Reading Rudin(pg. 22) the question 6 is about exponential function construction.
- Fix $b>1.$
(a) If $m,n,p,q$ are integers, $n>0,q>0,$ and $r=m/n=p/q,$ prove that $$(b^m)^{1/n}=(b^p)^{1/q}.$$ Hence it makes sense to define $b^r=(b^m)^{1/n}.$
(b) Prove that $b^{r+s}=b^rb^s$ if $r$ and $s$ are rational.
(c) If $x$ is real, define $B(x)$ to be the set of all numbers $b^t,$ where $t$ is rational and $t\le x.$ Prove that $$b^r= \sup B(r)$$ when $r$ is rational. Hence it makes sense to define $b^x=\sup B(x)$ for every real $x.$
My question is how can i prove item c) and more important Why does it make sense to define $B(x)$? It goes straight from rational to real, so that already takes the irrational, right? and it makes sense for example $2^{\pi}$, or any irrational exponent. Any help with the construction, thanks in advance.
Best Answer
We are updating the definition of $b^x$ ($b>1$ fixed).
Originally, $b^x$ is defined for nonnegative integers $x=n\in\mathbb{N}$: $$ b^n:=\underbrace{b\cdot b\cdots b}_{n \textrm{ factors}} $$ where we use the convention that $b^0=1$.
We also define for positive integer $n$, $$ b^{-n}:=\frac{1}{b^n}. $$
By existence of $\sqrt[n]{b}$ for positive integer $n$, (see Rundin), we are allowed to define $$ b^{1/n}:=\sqrt[n]{b},\quad,n=1,2,3,\cdots $$
Then by Problem 6(a) in Rudin, we can define $$ b^r:=(b^m)^{1/n},\quad r=\frac{m}{n},\quad m\in\mathbb{Z},n\in\mathbb{Z}\setminus\{0\}. $$
For $r\in\mathbb{Q}$, observes from Problem b(c) that $$ b^r=\sup B(r) \tag{1} $$ where the set $B(r):=\{b^t\mid t\le r,t\in\mathbb{Q}\}$.
Now for general $x\in \mathbb{R}$, we define $$ b^x:=\sup B(x) \tag{2} $$ where the set $B(x):=\{b^t\mid t\le x,t\in\mathbb{Q}\}$. This definition makes sense because (1) is true. It would not make sense if (1) is not true, because $\mathbb{Q}\subset\mathbb{R}$ and by (2) means (1) must be true.
To prove (1), i.e., the statement in Problem (c), simply follow the definition and show the following: