6
$\begingroup$

this is something that often comes up in both Physics and Mathematics, in my A Levels. Here is the crux of the problem.

So, you have something like this :

$A \propto B$ which means that $A = kB \tag{1}$

Fine, then you get something like :

$A \propto L^2$ which means that $A = k'L^2 \tag{2}$

Okay, so from $(1)$ and $(2)$ that they derive :

$$A \propto BL^2$$

Now how does that work? How do we derive from the properties in $(1)$ and $(2)$, that $A \propto BL^2$.

Thanks in advance.

$\endgroup$
6
  • 2
    $\begingroup$ This is not true in general. For exampe, if $r,A,V,m$ are radius, surface area, volume, and mass of a ball, then $m\propto V$ (this ways introducing the density as factor) and also $m\propto A^{3/2}$ and $m\propto r^3$. But that doesn't imply that $m=kVA^{3/2}r^3$. $\endgroup$ Jul 1, 2013 at 16:27
  • 1
    $\begingroup$ @HagenvonEitzen: It works when all factors are independent of each other. $\endgroup$
    – Aang
    Jul 1, 2013 at 17:29
  • 1
    $\begingroup$ @Avatar Yes, but that must be formulated as premise about the "state space" or whatever $\endgroup$ Jul 1, 2013 at 18:57
  • $\begingroup$ @HagenvonEitzen: I agree. $\endgroup$
    – Aang
    Jul 2, 2013 at 3:10
  • $\begingroup$ In the general case, the implication $(A \propto B$ and $A \propto C) ⟹ A \propto BC$ is false as clearly pointed out in Hagen Von Eitzen's comment above and in @ZarifMuhtasim answer below. It is necessary to add the assumption that $A$ and $B$ are independent to make the implication $(A \propto B$ and $A \propto C) ⟹ A \propto BC$ true. $\endgroup$
    – Ramiro
    Dec 5, 2021 at 12:34

2 Answers 2

7
$\begingroup$

Actually, you are right. We can't prove it.

The following proposition is incorrect: $(A∝B$ and $A∝C) ⟹ A∝BC$

Let's assume that the above proposition is true. If we can prove that this proposition leads to a contradiction, for any example, then the proposition is incorrect. Here is a counter-example:
Let's say, $A = 4B$ and $B = 3C$
Therefore, $A = 4(3c) = 12C$
Therefore, $A∝B$ and $A∝C$

Given, the proposition is true, the above statement implies A∝BC
Therefore, $A = kBC = k(3C)C = 3kC^2$
Therefore, $A∝C^2$
Which is a contradiction. Both $A∝C$ and $A∝C^2$ can't be true.

But here is what your Physics and Mathematics books tell you.
$(A=k'B$ and $k'=k''C) ⟹ A∝BC$
And we can prove it very easily
$A=k'B=(k''C)B=k''BC⟹ A∝BC$

$\endgroup$
5
  • $\begingroup$ k’=k’’C is not given as an assumption in the question. So have you assumed this yourself because it is necessary to solve this question $\endgroup$
    – aaksaksyk
    Dec 3, 2021 at 14:36
  • $\begingroup$ @aaksaksyk yes. $\endgroup$ Dec 4, 2021 at 15:07
  • $\begingroup$ @aaksaksyk Without any additional assumption, the implication $(A∝B$ and $A∝C) ⟹ A∝BC$ is false, as Zarif has pointed out. $\endgroup$
    – Ramiro
    Dec 4, 2021 at 21:14
  • $\begingroup$ Are you saying that $A \propto C$ and $A \propto C^2$ can both never be true? If you let $A = k_1 C$ and $A = k_2 C^2$, this implies that either $C = 0$ or $C = \frac{k_1}{k_2}$. So while it is not always true, it looks like it can be true under very specific conditions. $\endgroup$
    – Mailbox
    Feb 11, 2023 at 18:24
  • 1
    $\begingroup$ @Mailbox $(A∝B$ and $A∝C) ⟹ A∝BC$ given B is dependent on C will not hold for an infinite number of cases except the two you mentioned. If it doesn't hold for even one case, the statement would be considered invalid. So yes, it is true under very specific conditions, but the fact that it is not always true makes the statement false. $\endgroup$
    – Tca
    Sep 12, 2023 at 14:49
3
$\begingroup$

Suppose a variable $A$ depends on two independent factors $B,C$, then

$A\propto B\implies A=kB$, but here $k$ is a constant w.r.t. $B$ not $C$, in fact, $k=f(C)\tag{1}$

Similarly, $A\propto C\implies A=k'C$ but here $k'$ is a constant w.r.t. C not $B$, in fact, $k'=g(B)\tag{2}$

From $(1)$ and $(2)$,

$f(C)B=g(B)C\implies f(C)\propto C\implies f(C)=k''C$

Putting it in $(1)$ gives,

$A=k''CB\implies A\propto BC\tag{Q.E.D.}$

$\endgroup$
12
  • $\begingroup$ How do we know that k=f(C) or k′=g(B) is a fact Also, f(C)B=g(B)C which means that f(C)∝C… I didn't get this part... I mean wouldn't that mean that g(B)/B is a constant. Sorry to be a little thick here, but I would deeply appreciate a deeper explanation. $\endgroup$ Jul 1, 2013 at 16:54
  • $\begingroup$ As i said $k$ is constant w.r.t. $B$ not $C$ therefore it depends on $C$ and hence a function of $C$. $\endgroup$
    – Aang
    Jul 1, 2013 at 16:57
  • $\begingroup$ Oh... now I get that part :) but could you please tell me why f(c) is proportional to C? $\endgroup$ Jul 1, 2013 at 17:12
  • 1
    $\begingroup$ $\frac{f(C)}{C}=\frac{g(B)}{B}$. Since L.H.S. depends only on $C$ and R.H.S. depends only on $B$. Since $B,C$ are independent, therefore, L.H.S.=R.H.S. must be equal to some constant. $\endgroup$
    – Aang
    Jul 1, 2013 at 17:22
  • 1
    $\begingroup$ This is the correct answer, but it does need the additional assumption that $B$ and $C$ are independent. $\endgroup$ Dec 3, 2021 at 14:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .