Why a(pt=proportional to) pt b and a pt c implies a pt bc?

[Rishabh Narula]

hmm.very thankful to everyone who answered but i think where I am actually getting stuck is nothing high level.im just not getting why a proportional to b and a proportional to c implies a proportional to bc .i get it that if b changes by factor x then a changes by factor x and at the same time if c changes by y then that a.x would change by y resulting a net change a into a.x.y and also i mean yeah when you see a proprtional to bc it is evident that b,c getting changed by x,y makes a change by xy.but i want some more derived proof or something i guess.

 

[blue_leaf77]

$a$ proportional to $b$ means $a=K_1b$ and $a$ proportional to $c$ means $a=K_2c$. Combining the last two equations, you get $K_1b=K_2c$ or $\frac{K_1}{K_2}=\frac{c}{b}$ valid for all $b$ and $c$. This equation is satisfied by $K_1=K_3c$ and $K_2=K_3b$ with $K_3$ being non-zero. Therefore $a=K_3bc$.

 

[Rishabh Narula]

$a$ proportional to $b$ means $a=K_1b$ and $a$ proportional to $c$ means $a=K_2c$. Combining the last two equations, you get $K_1b=K_2c$ or $\frac{K_1}{K_2}=\frac{c}{b}$ valid for all $b$ and $c$. This equation is satisfied by $K_1=K_3c$ and $K_2=K_3b$ with $K_3$ being non-zero. Therefore $a=K_3bc$.

$a$ proportional to $b$ means $a=K_1b$ and $a$ proportional to $c$ means $a=K_2c$. Combining the last two equations, you get $K_1b=K_2c$ or $\frac{K_1}{K_2}=\frac{c}{b}$ valid for all $b$ and $c$. This equation is satisfied by $K_1=K_3c$ and $K_2=K_3b$ with $K_3$ being non-zero. Therefore $a=K_3bc$.

this helped a lot.thank you for your time.