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.
Asked
Active
Viewed 43 times
2
-
3The 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 $a^2=kl(bc)$. – André Nicolas Jul 02 '16 at 00:31
1 Answers
1
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 $a\propto c$.
André Nicolas
- 507,029
-
This proof (and the OP's) begins by implicitly assuming that $b$ and $c$ depend on each other (for example, $b=3c$). On the other hand, that well-known result quoted at the beginning of the OP assumes that $b$ and $c$ are independent of each other. – ryang Jun 27 '23 at 14:42