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I have a slight confusion regarding combining proportionalities. There is a question related to this topic, the link of the question concerned is as follows. I was led to this question for I was pondering the same thing as the OP in that question.

How does one combine proportionality?

Aang has posted a nice answer. However, I have a query.

When you take A=kB and k=f(C),aren't you already assuming that A is proportional to CB? When we are regarding A's change with B, we are neglecting the effects of C, which is to say we are assuming C to be constant at some value, which C can take. A is proportional to B, hence we take the proportionality constant, k. As, A depends on both B and C, it makes sense to say that A is proportional to B and the proportionality constant involved is equal to the function f(C) which represents all values C can take. But, in saying so, aren't we assuming that A is proportional to CB, as the proportionality constant is equal to C at some value?

Have I missed something? Thanks for any help.

Community
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SNB
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  • I know this question is better suited as a comment in the question linked, itself. But, I don't have the rep necessary to post a comment. – SNB Dec 23 '16 at 17:32
  • 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 12:57

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Not sure if that answers your question but if $A$ is proportional to $B$ and also proportional to $C$ then mathematical $A$ is bilinear in $B$ and $C$. This means it is linear in $B$ for a fixed $C$ and linear in $C$ for a fixed $B$. So in that sense it is quite natrual to do what you say, namely, consider one of them constant when looking at the other.

edit (added for clarification): When you study the proportionality of $A$ and $B$, you treat $C$ as a constant. This means $A=kB$ where $k(C)$ could be any function. Thus you're not assuming $A$ is proportional to $CB$. Rather, you say that for every $C$ there is a constant $k$ such that $A=kB$, Note that $A$ and $B$ vary, $k$ does not.

Florian
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  • Yes, it makes sense. But, when we represent the constant involved as a function of C, aren't we already assuming that A is proportional to the multiple CB, (not C and B individually, but C AND B)which we are trying to prove? – SNB Dec 24 '16 at 18:07
  • No. When you study the proportionality of A and B, you treat C as a constant. This means $A=kB$ where $k(C)$ could be any function. Thus you're not assuming A is proportional to $CB$. Rather you say that for every $C$ there is a constant $k$ such that $A=kB$, Note that A and B vary, k does not. – Florian Dec 25 '16 at 08:24
  • I understand it now. Much obliged :) Pls add the crux of your comment to the answer and I'll choose it as the accepted answer – SNB Dec 25 '16 at 17:16
  • I did. You're welcome. – Florian Dec 25 '16 at 22:29