# Why variables in directly proportinality are multipiled

by 22990atinesh
Tags: directly, multipiled, proportinality, variables
 P: 58 Why variables (RHS) in directly proportionality are always multiplied. Suppose the newton 2nd law ##{F}\propto{m}## ##{F}\propto{a}## ##{F}\propto{m*a}##
 Homework Sci Advisor HW Helper Thanks P: 12,463 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##
P: 58
 Quote by Simon Bridge 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##
If
##Q \propto a##
##Q \propto b##

then you also know that

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

P: 128
Why variables in directly proportinality are multipiled

 Quote by 22990atinesh If ##Q \propto a## ##Q \propto b## then you also know that ##Q\propto (a+b)## this can aslo be true why multiply.
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 $Q \propto ab$.
P: 58
 Quote by chogg 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 $Q \propto ab$.
I didn't understand what are u trying to say.
 P: 128 Proportional to $x$ means the same as Equal to $x$, times some constant. Right?
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P: 21,215
 Quote by chogg 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 $Q \propto ab$.
 Quote by 22990atinesh I didn't understand what are u trying to say.
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.
 P: 128 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.
 Homework Sci Advisor HW Helper Thanks P: 12,463 @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?
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P: 21,215
 Quote by Simon Bridge You seem to be confused about what "directly proportional to" means
Who is "you" here?
 Quote by Simon Bridge Here is the definition: If ##Q \propto P## then ##Q=kP## where ##k## does not depend on ##P##. 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##
I agree, and this is pretty much what I said in post 7.
 Quote by Simon Bridge [Note: this is pretty much the argument in post #4]
 Homework Sci Advisor HW Helper Thanks P: 12,463 @Mark44: so noted - post #9 edited to reflect your comments :)
P: 58
 Quote by Simon Bridge @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?
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##
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P: 12,463
 Quote by 22990atinesh 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##
##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)##?
P: 58
 Quote by Simon Bridge ##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)##?
Still didn't understand. I dont want a rigorous proof. I just want a simple explanation.
P: 128
 Quote by 22990atinesh Still didn't understand. I dont want a rigorous proof. I just want a simple explanation.
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.
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P: 12,463
 Still didn't understand. I dont want a rigorous proof. I just want a simple explanation.
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.
P: 58
 Quote by chogg 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.
Thanx for example Chogg. This simple example cleared my doubt.

 Quote by Simon Bridge 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.
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
 P: 128 Very happy I could help! :)

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