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 $x^n < y ≤ x^{n+1}$.


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 $n = m_0 - m_1$.

  • $\begingroup$ 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.) $\endgroup$ Jun 6, 2023 at 18:38
  • $\begingroup$ What about $y=1/2$? Am I missing something? $\endgroup$ Jun 6, 2023 at 18:49
  • $\begingroup$ @MichaelHoppe If y = 1/2, then we may let x = 2 and n = -2. Then 1/4 < 1/2 <= 1/2. $\endgroup$
    – Beerus
    Jun 6, 2023 at 19:12
  • $\begingroup$ @GregMartin Thank you so much! $\endgroup$
    – Beerus
    Jun 6, 2023 at 19:13


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