In Rudin's Principles of Mathematical Analysis, 3e, problem 6a on page 22, the following problem is stated:
Fix $b>1$. If $m,n,p,q$ are integers, $n>0, q>0$, and $r=m/n=p/q$, prove that $(b^m)^{1/n}=(b^p)^{1/q}$. Hence it makes sense to define $b^r=(b^m)^{1/n}$.
What puzzles me is how does proving that $(b^m)^{1/n}=(b^p)^{1/q}$ imply that $b^r=(b^m)^{1/n}$ is well-defined? I don't quite see the connection.