Suppose $\{v_1,v_2,v_3\}$ $\subseteq$ $\mathbb{R}^{10}$ is a set of linearly independent vectors, and $w\in\mathbb{R}^{10}$ is not in span $\{v_1,v_2,v_3\}$. Prove $\{v_1,v_2,v_3,w\}$ is linearly independent.
Proof by Contradiction:
Let $U=\{v_1,v_2,v_3,w\}$
Assume $U$ is linearly dependent.
Then if $x_1v_1+x_2v_2+x_3v_3+x_4w=0$ , $x_1,x_2,x_3,x_4\in \mathbb{R}$ has a solution, then at leat one of the $x$'s must not equal to zero. Without loss of generality, assume $x_4\neq0$.
Then,
$w= -(\frac{x_1}{x_4}v_1+\frac{x_2}{x_4}v_2+\frac{x_3}{x_4}v_3)$
Therefore, $w$ is a linear combination of $\{v_1,v_2,v_3\}$. Then $w$ is in span $\{v_1,v_2,v_3\}$. This is a contradiction to the given statement.
Hence, $U=\{v_1,v_2,v_3,w\}$ is linearly independent.
Does this make sense?
Yes it is correct, the only observation is that we are allowed assume $x_4\neq 0$ since $v_i$ are linearly independent and then at least two of the x coefficient ($x_4$ and one other) must be $\neq 0$.