I asked this question earlier and got a very useful hint, but I still can't get the full solution. Here is the original question:
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?
The hint I got, courtesy of Greg Martin:
Try this one: let $S$ be the sup and $I$ be the inf. If $S<I$, then set $\epsilon =I−S$. Choose $p<x$ rational such that $b^p > S-\frac\epsilon{100}$ (say) and $q>x$ rational such that $b^q<I+ \frac \epsilon{100}$. Then with $r=\frac12(p+q)$, use some inequalities to show that $b^r>S$ and $b^r<I$, contradiction.
This was my reply:
I have the inequality $S-\frac\epsilon{100}<b^p<b^r<b^q<I+\frac\epsilon{100}$, and I'm trying to show that for all $n$, $S-\frac\epsilon{n}<b^r<I+\frac\epsilon{n}$. This would show that $S<b^r<I$. But $p$ and $q$ depend on $n$, and $r$, being their midpoint, depends on $p$ and $q$, and hence $n$. How do I proceed from here? That is, how do I show that $r$ doesn't change when $n$ changes?
Can anybody help? Thank you.
The link to the question: Trying to show that $\sup \{b^p:p\in Q,\;0<p<x\} = \inf\{b^q:q\in Q,\;x<q\}$
We should not choose $n=100$ but we will choose $n=\lceil{I\over S}\rceil\geq {I\over S}$.
$${b^{{1\over2}(p+q)}<S^{1\over2}\cdot(I+{\epsilon\over n})^{1\over2}}<(I-{\epsilon\over n})^{1\over2}\cdot(I+{\epsilon\over n})^{1\over2}<(I^2)^{1\over2}=I$$
$${b^{{1\over2}(p+q)}>(S-{\epsilon\over n})^{1\over2}\cdot I^{1\over2}}=(S-{\epsilon\over n})^{1\over2}\cdot (S+{\epsilon})^{1\over2}=(S^2+{n-1\over n}\epsilon S-{1\over n}\epsilon^2)^{1\over2}\geq (S^2)^{1\over2}=S$$
The last inequality comes from
$${n-1\over n}\epsilon S-{1\over n}\epsilon^2={\epsilon\over n}((n-1)S-\epsilon)={\epsilon\over n}(nS-S-I+S)={\epsilon\over n}(nS-I)\geq 0$$