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Pf:

Let $x > 0$. By the Archimedean property, $\exists M \in \mathbb{Z}^{+}$ s.t. $\frac{1}{x} < M$.

Likewise, $\exists n \in \mathbb{Z}^{+}$ s.t. $x < n$.

Let $N = n + M$. Because $n, M > 0 \Rightarrow n + M > 0$ and $(n + M) \in \mathbb{Z}$ because the integers are closed under addition. Thus, $N \in \mathbb{Z}^{+}$.

We have that $$x < n < N$$ and $$\frac{1}{x} < M < N \Rightarrow \frac{1}{N} < \frac{1}{M} < x.$$ Thus, $$\frac{1}{N} < \frac{1}{M} < x < n < N$$ $$\Rightarrow \frac{1}{N} < x < N.$$ Because $N \in \mathbb{Z}^{+}$, we conclude that $\exists N \in \mathbb{Z}^{+}$ s.t. $\frac{1}{N} < x < N$.

Faith Alone
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    Looks good. You switched between $m$ and $M$ a few times though. The last few lines are also a bit verbose, but that's just a matter of personal preference. You could probably just end with something like "Thus $1/N<x<N$, as desired." – pancini Feb 10 '24 at 06:06
  • https://math.stackexchange.com/questions/tagged/solution-verification: For posts looking for feedback or verification of a proposed solution. "Is my proof correct?" is off topic (too broad, missing context). Instead, the question must identify precisely which step in the proof is in doubt, and why so. – Anne Bauval Feb 10 '24 at 08:07

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