The set of all positive real numbers, with addition defined by $x + y = x * y$ and scalar multiplication defined by $c * x = x ^ c$. How is this a vector space? Doesn't the axiom stating that $(c+d)u = cu + du$ fail?
example: $u = 2$ (since two is a positive real number, it is in the set). $c = 3$ and $d = 4$ ( these two are just scalars).
$(c+d)u = (3+4)(2) = 2^7$
$cu + du = 2^3 + 2^4$, which does not equal $2^7$, so the axiom fails.
How is this a vector space if an axiom fails? Or does this axiom somehow not fail? Any help is appreciated.
Best Answer
It definitely is a vector space. You seem to be getting confused about the two $+$ and two $*$ operations, as well as the fact that vectors and scalars have some overlap. It would be better to give the operations their own names:
\begin{align*} u \oplus v &= uv \\ \lambda \odot u &= u^\lambda. \end{align*}
The distributivity law that you're trying to verify is as follows:
$$(\lambda + \mu) \odot u = (\lambda \odot u) \oplus (\mu \odot u).$$
Please note the scalar $+$ and the vector $\oplus$, and where they belong. Don't forget that $\lambda$ and $\mu$ are scalars, and so they must be added by regular addition on $\mathbb{R}$.
When simplifying this rule, we get,
$$u^{\lambda + \mu} = u^\lambda \cdot u^\mu,$$
which is a well-known exponential law.