# If $x, y$ are rationals with $y > 0$ and $x > 1$, then there exists a positive integer $n$ such that $x^n > y$

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!

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

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 $$x^{n+1}=(1+h)^{n+1}\geq (n+1)h>nh>y.$$
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.