# Math Help - Set

1. ## Set

Hi everybody,

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

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

Can you help me please???

2. Originally Posted by bhitroofen01
Hi everybody,

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

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

Can you help me please???
Take some $x = (x_1,x_2,x_3), ~ y = (y_1,y_2,y_3) \in E, ~ \alpha \in \mathbb{R}$, then you get that:

$x = (x_1,x_1,2x_1), ~ y = (y_1,y_1,2y_1) \Rightarrow x+y = (x_1+y_1, x_1+y_1,2(x_1+y_1))$. Now let $t = x_1+y_1 \Rightarrow x+y = (t,t,2t) \Rightarrow x+y \in E$

$\alpha x = \alpha(x_1,x_1,2x_1) = (\alpha x_1, \alpha x_1, 2\alpha x_1)$. Now let $k=\alpha x_1 \Rightarrow \alpha x = (k,k,2k) \Rightarrow \alpha x \in E$

3. OK.
And how to show that (E;+;.) is a vector space, it's clear that (E;+) is an abelian group but i don't know how to continue???