# Trying to show that $\sup \{b^p:p\in Q,\;0<p<x\} = \inf\{b^q:q\in Q,\;x<q\}$

I’m trying to show that for $b>1$, $x>0$ and $x$ irrational, that $$\sup \{b^p:p\in Q,\;0<p<x\} = \inf\{b^q:q\in Q,\;x<q\}$$ I know this follows immediately if we define $b^x=\exp(x\log(b))$ and use the fact that the exponential function is strictly increasing and continuous over $\mathbb R$, but I’m not allowed to do that. In fact, I only have that $b^{m/n}=(b^{1/n})^m$, where $b^{1/n}$ means the positive $n^{th}$ root of $b$; $b^x$ is not yet defined for irrational $x$.

I know that the supremum and infimum both exist because if $b>1$ and $p<q$, then $b^p<b^q$. Also, by assuming that the inf $<$ sup, I got a contradiction. The part I’m stuck on is deriving a contradiction from assuming that sup $<$ inf. Does anybody have any ideas?

Thank you.