Fix $b>1$. Show the following:
(a) If $m,n,p,q$ are integers, $n>0, q>0$, and $r = m/n = p/q$, prove that $(b^{m})^{1/n} = (b^p)^{1/q}$. So $(b^r) = (b^m)^{1/n}$.
(b) Prove that $b^{r+s} = b^{r}b^{s}$ for $r,s \in \mathbf{Q}$.
(c) If $x$ is real, define $B(x)$ to be the set of all numbers $b^t$, there $t$ is rational and $t \leq x$. Prove that $b^r = \sup B(r)$ where $r$ is rational. So $b^x = \sup B(x)$ for every real $x$.
(d) Show that $b^{x+y} = b^{x}b^{y}$ for all real $x$ and $y$.
For (a), we know that $mq = np$. Thus $b^{mq} = b^{np}$. Then $(b^{m})^{q} = (b^{n})^{p}$. Or $(b^{m})^{q} = (b^{p})^{n}$. So the result follows if we note that $m = np/q$ and $n = mq/p$?
For (b) just start with $b^{\frac{mq+pn}{qn}}$?
For (c) and (d), we just need to show that $B(x)$ is non-empty. Then use a proof by contradiction?