If $x \in \mathbb{Q}$ and $b \in \mathbb{R}$ with $b > 1$ and $x > 0$, then $b^x \geq 1$

Question

The following problem is motivated from Baby Rudin, Chapter 1, Problem 6(c): if $x \in \mathbb{Q}$ with $b > 1$ and $x \geq 0$, then $b^x \geq 1$. I would prefer not to use techniques such as differentiation and sequences as it's still very early on within the textbook.

The case for $\mathbb{N}$ is simple enough, but I'm not sure how to prove that $b^x$ is monotonically increasing with what's given so far from Rudin.

Answer 1

If $x=\frac mn$, with $m,n\in\Bbb N$, then$$b^x=b^{m/n}=\sqrt[n]{b^m}.$$Can you take it from here?

Answer 2

Note that $f(b)=b^{m/n}=(b^m)^{1/n}$ is a composition of increasing functions so $f(b)\ge f(1)=1$ for $b\ge 1$.

Answer 3

Simple enough.

Let $x = \frac mn; m,n\in \mathbb N$ and $b^x =b^{\frac mn}=(b^m)^{\frac 1n}$. (That is to say: the unique positive integer $c$ so that $c^n = b^m$.)

Note as $b> 1$ then $b^{k+1}=b*b^k > 1*b^k=b^k$ so by induction $1 < b< b^2 <.....< b^k$ for all natural $k$.

And conversely if $0< c < 1$ we have $1 > c > c^2 > .......> c^k>0$ for all natural $k$.

And of course if $d =1$ then $d^k =1$ for all $k$.

so if $b^x \le 1$ then $(b^x)^n \le 1$. But by definition $(b^x)$ is the unique $c$ so that $c^n = b^m$ and $b^m > 1$ so that is a contradiction. By the way if $b^x =1$ then $b = 1$ so that inequality can be a strict inequality.