If X∝A and X∝B , then why X∝AB ?? Give me the Mathematical Explanation....


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Question Number 11701 by Nayon last updated on 30/Mar/17

$I f X \propto A a n d X \propto B,$ $t h e n w h y X \propto A B ? ?$ $G i v e m e t h e M a t h e m a t i c a l$ $E x p l a n a t i o n . . . .$
Commented by Nayon last updated on 30/Mar/17

$m r w 1 p l s s o l v e$
Commented by mrW1 last updated on 01/Apr/17

$X \propto A m e a n s X a n d A a r e p r o p o r t i o n a l,$ $i . e . \frac{X}{A} = k_{1} = c o n s t a n t \neq 0$ $b u t X a n d A a r e n o t c o n s t a n t!$ $I n t h e s a m e w a y : \frac{X}{B} = k_{2} = c o n s t a n t \neq 0$ $\frac{X}{A B} = \frac{X}{\frac{X}{k_{1}} \times \frac{X}{k_{2}}} = \frac{k_{1} k_{2}}{X} = \frac{c o n s t a n t}{n o c o n s t a n t} \neq c o n s t a n t$ $\Rightarrow X a n d A B a r e n o t p r o p o r t i o n a l!$ $i . e . X \propto A B i s n o t t r u e!$ $E x a m p l e :$ $A = 2 X$ $B = 3 X$ $\frac{X}{A} = \frac{1}{2} \Rightarrow X \propto A$ $\frac{X}{B} = \frac{1}{3} \Rightarrow X \propto B$ $b u t$ $\frac{X}{A B} = \frac{X}{6 X^{2}} = \frac{1}{6 X} \neq c o n s t a n t$ $\Rightarrow X n o t \propto A B!$
Commented by Nayon last updated on 03/Apr/17

$B u t t h i s f o r m u l a w e u s e i n p h y s i c s$
	Answered by linkelly0615 last updated on 30/Mar/17

	$X \propto A \Rightarrow X = k_{1} A, k_{1} \neq 0$ $i s t h e m e a n i n g ‘ ‘ X \propto A^{″} l i k e t h i s ? ?$
	Commented by linkelly0615 last updated on 30/Mar/17

	$I f m y v i e w i s r i g h t . . .$ $t h e n i t s h o u l d b e w r i t t e n a s ‘ ‘ X^{2} \propto {A B}^{″} .$
	Commented by Nayon last updated on 30/Mar/17

	$n o, i t i s X \propto A B$
	Commented by Nayon last updated on 30/Mar/17

	$G i v e m e t h e E x a t R e a s o n .$
	Answered by linkelly0615 last updated on 30/Mar/17

	$W e l l . . .$ $i f A a n d B a r e c o n s t a n t s$ $t h e n,$ $X \propto A \Rightarrow \exists k_{1} \in c o n s t a n t, X = {A k}_{1}, k_{1} \neq 0$ $X \propto B \Rightarrow \exists k_{2} \in c o n s t a n t, X = {B k}_{2}, k_{2} \neq 0$ $∴ X = {A k}_{1} = {B k}_{2}$ $\Rightarrow \frac{X}{A B} = \frac{k_{1}}{B} = \frac{k_{2}}{A} \in c o n s t a n t$ $\Rightarrow X \propto A B$
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