Here's the problem
Fix b>1. If $x$ is real, define $B(x)$ to be the set of all numbers $b^t$ where $t$ is rational and $t\leq x$. Prove that $b^r=\sup B(r)$ when $r$ is rational.
So I basically know exactly what I need to do. We know that $b^r\in B(r)$ so all that's left is to show that for any $t\leq r$, $b^t\leq b^r$. Seems simple enough but I just can't figure out how to prove this without begging all kinds of questions. All I need is a lemma that says for rationals $r,r'$, and a real number $b>1$, $b^r implies $r
By best attempt so far has been this: suppose $b^r. We can write $b^{r'}$ as $b^{r'-r+r}=b^{r'-r}b^r$ (I proved this in a previous problem). Now dividing both sides by $b^r$ gives $b^{r-r}. Now I really would like to take that and say that it implies that $r-r
Any ideas? Maybe a different approach I could take?