General linear independence proof Q: Suppose $\{\vec v_1, \vec v_2, ..., \vec v_k\}$ is linearly independent and $\vec w \notin \left<\vec v_1, \vec v_2, ..., \vec v_k\right>$. 
Proving that $\{\vec v_1, \vec v_2, ..., \vec v_k, \vec w\}$ is linearly independent.
I have chosen to prove by contradiction (maybe, I'm stuck right now so maybe this wasn't the right choice).
So suppose $\{\vec v_1, \vec v_2, ..., \vec v_k, \vec w\}$ is linearly dependent. This means there are scalars such that:
$\{\,c_1\vec v_1+c_2\vec v_2+...+c_k\vec v_k + c_l\vec w_l=\vec 0\mid c \in R\,\}$
But this would mean $\vec w$ is in our span. I think I am missing some steps and may not even be coming to the correct conclusion. 
EDIT: so I can state that $c_l \neq 0$ bc otherwise $ \vec w$ would be the zero vector and any space with the zero vector in the span cannot be linearly independent? 
Beyond this, I've been trying a calculation to prove a vector is in a span. 
Q: In $R^5$ let $S=\left<(1,0,1,0,1), (2,1,0,1,1), (4,1,2,1,3), (0,0,1,1,1)\right>$. I need to determine if $(1,2,3,4,5)$ and $(2,1,1,1,1)$ are in $S$. 
I made an augmented matrix with $(1,2,3,4,5)$ but ended up with three rows giving me difference answers as to what $x_4$ would be? Am I messing up my math?
 A: What you know is that there are scalars $c_1,\dots,c_k,c_ld$ such that $c_1\vec{v}_1+\cdots+c_k\vec{v}_k+c_l\vec{w}=\vec{0}$ and not all of the scalars are zero. You have to consider the cases $c_l=0$ and $c_l\neq 0$ separately. In particular, if $c_l=0$, you cannot rearrange the equation to show that $\vec{w}$ is in the span of $\vec{v}_1,\dots,\vec{v}_k$. But in that case, you can say something about the linear (in)dependence of $\vec{v}_1,\dots,\vec{v}_k$.
A: Since you write $R$ for your ring of scalar, I am assuming that you are working in more general context than a field, for example including $\mathbb{Z}$. In that case the claim is wrong, for example, $1\notin\langle 2\rangle$ but $\{2,1\}$ is not linearly independent.
However, if you are indeed assuming a field. Then what you need to do next is as follow:
First $c_{l}\not=0$ since otherwise you will have a nontrivial linear combination not involving $w$, contradicting a hypothesis that the rest are linearly independent.
Then since you are in a field, $c_{l}$ is invertible, so by dividing by $c_{l}$ and move $w$ to the other side, you can write $w$ as a linear combination of the rest, also violating the span condition.
