Question: Say we have scalar potential in a form 
$$
U = A \ln (\vec{a} \times \vec{r})^2 e^{-\vec{b} \cdot \vec{r}}.
$$
How would one calculate gradient $\vec{E}=-\nabla U$ of such potential? 

A is a constant, $\vec{a}$ is a constant vector and $\vec{r}=(x, y, z)$.

Answer: 
Pick two points in $(x\_1,x\_2), (y\_1,y\_2) \in A^c$.

At least one of $x\_1,x\_2$ and at least one of $y\_1,y\_2$ are irrational.

Suppose $x\_1, y\_1$ are irrational. Then $(x\_1,s), (y\_1,t) \in A^c$ for all $s,t \in \mathbb{R}$. Then the path $(x\_1,x\_2) \to (x\_1,\sqrt{2}) \to (y\_1,\sqrt{2}) \to (y\_1,y\_2)$ lies entirely in $A^c$.

Suppose $x\_1,y\_2$ are irrational. Then the path $(x\_1,x\_2) \to (x\_1,y\_2) \to (y\_1, y\_2)$ lies entirely in $A^c$.

The other combinations follow the same pattern.
Consider two points $P\_1 = (x\_1, y\_1)$ and $P\_2 = (x\_2, y\_2)$ in $A^c$. For each point, at least one of the coordinates is irrational.

Say, for example, that $y\_1 \in \mathbb{R} \setminus \mathbb{Q}$. Then, you can travel horizontally along the line $y = y\_1$, all the while, staying within $A^c$. Similarly, if $x\_1 \in \mathbb{R} \setminus \mathbb{Q}$, then the vertical line $x = x\_1$ lies entirely in $A^c$.

A polygonal path connecting $P\_1$ to $P\_2$ can be found entirely in $A^c$ by zig-zagging horizontally and vertically, all the while maintaining at least one irrational coordinate.