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Its known that if one variable is proportional to two others than it is also proportional to their product. $$\forall a,b,c\in ℝ:a\propto b\wedge a\propto c\Rightarrow a\propto b\cdot c$$ I think i`ve proven that if one variable is proportional to two others than its square is proportional to their product. But these two theorems would be in contradiction to each other so my proof have to contain a mistake. But where is the mistake? $$Assumption:\forall a,b,c\in ℝ:a\propto b\wedge a\propto c\Rightarrow { a }^{ 2 }\propto b\cdot c$$ $$a\propto b\wedge a\propto c\Rightarrow b=m\cdot a \wedge c=n\cdot a\Rightarrow b\cdot c=m\cdot n\cdot { a }^{ 2 }\Rightarrow { a }^{ 2 }\propto b\cdot c $$ Where $m$ and $n$ are parameters.

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    $\begingroup$ The assertion in the first displayed line is not correct. For your correct assertion about $a^2$, I would mildly prefer $a=kb$, $a=lc$ so $a^2=kl(bc)$. $\endgroup$ – André Nicolas Jul 2 '16 at 0:31
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There is no mistake. If the variable $a$ is proportional to the variable $b$, then $a=kb$ for some constant $k$. Similarly, $a=lc$ for some constant $l$. Thus $a^2=(kl)bc$, and therefore $a^2$ is proportional to $bc$. If $a$, $b$, and $c$ are positive, which they usually are, we can conclude that $a$ is proportional to $\sqrt{bc}$.

The assertion in the first displayed line is not correct. Perhaps you had the following correct result in mind. If $a\propto b$ and $b\propto c$, then $a\propto c$.

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