Prove that $\mathbb{R}_{\neq0}$ is not a real vector space Assume a set of $\mathbb{R}_{\neq0}=\{a \in \mathbb{R} \mid a \neq 0\}$, where addition of elements in $\mathbb{R}_{\neq0}$ is the product in scalar $ab$. Prove that this is not a real vector space.
I have made the assumption that the scalar product for the elements is the power of it, $a^k$, $a \in \mathbb{R}_{\neq0}$, $k \in \mathbb{R}$.
And then I went through all the definitions for vector space and this vector space fulfills it. I've seen a rule somewhere that says if $ka=kb$, where $a,b\in\mathbb{R}_{\neq0}$ and $k\in\mathbb{R}$, then $a=b$. If I put $k=2$ and $a=1$, $b=-1$ then it doesn't satisfy the rule, but is that rule even correct?
 A: To make $V=\mathbb R_{\ne0}$ a vector space, we need two operations: An addition $\oplus\colon V\times V\to V$, which we are given by the problem statement: $v\oplus w=v\cdot w$; and a scalar multiplication $\odot\colon \mathbb R\times V\to V$, which we are not given. However, we can see that no matter how we try to define $\odot$, we run into trouble. Let us first study $\oplus$ further: $V$ should be an abelian group und $\oplus$ and we easily check that it indeed is. The neutral element is $1$ and hence we need $x\odot 1=1$ for all $x\in\mathbb R$. Apart from that we have $2\odot(-1)=(1+1)\odot(-1)=(1\odot(-1))\oplus(1\odot(-1))=(-1)\oplus(-1)=1$, but then $-1=1\odot(- 1)=(\frac12\cdot 2)\odot(-1)=\frac12\odot(2\odot(-1))=\frac12\odot 1=1$, contradiction.
A: Oh, I didn't see that part of the scalar multiplication. Ok. 
But then we might have some serious problems: what is 
$$\;\frac12\cdot (-1):=(-1)^{1/2}=\sqrt{-1}\;\;\text{within}\;\;\Bbb R^*\;?$$
A: Hint: Here is a proof of the rule you cited. $k\in\mathbb R$ so its inverse $k^{-1}\in\mathbb R$ is also a scalar. Left multiply by $k^{-1}$ on both sides to get $k^{-1}(ka)=k^{-1}(kb)$. By "associativity" we have that $(k^{-1}k)a=(k^{-1}k)b$ and by definition $1a=1b$. Then since $1$ is an "identity" for scalar multiplication, this gives $a=b$.
Since you showed an example for which this rule fails, there should be some step in the proof which is not justified for this space. Which is it?
