Which mathematical property allows us to combine proportional relationships? Coulumb's law states that $$F \propto q_1 \cdot q_2 \tag{1} $$ and $$ F \propto \frac{1}{r^2} \tag{2} $$
Why can we combine these two proportions into $$ F \propto \frac{q_1.q_2}{r^2}?$$
What property allows that? If we combine them by the transitive property, shouldn't the final relation be $ q_1 \cdot q_2 \propto \frac{1}{r^2}$? 
 A: your description is imprecise
∝1.2 for fixed r
 and 
∝1/^2 for fixed q1,q2
and r independent of q1 and q2
only then you can deduct Coulombs law
A: Let us say we want to define electrostatic force between two charges. What would it be a function of? We make the simplest guess that it must be related to the charges $q_1,q_2$ and the distance $r$ between the two, $F\left(q_1,q_2,r\right)$. Now to get any functional form we need a controlled experiment where we measure the force by varying any one of the three parameters at a time while keeping the others fixed. In doing these experiments we find the following,


*

*$F\left(q_1,q_2,r\right)\propto q_1q_2\big|_{\text{fixed r}}$

*$F\left(q_1,q_2,r\right)\propto \frac{1}{r^2} \big|_{\text{fixed }q_1q_2} $
Due to the fact that these are true only when the other parameters are fixed, we can not directly “combine” the two proportionalities in any meaningful way. To do so, we need to do the following. 
Notice that the first proportionality is true for a fixed $r$. Thus in general the constant of proportionality must be some number $\alpha_1$ multiplied by some function of $r$, say $f(r)$. Similarly the second constant of proportionality must be $\alpha_2~g(q_1q_2)$. Now that we have converted our proportionalities into equalities, we can use transitivity to obtain
$$\alpha_1f(r)q_1q_2=F=\frac{\alpha_2g(q_1q_2)}{r^2}\\
\alpha_1f(r)r^2= \frac{\alpha_2g(q_1q_2)}{q_1q_2}
$$
Since $\alpha$’s are numbers, the LHS is purely a function of $r$ and the RHS purely a function of $q_1q_2$, the only way the equation will hold true is if both sides equal a constant (can be set to 1 without loss of generality). This can only happen if $f(r)=1/r^2$ and $g(q_1q_2)=q_1q_2$ and $\alpha_1=\alpha_2$. This finally gives us (using either of the proportionalities)
$$F\propto \frac{q_1q_2}{r^2}$$
A: The first two formulas are useful because they correspond directly to basic, practically easy, experimental procedures. That makes them easily justified by experiments, and that makes them useful as "axioms" to derive others, "theorems".
The first formula F∝q1.q2 corresponds to an experiment where we keep r constant, vary q and measure F. It is easy to keep r constant in an experiment, because there are rigid bodies (at least approximately) and because r is easy to measure.
The second formula F ∝ 1/r2 corresponds to an experiment where we keep the q's constant, vary r and measure F. It is easy to keep the q's constant in an experiment, because there is a law of charge conservation, and because many materials are electric insulators.
The fourth formula q1.q2 ∝ 1/r2 corresponds to an experiment where we keep F constant while varying r. But that is a less easy experimental procedure. There is no law of force conservation, there are no naturally occurring constant forces to help us.
The fourth formula q1.q2 ∝ 1/r2 is in a sense just as true as the first two formulas. But it is not very useful, and that is so because it does not correspond to a basic, practically easy, experimental procedure.
Finally, the third formula, Coulomb's law, is very different from the others: it combines all quantities F, r, q1 and q2. So deriving Coulomb's law from the first two is much more valuable than deriving the fourth formula q1.q2 ∝ 1/r2, just because that formula does not contain F. (And F is a physical quantity that really matters. One may prefer another quantity, such as field strength or potential energy, over force. But one can not make it go away.)
