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I won't reproduce the proof here, but here is a link to another MathSE question about the same theorem. I've read Brian M. Scott's answer, but I think I'm missing something — I still don't understand why $m_1$ and $m_2$ are necessary for the proof; why can't we just pick an $m$ like in the following examples?

$$ \text{Let } m = \begin{cases} 5 &\text{if } nx = 4.6 \\ 2 &\text{if } nx = \sqrt{2} \\ -6 &\text{if } nx = -7 \\ \vdots &\text{etc.} \end{cases} , $$

i.e. let $m$ be the next largest integer. I believe this method will always satisfy $m - 1 \leq nx < m$.

Now I'm guessing that the reason we can't directly pick an $m$ without the help of $m_1$ and $m_2$ has something to do with rigour; I think I would be assuming things about $\mathbb{R}$ that Rudin wouldn't want me to just yet, but I can't quite put my finger on what exactly.

Jeremy Lindsay
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    Which theorem or proposition are you using to construct $m$? What's etc mean? The reader just learned what + and < meant, what's this funny dot between 4 and 6? and why is there a roof on 2? – Calvin Khor Aug 27 '20 at 12:38

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Because at that point we don’t know that there is a next larger integer. As I said in that answer, we could use the well-ordering principle for the natural numbers to show that such an integer exists if we knew that $nx$ was positive, but we don’t: $n$ is positive, but $x$ can be any real number. The argument using $m_1$ and $m_2$ establishes that there is such an $m$ even if $nx$ is negative. This part of the proof of could just as well have been pulled out as a separate lemma or proposition: for any $x\in\Bbb R$ there is in integer $m$ such that $m-1\le x<m$.

Brian M. Scott
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  • I should've asked a more specific question. I feel like the $m_1, m_2$ method is just a more complicated way to note that $\mathbb{Z}$ is unbounded in both the positive and negative directions — a corollary from part a), I suppose. Then just choose $m$ and $m - 1$ from all of $\mathbb{Z}$ rather than from the subset in your other answer. Is this method not simpler/more elegant? – Jeremy Lindsay Aug 27 '20 at 22:26
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    @JeremyLindsay: That’s irrelevant, since it’s not justifiable on the basis of what has been developed at that point in the text. I don’t see any legitimate approach that’s any simpler than the one that Walter used. (And in my experience — his wife Mary Ellen was my thesis adviser — he was not given to unnecessary complexity.) – Brian M. Scott Aug 28 '20 at 02:31