# Proving that $c(\vec x +' \vec y) = c \vec x + c \vec y$

I'm trying to prove an axiom of a vector space - namely, distributivity of scalar multiplication over addition. In the answer section of the book, it says that the property holds. When I do it, it doesn't seem that it does. This is what I have:

The Property

In order for a set to be a vector space it has to satisfy,

$$\forall \vec x, \vec y \in V: \forall c \in \Bbb R,\ c(\vec x + \vec y) = c \vec x + c \vec y$$

For my problem, vector addition has been defined as, $$(x_1,x_2)\ +' \ (y_1, y_2) = \ (x_1 + 2y_1, 3x_2 - y_2)$$

What I Have

Let $c$ be a scalar s.t. $c \in \Bbb R$. Then, $$c(\vec x \ +' \vec y) = c((x_1, x_2) \ +' \ (y_1, y_2)) = c(x_1 +2y_1, 3x_2 - y_2) = (c(x_1 + 2y_1) \ + \ c(3x_2 - y_2)) \\ = (cx_1 + c2y_1, c3x_2 - cy_2) = \ (cx_1, c3x_2) \ +' \ (c2y_1, cy_2) = c(x_1, 3x_2) \ +' \ c(2y_1, y_2)$$

Which is not equal to $c \vec x + c \vec y$. So it seems the property doesn't hold for this definition of vector addition. What am I doing wrong? Thanks in advance.

• @Romaion yes, the scalars $2$ and $3$ should disappear. Try simplifying $c(\vec{x} \ +' \ \vec{y})$ and $c\vec{x} \ +' \ c\vec{y}$ from both ends and see what you get. Sep 17, 2017 at 16:37
$\newcommand{\vect}{{\bf #1}}$