Gradient vector $\nabla F = (z_x, z_y, -1)$ is normal to the integral surface? I have an integral surface $z = z(x, y)$.
Writing this integral surface in implicit form, we get
$$F(x, y, z) = z(x, y) - z = 0$$
I am then told that the gradient vector $\nabla F = (z_x, z_y, -1)$ is normal to the integral surface $F(x, y, z) = 0$.
First of all, how was this calculated? I understand how the gradient is calculated, but I don't understand how it was calculated in this case?
And lastly, where did the $-1$ come from and why? Couldn't they also have had $\nabla F = (z_x, z_y, 1)$, where this would just be the normal vector in the other direction? Why and how did they pick the $-1$ direction instead?
I apologise. My vector calculus understanding is not particularly strong, and I strive to improve it.
Thank you for any help.

Let us start with an example.
$$ z=x^2+y^2$$
$$ F(x,y,z)=x^2+y^2-z$$
$$\nabla F = (z_x, z_y, -1)=< 2x,2y,-1>$$
If a point is given, for example $P(1,2,5)$ Then at that point you have two normal vector to the surface.
Upward normal $$< -2x,-2y,1> = <-2,-4,1>$$
Downward normal $$< 2x,2y,-1> = <2,4,-1>$$
