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I have followed one question in this forum about the same topic. That question was too old so I thought I won't get any response even if I comment on that. That is why I am asking sort of a similar question but pointing out the part where I am facing difficulty.

Let $A \propto B$ when $C$ is constant. And $A \propto C$ when $B$ is constant. Again when $A$ is constant, $B \propto C$, but we are ignoring this for a while. I understood that for the first case $A=KB$ where $K=f(C)=mC$ since $A \propto C$ too. So the for the second case $A=LC$ where $L=g(B)=nB$. Now following the first answer to this question:

How does one combine proportionality?

How can we simply assure that at any moment for a certain value of $A$, $f(C)B=g(B)C$ instead of $f(C')B=g(B')C$?

MSKB
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  • In the general case, the implication $(A \propto B$ and $A \propto C) \Rightarrow A \propto BC$ is false as clearly pointed out in Hagen Von Eitzen's comment and @ZarifMuhtasim answer to the question

    https://math.stackexchange.com/questions/433754/how-does-one-combine-proportionality

    . It is necessary to add the assumption that $A$ and $B$ are independent to make the implication $(A \propto B$ and $A \propto C) \Rightarrow A \propto BC$ true.

     – Ramiro Dec 05 '21 at 15:54
  • When you formulate your question as : "If $A$ is proportional to $B$ when $C$ is constant and $A$ is proportional to $C$ when $B$ is constant then how is $A$ is proportional to $BC$?" , then you are assuming $B$ and $C$ to be independent, and so the implication $(A \propto B$ and $A \propto C) \Rightarrow A \propto BC$ is true. – Ramiro Dec 05 '21 at 15:58
  • Does this answer your question? Does $f(a,b)$ being directly proportional to $a$ and $b$ separately imply that $f(a,b)$ is directly proportional to $ab?$ – ryang Jun 25 '23 at 15:00

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Let's hold everything other than $B, C$ constant and treat $A$ as a function only of $B$ and $C$.

Let's compute $A(B, C) / A(B', C')$.

We have $\frac{A(B, C)}{A(B', C')} = \frac{A(B, C)}{A(B', C)} \frac{A(B', C)}{A(B', C')} = \frac{B}{B'} \frac{C}{C'} = \frac{BC}{B'C'}$.

Therefore, $A(B, C)$ is proportional to $BC$. In particular, we have $\frac{A(B, C)}{A(1, 1)} = \frac{BC}{1 \cdot 1}$, hence $A(B, C) = A(1, 1) BC$.

Mark Saving
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  • Maybe I could have proved this thing quite easily since I have already mentioned A=aCB, so from here on I could have said that A is proportional to BC. Isn't it? – MSKB Aug 19 '21 at 16:05
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    @MSKB As soon as you show $A = aCB$, that's enough to know that $A$ is proportional to $CB = BC$. – Mark Saving Aug 19 '21 at 16:31
  • A little bit clarification is required still that is the answer to the question linked above correct? – MSKB Aug 19 '21 at 17:07
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    @MSKB I believe it is, but it's phrased confusingly. – Mark Saving Aug 19 '21 at 17:23