Vector space with x+y=xy, cx=x^c If $x+y=xy$ and $cx=x^c$ ($c$ is a real number) where $x,y$ posivitive. I want to prove that this can be a vector space for all $x,y>0$. Isn't there a problem with the axiom that states "For every $v \in V$, there exists an element $−v \in V$, called the additive inverse of $v$, such that $v + (−v) = 0$" because by the way that addition is defined $x+y=xy$ for every positive number x theatre element will be zero so that $x+0=x0=0$. Except if the "zero" is actually $1$ because $x+1=x$ and the additive invent will be $1/x$. 
$x+1/x=1$. Can somebody clear things a bit?
 A: Perhaps using sum symbols for other operations can be confusing, so let's not do that.
Consider the set $V=\mathbb{R}^+_0$, i.e. the set of strict positive real numbers, and we now endow it with two operations:


*

*$\square: V\times V\rightarrow V:(x,y)\mapsto x\square y:=xy$

*$\cdot:\mathbb{R}\times V\rightarrow V:(c,x)\mapsto x^c$


We claim that $(\mathbb{R},V,\square,\cdot)$ is a real vector space. We'll check all axioms:


*

*$(V,\square)$ is a commutative group: The neutral element of $\square$ is $1$, indeed, $1\square x=x=x\square 1$ for all $x\in V$. The inverse of $x\in V$ is given by $\frac{1}{x}$. Clearly $\square$ is associative and commutative. This shows that $(V,\square)$ is an abelian group (often the neutral element is denoted by $0$ for such groups, but we'll not do that here as the neutral element is $1$).

*The operation $\cdot$ is a scalar product: $$c\cdot(x\square y):=c\cdot(xy)=(xy)^c=x^cy^c=(x^c)\square (y^c)=(c\cdot x)\square (c\cdot y)$$
$$(c+d)\cdot x= x^{c+d}=x^cx^d=(c\cdot x)\square (d\cdot x)$$
$$(cd)\cdot x=x^{cd}=(x^d)^c=c\cdot (x^d)=c\cdot(d\cdot x)$$
$$1\cdot x=x^1=x$$


Since all of this seems to work, we conclude that $(\mathbb{R},V,\square, \cdot)$ is a real vector space with the given operations.
So why was this confusing to you? It simply has to do with common notational conventions in mathematics. We often write $0$ for the neutral element in an abelian group, especially for the underlying group of a vector space. But even if you consider $\mathbb{R}^2$ as a vector space, $0$ actually denotes the vector $(0,0)$. The example above is more extreme in the sense that the actual neutral element $0$ is given by $1$.
A: In cases like this, it's a good idea to keep track of where each concept comes from. 
For instance, the number $1$ is called that way because of how it behaves with respect to multiplication in $\Bbb R$, while $0$ is named for how it behaves with respect to regular addition.
On the other hand, in any vector space, the vector $\vec0$ is named that way because of how it behaves with respect to the defined vector addition. In this vector space, vector addition is regular multiplication, so $\vec0$ is $1$.
If you, for each operation or element named for some property it has like $1$ or $0$, are very clear on which structure it's part of (for instance by writing $\vec 0$ instead of $0$ for the zero vector, writing $\vec+$ instead of $+$ for the vector addition, $\vec\cdot$ for the scalar multiplication and $\vec -$ for vector inversion), it's a lot easier to see what's going on.
