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?
$$
\newcommand{\Vec}[1]{\overrightarrow{#1}}
$$

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. 
