0
$\begingroup$

I have problems with understanding how to operate with proportionality the first problem is with a question above.

Say we have two relations A proportional to B and C is proportional to D then if we divide the two relations why do we get A upon C is equals to B upon D and not A/C = KB/QD k and q being some numbers

Another question is why can a term be substituted in one relation from another for example:-

   a is proportional to p
   a + (n-1)x is proportional to q
   Then why is
   p + (n-1)x proportional to q

Also, Why can we substitute relations in equations

I don't seem to understand how proportionality works if there are any other rules for applications that I've missed can you also please tell me those?

$\endgroup$
  • 3
    $\begingroup$ No wonder you don't understand this. The claim is false in general. $\endgroup$ – Hagen von Eitzen Oct 1 '18 at 13:45
0
$\begingroup$

The relation "$a$ is proportional to $b$" actually represents an equation already, namely: $a=k\cdot b$ for some number $k$. So all of the rules of equal signs can be used, including transitivity and substitution.

$\endgroup$
  • $\begingroup$ Then why in the relations given in question, a is substituted by p and not kp. $\endgroup$ – user25614 Oct 1 '18 at 13:52
  • $\begingroup$ Say we have two relations A proportional to B and C is proportional to D then if we divide the two relations why do we get A upon C is equals to B upon D and not A/C = KB/QD k and q being some numbers $\endgroup$ – user25614 Oct 1 '18 at 13:54
0
$\begingroup$

Saying that one variable is proportional to another is simply another way os saying that there is a linear relationship between them. In other words the following statements:

  1. $a$ is proportional to $b$

  2. $a=kb$ for some constant $k$

  3. $\frac{a}{b}=k$ for some constant $k$

all say the same thing.

If $a$ is proportional to $b$ and $a$ is also proportional to $c$, then we have $a=k_1b$ and $a=k_2c$. So $b=\frac{k_2}{k_1}c$. But since $k_1$ and $k_2$ are constants then so is $\frac{k_2}{k_1}$. So we can conclude that $b$ is proportional to $c$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.