Show that Axioms 7, 8, and 9 hold. 

I'm having trouble seeing how axiom 7 holds since ku makes the first element a 0 but not kv.. also I'm not sure what the m is in axiom 8 and 9..
 A: Do the subtitutions mechanically:
$$\textbf{u}+\textbf{v}=(u_1+v_1,u_2+v_2),\\\textbf{v}+\textbf{u}=(v_1+u_1,v_2+u_2)$$
$$\textbf{u}+(\textbf{v}+\textbf{w})=\textbf{u}+(v_1+w_1,v_2+w_2)=(u_1+(v_1+w_1),u_1+(v_2+w_2)),\\(\textbf{u}+\textbf{v})+\textbf{w}=(u_1+v_1,u_2+v_2)+\textbf{w}=((u_1+v_1)+w_1,(u_1+v_2)+w_2))$$
$$\textbf{0}+\textbf{u}=(0+u_1,0+u_2),\\\textbf{u}+\textbf{0}=(u_1+0,u_2+0),\\\textbf{u}=(u_1,u_2).$$
$$\textbf{u}+(-\textbf{u})=(u_1+(-u_1),u_2+(-u_2)),\\(-\textbf{u})+\textbf{u}=((-u_1)+u_1,(-u_2)+u_2),\\\textbf{0}=(0,0).$$
$$k(\textbf{u}+\textbf{v})=k(u_1+v_1,u_2+v_2)=(0,k(u_2+v_2)),\\ k\textbf{u}+k\textbf{v}=(0,kv_1)+(0,kv_2)=(0,kv_1+kv_2).$$
$$(k+m)\textbf{u}=(0,(k+m)u_1),\\k\textbf{u}+m\textbf{u}=(0,ku_1)+(0,mu_1)=(0,ku_1+mu_1).$$
$$k(m\textbf{u})=k(0,mu_1)=(0,k(mu_1)),\\(km)\textbf{u}=(0,(km)u_1).$$
$$1\textbf{u}=(\color{red}0,u_2),\\\textbf{u}=(\color{red}{u_1},u_2).$$
A: The definition of scalar multiplication given in the question is for a general scalar $k$ and vector $\vec u$. And in order to show that the axioms hold, you must use general values, not the specific values given in the question. So $k\vec u = (0, ku_2)$ and $k\vec v = (0, kv_2)$. Then according to the addition defined, $\vec u + \vec v = (u_1 + v_1, u_2 + v_2 \Rightarrow k(\vec u + \vec v) = (0, k(u_2 + v_2))$, and $k\vec u + k\vec v = (0, ku_2) + (0, kv_2) = (0, k(u_2 + v_2)) = k(\vec u + \vec v)$.
For axioms $8$ and $9$, use another scalar $m$.
