Prove that $b^{r+s} = b^{r}b^{s}$ if r and s are rational. Also $b > 1, b \in \mathbb{R}$
My question has to do with a specific aspect of this proof. I'm taking the question from Principles of Mathematical Analysis by Rudin.
So say we have defined $r = \frac{p}{q}$ and $s = \frac{m}{n}$, where $p, q, m, n \in \mathbb{Z}$
To prove this statement I will end up showing that both sides are equivalent to a particular expression. That expression being: $$b^{np}b^{mq}$$
In proving from the side where $(b^{r+s})^{nq}$, I arrive at the following expression: $$b^{pn + mq}$$
In all of the solutions that I have seen, they make the jump from $$b^{pn + mq} = b^{pn} b^{mq}$$
Of course this is how the properties of exponents behave, but it is this exact property that I am trying to prove, so how is making that jump to the final step even possible?
It is not the exact property you are trying to prove. In the latter the exponents are integers; and it can be proven easily by induction (or almost by definition) if you prefer, as it is just counting $b$'s. That is, $b^{m+n}$ means to multiply $m+n$ instances of $b$; while $b^mb^n$ means to multiply $m$ instances of $b$ with $n$ instances of $b$ for a total of $m+n$ $b$'s.