If scalar and vector both are non zero then resulting vector is again non zero by scalar multiplication It seems easy but I am stuck a bit. Let V be a vector space over field F. Take x be non zero vector and a,b be different scalars then I want to show that ax and bx are different.
My try: let ax=bx . Multiplying both sides by inverse of a(say a'), we will get 
              (1)x= x = (a'b)x. Now from here I can take things ahead if I show given any nonzero scalar c and non zero vector x we have cx again non zero vector. It seems obvious but I am just stuck to prove this. Thanks.
 A: Let $x\in V$ with $x\ne 0$, and let $a,b\in F$ with $a\ne b$.

Suppose $ax=bx$.
\begin{align*}
\text{Then}\;\;&ax=bx\\[4pt]
\implies\;&ax-bx=0\\[4pt]
\implies\;&(a-b)x=0\\[4pt]
\implies\;&cx=0,\;\text{where}\;c=a-b\\[4pt]
\implies\;&c^{-1}cx=c^{-1}0\qquad\text{[$\,c^{-1}\;$exists since$\;a-b\ne 0\,$]} \\[4pt]
\implies\;&x=0\\[4pt]
\end{align*}
contradiction.

Therefore $ax\ne bx$.
A: There is no reason to involve $bx$ or $c$ in this. Assume $a\cdot x = 0$ (because "scalar multiplication that becomes zero" is the actual thing that this problem is about), and to basically the same argument from there. You get something along the lines of $$0 = a^{-1}\cdot 0 = a^{-1}a\cdot x = 1\cdot x = x$$ which is a contradiction, and you're done.
A: You just have stepped into the correct way. Let me explain further, let $V$ be a vector space over $\Bbb R$ and $x\in V$ be a non-null vector. Let $a,b\in \Bbb R$ be non-zero scalars.
Suppose that $ax=bx\implies x=a^{-1}bx\implies (1-a^{-1}b)x=0\implies 1-a^{-1}b=0\implies a^{-1}b=1$ if not then $(1-a^{-1}b)^{-1}$ exists and so $x=0(null-vector)$ which is not possible so, $a^{-1}b=1\implies a=b$. Hence if $a\ne b$ then $ax\ne bx$ for all non null $x\in V$.
