Basic question about torques In the first pages of the book Introductory classical mechanics by David Morin, there is a problem:

Now, let me list my questions:


*

*why is a reasonable assumption to suppose that there is a function $f$ such that $(1.8)$ and $(1.9)$ hold? and also, in the statement of the problem, it is written that the system is motionless, so, doesn't that already says that the net torque about any axis is $0$ (which would trivially prove Claim 1.1)? 

*why if $f(a)+f(b)=f(a+b)$ then $f(x)=Ax$ for some constant $A$?, I mean, $a$ and $b$ are some particular values and also we don't even know what the domain of $f$ is. Of course one can say "well, the condition $f(a)+f(b)=f(a+b)$ will hold regardless of the values of $a$ and $b$ because the statement of the problem is not assuming anything about them, just maybe that they are non-negative (because they represent distances)", but still in that case, one cannot necessarily conclude that $f(x)=Ax$ for some constant $A$
(moreover, they also assume that $A\neq0$).


I really appreciate your help and advices!
 A: The very basic foundation of any physics proof is the condition. First you define a condition and then you prove your claim using physics and mathematical laws to say that it is correct for that condition. It simply doesn't work the other way. Your claim cannot be correct by the condition you yourself applied.
Here your condition is that the body is motionless and your claim is that it's net torque is  $0$ for your condition. 
Let's start from the very basics used to solve this claim.
Firstly for a system to be in complete equilibrium it should be in dynamic as well as static equilibrium, which means that both force balance and torque balance should take place.
Hence from force balance we get:
$$F_1 + F_2 = F_3$$
Now let's focus on dynamic equilibrium. Let's balance the torque from a reference point which is considered in your book. Lets's consider our reference point where $F_1$ is acting.
The torque at the reference point due to $F_1$ is $0$ because the distance from the reference point to $F_1$ is $0$. The torque due to $F_3$ is $F_3\times a$ in the clockwise sense. The torque due to $F_2$ is $F_2\times (a+b)$ in counter clockwise sense. (Remember we chose the reference point where $F_1$ is acting.)
So to keep the stick in dynamic equilibrium torque due to $F_3$ should be equal to torque due to $F_2$. Hence
$$F_3\times a = F_2\times(a+b)$$
........Edit.......
The term $f(x)$ included in the torque equation in place of $a$ and $a+b$ is just a  way to tell that we still don't know the correct relationship between force and distance from the reference point. Later they prove the function to be linear using $$F_1 + F_2 = F_3$$ and then cancel out the common constant coming on both sides of eqn that we just proved.
Proving $f(x)$ to be linear
Since we can actually balance torque from any point on the stick, we would consider 2 reference points from where we would balance the torque on the stick.
Since at this point we don't know the relationship between force and distance we would consider function $f(a)$ as function of distance $a$ and $f(b)$ as function of distance $b$ and $f(a+b)$ as function of total length of rod $(a+b)$.
Now from one reference point we get $$F_1\times f(a+b) = F_3\times f(b)$$
From second reference point we get $$F_2\times f(a+b) = F_3\times f(a)$$
Adding these two eqns $$(F_1+F_2)\times f(a+b) = F_3\times (f(a)+f(b))...(1)$$
We know that $$F_1+F_2 = F_3...(2)$$
Dividing eqn (1) by (2) we get $$f(a+b) = f(a)+f(b)...(3)$$
Let $$f(x) = C\times x$$ where C is constant
Then, $$f(a) = C\times a$$
$$f(b) = C\times b$$
$$f(a+b) = C\times (a+b)$$

These three relations satisfy (3)
Therefore we now know that $f(x)$ is a linear function.
