Show that $ \forall x \neq 0 \in V$ (where V is a vector space) , if $a \neq b$, then $a*x \neq b*x $? This must be proven using only the 8 basic axioms that make up a vector space. So far, I've tried to create the argument: 
Let $x = (x_1,...,x_n)$ and $a,b \in \Bbb R$
then $a*x = a (x_1,...,x_n)$
=>     $a*x = (ax_1,....,ax_n)$
Note that $ \exists -x $ such that $a*x + a*(-x) = 0 $
This additive inverse is unqiue.
This is as far as I've gotten, I'm not really sure how to connect the above to proving the required statement. Any help is appreciated!
 A: In what follows, I've tried to cite the axioms as I go along; I think I got it mostly right, but may have missed in a spot or two; but you should be able to get the idea . . .
Not sure exactly what eight axioms you use; I take my eight from this wikipedia entry, thus:
If
$a \ne b, \tag 1$
then
$a -b \ne 0, \tag 2$
so $(a - b)^{-1}$ exists in $\Bbb R$; then if
$ax = bx, \tag 3$
we have, by distributivity of scalar multiplication with respect to field addition and the vector additive inverse axiom:
$(a - b)x = ax - bx = ax + (-bx) = ax + (-ax) = 0; \tag 4$
whence
$x = 1x = ((a - b)^{-1}(a - b))x = (a - b)^{-1}((a - b)x) = (a - b)^{-1}(0), \tag 5$
where we have used the axioms identity of scalar multiplication, compatibility of scalar multiplication, and our equation (4).  To finish this off, we have
$(a - b)^{-1}(0) = (a - b)^{-1}(0 + 0) = (a - b)^{-1}(0) + (a - b)^{-1}(0), \tag 6$
by the axioms of identity element of vector addition (0 + 0 = 0!) and distributivity of of scalar multiplication with respect to vector addition.  From (6), using the vector additive inverse axiom (e.g., to add $-(a - b)^{-1}x$ to each side), and vector associativity (to go from $(v + v) + (-v)$ to $v + (v + (-v))$), and once again the vector identity axiom,
$(a - b)^{-1}(0) = 0, \tag 7$
so combining (7) with (5) we obtain
$x = 0, \tag 8$
a contradiction.  So 
$ax \ne bx, \tag 9$
which was required to be proved.
NB: Reviewing the above, I realize how one really has to pick nits to get all the axioms cited where used.  Proofs like this should really be set up in Principia Mathematica format.  End of Note.
A: Let $x\neq \vec{0}$ and $a\neq b$. Then: $ax\neq bx$
By reduction to the absurd: Let us suppose $ax=bx\rightarrow ax+(-bx)=bx+(-bx)\rightarrow$
$$ax+(-1)bx=\vec{0}\rightarrow ax+(-b)x=\vec{0}\rightarrow [a+(-b)]x=\vec{0}\xrightarrow{\text{It could be proved}} \left\{ \begin{array}{lcc}
             a+(-b)=0 \\
             \\ \vee \\
             \\ x=\vec{0}
             \end{array}
   \right.\rightarrow \left\{ \begin{array}{lcc}
             a=b \\
             \\ \vee \\
             \\ x=\vec{0}
             \end{array}
   \right.$$
(Contradiction) $\rightarrow ax\neq bx$
