TITLE: Show that if some nontrivial linear combination of vectors $\vec{u}$ and $\vec{v}$ is $\vec{0}$, then $\vec{u}$ and $\vec{v}$ are parallel.
QUESTION [7 upvotes]: I've never been that great at writing proofs, but I'm getting a bit better. I think I have the answer correct, but I don't know if I'm missing anything. My logic seems right but there may be some minute detail that I'm leaving out. Can anybody give any feedback on this? Thanks.

$\vec{0}$ being a nontrivial linear combination of $\vec{u}$ and $\vec{v}$ implies that there exists a non-zero $a$ or $b$ such that $a\vec{u}=-b\vec{v}$. Without loss of generality, assume $a\neq 0$. Then divide by $a$ and the equality holds: $\vec{u}=-\frac{b}{a}\vec{v}$. And since $-\frac{b}{a}\vec{v}$ is a scalar multiple of $\vec{u}$, it remains that $\vec{u}$ and $\vec{v}$ are parallel.
More rigorous proof:
\begin{align*}
\vec{0}=a\vec{u}+b\vec{v}&\Longrightarrow a\neq 0\vee b\neq 0&&\text{Given}\\
&\Longrightarrow a\vec{u}=-b\vec{v}\\
&\Longrightarrow \vec{u}=-\frac{b}{a}\vec{v}&&\text{WLOG assume $a\neq 0$}\\
&\Longrightarrow \vec{u}\text{ and }\vec{v}\text{ are parallel.}
\end{align*}

REPLY [3 votes]: Almost perfect.
But, we don't have $a\ne 0$ and $b\ne 0$, only $a\ne 0$ or $b\ne 0$. So, one side might be $0$, but then the other is again parallel to it.