Proof that $\mathbb{R}^+$ is a vector space I was doing some beginner linear algebra tasks and stumbled upon this one:
Proove that $\mathbb{R}^+$ is a vector space over field $\mathbb{R}$ with binary operations defined as $a+b = ab$ (where $ab$ is multiplication in $\mathbb{R}$ and $\alpha *b =b^\alpha$, where $b \in \mathbb{R}$ and $\alpha \in \mathbb{R} $.
It's easy to prove that $(\mathbb{R}^+,+)$ is an Abelian group and i will leave that part of proof out. However, when proving the following property of vector spaces, there seems to be a problem: 
$\alpha (x+y) = \alpha x +  \alpha y$ ( where $x,y \in \mathbb{R}^+$ and $\alpha \in \mathbb{R}$)
By definition:
$$\alpha (x+y) =(x+y)^\alpha $$
and
$$\alpha x + \alpha y = x^\alpha + y ^\alpha $$
In general case: 
$(x+y)^\alpha \ne x^\alpha + y^\alpha$
so this appears not to be a vector space, but even the solution in textbook states it is ( this property proof is completely omitted). Could this be author's error or did I make a mistake? 
If the mistake is mine, I would like to ask and additional question ,which should probably be posted in a separate thread: How would i find one base of this vector space. By defintion, I need to find a positive real number who's linear combination would generate all  positive real numbers. This is quite simple but should i use this vector space operations to form linear combinations or general multiplication and addition i.e. would a linear combination of $a \in \mathbb{R}^+$ be $b=5a$ or would that be $b=a^5$. If it's the latter, is it safe to assume that any positive number other than 1 is a base vector ?

Remember what addition means! Note:
$$\alpha(x+y)=x^{\alpha}+y^{\alpha}=x^{\alpha}y^{\alpha}=\alpha x+\alpha y$$
As for a basis, you can take any (non-zero) vector (also, what is $\vec{0}$ in this space? Hint: it's not $0$). For example, $2$ can be a basis vector. Every nonzero number $x$ can be reached by $\pm \log_2(x)$.
