I have to prove the next proposition and I think I have to do it reducing it to the absurd but I don't know how to do it.
Being $V$ a finite vectorial space and $v_1,v_2\in V$ ($v_1 \neq v_2$). Prove that $\exists \phi\in V^*$ where $\phi(v_1)\neq\phi(v_2)$
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Let $v_1$ and $v_2$ be distinct vectors of $V$ and let $\Omega = (a_1,a_2...,a_n)$ be an orthogonal basis of V. So there are unique vector combinations of both $v_1$ and $v_2$. So define in particular $\phi$ takes vector to his linear combination on $R^n$. You can check as exercise if $\phi$ is well-defined and $1-1$.
Obviously we have to assume that $v_1 \neq v_2$. Let $x_1=v_1-v_2$. Extend $\{x_1\}$ to a basis $\{x_1,x_2,..,x_n\}$ and define $\phi (\sum a_ix_i)=a_1$. Then $\phi \in V^{*}$ and $\phi (x_1)=1$ so $\phi (v_1)=\phi (v_2)+1$.