$x$ with the operations $x+x´=xx´$ vector space? The set of all positive real numbers $x$ with the operations $x+x´=xx´$ and $kx=x^k$. Determine if this set is a vector space.
My book solutions claim this set is a vector space however axiom 4)(u+0=u) seems to fail since $0+x=x+0=x\times 0=0$ 
What am I doing wrong?
 A: Note that in a vector space, "$0$" doesn't usually refer to the number $0$ (in fact, note that the number $0$ isn't even an element of our set).  Rather, a set satisfies axiom 4 if there exists a vector $x_0$ such that $x + x_0 = x$ for all $x$ (and confusingly, it is common to label this vector as $0$).
In the context of the positive real numbers multiplication, we can take $x_0 = 1$, since we find that
$$
x + x_0 = x(1) = x
$$
A: We can check all of the axioms. Let's use $*$ for the operation between vectors.
Associativity: $u*(v*w)=(u*v)*w$
In our case we need $u(vw)=(uv)w$ which is true.
Commutativity: $u*v=v*u$
In our case we need $uv=vu$ which is true.
Identity: $I * u = u* I$
In our case we need $Iu=uI$ and this is true for $I=1$.
Inverse: For every $u$, there is a $v$ with $u*v = I = v*u$
In our case we need $uv=1=vu$ and this is true for $v=\frac{1}{u}$
Comparability: For scalars $\lambda$ and $\mu$ we have $\lambda(\mu u)=(\lambda\mu) u$
In our case we need $\left(u^{\lambda}\right)^{\mu} = u^{\lambda \mu}$, which is true.
Distributivity 1: For scalar $\lambda$ and vectors $u$ and $v$ we need $\lambda(u*v) =\lambda u * \lambda v$
In our case we need $(uv)^\lambda = u^{\lambda}v^{\lambda}$ which is true.
Distributivity 2: For scalars $\lambda$ and $\mu$ and vector $v$ we need $(\lambda+\mu)u = \lambda u * \mu u$
In our case we need $u^{\lambda+\mu} = u^{\lambda}u^{\mu}$, which is true.
