5

I've commonly seen the following in physics and math textbooks, but never understood how it is mathematically deduced:

$A \propto B$ $\space$ and $\space$ $A \propto C \space\space\space \implies \space\space\space A \propto BC.$

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

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?

ryang
  • 38,879
  • 14
  • 81
  • 179
chevestong
  • 325
  • 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. –  Sep 02 '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 $A = n(B)C$) – chevestong Sep 02 '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. –  Sep 02 '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 – chevestong Sep 02 '12 at 22:51
  • 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 23 '23 at 19:23

3 Answers3

2

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, $A \propto BC$.

1

Here’s a simpler explicit proof.

Let A be k when B and C are 1: $(k,1,1)$

Then, if we keep C constant and multiply B by $b$, A goes up by b: $(kb,b,1)$.

If we keep B constant now and scale C by $c$, A goes up by $c$: $(kbc,b,c)$

Thus, $a=kbc$ in general, so A is proportional to BC.

Eric
  • 6,348
0

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 $$BC \propto A$$ $\blacksquare$

Gautam Shenoy
  • 10,318