definition of rational powers of real numbers

Question

Suppose that $b\gt1$ and x is a real number. Rudin defines $B(x)$ to be the set of all numbers $b^{t}$, where $t$ is a rational number and $t\le x$. I want to prove that if $r$ is a rational number then $b^{r} = \sup B(r)$ by showing that $b^{r_1}\lt b^{r_2}$ whenever $r_1\lt r_2$ and $r_1$,$r_2$ are rational. What is the idea behind proving it?

Answer 1

You need to know/show the following three facts:

1. $(b^r)^n=b^{rn}$ for $r$ rational and $n$ an integer.
2. If $n<m$ are integers and $b>1$ then $b^{n}<b^m$.
3. Given integer $n>0$ and real $x,y>0$ then $x^n<y^n$ implies $x<y$.

Then if $r_1=p_1/q_1$ and $r_2=p_2/q_2$ with $q_1,q_2>0$ and $r_1<r_2$ we first have that $p_1q_2<p_2q_1$ and thus, by (2):

$$b^{p_1q_2}<b^{p_2q_1}$$

But this means, by (1) that:

$$\left(b^{r_1}\right)^{q_1q_2}<\left(b^{r_2}\right)^{q_1q_2}$$

Finally, since $q_1q_2$ is a positive integer, we get by (3):

$$b^{r_1}<b^{r_2}$$

Answer 2

You can suppose $r_1,r_2$ have the same denominator $d$: $r_1=\dfrac{n_1}d,\enspace r_2=\dfrac{n_2}d$.

Now the function $b^d$ is defined, and increasing and $$b^{r_1}<b^{r_2}\iff (b^{r_1})^d=b^{n_1}<(b^{r_2})^d=b^{n_2}\iff b^{n_2-n_1}>1$$ Ultimately, it all amounts to proving that, if $n\ge 1$, then $\; b^n>1$, which results from a trivial induction.