# Why variables in directly proportinality are multipiled

1. ### 22990atinesh

127
Why variables (RHS) in directly proportionality are always multiplied.

Suppose the newton 2nd law

##{F}\propto{m}##

##{F}\propto{a}##

##{F}\propto{m*a}##

Last edited by a moderator: Apr 18, 2014
2. ### Simon Bridge

14,479
If you know that for variables Q, a, and b, if you know ##Q \propto a##, and that ##Q \propto b## then you also know that ##Q\propto ab## ...

This is because that is what "directly proportional to" means.

Similarly if ##Q\propto a## and ##Q\propto 1/b## then ##Q\propto a/b##

3. ### 22990atinesh

127
If
##Q \propto a##
##Q \propto b##

then you also know that

##Q\propto (a+b)##
this can aslo be true why multiply.

Last edited by a moderator: Apr 18, 2014
4. ### chogg

131
This isn't true.

The constant of proportionality can't depend on whatever's on the right side. So, $Q \propto a$ means $Q = C_1 \cdot f(b) \cdot a$ (where $C_1$ doesn't depend on $a$ or $b$).

Similarly, $Q \propto b$ really means $Q = C_2 \cdot f(a) \cdot b$, for some (possibly different) constant $C_2$.

The only way this can be true simultaneously is if $Q = C \cdot a \cdot b$ for some constant $C$ -- or, more simply, if $Q \propto ab$.

5. ### 22990atinesh

127
I didn't understand what are u trying to say.

Last edited by a moderator: Apr 18, 2014
6. ### chogg

131
Proportional to $x$
means the same as
Equal to $x$, times some constant.

Right?

### Staff: Mentor

I didn't either, especially the parts about f(a) and f(b).
##Q \propto a## means Q = ka for some constant k. If you form the ratio Q/a, you get (ka)/a, or k, the constant of proportionality.

BTW, using "textspeak" such as "u" for "you" isn't allowed here.

8. ### chogg

131
Let me start by answering your original question in a different way.

Suppose $Q$ is some function of $a$ and $b$. I'll write it as $Q(a, b)$ to emphasize this.

To say $Q(a, b) \propto a$ means that $Q(ka, b) = kQ(a, b)$ for any constant $k$. In words: if you scale up $a$, you scale up $Q$ by the same amount, because $Q$ is proportional to $a$.

You said that $Q(a, b) = a + b$ satisfies $Q(a, b) \propto a$. Let's check!
\begin{align} Q(ka, b) &= ka + b \\ kQ(a, b) &= ka + kb \\ &\ne Q(ka, b) \end{align}
Therefore, $a + b$ is not proportional to $a$.

---

Now as to my apparently-hard-to-understand notation: the $f(a)$ notation just means "any function of $a$". Note that Mark44's constant $k$ could well depend on $a$! For example, if $Q(a, b) = \sin(a)b$, then $Q(a, b) \propto b$ is true. I used the $f(a)$ notation to emphasize this.

9. ### Simon Bridge

14,479
@22990atinesh (OP): You seem to be confused about what "directly proportional to" means
Here is the definition:

If ##Q \propto P## then ##Q=kP## where ##k## does not depend on ##P##.

(I suspect you've got the first part but not the second part.)

Applying this definition:

(1) ##Q \propto a## means ##Q = k_1 a## where ##k_1## does not depend on ##a##
(2) ##Q\propto b## means ##Q = k_2 b## where ##k_2## does not depend on ##b##

Now consider:

(3) ##Q \propto ab## means that ##Q= k ab## and we can see from (1) and (2) that ##k_1=k/b## does not depend on ##a## and ##k_2=k/a## does not depend on ##b## - so if (1) and (2) are both true, then (3) is also true.

(4) ##Q \propto a+b## means that ##Q=k(a + b)## and we can see from (1) and (2) that: $$k_1=\frac{k(a+b)}{a},\; k_2=\frac{k(a+b)}{b}$$... these expressions are saying that the only way (4) is true is if ##k_1## depends on ##a## and ##k_2## depends on ##b## - which contradicts the definition of "directly proportional to" used to make (1) and (2).

In other words, if (1) and (2) are both true, then (4) is false.

[Note: this is pretty much the argument first appearing in this thread in post #4 (and repeated since)]

You can try this reasoning process yourself for:

(5) ##Q\propto f(a,b)## ... where ##f## is an arbitrary function of ##a## and ##b## together...

... if (1) and (2) are both true, what form(s) can ##f## take so that (5) is also true?

Last edited: Apr 17, 2014

### Staff: Mentor

Who is "you" here?
I agree, and this is pretty much what I said in post 7.

11. ### Simon Bridge

14,479
@Mark44: so noted - post #9 edited to reflect your comments :)

12. ### 22990atinesh

127
How did ##k_2=k/a## and ##k_1=k/b## come in point (3).
It must be ##k_2=k*a## and ##k_1=k*b##

Last edited by a moderator: Apr 18, 2014
13. ### Simon Bridge

14,479
##Q=k_1a## and ##Q= kab## true at the same time means that ##k_1 a = kab## or ##k_1=kb## ... so well done: but the argument holds: k_1 does not depend on a - in accordance with the definition given in (1).

Can you say the same for ##Q=k(a+b)##?

14. ### 22990atinesh

127
Still didn't understand. I dont want a rigorous proof. I just want a simple explanation.

Last edited by a moderator: Apr 18, 2014
15. ### chogg

131
Well, several people have already supplied one. But let's try plugging in numbers; maybe examples will help!

You said that $Q = a + b$ can satisfy $Q \propto a$. If that's true, then for any values of $a$ or $b$, if we scale up $a$, we scale up $Q$ by the same amount.

Let's say $a = 1$ and $b = 2$. This means that $Q = 3$.

Now let's double $a$, and try predicting what happens to $Q$ in two ways.
• Using $Q \propto a$, when we double $a$, we double $Q$. Therefore, we expect $Q = 6$.
• Using $Q = a + b$, we can just plug in the values. We actually find $Q = 4$.

4 is not the same as 6. Therefore, we were wrong when we said $Q \propto a$ is true when $Q = a + b$.

Proportional means multiply.

Last edited by a moderator: Apr 18, 2014
1 person likes this.
16. ### Simon Bridge

14,479
I am at a loss: what education level are you at?

Do you understand that the definition of ##Q\propto P## is
##Q=kP## where k is a number that does not depend on P?​

I'm afraid that is as simple as it gets.

17. ### 22990atinesh

127
Doubt Cleared

Thanx for example Chogg. This simple example cleared my doubt.

Simon Bridge I appreciate your efforts to explain. But your explanation was a little bit theoretical. If you would have said the same thing with an example, I would have gotten it earlier. Anyway thanx for help

Last edited by a moderator: Apr 18, 2014
18. ### chogg

131
Very happy I could help! :)