Why variables in directly proportinality are multipiled

[22990atinesh]

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

Suppose the Newton 2nd law

${F}\propto{m}$

${F}\propto{a}$

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

 

[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 $Q\propto a/b$

 

[22990atinesh]

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.

 

[chogg]

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 $Q \propto ab$.

 

[22990atinesh]

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.

 

[chogg]

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

Right?

 

[Mark44]

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.

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.

 

[chogg]

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.

 

[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?

 

[Mark44]

You seem to be confused about what "directly proportional to" means

Who is "you" here?

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.

[Note: this is pretty much the argument in post #4]

 

[Simon Bridge]

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

 

[22990atinesh]

@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 $k_1=k*b$

 

[Simon Bridge]

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 $Q=k(a+b)$?

 

[22990atinesh]

$Q=k_1a$ 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 don't want a rigorous proof. I just want a simple explanation.

 

[chogg]

Still didn't understand. I don't 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.

 

[Simon Bridge]

Still didn't understand. I don't 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.

 

[22990atinesh]

Doubt Cleared

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.

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 thanks for help

 

[chogg]

Very happy I could help! :)