How to prove $\exists n \in \mathbb{N}$ such that $a < b^n$ with $a > 0$ and $b>1$, with $a,b \in \mathbb{Q}$

Question

A friend gave me this problem and I can't solve it. I tried to fool around with the Archimedean property but with no success. Any hints? Thanks a lot in advance!

Answer 1

If $b>1$, one can write $b=1+r$ for some $r>0$. Thus, $b^m = (1+r)^m \geq 1 + mr$ for any $m\in\mathbb{N}$. Choose $n\in\mathbb{N}$ such that $nr > a$. We have then $$b^n \geq 1+nr > 1 + a > a.$$

Answer 2

This is basically the proof for infinite divergence of $\lim b^n$ as $n\to\infty$, wherein you let $M=a\in\mathbb{Q}$, instead of reals.

Let $a>0$ be some real number. And notice that the natural logarithm is a continuous but increasing function, therefore since $a<b^n$ we also have $\ln(a)<n\ln(b)$ or more preferably written $\ln(a)/\ln(b)<n$.

So this is what we do for a proof.

For any $a>0$ in the reals, we choose an $N\in\mathbb{N}$ such that $N> \ln(a)/\ln(b)$. For all $b>1$ and for all $n\ge N$ we observe $b^n\ge b^N >b^{\ln(a)/\ln(b)}=a$. Thus $b^n>a$.

Stipulating that $a,b\in\mathbb{Q}$ doesnt make it less true, because $\mathbb{Q}\subset\mathbb{R}$.

So $\forall n\ge N$ when you define $N$ in this way ensures the inequality holds. Coincidentally, this proves divergence of $\lim b^n$. Because it doesnt matter how big you make $a$, you can always find an $N$ that makes $b^n>a$.