Theorem: Cartesian Definition of Gradient defines Gradient Operator

Let $R$ be a region of Cartesian $3$-space $\R^3$.
Let $\map F {x, y, z}$ be a scalar field acting over $R$.
Let $\tuple {i, j, k}$ be the standard ordered basis on $\R^3$.
Let $\grad F$ be defined according to the Cartesian $3$-space definition of the gradient of $F$:
{{begin-eqn}}
{{eqn | l = \grad F
      | o = :=
      | r = \nabla F
}}
{{eqn | r = \paren {\mathbf i \dfrac \partial {\partial x} + \mathbf j \dfrac \partial {\partial y} + \mathbf k \dfrac \partial {\partial z} } F
      | c = {{Defof|Del Operator}}
}}
{{eqn | r = \dfrac {\partial F} {\partial x} \mathbf i + \dfrac {\partial F} {\partial y} \mathbf j + \dfrac {\partial F} {\partial z} \mathbf k
      | c = 
}}
{{end-eqn}}
Then $\grad F$ is a gradient operator as defined by the geometrical representation:
:$\grad F = \dfrac {\partial F} {\partial n} \mathbf {\hat n}$
where:
:$\mathbf {\hat n}$ denotes the unit normal to the equal surface $S$ of $F$ at $A$
:$n$ is the magnitude of the normal vector to $S$ at $A$.

Proof:
The vector rates of increase of $F$ in the directions of the $3$ axes are:
:$\dfrac {\partial F} {\partial x} \mathbf i$, $\dfrac {\partial F} {\partial y} \mathbf j$, $\dfrac {\partial F} {\partial z} \mathbf k$
Their sum will be a vector with the magnitude and direction of the most rapid rate of increase of $F$.
It remains to show that this expression:
:$\dfrac {\partial F} {\partial x} \mathbf i + \dfrac {\partial F} {\partial y} \mathbf j + \dfrac {\partial F} {\partial z} \mathbf k$
is equivalent to:
:$\dfrac {\partial F} {\partial n} \mathbf {\hat n}$
Let us take the dot product of both sides of the gradient equation with the position vector $\d \mathbf r$ of an arbitrary point $A$ on an equal surface $S$ of $F$.
Thus:
{{begin-eqn}}
{{eqn | l = \paren {\grad F} \cdot \d \mathbf r
      | r = \dfrac {\partial F} {\partial n} \mathbf {\hat n} \cdot \d \mathbf r
      | c = 
}}
{{eqn | r = \dfrac {\partial F} {\partial n} \d r \cos \theta
      | c = where $\d r = \norm {\d \mathbf r}$: {{Defof|Dot Product}}
}}
{{eqn | r = \dfrac {\partial F} {\partial n} \d n
      | c = 
}}
{{eqn | r = \d F
      | c = as $\dfrac {\partial F} {\partial n}$ is the total rate of change of $F$ with respect to $n$
}}
{{end-eqn}}
In Cartesian coordinates:
{{begin-eqn}}
{{eqn | l = \d F
      | r = \dfrac {\partial F} {\partial x} \d x + \dfrac {\partial F} {\partial y} \d y + \dfrac {\partial F} {\partial z} \d z
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {\grad F} \cdot \d \mathbf r
      | r = \dfrac {\partial F} {\partial x} \d x + \dfrac {\partial F} {\partial y} \d y + \dfrac {\partial F} {\partial z} \d z
      | c = 
}}
{{eqn | r = \paren {\dfrac {\partial F} {\partial x} \mathbf i + \dfrac {\partial F} {\partial y} \mathbf j + \dfrac {\partial F} {\partial z} \mathbf k} \cdot \paren {\d x \mathbf i + \d y \mathbf j + \d z \mathbf k} 
      | c = 
}}
{{eqn | r = \paren {\nabla F} \cdot \d \mathbf r
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \grad F = \nabla F
      | r = \dfrac {\partial F} {\partial x} \mathbf i + \dfrac {\partial F} {\partial y} \mathbf j + \dfrac {\partial F} {\partial z} \mathbf k
      | c = 
}}
{{end-eqn}}
Thus the operations $\grad$ and $\nabla$ applied to a point in a scalar field are identical.
{{qed}}