7
$\begingroup$

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 $k′=g(B).$

$\endgroup$
19
  • 1
  • 1
    $\begingroup$ Does this answer your question? How does one combine proportionality? $\endgroup$ Dec 2, 2021 at 17:59
  • 1
    $\begingroup$ @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. $\endgroup$
    – aaksaksyk
    Dec 3, 2021 at 9:43
  • 2
    $\begingroup$ 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. $\endgroup$
    – David K
    Dec 3, 2021 at 12:37
  • 1
    $\begingroup$ @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 $M(V,L) \propto VL^3$. 3. 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. $\endgroup$
    – Ramiro
    Dec 4, 2021 at 17:44

3 Answers 3

8
$\begingroup$

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

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

$\endgroup$
0
4
$\begingroup$

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 $xy.$

$\endgroup$
0
1
$\begingroup$

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 $f(a,b)=ab\cdot f(1,1)$.

$\endgroup$
0

You must log in to answer this question.

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