I am looking for help in proving that given $x>1$, $A$ $\in \mathbb{Q}$, there exists an integer $n$ such that $x^n>A$ as long as $x>1$.

Some rules:

The proof may not use the least upper bound property. It may assume any of the definitions that allow $\mathbb{Q}$ to be an ordered field. The reason is that I am trying to use this fact to prove that Dedekind cuts (special subsets of $\mathbb{Q}$) can be constructed to form the real field in the first place.

My understanding:

My intuition tells me that if this weren't true, we should find a contradiction with the archemedian property of the rational numbers. I cannot find the contradiction, however.

EDIT: My (dead-end at the moment) approach is as follows:

Let $x>1 \in \mathbb{Q}$ and let $A \in \mathbb{Q}$ be an arbitrary rational number. I claim that there exists $n \in \mathbb{Z}^+$ such that $x^n>A.$ Suppose this were not true and that there existed some $A \in \mathbb{Q}$ such that $x^n \leq A$ for all $n \in \mathbb{Z}^+$. By the Archemidean property of $\mathbb{Q}$, there exists some $m \in \mathbb{Z}$ such that $mx^n > A$.

This is where I am stuck. It looks like you should just be able to do inequality gymnastics of some sort to force a contradiction. I have tried one approach which is as follows:

Since by our contradiction assumption, $x^n \leq A$, the $m$ obtained from the Archemedian property has the bound $m>1$, and by Well-Ordering there must be a minimum such $m$ for every $x^n$, call it $m_n$. Let $m_{\alpha}$ be the lowest such integer for all positive integer powers of $x$. Something something contradiction? It would help if I could find a place to use $x>1$, which implies $x^n < x^{n+1}$ for all $n \in \mathbb{Z}$ and would therefore imply that $m_n \geq m_{n+1}$. I don't even know if this is a fruitful approach, however.

  • $\begingroup$ Hint: take nth root on both sides and get the familiar limit $n^{1/n} \to 1$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 2 '18 at 2:29
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    $\begingroup$ I would prefer an approach that doesn't utilize any sort of theory of limits. I will try to update so people can better see the type of assumptions I am allowing/disallowing since it is admittedly a touch murky. $\endgroup$ – Bud Feb 2 '18 at 2:34

Let $A>0$, or it is easy. Write $x = 1+ \delta$ for some $\delta \in \mathbb Q$, $\delta >0$.


$$x^n = (1+ \delta)^n > 1+ \delta n.$$

Now write $A = p/q, \delta = r/s$. So if $n = sp$, then

$$1+\delta n > \delta n = (r/s)sp =rp \ge p \ge p/q = A.$$

  • $\begingroup$ Precisely the sort of argument I was looking for. Thanks so much!!! $\endgroup$ – Bud Feb 2 '18 at 2:56
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    $\begingroup$ Good old Bernoulli! Gets 'em every time (almost). $\endgroup$ – marty cohen Feb 2 '18 at 5:00

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