Show that : $\displaystyle b\neq 0$ and $\displaystyle a=bx+cy \Rightarrow (b,c)\leq (a,b)$
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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?) $\displaystyle \Rightarrow (b,c) | (a,b) \Rightarrow (b,c) \leq (a,b)$
I think... $\displaystyle (a,b)\geq 0 , (b,c)\geq 0$ $\displaystyle (b,c)|a \Rightarrow(b,c)|a $ and $\displaystyle (b,c)|b\Rightarrow (b,c)|(a,b)\Rightarrow (b,c)\leq (a,b)$
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 Also $\displaystyle (b,c)|a$ since $\displaystyle a = bx+cy$
I can understand...... Thanks a million...
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