Exponents and Suprema 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? 
 A: I've been working on this exercise today as well. I managed to prove (I think) (a), but it's a bit complicated. Let's see:
Lemma 1: $(a^x)^{1/x} = a$,  for all $a>0,x\in\mathbb{Q}$. This follows from the uniqueness of positive roots.
Lemma 2: $(a^x)^{1/y} = (a^{1/y})^x$, for all $a>0\in\mathbb{Q}$, $x,y\in\mathbb{Z}$. This follows from Theorem 1.2.1 from Rudin's.
Lemma 3: $(a^x)^y = a^{xy}$, for all $a\in\mathbb{Q}$, $x,y\in\mathbb{Z}$, since $\prod_1^y\prod_1^x a = \prod_1^{xy} a$
To prove (a), we start from $b = (b^{1/q})^q$ because of (1).
Then we raise both sides to the $m$th power to get $b^m = ((b^{1/q})^q)^m= (b^{1/q})^{qm}$, because of (3).
Since $m/n = p/q$, then $qm = pn$. Then $b^m = (b^{1/q})^{pn}=((b^{1/q})^p)^n$, once again because of (3).
Then $b^m = ((b^p)^{1/q})^n$ because of (2). Taking the $n$th root to both sides we get $(b^m)^{1/n} = (b^p)^{1/q}$, which is what we wanted to prove.
To prove (b), consider $b^{mq+np}=b^{mq}b^{np}$ (which holds because $m,n,p,q\in\mathbb{Z}$). This is equivalent to $(b^{m/n})^{nq}(b^{p/q})^{qn}$ because of the previously mentioned lemmas.
Taking the $nq$th root to both sides, we end up with $b^{\frac{mq+np}{nq}} = b^{m/n}b^{p/q}$, which is what we wanted to prove.
(c) is proved easily since $b^r \geq x\in B(r)$ and $b^r\in B(r)$, therefore $b^r=\sup(B(r))$
I haven't proved (d) yet but will edit this if I do.
Anyway, I'm not sure about this so feedback is really appreciated.
