newtonian mechanics - Defining gravitational potential

I recently came across the definition of gravitational potential where .....

Suppose a particle of mass $m$ is taken from a point $A$ to $B$. Let $U(A)$ and $U(B)$ denote the gravitational potential energy when the mass $m$ is at point $A$ and point $B$ respectively ..Also $V(B) - V(A)$ is the change in potential, then $$V(B) - V(A) =\frac{U(B) - U(A)}{m}$$

My teacher then said that if we take $A$ to be the reference point then $V(A)$ becomes equal to $0$ while I have no problem with this but why can't we also say that $U(A)$ is also equal to zero? After all we are taking $A$ as the origin? But my teacher says that we cant take $U(A)$ as equal to zero. Can anyone please explain why?

My book(Concepts of Physics by HC verma) also says the same thing.

Answer

Your teacher is wrong. The gravitational potential $V(x)$ is generally defined as potential energy per unit mass i.e. $V(x) \equiv \dfrac{U(x)}{m}$. So for the points where $U(x)$ is zero, $V(x)$ is zero and vice-versa by definition.

EDIT: After you added the comment and a snapshot of the book, I realized your book has defined Gravitational Potential in a different manner as $V(B)-V(A)=\dfrac{U(B)-U(A)}{m}$ than the standard $V(r)=U(r)/m$. This allows the room to choose different references for both $V(r)$ and $U(r)$. So to solve your confusion, the reference of $U(r)$ might be different than that of $V(r)$ so $U(A)$ need not be equal to $mV(A)$.