\nabla} \times \vec{V}) = 0 $$ দেখানো হলো।