Does $f(a,b)$ being directly proportional to $a$ and $b$ separately imply that $f(a,b)$ is directly proportional to $ab?$

Question

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 $k′=g(B).$

Answer 1

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

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

Answer 2

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, $$z\propto xy.$$

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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 $xy.$

Answer 3

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