Results 1 to 3 of 3

Math Help - Set

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    40

    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???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by bhitroofen01 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2010
    Posts
    40
    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???
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum