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Possible Duplicate:
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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    @ Alraxite Oh, I missed the fact that one is proportional while the other is constant.2012-12-12

1 Answers 1

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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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    ... 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.)2012-12-12