(ক) প্রমাণ করতে হবে যে, $\vec{\nabla} \times (\vec{\nabla} \phi) = \vec{0}$।
আমরা জানি, গ্রেডিয়েন্ট $\vec{\nabla} \phi = \hat{i} \frac{\partial \phi}{\partial x} + \hat{j} \frac{\partial \phi}{\partial y} + \hat{k} \frac{\partial \phi}{\partial z}$।
এখন কার্লের সংজ্ঞা অনুযায়ী,
$$
\begin{aligned}
\vec{\nabla} \times (\vec{\nabla} \phi) &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z} \end{vmatrix} \\
&= \hat{i} \left( \frac{\partial^2 \phi}{\partial y \partial z} - \frac{\partial^2 \phi}{\partial z \partial y} \right) - \hat{j} \left( \frac{\partial^2 \phi}{\partial x \partial z} - \frac{\partial^2 \phi}{\partial z \partial x} \right) + \hat{k} \left( \frac{\partial^2 \phi}{\partial x \partial y} - \frac{\partial^2 \phi}{\partial y \partial x} \right)
\end{aligned}
$$
যেহেতু আংশিক অন্তরীকরণের ক্রম পরিবর্তনশীল (অর্থাৎ $\frac{\partial^2 \phi}{\partial y \partial z} = \frac{\partial^2 \phi}{\partial z \partial y}$ ইত্যাদি),
$$
\begin{aligned}
\vec{\nabla} \times (\vec{\nabla} \phi) &= \hat{i}(0) - \hat{j}(0) + \hat{k}(0) \\
&= \vec{0}
\end{aligned}
$$
প্রমাণিত।
(খ) প্রমাণ করতে হবে যে, $\vec{\nabla} \cdot (\vec{\nabla} \times \vec{A}) = 0$।
ধরি, $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$।
কার্লের সংজ্ঞা অনুযায়ী,
$$
\begin{aligned}
\vec{\nabla} \times \vec{A} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix} \\
&= \hat{i} \left( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \right) - \hat{j} \left( \frac{\partial A_z}{\partial x} - \frac{\partial A_x}{\partial z} \right) + \hat{k} \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right)
\end{aligned}
$$
এখন ডাইভারজেন্সের সংজ্ঞা অনুযায়ী,
$$
\begin{aligned}
\vec{\nabla} \cdot (\vec{\nabla} \times \vec{A}) &= \frac{\partial}{\partial x} \left( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \right) - \frac{\partial}{\partial y} \left( \frac{\partial A_z}{\partial x} - \frac{\partial A_x}{\partial z} \right) + \frac{\partial}{\partial z} \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right) \\
&= \frac{\partial^2 A_z}{\partial x \partial y} - \frac{\partial^2 A_y}{\partial x \partial z} - \frac{\partial^2 A_z}{\partial y \partial x} + \frac{\partial^2 A_x}{\partial y \partial z} + \frac{\partial^2 A_y}{\partial z \partial x} - \frac{\partial^2 A_x}{\partial z \partial y}
\end{aligned}
$$
পদগুলো কাটাকাটি করলে পাই,
$$
\begin{aligned}
\vec{\nabla} \cdot (\vec{\nabla} \times \vec{A}) &= 0
\end{aligned}
$$
প্রমাণিত।