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For example, in physics, if $$\text{F} \propto m_1m_2$$ and $$\text{F} \propto \frac{1}{r^2},$$ then $$\text{F} \propto (m_1m_2)\left(\frac{1}{r^2}\right)= \frac{m_1m_2}{r^2}.$$

This property (combining proportionality) intuitively makes sense, but I have never seen it formally written in a textbook.

Could someone please rigorously prove this property (and fully specify its conditions?), or give a counterexample?

P.S. I know that this question has been answered, including here, but I do not understand the explanations: e.g., I don’t understand how $k=f(C)$ or $k′=g(B).$

aaksaksyk
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    Relevant and/or related: How does one combine proportionality? AND Proportionality AND If $A$ is proportional to $B$ when $C$ is constant and $A$ is proportional to $C$ when $B$ is constant then how is $A$ is proportional to $BC$? – Dave L. Renfro Dec 02 '21 at 14:54
  • As far as I know there only exist very basic hand waving mathematical/physical analogies and crude approximations to sort of classically link the law of gravitation to coulombs law. These work effectively in a pied-piper sort of way to stop students thinking about issues you or I might consider important. The basic problem (as far as I see it) is no one really understands what mass and charge are (even non-classically) and if/how they related to each other at some more fundamental level. – James Arathoon Dec 02 '21 at 15:28
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    Does this answer your question? How does one combine proportionality? – Adrian Keister Dec 02 '21 at 17:59
  • @AdrianKeister: Your thinking about the process of relating two, or more, independently measurable variables. Gravitational Force, $F$, is not a measurable quantity; its existence is entirely inferred from the measured gravitational acceleration with reference to Newton's absolute space (involving gravitational mass), together with Newton's 3 Laws (involving inertial mass). Then we have to assume that gravitational mass is equal to inertial mass, or at the very least directly proportional to one another. – James Arathoon Dec 02 '21 at 20:16
  • Mach and Einstein amongst others rightly pointed out that absolute space is also not a measurable thing. Today we talk about acceleration with reference to the centre of mass of a gravitationally interacting system. I am not entirely convinced about the necessity of assuming absolute time, but some people are. – James Arathoon Dec 02 '21 at 20:18
  • @JamesArathoon I think you're way overthinking this. I agree that the independence of the variables on the RHSs is important, but all the physics stuff you're mentioning is at a much different level than the OP is asking. – Adrian Keister Dec 02 '21 at 20:19
  • @AdrianKeister that posts asks the same question I have but I don’t understand the answer that is given. – aaksaksyk Dec 03 '21 at 03:39
  • @aaksaksyk: Can you provide some context as to what level of maths/physics you are studying and why you have picked this problem rather than simpler measurement examples first? - i.e. not one that is so theory-laden as gravitational theory (where inferences made rely on a sophisticated mixture of measurement and theoretical constructs often involving unobservable elements / free parameters). – James Arathoon Dec 03 '21 at 09:27
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    @JamesArathoon there isn’t a specific reason as to why a chose the law of gravitation as an example to express my problem over measurement example. I just chose the law of gravitation because it’s the most popular application of property of proportionality that I want to know about . I don’t want a physics answer to my question. I only want to know how 2 proportional relationships can be combined. And I am in 12th grade. – aaksaksyk Dec 03 '21 at 09:43
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    Can you point to any specific deficiency in the explanations given in the answers to the other question? Why do you doubt them? How do they not answer your question? This could help someone understand what kind of explanation would work better for you. – David K Dec 03 '21 at 12:37
  • @DavidK I don’t understand how k=f(C) or k′=g(B) in other question – aaksaksyk Dec 03 '21 at 12:50
  • Let us continue this discussion in chat. – aaksaksyk Dec 03 '21 at 16:08
  • The answers below so far involve specially picked/designed functions tailored to the task at hand. Mathematics does not know about the form of final physical theories, only how raw data (without theoretical baggage) may be fitted to arbitrary polynomial equations of well defined order (and perhaps the products of these polynomials also). Perhaps partial differentiation is more useful in a general sense here. – James Arathoon Dec 04 '21 at 03:34
  • @ryang If $\rho$ is the density, $M(V,L) = \rho V$. The fact that $L$ does not appears in right-hand size does not prevent listing it on the left-hand site. From a physical point of view, the mass $M$ depends on $V$ and $L$. – Ramiro Dec 04 '21 at 17:07
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    @ryang 1. In Mathematics, nothing prevents from writing $M(V, L)= \rho V$. 2. In Physics, even using mathematical notation, we would say $M$ is directly proportional to $V$ and $M$ is directly proportional to $L^3$, and using notation: $M(V,L) \propto V$ and $M(V,L) \propto L^3$. And we can not conclude $M(V,L) \propto VL^3$.
    1. In Physics, in general, it may not be obvious if the variables are independent or not. So, clearly stating that the variables are mutually independent is very necessary, specially in Physics.
     – Ramiro Dec 04 '21 at 17:44
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    @ryang Here is a pure mathematical example. Consider $\pi_1$ : $\Bbb R^2 \to \Bbb R$ defined as $\pi_1(x,y) =x$. Now consider the curve $C$ in $\Bbb R^2$ given by $y=2x$. So $C= {(x,y) \in \Bbb R^2 : y=2x }$. Let $f$ be the restriction of $\pi_1$ to $C$. So $f : C \to \Bbb R$, $f(x,y)= x$. We also have $f(x,y)= \frac{1}{2}y$ So, $f(x,y)\propto x$, $f(x,y)\propto y$, but it is not true that $f(x,y)\propto xy$. – Ramiro Dec 04 '21 at 18:18
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    @ryang OP did not specify the domain of $f$. NOTHING prevents the domain of $f$ to be a sub-manifold of $\Bbb R^2$. – Ramiro Dec 04 '21 at 18:34
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    The question is: "Does $f(a,b)$ being directly proportional to $a$ and $b$ separately imply that $f(a,b)$ is directly proportional to $ab?$"

    The correct answer depends on what is the domain of $f$.

