1. ## Set

Hi everybody,

E={(x;y;z)$\displaystyle \in \mathbb{R}^3$/ x=y=2z}

I must show that $\displaystyle \forall \alpha \in \mathbb{R}\$and $\displaystyle \forall$ (x;y;z) and (x';y';z') from E: (x;y;z)+ (x';y';z') $\displaystyle \in$ E and $\displaystyle \alpha$.(x;y;z)$\displaystyle \in$ E.

2. Originally Posted by bhitroofen01
Hi everybody,

E={(x;y;z)$\displaystyle \in \mathbb{R}^3$/ x=y=2z}

I must show that $\displaystyle \forall \alpha \in \mathbb{R}\$and $\displaystyle \forall$ (x;y;z) and (x';y';z') from E: (x;y;z)+ (x';y';z') $\displaystyle \in$ E and $\displaystyle \alpha$.(x;y;z)$\displaystyle \in$ E.

Take some $\displaystyle x = (x_1,x_2,x_3), ~ y = (y_1,y_2,y_3) \in E, ~ \alpha \in \mathbb{R}$, then you get that:
$\displaystyle x = (x_1,x_1,2x_1), ~ y = (y_1,y_1,2y_1) \Rightarrow x+y = (x_1+y_1, x_1+y_1,2(x_1+y_1))$. Now let $\displaystyle t = x_1+y_1 \Rightarrow x+y = (t,t,2t) \Rightarrow x+y \in E$
$\displaystyle \alpha x = \alpha(x_1,x_1,2x_1) = (\alpha x_1, \alpha x_1, 2\alpha x_1)$. Now let $\displaystyle k=\alpha x_1 \Rightarrow \alpha x = (k,k,2k) \Rightarrow \alpha x \in E$