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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.

Rajdeep Sindhu
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Nafiul Islam
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2 Answers2

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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
$A=k'B=(k''C)B=k''BC⟹ A∝BC$

Zarif Muhtasim
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  • 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 – aaksaksyk Dec 03 '21 at 14:36
  • @aaksaksyk yes. – Zarif Muhtasim Dec 04 '21 at 15:07
  • @aaksaksyk Without any additional assumption, the implication $(A∝B$ and $A∝C) ⟹ A∝BC$ is false, as Zarif has pointed out. – Ramiro Dec 04 '21 at 21:14
  • 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. – Mailbox Feb 11 '23 at 18:24
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    @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. – Klg Sep 12 '23 at 14:49
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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,

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

Aang
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  • 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. – Nafiul Islam Jul 01 '13 at 16:54
  • As i said $k$ is constant w.r.t. $B$ not $C$ therefore it depends on $C$ and hence a function of $C$. – Aang Jul 01 '13 at 16:57
  • Oh... now I get that part :) but could you please tell me why f(c) is proportional to C? – Nafiul Islam Jul 01 '13 at 17:12
  • $f(C)B=g(B)C\implies$ $f(C)=\frac{g(B)}{B}C$. Now as you said $\frac{g(B)}{B}$ is a constant $\implies f(C)=$ constant*$C$ – Aang Jul 01 '13 at 17:14
  • This is the part I didn't get... how is g(B)/B a constant? – Nafiul Islam Jul 01 '13 at 17:18
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    $\frac{f(C)}{C}=\frac{g(B)}{B}$. 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. – Aang Jul 01 '13 at 17:22
  • let us continue this discussion in chat – Nafiul Islam Jul 01 '13 at 17:23
  • I am glad, it helped :) – Aang Jul 01 '13 at 17:26
  • What do $f(C)$ and $g(B)$ mean here? Functions? – Rajdeep Sindhu May 29 '20 at 08:38
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    This is the correct answer, but it does need the additional assumption that $B$ and $C$ are independent. – Adrian Keister Dec 03 '21 at 14:40
  • Attention: This answer is correct only if we assume that $B$ and $C$ do not depend on each other (that is, $B$ and $C$ are independent). Otherwise, it is easy to build a counterexample as @ZarifMuhtasim pointed out. Consider that $A= 2B$ and $B=3C$. Clearly $A=6C$. So $A \propto B$ and $A\propto C$, but it is not true that $A \propto BC$. Following the steps of this answer in this counterexample, we can get to $f(C)B=g(B)C$ where $f(C)=1$ and $g(B)=3$. So we can not conclude that $f(C)\propto C$. – Ramiro Dec 04 '21 at 13:58
  • In summary: This answer used as a key step that $f(C)B=g(B)C\implies f(C)\propto C$ which may not be true if $B$ and $C$ are not independent. – Ramiro Dec 04 '21 at 14:00