Definition of $b^x$ and its properties. (Walter Rudin's "Principles of Mathematical Analysis")

Question

I am reading Walter Rudin's "Principles of Mathematical Analysis".

There is the following theorem in this book:

p.57
Theorem 3.20(a)
If $p > 0$, then $\lim_{n\to\infty} \frac{1}{n^p}=0$.

Take $n > (\frac{1}{\epsilon})^{\frac{1}{p}}$. Then $n^p > ((\frac{1}{\epsilon})^{\frac{1}{p}})^p = \frac{1}{\epsilon}$. So $\epsilon > \frac{1}{n^p}$. To prove $((\frac{1}{\epsilon})^{\frac{1}{p}})^p = \frac{1}{\epsilon}$, I think we need the property $(a^x)^y = a^{x y}$. And Rudin didn't write this property on p.22 ex6.

On p.22 Exercise 6, Rudin defined $b^x$ for $b \in \{y \in \mathbb{R} | y > 1\}, x\in\mathbb{R}$.
And the reader proves that $b^{x+y} = b^x b^y$ for all $x, y \in \mathbb{R}$.

But Rudin didn't define $b^x$ for $b \in \{y \in \mathbb{R} | 0 < y \leq 1\}, x\in\mathbb{R}$.

And Rudin didn't write other properties of $b^x$.
For example, Rudin didn't write $(b^x)^y = b^{xy}$ for all $x, y \in \mathbb{R}$.

I am disappointed and sad.
Walter Rudin's "Principles of Mathematical Analysis" isn't perfect.

I can guess $b^x$ is defined as $(\frac{1}{b})^{-x}$ for $b \in \{y \in \mathbb{R} | 0 < y \leq 1\}, x\in\mathbb{R}$.

Isn't Walter Rudin's "Principles of Mathematical Analysis" self-contained?

Is there a self-contained analysis book in the world?

Answer 1

You can easily extend the definition of $b^x$ for $0 < b \leq 1$ by defining $$b^x := \frac{1}{(1/b)^x}$$ and show the property $$b^{x+y} = b^x b^y$$ still holds.