Rudin proves $b^{x+y} = b^x b^y$ for $x$, $y \in \mathbb{R}$ and $b>1$. But what about $b\leq 1$?

Question

The title pretty much sums up my question. Rudin only has the exercise for $b>1$, but as I recall it is true that $b^{x+y} = b^x b^y$ for $x$, $y \in \mathbb{R}$ and for each $b\in \mathbb{R}$. So what would the rest of the proof look like? That is for $b\leq 1$?

Answer 1

The proof for $b = 1$ is trivial, since $b^x$ is always $1$.

For the case of $0<b<1$, it suffices to note that $b^x = (b^{-1})^{-x}$ and apply the established result to the base $b^{-1}>1$.

For the case $b \leq 0$, the exponentiation $b^x$ is no longer defined as a real-valued function for arbitrary values $x \in \Bbb R$.