Let $V$ be a vector space and $\dim(V) ≥ 2$. How to prove with that information that there exists three points that are not on the same line?
I understand how to normally prove that in a vector space there exists three points that are not on the same line, but what does the $\dim(V)$ have to do with this?
The comment by alxchen already gives a short, very good proof. I am offering an alternative. I shall prove that, for any vector space $V$ over a field $K$, any three elements of $V$ are always collinear if and only if $$\dim_K(V)\in \{0,1\}\,.$$
The converse is trivial, so I am proving the direct implication. Suppose that $V$ is a $K$-vector space in which three arebitrary elements of $V$ are on the same line. If every element of $V$ is the zero vector, then $\dim_K(V)=0$. Now, suppose that $V$ contains a nonzero vector $u$, i.e., $\dim_K(V)=1$. Now, let $v\in V$ be arbitrary. Since $0$, $u$, and $v$ are on the same line and $u\neq 0$, there exists $\lambda\in K$ such that $$v-0=\lambda(u-0)\,.$$ That is, $v=\lambda u$. Ergo, every element of $V$ is a scalar times $u$, whence $V$ is spanned by a nonzero element $u$. This shows that $\dim_K(V)=1$.