    Case 1. If the domain of $f$ is $\Bbb R^2$, then, since $f(a,b)\propto a$ and $f(a,b)\propto b$, we have $$f(a,b) = a f(1,b) = ab f(1,1)$$ So $f(a,b)\propto ab$.

    Case 2. If the domain of $f$ is a submanifold of $\Bbb R^2$, then we may have $f(a,b)\propto a$ and $f(a,b)\propto b$ without having $f(a,b)\propto ab$. For instance: (continue)

     – Ramiro Dec 05 '21 at 15:37
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    Consider the curve $C$ in $\Bbb R^2$ given by $y=2x$. So $C= {(x,y) \in \Bbb R^2 : y=2x }$. Let $\pi_1$ : $\Bbb R^2 \to \Bbb R$ be defined as $\pi_1(x,y) =x$ and let $f$ be the restriction of $\pi_1$ to $C$. So $f : C \to \Bbb R$, $f(x,y)= x$. We also have $f(x,y)= \frac{1}{2}y$. So, $f(x,y)\propto x$, $f(x,y)\propto y$, but it is not true that $f(x,y)\propto xy$.

    Remark: In Physics, it is common to refer to case 1 by saying that $a$ and $b$ are independent, and to refer to case 2 by saying that $a$ and $b$ are dependent.

     – Ramiro Dec 05 '21 at 15:37

3 Answers3

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Let $F$ be a function depending on two variables $x$ and $y$. Assume that $F$ is proportional to $x$. This means that the value of $\frac{F(x,y)}{x}$ only depends on the value of $y$. Thus for non-zero $x$ and $y$, $\frac{F(x,y)}{x}=\frac{F(1,y)}{1}$, meaning $$\frac{F(x,y)}{x}=F(1,y).$$

Further, assume that $F$ is proportional to $y$. Analogously, we acquire for non-zero $x$ and $y$, $$\frac{F(x,y)}{y}=F(x,1).$$

To prove that $F$ is proportional to $xy$, it suffices to show that $\frac{F(x,y)}{xy}$ is constant. Indeed, combining both formulas above, we obtain

$$\frac{F(x,y)}{xy}=\frac{F(1,y)}{y}=F(1,1).$$

Zuy
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Theorem

Let $x$ and $y$ be independent of each other. Then $$z\propto xy$$ iff $$\:z\propto x\quad\text {and}\quad z\propto y.$$

Proof

  1. Suppose that $\,z\propto xy,$ i.e., $z$ is jointly proportional to $x$ and $y.$

    Then there exists a non-identically-zero expression $w$ that is independent of $xy$ such that $$z=w(xy).$$ Therefore, by the boldfaced condition, there exists a non-identically-zero expression $w$ that is independent of $x$ and $y$ such that $$z=(wy)x=(wx)y.$$ Thus, there exists a non-identically-zero expression $u$ independent of $x$ such that $$z=ux,$$ and there exists a non-identically-zero expression $v$ independent of $y$ such that $$z=vy;$$ that is, $$z\propto x\quad\text {and}\quad z\propto y.$$

  2. Suppose that $\,z\propto x\:$ and $\:z\propto y.$

    Then there exist non-identically-zero expressions $u$ independent of $x$ and $v$ independent of $y$ such that $$ux=z=vy\\\frac uy=\frac vx.$$ Therefore, denoting $\dfrac uy$ by $w,$ by the boldfaced condition, $w$ is independent of $x$ and $y.$

    Hence, there exists a non-identically-zero expression $w$ that is independent of $xy$ such that \begin{align}z&=(wy)x\\&=w(xy);\end{align} that is, $$z\propto xy.$$

When variables $x$ and $y$ depend on each other, the theorem does not apply:

  1. Let $z=xy$ and $x=y.$

    Then $z=x^2.$

    Thus, $z$ is directly proportional to $xy,$ but not to $x.$

  2. Let $x=y=z.$

    Then $z=\frac12(x+y).$

    Thus, $z$ is directly proportional to each of $x$ and $y,$ but not to $xy.$

ryang
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The condition $f(a,b) \propto a$ means that, for every $b$, there is a constant $k_b$ such that $f(a,b)=a\cdot k_b$. Thus, $k_b= f(1,b)$ and then $f(a,b)= a \cdot f(1,b)$.
Similarly, $f(a,b)=b\cdot f(a,1)$.
You can substitute any of the two expressions in the other to find $f(a,b)=ab\cdot f(1,1)$.

Matheus Barros
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