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!
<br /> \begin{align}<br /> Q(ka, b) &amp;= ka + b \\<br /> kQ(a, b) &amp;= ka + kb \\<br /> &amp;\ne Q(ka, b)<br /> \end{align}<br />
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! :)