Is this proportionality proof erroneous?

Question

If $a$ is directly proportional to $b$ and also directly proportional to $c$,

is it true that $a$ is directly proportional to $bc?$ (It seems like it is true.)

Here is what I did, and I have a feeling I have made a pretty big mistake somewhere:

From the given info, we have $$\frac{a}{b}=x$$ for some constant $x,$ and $$\frac{a}{c}=y$$ for some constant $y.$ Multiplying the two equations gives $${a}^2=bc(xy),$$ where $xy$ is a constant.

So, ${a}^2$ is directly proportional to $bc.$

Update: Now for my real doubt. My physics textbook states the following:

Let two objects $A$ and $B$ of masses $M$ and $m$ lie at a distance $d$ from each other. Let the force of attraction between two objects be $F$. According to the universal law of gravitation, the force between two objects is directly proportional to the product of their masses; that is, $F ∝ M m.$ And the force between two objects is inversely proportional to the square of the distance between them; that is, $F ∝\frac1{d^2}.$ Combining the proportionalities we get $F ∝ \displaystyle\frac{Mm}{d^2}$.

So, does it mean that my textbook is wrong? Doesn't the combination of the two proportionalities result in $F^2$ being directly proportional to $\displaystyle\frac{Mm}{d^2}?$

Answer 1

Revised and updated answer

I did not examine carefully your assertion that ${a}^2$ is directly proportional to $bc$

Actually, a is jointly proportional to $b$ and $c$ which means

$a \propto bc$

Here are two examples:

Distance travelled varies directly with speed for a given time.
Distance travelled varies directly with time for a given speed.
Distance travelled varies jointly with speed and time, distance $\propto$ speed$\times time$

To take another example with both direct variation, and inverse variation,

Time taken for a journey varies directly with the distance for a given time
Time taken for a journey varies inversely as the speed, fpr a given time
Time $\propto$ distance/speed

[ It just happens that in the above two examples, the constant of variation is $1$ ]

Added: Proof for the original example

Let $a$ change to $a_1$ while $c$ is unchanged, and $b$ changes to $b'$, then $a/a_1 = b/b'$

Then let $a_1$ change to $a_2$ while $b'$ is unchanged, and $c$ changes to $c'$, then $a_1/a_2 = $c/c'$

Multiplying the two, $\frac{a}{a_2} = \frac{bc}{b'c'}$
so $a\propto bc$

Answer 2

The following argumentation has been used in the good old days to teach us about the ideal gas laws of Boyle and Gay Lussac.

1. Experiments reveal that the pressure $p$ is proportional to the amount of gas $n$
   if temperature $T$ and volume $V$ are held constant. Thus: $\, p \sim n$ .
2. Experiments reveal that the pressure $p$ is inversely proportional to the volume $V$
   if temperature $T$ and amount of gas $n$ are held constant. Thus: $\, p \sim 1/V$ .
3. Experiments reveal that the pressure $p$ is proportional to the temperature $T$
   if volume $V$ and amount of gas $n$ are held constant. Thus: $\, p \sim T$ .
4. Experiments reveal that the volume $V$ is proportional to the temperature $T$
   if pressure $p$ and amount of gas $n$ are held constant. Thus: $\, V \sim T$ .

Then the book (*) tells us that from (2) it follows that $p_1\cdot V_1 = p_2\cdot V_2$ : Boyle's law.
A next step is to introduce absolute temperature $T$ (Kelvin) instead of relative temperature.
Then from (3) we have $p_1/T_1 = p_2/T_2$ : pressure law of Gay Lussac.
The book proceeds with (4) and the volume law of Gay Lussac : $V_1/T_1 = V_2/T_2$.
And then comes a clue. What happens if we change the volume $V$ as well as the temperature $T$ ?
The book says that we should do this in two subsequent steps:

1. Keep the volume $V$ constant and raise the temperature $T$.
   Then we have according to Gay Lussac: $\;p_1/T_1 = p_2'/T_2 \quad \Longrightarrow \quad p_2' = p_1\cdot T_2/T_1$
2. Keep the temperature $T$ constant and change the volume $V$.
   Then we have according to Boyle: $\;p_2'\cdot V_1 = p_2\cdot V_2 \quad \Longrightarrow \quad p_1\cdot T_2/T_1 \cdot V_1 = p_2\cdot V_2$

It follows that: $$ \frac{p_1 \cdot V_1}{T_1} = \frac{p_2 \cdot V_2}{T_2} \quad \Longrightarrow \quad p\cdot V = C\cdot T $$ Where $C$ is a constant that depends only on the amount of gas $n$.

