I remember of learning the rule of three in elementary school. These days I had a random thought and formulated it this way:
We have the values $x,f(x),x+h$ and we want to know the value of $f(x+h)$ in terms of the previous values. We write:
$$\cfrac{x}{x+h}=\cfrac{f(x)}{f(x+h)}$$
This yields:
$$f(x+h)=\cfrac{f(x)(x+h)}{x}\tag{1}$$
I used the symbol of function on purpose: When using a rule of three, we are assuming that there is a function that behaves in a certain way. The problem is that for certain functions, it would miss tremendously, for example: Suppose the unknown function is $f(x)=x^2$, a numerical example would be:
$$\cfrac{2}{4}=\cfrac{4}{x} \quad \quad\quad \quad\quad \quad 2x=16\quad \quad\quad \quad\quad \quad x=8$$
But $4^2=16$. My guess is that $f$ must be linear or affine but I'm not sure on how to see it. I did the following: Rewriting $(1)$:
$$f(x+h)x=f(x)(x+h)$$
As multiplication is well-defined and none of these quantities is $0$, then:
$$f(x+h)= f(x) \quad \quad \quad \quad x=x+h$$
Which implies that $h=0$, or: $$f(x+h)= x+h \quad \quad \quad \quad x=f(x)$$
But this result left me confused. From gut feeling, this doesn't seems right. What have I missed?
You are wrong when you say :
Indeed, let us consider the equality $ 3 \times 4 = 6 \times 2$. From what you wrote one could conclude that $3 = 4$ or $6 = 2$. I guess you where thinking about the following (true) statement:
However, it doesn't work if the right-hand term is not $0$.
Now, for your problem. The rule of three is related to the idea of proportion, and as such to linearity, and hence is only true for linear functions.
Let us assume $f$ is a function satisfying the "rule of three". We have $\frac{x}{x+h} = \frac{f(x)}{f(x+h)}$, which gives $\frac{f(x)}{x} = \frac{f(x+h)}{x+h}$, for any $x$ and $h$. Which means that $\frac{f(x)}{x}$ is constant, ie $\frac{f(x)}{x} = a$ for some a.
All in all, we get that if $f$ follows the rule of three, then $f(x) = ax$.