Which Axiom does this definition fail? 
Consider the set of vectors $\renewcommand{\vec}[1]{\mathbf{#1}}\vec{u}=(x,y,z)$ in $\mathbb{R}^3$ with standard addition, but scalar multiplication defined by $r\vec{u}=(r^2x,r^2y,r^2z)$. This is not a vector space. Determine which vector space axiom fails and exhibit a counterexample to demonstrate the failure of that axiom.

I have attempted this and have come to believe it is the $r(s\vec{v}) = (rs)\vec{v}$ axiom that it fails.
I am new to this subject and am hoping for some guidance/confirmation. 
 A: (r+s)u = ${(r+s)^2x,(r+s)^2y,(r+s)^2z)}$ = {$(r^2+2rs+s^2)x,(r^2+2rs+s^2)y,(r^2+2rs+s^2)z$} $\ne$ ru+su = $(r^2x,r^2y,r^2z)+(s^2x,s^2y,s^2z)$
Therefor,
(r+s)u $\ne$ ru+su
A: Your initial question has been answered already, so let me explain how to think about this problem—at least, how I thought about the problem.
You can see that $r\pmb{v}$ is never going to be a negative multiple of $\pmb{v}$, but we know that for vector spaces, $(-1) \cdot \pmb{v} = - \pmb{v}$. So now you need to either look up or try your best to remember the proof that $(-1) \cdot \pmb{v} = - \pmb{v}$ to see what the key axioms that are used in that proof.
Because you have the standard addition on $\mathbf{R}^3$, all the axioms that refer just to the vector addition and not the scalar multiplication will always be satisfied. So what you are left with is:


*

*$r(s\pmb{v}) = (rs)\pmb v$

*$(r + s)\pmb v = r \pmb v + s \pmb v$

*$r(\pmb v + \pmb w) = r\pmb v + r \pmb w$

*$1 \pmb v = \pmb v$
The essential property of the number $(-1)$ is the equation $1 + (-1) = 0$ which involves both $1$ and $+$ (scalar addition). Of the 4 axioms, the fourth mentions $1$ and you can check that easily, and the second uses the scalar addition. So because the essential equation $1 + (-1) = 0$ uses scalar addition, that tells us to look at the second axiom.
