2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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

Then

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

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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.