Divisibility (gcd) 3

Printable View

December 16th 2008, 12:36 AM
Sea

Divisibility (gcd) 3

Show that : $b\neq 0$ and $a=bx+cy \Rightarrow (b,c)\leq (a,b)$
December 16th 2008, 05:17 AM
Isomorphism

Quote:

Originally Posted by Sea

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

$(b,c) | a$ (Why?) $\Rightarrow (b,c) | (a,b) \Rightarrow (b,c) \leq (a,b)$
December 16th 2008, 11:03 AM
Sea

I think...

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

$(b,c)|a \Rightarrow(b,c)|a$ and $(b,c)|b\Rightarrow<br /> (b,c)|(a,b)\Rightarrow (b,c)\leq (a,b)$
December 16th 2008, 11:21 AM
Isomorphism

Quote:

Originally Posted by Sea

I think...

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

$(b,c)|a \Rightarrow(b,c)|a$ and $(b,c)|b\Rightarrow<br /> (b,c)|(a,b)\Rightarrow (b,c)\leq (a,b)$

Right (Clapping)

Also $(b,c)|a$ since $a = bx+cy$
December 16th 2008, 11:32 AM
Sea

I can understand...:)...

Thanks a million...

All times are GMT -8. The time now is 05:24 PM.

Copyright © 2005-2013 Math Help Forum. All rights reserved.

Copyright © 2005-2013 Math Help Forum. All rights reserved.