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I have difficulty proving the following statement:

Let $x$, $y \in \mathbb{Q}$, $y > 0$, and $x > 1$. Then there exists a positive integer n such that $x^n > y$.

What I have done so far is the following:

Suppose $y \leq 1$. Then take $n = 1$ and we are done.

Suppose $y > 1$. Then let $x = \frac{a}{b}$ and $y = \frac{c}{d}$, where $a$, $b$, $c$, $d \in \mathbb{N}$ and $b > 0$, $d > 0$. Since $x$, $y > 1$, we have $a > b > 0$ and $c > d > 0$.

However I failed to construct an inequality to prove $y = \frac{c}{d} \leq \dots \leq (\frac{a}{b})^n = x^n$.

I would really appreciate it if someone could help!

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    $\begingroup$ Bernoulli inequalty and Archimedean property $\endgroup$
    – ajotatxe
    May 27, 2023 at 18:14
  • $\begingroup$ @ajotatxe Could you please elaborate it? [(a/b)^n >= 1 + n(a/b - 1) >= ...?] Thank you so much! $\endgroup$
    – Beerus
    May 27, 2023 at 18:26
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    $\begingroup$ $x=1+h$ for $h>0$ and $x^n\ge nh$ follows $\endgroup$
    – FShrike
    May 27, 2023 at 18:41
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    $\begingroup$ @FShrike I really appreciate it! $\endgroup$
    – Beerus
    May 27, 2023 at 18:47

2 Answers 2

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Since $x>1$ there must exist a positive real number $h$ such as $x=1+h$. By the Archimedian propertie of real numbers given a real number $y$ and a positive real number $h$ there exists a positive integer $n$ such as $nh>y$. Now from the Bernoulli inequalty $$x^{n+1}=(1+h)^{n+1}\geq (n+1)h>nh>y.$$

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  • $\begingroup$ Thank you so much! $\endgroup$
    – Beerus
    May 27, 2023 at 18:48
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Outline of an alternative approach:

Suppose $x$, $y \in \mathbb{R}$, $y > 0$, and $x > 1$. Define $U:= \{x^n:n\in\mathbb{N}\}.$ If there does not exist a positive integer $n$ such that $x^n > y,$ then there exists $\ t>1\ $ such that $\ t:=\sup\{x^n:n\in\mathbb{N}\}.\ $ But then, there exists $s\in \left( \frac{t+1}{2} \ ,\ t\right]\ $ such that $s\in U,\ $ i.e., $\ s = x^N$ for some $N\in\mathbb{N},$ and so $s^2 = x^{2N}\in U.\ $ But $\ s^2 > \frac{(t+1)^2}{4} = \frac{t^2 + 2t + 1}{4} > t,\ $ because $\ t^2 +2t + 1 > 4t\ $ because $\ (t-1)^2 > 0\ $ because $t>1.\ $ Since $\ x^{2N}\in U\ $ and $\ x^{2N} > t,\ \sup\{x^n:n\in\mathbb{N}\}> t,\ $ a contradiction. This contradiction means our assumption that there does not exist a positive integer $n$ such that $x^n > y$ is false.

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