# Divisibility (gcd) 3

• Dec 16th 2008, 12:36 AM
Sea
Divisibility (gcd) 3
Show that : $\displaystyle b\neq 0$ and $\displaystyle a=bx+cy \Rightarrow (b,c)\leq (a,b)$
• Dec 16th 2008, 05:17 AM
Isomorphism
Quote:

Originally Posted by Sea
Show that : $\displaystyle b\neq 0$ and $\displaystyle a=bx+cy \Rightarrow (b,c)\leq (a,b)$

$\displaystyle (b,c) | a$ (Why?) $\displaystyle \Rightarrow (b,c) | (a,b) \Rightarrow (b,c) \leq (a,b)$
• Dec 16th 2008, 11:03 AM
Sea
I think...

$\displaystyle (a,b)\geq 0 , (b,c)\geq 0$

$\displaystyle (b,c)|a \Rightarrow(b,c)|a$ and $\displaystyle (b,c)|b\Rightarrow (b,c)|(a,b)\Rightarrow (b,c)\leq (a,b)$
• Dec 16th 2008, 11:21 AM
Isomorphism
Quote:

Originally Posted by Sea
I think...

$\displaystyle (a,b)\geq 0 , (b,c)\geq 0$

$\displaystyle (b,c)|a \Rightarrow(b,c)|a$ and $\displaystyle (b,c)|b\Rightarrow (b,c)|(a,b)\Rightarrow (b,c)\leq (a,b)$

Right (Clapping)

Also $\displaystyle (b,c)|a$ since $\displaystyle a = bx+cy$
• Dec 16th 2008, 11:32 AM
Sea
I can understand...:)...

Thanks a million...