Question about exponential function construction

Question

Reading Rudin(pg. 22) the question 6 is about exponential function construction.

6. 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.

Answer 1

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.

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To prove (1), i.e., the statement in Problem (c), simply follow the definition and show the following:

• show $b^r\leq \sup B(r)$ by observing that $b^r\in B(r)$;
• show that $b^r\geq \sup B(r)$ by observing that $b^r\geq b^t$ for any $t\in\mathbb{Q}$ with $t\leq r$.

Answer 2

What is wrong with the definition of $B(x)$? There were irrational numbers in this context before its introduction; for instance, $2^{1/2}$ is irrational.

The set $B(x)$ is defined as the set of all real numbers of the form $b^t$, where $t$ is rational and $t\leqslant x$. It is a set of real numbers, which has an upper bound. And then Rudin defines $b^x$ as $\sup B(x)$. There is no problem here (except that one needs to prove that the set $B(x)$ has un upper bound).