# Proof: Let $x$, $y \in Q$, $y > 0$, $x > 1$. Then there is an integer $n$ such that $x^n < y ≤ x^{n+1}$.

I want to know if my proof of the following result is correct:

Let $$x$$, $$y \in Q$$, $$y > 0$$, $$x > 1$$. Then there is an integer $$n$$ such that $$x^n < y ≤ x^{n+1}$$.

Proof:

Suppose I have proved that "let $$x$$, $$y \in Q$$, $$y > 0$$, $$x > 1$$. Then there is a positive integer $$n$$ such that $$x^n > y$$."

By the above result, there is a positive integer $$m_0$$ such that $$x^{m_0} > y$$. Consider the set $$S = \{n \in \mathbb{N}: y > x^{m_0 - n}\}$$. Again, by the above result, there exists a positive integer $$m$$ such that $$1/y < x^m$$. Then $$x^{-m} < y < x^{m_0}$$. Thus, $$m_0 + m \in S$$ where $$m_0 + m \in \mathbb{N}$$. By the well-ordering principle, $$S$$ has a minimal element $$m_1$$. Then I claim: $$x^{m_0 - m_1} < y \leq x^{m_0 - m_1 + 1}$$.

Indeed, $$m_1 \neq 0$$ since $$0 \notin S$$. Thus, $$m_1 - 1 \in \mathbb{N}$$, and so if $$y > x^{m_0 - m_1 + 1}$$, then $$m_1 - 1 \in S$$, contradicting the minimality of $$m_1$$. Now we can conclude by letting $$n = m_0 - m_1$$.

• I glanced over the proof and it seems like a good approach. Depending on what you have at your disposal, there are probably easier ways—one can simply take $n=\lceil \frac{\log y}{\log x} \rceil-1$ for example. (The proof of the well-definedness of $\lceil\cdot\rceil$ already uses the inductive property of the integers in a way similar to this proof, so we're not cheating, just avoiding repeating arguments.) Jun 6, 2023 at 18:38
• What about $y=1/2$? Am I missing something? Jun 6, 2023 at 18:49
• @MichaelHoppe If y = 1/2, then we may let x = 2 and n = -2. Then 1/4 < 1/2 <= 1/2. Jun 6, 2023 at 19:12
• @GregMartin Thank you so much! Jun 6, 2023 at 19:13