Show that if $r\overrightarrow {X}=\overrightarrow {0}$, then either $r=0$ or $\overrightarrow {X}=\overrightarrow{0}$ 
Suppose $\overrightarrow{X}\in{\bf R}^2$ and $r\in{\bf R}$. Show that if $r\overrightarrow {X}=\overrightarrow{0}$, then either $r=0$ or $\overrightarrow {X}=\overrightarrow{0}$.


[Attempt:] 
Let $\overrightarrow {X}=\left( \begin{matrix} a\\ b\end{matrix} \right)$. Assume $r\neq 0$. Then,
$r\overrightarrow {X}=\overrightarrow {0}$,
$\overrightarrow {X}=\dfrac {1} {r }\overrightarrow {0}$,
$\left( \begin{matrix} a\\ b\end{matrix} \right)=\dfrac {1} {r }\left( \begin{matrix} 0\\ 0\end{matrix} \right)$,
$\left( \begin{matrix} a\\ b\end{matrix} \right)=\left( \begin{matrix} \dfrac {1} {r }0\\ \dfrac {1} {r }0\end{matrix} \right)$,
$\left( \begin{matrix} a\\ b\end{matrix} \right)=\left( \begin{matrix} 0\\ 0\end{matrix} \right)$.
Thus, we have $a=0$ and $b=0$, that is, $\overrightarrow {X}=\overrightarrow {0}$.
Now, assume $\overrightarrow {X}\neq\overrightarrow {0}$. We will show that $r=0$. Then,
$r\overrightarrow {X}=\overrightarrow {0}$,
$\left( \begin{matrix} r a\\ r b\end{matrix} \right)=\left( \begin{matrix} 0\\ 0\end{matrix} \right)$. Since $a\neq 0$ and $b\neq 0$, we obtain $r=0$.
So, we are done.
Can you check my proof?
$$
\newcommand{\Vec}[1]{\overrightarrow{#1}}
$$
 A: This holds in abstracto for any vector space $\mathbf V$ over any field $\Bbb F$, straight from the axioms.  For if 
$r \vec X = 0 \tag 1$
for $0 \ne r \in \Bbb F$ and $\vec X \in \mathbf V$, then $\exists r^{-1} \in \Bbb F$ and so
$\vec X = 1_{\Bbb F} \vec X = (r^{-1}r)\vec X = r^{-1}(r \vec X) = r^{-1} (0) = 0; \tag 2$
the component representation of $\vec X$ is not needed here.
P S.  The OP's proof liiks fine, except one only needs $(a = 0) \vee (b = 0)$ near the end, not $(a = 0) \wedge (b = 0)$.
A: This is an axiom for vector spaces in general, regardless of vector space dimension.
But if you must prove it, I think there's an easier way, by working in reverse:
Consider both cases.
Let $r=0$

Then $r\vec X=0\vec X=(0+0)\vec X=0\vec X+0\vec X=\color{red}{\vec 0+\vec 0=\vec 0}$

Let $\vec X=\vec 0$

Then $r\vec X=r\vec 0=r(\vec 0+\vec 0)=r\vec 0+r\vec 0=\color{red}{\vec 0+\vec 0=\vec 0}$

The $\color{red}{red}$ is where I used the fact given in the original expression
A: Your proof is correct. As long as you have shown that $\vec{X}=\vec{0}$ assuming $r\neq 0$, all the rest of your work are redundant. 