It is supposed that the same procedure may be applied for Newton's law of gravitation:

1. Experiments reveal that the force $F$ is proportional to one of the two masses $m$
   if the other mass $M$ as well as the distance $r$ between the masses is held constant: $\, F \sim m$ .
2. The same argument hold for the other mass if mass $m$ and distance $r$ are held constant.
   Thus: $\, F \sim M$ .
3. Experiments reveal that the force $F$ is inversely proportional to the distance $r$ squared
   if the mass $M$ as well as the mass $m$ are held constant. Thus: $\, F \sim 1/r^2$ .

Now we are going to derive Newton's law in three steps. (Don't get confused by the naming!)

1. Keep the distance constant and change only the mass $m$ : $ F_1/m_1 = F_2/m_4 \quad \Longrightarrow \quad F_2 = F_1 m_4/m_1 $
2. Keep the distance constant and change only the mass $M$ : $ F_2/M_1 = F_3/M_4 \quad \Longrightarrow \quad F_3 = F_1 m_4 M_4/(m_1 M_1) $
3. Keep the masses constant and change only the distance $r$ : $ F_3 r_1^2 = F_4 r_4^2 \quad \Longrightarrow \quad F_4 = F_1 m_4 M_4 r_1^2/(m_1 M_1 r_4^2) $

It follows that: $$ \frac{F_4 r_4^2}{m_4 M_4} = \frac{F_1 r_1^2}{m_1 M_1} \quad \Longrightarrow \quad F = \mbox{constant} \times \frac{m M}{r^2} $$ (*) Reference:

DR. SCHWEERS EN DRS. P. VAN VIANEN
NATUURKUNDE
op corpusculaire grondslag
DEEL I
voor de onderbouw van het v.h.m.o.
ZEVENDE DRUK (1960)
L.C.G. MALMBERG 's-HERTOGENBOSCH

Answer 3

From your first two identities,

$$a=bx=cy.$$

These equalities can hold simultaneously iff $b$ and $c$ are also proportional to each other ! Then $a^2=bcxy$ indeed shows that $a^2$ is proportional to $bc$.

But the usual way to work this out is "with $c$ fixed, $a=bx$, and with $b$ fixed $a=cy$: proportionality works independently for $b$ and $c$.

For this to be possible, you must have

$$a=bx_c=cy_b=bcz$$ or if you prefer, both "constants" are proportional to the other variable

$$x_c=cz,y_b=bz.$$

Answer 4

Consider that $F$ is a function of $M,m,$ and $d.$ That is $F=F(M,m,d).$ Pick some particular non-zero values $M_0,m_0,d_0.$ We have $$F(M,m,d)=F(M,m,d_0)(d/d_0)^{-2}$$ because $F(M,m,d)$ is inversely proportional to $d^2.$ We have $$F(M,m,d_0)=F(M,m_0,d_0)(m/m_0)$$ because $F(M,m,d_0)$ is proportional to $M m,$ and because $M m/(M m_0)=m/m_0.$ We have $$F(M,m_0,d_0)=F(M_0,m_0,d_0)(M/M_0)$$ because $F(M,m_0,d_0)$ is proportional to $M m_0,$ and because $M m_0/(M_0 m_0)=M/M_0.$ Altogether we have $$F(M,m,d)=(M m/d^2)K_0$$ for all $M,m,d$, where $K_0=d_0^2/M_0 m_0.$

Answer 5

1. If $a$ is directly proportional to $b$ and also directly proportional to $c$,

   we have $\dfrac{a}{b}=x$ for some constant $x,$ and $\dfrac{a}{c}=y$ for some constant $y.$

   It is tacit that each described variation occurs while the other variable is held constant. As such, neither $x$ nor $y$ is a constant: $x$ is a function of $c,$ while $y$ is a function of $b.$

   $${a}^2=bc(xy),$$ where $xy$ is a constant.

   So, ${a}^2$ is directly proportional to $bc.$

   Actually, $$a^2=bc\times f(b,c),$$ so, it is invalid to conclude that $a^2\propto bc.$

2. If $a$ is directly proportional to $b$ and also directly proportional to $c$,

   is it true that $a$ is directly proportional to $bc?$

   Yes, on the condition that $b$ and $c$ are independent of each other.

   On the other hand, suppose that $$a=b+c\\b=c$$ so that the above condition is not satisfied; in this case, $a$ is directly proportional to $b$ and to $c$ (since $a=2b=2c$) but not to $bc.$

3. $F ∝ M m$ and $F ∝\dfrac1{d^2}.$ Combining the proportionalities we get $F ∝ \dfrac{Mm}{d^2}$.

   Doesn't the combination of the two proportionalities result in $F^2$ being directly proportional to $\displaystyle\frac{Mm}{d^2}?$

   As explained in the previous point, it is $F,$ not $F^2,$ that is directly proportional to $\displaystyle\frac{Mm}{d^2}.$