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Do the rules of algebra apply when you’re working with proportionalities?

For example, let $P$ be pressure, $\rho$ be density, and $m$ be mass. Given that $$P \propto \rho \quad\text{and}\quad \rho \propto m,$$ it would be logical to conclude that $$P \propto \rho \propto m$$ and $$P \propto m.$$

However, if I isolate $\rho$ on one side of the proportionality, I would, logically, get $$Pm \propto \rho,$$ since that is the only way both proportionalities—$P \propto \rho$ and $m \propto \rho$ are expressed. Clearly, this does not seem to follow much algebraic sense. How about if I wrote the proportionalities as equalities? $$P=\alpha\rho=\beta m,$$ where $\alpha$ and $\beta$ are proportionality constants. From here, how would you use algebra to obtain $$Pm=\gamma \rho$$ (whatever $\gamma$ is equal to in terms of $\alpha$ and $\beta$)?

ryang
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lightweaver
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3 Answers3

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I'm not entirely sure where you are coming from on this, so tell me if I am answering a different question here.

Proportionality is defined as follows. $P\propto \rho$ means $P=c\rho$ for a constant $c$. It must mean that and that is all it ever means. It is literally a shorthand way of writing $P=c\rho$.

Thus, you would not logically get $Pm\propto \rho$. This would imply that $Pm=c\rho$ for some constant $c$, or $P=\frac{c}{m}\rho$, but you already said that $P\propto \rho$. Since $m$ is not a constant (at least, not mathematically, although mass usually stays constant), we have a contradiction.

Going on from that, your last line would also give a contradiction. You write $Pm=\gamma m$, which implies of course that $P=\gamma$, a constant.

Hopefully that helps.

pancini
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  • I apologize for a mistake. I meant $Pm=\gamma \rho$, not $\gamma m$. And what of the proportionality between mass and density then? The statement $Pm\propto \rho$ isn’t false, is it? – lightweaver Dec 29 '15 at 07:31
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Your two proportionalities can be expressed as standard linear equations - you have constants of proportionality $A$ and $B$ with $$P=A\rho$$ and $$\rho=Bm$$ and from these you obtain $$P=(AB)m$$ with constant of proportionality $AB$.

You would also have $$P\cdot Bm=A\rho\cdot \rho$$ or $$Pm=\frac AB\rho^2$$ combining the two linear relationships to get a quadratic expression.

To answer a question you have asked, but probably don't mean, some mathematicians study algebraic geometry with projective co-ordinates which gives a systematic algebra of proportion in a much wider (and more abstract) context.

Mark Bennet
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let $P$ be pressure, $\rho$ be density, and $m$ be mass. Given that $$P \propto \rho \quad\text{and}\quad \rho \propto m,$$ it would be logical to conclude that $P \propto m$

No, proportionality involving multiple independent variables is not transitive, because, for example, the value of $m$ is being fixed while $P$ and $\rho$ are varying against each other.

$$P=\alpha\rho=\beta m,$$ where $\alpha$ and $\beta$ are proportionality constants.

Instead: $$f(m)\times P=\rho=g(P)\times m,\tag1$$ where the functions $f$ and $g$ are not identically zero.

$Pm \propto \rho$ does not seem to follow much algebraic sense.

From $(1):$ $$\frac{f(m)}m=\frac{g(P)}P.$$ Since $m$ and $P$ are independent of each other, $\dfrac{f(m)}m$ and $\dfrac{g(P)}P$ must be a (nonzero) constant, say $k.$ Then, substituting $f(m)=km$ into $(1)$ gives $$kmP=\rho,$$ that is, $$Pm \propto \rho.$$

ryang
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