How does one combine proportionality? this is something that often comes up in both Physics and Mathematics, in my A Levels. Here is the crux of the problem.
So, you have something like this :
$A \propto B$ which means that $A = kB \tag{1}$
Fine, then you get something like :
$A \propto L^2$ which means that $A = k'L^2 \tag{2}$
Okay, so from $(1)$ and $(2)$ that they derive :
$$A \propto BL^2$$
Now how does that work? How do we derive from the properties in $(1)$ and $(2)$, that $A \propto BL^2$.
Thanks in advance.
 A: Actually, you are right. We can't prove it.
The following proposition is incorrect:
$(A∝B$ and $A∝C) ⟹ A∝BC$
Let's assume that the above proposition is true. If we can prove that this proposition leads to a contradiction, for any example, then the proposition is incorrect.
Here is a counter-example:
Let's say, $A = 4B$ and $B = 3C$
Therefore, $A = 4(3c) = 12C$
Therefore, $A∝B$ and $A∝C$
Given, the proposition is true, the above statement implies A∝BC
Therefore, $A = kBC = k(3C)C = 3kC^2$
Therefore, $A∝C^2$
Which is a contradiction. Both $A∝C$ and $A∝C^2$ can't be true.
But here is what your Physics and Mathematics books tell you.
$(A=k'B$ and $k'=k''C) ⟹ A∝BC$
And we can prove it very easily
$A=k'B=(k''C)B=k''BC⟹ A∝BC$
A: Suppose a variable $A$ depends on two independent factors $B,C$, then
$A\propto B\implies A=kB$, but here $k$ is a constant w.r.t. $B$ not $C$, in fact, $k=f(C)\tag{1}$
Similarly, $A\propto C\implies A=k'C$ but here $k'$ is a constant w.r.t. C not $B$, in fact, $k'=g(B)\tag{2}$
From $(1)$ and $(2)$,
$f(C)B=g(B)C\implies f(C)\propto C\implies f(C)=k''C$
Putting it in $(1)$ gives,
$A=k''CB\implies A\propto BC\tag{Q.E.D.}$
