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Proportional to 2 Separate Variables vs. Proportional to Product of 2 Variables

I guess some people may find this obvious, but I really don't. My question is:

If $A\propto B$ while $C$ is constant and $A\propto C$ while B remains constant, then why is $A\propto BC$ ?

Why $A$ has to be proportional to the product of $B$ and $C$ ?

Earlier I never really questioned this when it popped up in textbooks because for some reason I thought it was self-evident, but now I come to think of it, I really don't understand why this is true.

I guess people here could rigorously prove this which I would like to see but an intuitive explanation as to why this is true would also be really appreciated.

Thanks.

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marked as duplicate by Rahul, Asaf Karagila, Cameron Buie, Chris Eagle, Noah Snyder Dec 12 '12 at 12:23

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Can you just replace those with $A = kB$? If so, that doesn't seem right then. $ A = k_1 B$ and $A = k_2 C$. $A^2 = (k_1 k_2)BC$ so $A^2 = k_3 BC$ or $A^2 \propto BC$. Am I wrong? –  fhyve Dec 12 '12 at 6:59
    
@fhyve So what you just proved is $A \propto \sqrt{BC}$ ? –  Alraxite Dec 12 '12 at 7:06
    
There an explanation at the wiki page for the combined gas law. You might want to take a look. Also, I'm not sure why this question was down-voted. It seems perfectly fine. –  EuYu Dec 12 '12 at 7:12
    
@ Alraxite Oh, I missed the fact that one is proportional while the other is constant. –  fhyve Dec 12 '12 at 7:13

1 Answer 1

up vote 2 down vote accepted

Look at an example or two. Suppose, for instance, that you double $B$ and triple $C$, in that order, starting with a value $a$ for $A$. Doubling $B$ while holding $C$ fixed causes $A$ to double to $2a$, and then tripling $C$ while holding $B$ fixed causes $A$ to triple to $3\cdot2a=6a$; the net effect is to multiply $A$ by $2\cdot 3=6$. Making the changes in the opposite order first triples $A$, to $3a$, and then doubles the result, so that again you end up with $A=6a$. If $b$ and $c$ are the original values of $B$ and $C$, respectively, the new values are $2b$ and $3c$, so $BC$ has also increased by a factor of $2\cdot3=6$.

Clearly you can substitute any multipliers for $2$ and $3$ and see the same behavior.

The only point that may still not be intuitively obvious is that the outcome remains the same even when $B$ and $C$ are changed simultaneously. This is actually a consequence of the (possibly unstated) assumption that changes in $B$ and $C$ act independently on $A$.

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Thank you, that makes sense. Just one other thing: why on multiplying the two equations ($A = k_1B$, $A = k_2C$), we get a wrong result? This was done in the comments and also on the wikipedia page of combined gas law (also mentioned in the comments). Wikipedia didn't really explain why doing this is incorrect and I don't want to make another question, so I figured I'd just ask you. One could say that in both the equations we consider the other quantity to be constant and that's why we can't multiply them, but I kind of don't find it satisfactory. –  Alraxite Dec 12 '12 at 8:19
    
@Alraxite: Basically that is the reason: you get the wrong result because the two two-variable equations hold only when the third variable is held constant. Multiplying them is in a sense trying to combine two incompatible settings to infer something about a setting more general than either. Knowing that $A=kB$ when $C$ is fixed just tells you that the ratio $A/B$ is a function of $C$, say $f(C)$; it doesn’t tell you how it varies with $C$, i.e., what $f$ is. Now imagine holding $B$ fixed and varying $C$: multiplying $A$ by $r$ multiplies $A/B$ by $r$, so it multiplies $f(C)$ by $r$. ... –  Brian M. Scott Dec 12 '12 at 8:36
    
... This is possible only if $f(C)=kC$ for some constant $k$, so we now have $\frac{A}B=kC$, or $A=kBC$, as desired. (I didn’t think of this approach earlier.) –  Brian M. Scott Dec 12 '12 at 8:37

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