If $x$ is proportional to $y$ and $x$ is proportional to $z$, then is $x \propto yz$ or $x^2 \propto yz$?

Question

Suppose $x$ is proportional to $y$ and $x$ is proportional to $z$. Then we can write, \begin{align*} x &= k_{1}y \end{align*}

The constant $k_{1}$ is going to depend on $z$. So, $$k_{1} = f(z)$$ Since $x$ is proportional to $z$, $$k_{1} = f(z) = cz$$, for some constant $c$. So we have, $$x = czy $$ $$x \propto zy $$

But, since $x$ is proportional to $z$ we can write, $$x = k_{2}z$$ So multiplying the two equations we have,

$$x^{2} = k_{1}k_{2}yz$$ $$\frac{x^{2}}{yz} = k_{1}k_{2}$$ $$x^{2} \propto yz$$

These are two conflicting results. So what am I missing here? I have looked at this thread: How does one combine proportionality? and while it does give some insight, doesn't fully answer my question.

Edit: Ok, I think I understand it now. In the last part we had $x^2 = k_1k_2yz$, since we are considering $y$ and $z$ as things that can change, we must also consider the constants. We know that if $z$ changes, $k_1$ changes, and if $y$ changes, $k_2$ changes. That is, $k_1 = c_1z$ and $k_2 = c_2y$. So we have \begin{align} x^2 &= c_1c_2y^2z^2 \\ x^2 &\propto (yz)^2 \\ \end{align}

From this can I write $$x \propto (yz)$$ ?

Answer 1

The two results are both correct under different interpretations of what "$x$ is proportional to $y$ and $x$ is proportional to $z$" means.

One degree of freedom

One interpretation is that $x$, $y$, and $z$ are all related in a way that knowing any one of them determines the other two. Geometrically, the set of all points $(x,y,z)$ permitted by the relationship between them is a curve in space.

In this case, "$x$ is proportional to $y$" gives us a relationship $x = k_1 y$ that is absolute: if I tell you what $y$ is, you can immediately use $x = k_1y$ to find $x$. The same thing applies to the statement "$x$ is proportional to $z$": we have $x = k_2 z$, with some different constant.

In this case, $x^2 = k_1 k_2 yz$ is true but slightly misleading: it might make you think that you can choose any value of $y$ and $z$ and use the formula to find what $x^2$ is. In fact, from $x = k_1y$ and $x = k_2z$ we can determine that $z = \frac{k_1}{k_2} y$: if we choose a value of $y$, that determines both $x$ and $z$.

Still, it's true that $x^2$ is proportional to $yz$, in the sense that the ratio between $x^2$ and $yz$ is always the same.

Two degrees of freedom

Another interpretation is that $y$ and $z$ are separate inputs, independently free to take on any value we like. Geometrically, the set of points $(x,y,z)$ permitted by the relationship between them is a surface in space.

In this case, "$x$ is proportional to $y$" should really be said as "$x$ is proportional to $y$ when $z$ is held constant". We have $x = k_1 y$, but $k_1$ is a value that depends on $z$.

If $x$ is proportional to $y$ when $z$ is held constant, and $x$ is proportional to $z$ when $y$ is held constant, then:

• Because $\frac{x}{y} = k_1$, where $k_1$ does not depend on $y$, the quantity $\frac{x}{yz} = \frac{k_1}{z}$ also does not depend on $y$.
• Because $\frac{x}{z} = k_2$, where $k_2$ does not depend on $z$, the quantity $\frac{x}{yz} = \frac{k_2}{y}$ also does not depend on $z$.

We conclude that $\frac{x}{yz}$ does not depend on either $y$ or $z$: it is a true constant. Therefore $x$ is proportional to the product $yz$.

Answer 2

Explanation

Suppose x is proportional to y and x is proportional to z.

Mathematically, this means that $x=k_1y$ and $x=k_2z$, where $k_1$ and $k_2$ are constants. Multiply these two equations together: $x^2=k_1k_2yz$. Let $k_3=k_1k_2$. $x^2=k_3yz$, and since $k_3$ is a constant ($k_2$ and $k_1$ are constants), this means that $x^2\propto yz$.

Where you're going wrong

Like @David_Quinn mentioned, you are going wrong when you say that the constant $k_1$ is going to depend on $z$, or when you say that "we know that if $z$ changes, $k_1$ changes, and if $y$ changes, $k_2$ changes". Constants are constants; they don't change. The fact that $x$ is proportional to $z$ means that for any $x$, $z$ is equal to that $x$ divided by a constant $k_1$, and that constant has to be the same regardless of $x$ and $z$.

Testing with numbers

We can also arrive at the same conclusion by testing out numbers, although this is hardly rigorous. Let $x=1$, $y=2$, and $z=3$. This means that $k_1=2$ and $k_2=3$. If $x$ were to change to $2$, then, $z$ would change to $6$ and $y$ would change to $4$. $yz$ is $6$ when $x$ is $1$ and $24$ when $x$ is $2$. $x^2$ is $1$ when $x$ is $1$ and $4$ when $x$ is $2$. Since $\dfrac{24}{6}=4$, we can say that $yz$ might be proportional to $x^2$, but cannot be proportional to $x$. This is the same as what we get from our analysis above.

Edit

The other answer is talking about a case where $z\propto k_1$, and not $z\propto x$. This might be why you got confused and wrote $k_1=f(z)$. In your particular case, however, this is not true, so $A$ is not proportional to $BC$.