If S = {${v_1,...,v_n}$} is a set of vectors in $R^n$ such that no $v_i$ is a scalar multiple of $v_j$ with $i≠j$, then {${v_1,...,v_n}$} is linearly independent.
So far, I've used the contrapositive and assumed that the set is dependent and wrote it as a linear combination equalling 0. I'm stuck here.
The answer to that particular problem is that it isn't true.However we can find a result in that spirit: if there is no vector in the set that can be expressed as a linear combination of the rest then the set of vectors is independent.
And this in fact can be proved by contrapositive. Mainly what you want to prove is that if a set of vectors is dependent then one vector can be written as a combination of the others.
To prove this note that if the set of vectors is L.D then there is a non-trivial solution to $\alpha_1 v_1+\alpha_2 v_2+\dots\alpha_k v_k =0$ since the solution is non trivial one of the alphas is non-zero, suppose it is $\alpha_k$ . Then
$\alpha_1 v_1+\alpha_2 v_2+\dots\alpha_{k-1} v_{k-1} =-\alpha_k v_{k}$. Since $\alpha_k$ is not zero it has a multiplicative inverse and then $\frac{\alpha_1}{-\alpha_k}v_1+\frac{\alpha_2}{-\alpha_k}v_2+\dots \frac{\alpha_{k-1}}{-\alpha_k}v_{k-1}=v_k$