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

I've always seen the following in physics and math textbooks but never understood the process by which it was mathematically deducted:

$A \propto B$ $\space$ and $\space$ $A \propto C \space\space\space \rightarrow \space\space\space A \propto BC$

Could someone walk me through how this is done? This has been bothering me for a while now :P

Thanks

Update: Here's something I found that explains how this works. (Page 387; "Proof" section). Still, this proof takes the two statements one after the other. The author uses $x \propto y$ when $z$ is constant, and then takes care of $x \propto z$ when $y$ is constant, where it left off from the first (going from $x$ to $x'$ and then $x_1$). Is this the only way it can be done?

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Re: "Is this the only way it can be done?" Given that the premise $A\propto B$ tells you how $A$ changes with $B$ while holding everything else constant, and the same goes for $A\propto C$, I don't think you can get anywhere without only changing one of $B$ and $C$ at a time. If you like, you can prove that it doesn't matter whether you do $B$ and then $C$, or $C$ and then $B$, or $B$ halfway then $C$ then $B$ the rest of the way, or any other path, you will always get the same answer. – Rahul Sep 2 '12 at 21:20
@Rahul Narain - So in other words the constants of proportionality for each statement hold the "opposite" variable as a factor, correct? (ie. $A = m(C)B$ and $A = n(B)C$) – ProSteve037 Sep 2 '12 at 21:30
Pretty much, yes. The constant of proportionality with respect to $B$ is a function of $C$, and vice versa, as you wrote. – Rahul Sep 2 '12 at 21:32
Awesome. Thanks so much for clearing this up for me! :-) You should answer my question so I can +1 you and mark it as the answer haha – ProSteve037 Sep 2 '12 at 22:51

Putting my comments into an answer: If we say $A \propto B$ when $A$ also depends on other things, what we mean is that holding everything else fixed, $A$ increases linearly with $B$. So $A = mB$ for some $m$, and $m$ is constant relative to $B$, but may vary depending on the other things.

This is getting a bit wooly, so let's be more explicit. Let's say $A$ is a function of $B$, $C$, and $D$. Then $A \propto B$ means $A(B,C,D) = f(C,D)\cdot B$ for some function $f$. On the other hand, if $A \propto C$, then $A(B,C,D) = g(B,D)\cdot C$ for some other $g$. When you can put those together and go through some algebra, you'll find that $A(B,C,D) = h(D)\cdot BC$, that is, $A \propto BC$.

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This is based off of wikipedia's article on Combined Gas Law and some of my own inputs.

We have A $\propto B$ and $A \propto C$.

Thus we get (noting that proportionality is symmetric), $$B = k_C(C)A$$ $$C = k_B(B)A$$

where $k_C$ is a function of only C and $k_B$ that of B. The proportionality constants could depend on various other things, but for now, we focus on A, B, C only as all other parameters are considered fixed. Then we get $$A = \frac{B}{k_C(C)} = \frac{C}{k_B(B)}$$ $$\Rightarrow Bk_B(B) = Ck_C(C)$$

Now in the previous line, LHS is a function of B only, RHS is a function of C only!!!! Thus both are independent of B and C and therefore must be a constant. Call this constant K.

Then $k_B(B) = \frac{K}{B}$. Substitute back to get $$BC = KA$$ or $$BC \propto A$$ $\blacksquare$

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Temperature $T$ is proportional to pressure $P$ ( Charles Law $1$);

Temperature $T$ is proportional to Volume $V$ ( Charles Law $2$);

So, Temperature $T$ is prortional to product $P V$ ( Gas Law, $P V /T = \mathrm{const}$)

Narasimham

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Thanks for the response! I've been reading up on the Combined Gas Law after seeing this post and have come across multiple times that the constants in each of the laws are actually functions dependent on the other variable. In other words, $T \propto P \space\rightarrow\space T = c_1(V)P$ and $T \propto V \space\rightarrow\space T = c_2(P)V$ Will this always be the case though? Will the constants always depend on each other like they do here? – ProSteve037 Jul 31 '12 at 3:18

Assuming $A\propto B$ means $A=kB$ for some constant $k$, then we can also say that $A=mC$ for some constant $m \Rightarrow A=kmBC \Rightarrow A\propto BC$.

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Why do you multiply the two terms for $A$ together? And wouldn't $kmBC = A^2$? – ProSteve037 Jul 25 '12 at 9:25
I'm multiplying the $kB$ and $mC$ together. I'm sorry. I meant to write that $A=\ell BC$ for some $\ell=ckm$ some constant $c$. Sorry that this is confusing. Essentially, if you can relate one thing ($A$) to each of two other things ($B$ and $C$), by taking the least common denominator you should be able to relate $A$ to the product $BC$. Try using specific values to see what I mean. – chris Jul 26 '12 at 4:20
No worries! Thanks for the quick response. Here are the numbers I used $A = 12$, $B = 3$, and $C = 6$. (Thus $k = 4$ and $m = 2$). How would you determine that $\ell = c(4)(2)$? Will $\ell$ always be proportional to $k$ and $m$? – ProSteve037 Jul 27 '12 at 3:16
I'm also wondering if there is a geometric intuition behind this, can it be proven using lines and areas? – ProSteve037 Jul 27 '12 at 3:21