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 ?

Solutions Collecting From Web of "Proof that $\mathbb{R}^+$ is a vector space"

By definition $\alpha\odot(x\oplus y)=(xy)^\alpha$ and $\alpha\odot x\oplus\alpha \odot y=x^\alpha\oplus y^\alpha=x^\alpha y^\alpha$, and those two are the same. I denoted vector space addition and scalar multiplication by $\oplus$ and $\odot$ for distinguishability.

Edit: Though avid19 already answered this, any nonzero vector will provide a one-element basis for $\mathbb{R}^+$, however, in this case the zero vector is $1$, since $1\oplus x=1x=x$.

We can check this by the following. Let $g$ be a nonzero (eg. $\neq 1$) element of $\mathbb{R}^+$, and let $x$ be an arbitrary element of $\mathbb{R}^+$. And also let $\alpha\in\mathbb{R}$ be a scalar. In this case the equation $$ \alpha\odot g=x $$ is written as $$ g^\alpha=x. $$ Taking the $g$-base logarithm of both sides (remember $g$ and $x$ are larger than zero): $$ \alpha=\log_gx, $$ which by the properties of logarithm functions, always exists. Thus given a non $1$ element of $\mathbb{R}^+$, we can always find a scalar, which when multiplied together by vector space scalar multiplication , results in any desired vector, so $\{g\}$ is a generating set.

It is also linearly independent, since there is only one element in it, and it isn’t the zero vector, so $\{g\}$ is a basis.

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 $\pm \log_2(x)$.