You know that for a potential function (conservative force/fields) that
$\nabla U = \pm \vec{F}$
In math, we don't have that minus sign, we have only the plus one.
What does it mean if you get rid of the plus sign? I remember the minus signs tells us that the force is always in the opposite direction of the increasing potential function. Does math tell me there exists a potential function such that it's force also increases as potential goes up?
Mathematically, it just gives a duality with vector fields and scalar fields in multivariable calculus, associated with conservative vector fields and line integrals. As such, the $\pm$ is irrelevant, because it can be absorbed into the force vector. For physics, we take the sign convention to be negative, so that it agrees with the fact that the force is restoring the object it acts on to a lower energy configuration. Note that we could alternatively absorb the negative sign into the potential! It is all a matter of sign convention, and when you define potential and force in physics (as stated above), the negative sign appears in your equation.
Mathematically, we can write $$\mathbf{F}\cdot\mathrm{d}\mathbf{l}=\pm\mathrm{d}E,\qquad\mathrm{if}\ \mathbf{F}\equiv\pm\mathbf{\nabla}E.$$ Here $\mathbf{F}$ is a vector field, $\mathbf{l}$ is a position vector, and $E$ is a scalar field.
In physics, the work done by a force $\mathbf{F}$ over the infinitesimal displacement $\mathrm{d}\mathbf{l}$ equals the left-hand side in the above relation. Thus, under the condition as given above, the right-hand side is also true in physics. The scalar field $E$ is called the energy corresponding to the given (conservative) force. When evaluating the above relationship and a minus sign appears, the corresponding energy is called potential energy (as for weight and the restoring force on a spring, for example).
For Newton's second law, $\mathbf{F}\cdot\mathrm{d}\mathbf{l}=+\mathrm{d}\left(\frac{1}{2}mv^2\right)$. This is an example of $\mathbf{F}=+\mathbf{\nabla}E$.
