Theorem 1.20(b) on page 9 of Rudin's "Principles of Mathematical Analysis," 3rd edition. For those without the text handy:

1.20 Theorem

(a) If $x \in \mathbb{R}$ and $x > 0$, then there is a positive integer $n$ such that $nx > y$.

(b) If $x \in \mathbb{R}$, $y \in \mathbb{R}$, and $x < y$, then there exists a $p \in \mathbb{Q}$ such that $x < p < y$.

Also a picture to the page in question: http://i.imgur.com/bufiYkE.png

We are given that $x < y$, thus $y - x > 0$ is obvious to me. I quickly lose Rudin after this step. I understand that the archimedean property $(a)$ is being used for the next line, where it says $n(y - x) > 1$, however I have no clue where the number "$1$" came from.

Furthermore, he says to apply $(a)$ again, but I have no idea what it means to "apply" a theorem arbitrarily. He doesn't say what to apply it to, and if he meant to apply it on $n(y - x) > 1$, then I am even more confused with the following step. He "applies" $(a)$ to obtain positive integers $m_1$ and $m_2$ such that $m_1 > nx$ and $m_2 > -nx$. As far as I understand, the archimedean property says that for $x > 0$, there is a positive integer $n$ such that $nx > y$. I don't understand how $m_1 > nx$ and $m_2 > -nx$ follow this property. In $m_1 > nx$, the equality sign is reversed from the archimedean property definition, and for $m_2 > -nx$, there is a negative sign.

And then the next line, he says "hence" there is an integer $m$ (with $-m_2 \leq m \leq m_1$) such that $m - 1 \leq nx \leq m$. I don't understand how you can deduce from the previous lines ($m_1 > nx$ and $m_2 > -nx$) to arrive at this one. Where are all these $m$'s coming from? The only connection I see is that number $1$ from $n(y - x) > 1$ from the first step. I have no clue where these $m$'s appeared out of nowhere.

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    $\begingroup$ For your first question, as $y - x > 0$, there is $n \in \Bbb{N}$ such that $y - x > \frac{1}{n}$. $\endgroup$ – Michael Albanese Jul 13 '14 at 1:40
  • $\begingroup$ Take $x=1>0$ to prove that if $y$ is any real then there is an integer $n>y$. Then there is $m_1$ and $m_2$... $\endgroup$ – Thomas Andrews Jul 13 '14 at 1:44
  • $\begingroup$ You ask where the number $1$ came from. Think about what the theorem wants to establish: That whenever $x<y$, there is a rational in between. If this is true, then in particular it will be true when $x=0$, so we need to verify that whenever $y>0$, there is a rational $r$ with $0<r<y$. But if $r$ is rational, we can write it in the form $k/n$ for some integers $k,n$, with $n>0$. Now, if $0<k/n<y$, then certainly $k>0$, so $k\ge 1$, so we also have $0<1/n\le k/n<y$, and it would be enough to prove that if $y>0$ then there is an $n$ such that $0<1/n<y$. (Cont.) $\endgroup$ – Andrés E. Caicedo Jul 13 '14 at 2:29
  • $\begingroup$ This is the same as proving that if $y>0$, then there is an $n$ with $1<ny$. This is precisely what Rudin begins with, only that he begins not with a positive $y$, but with reals $x<y$, so that now it is $y-x$ that is positive, and therefore what we need is an $n$ with $n(y-x)>1$. It turns out that this particular case is actually useful to prove the general result (that between any two numbers there is a rational), and so he proves this first, and proceeds from there. $\endgroup$ – Andrés E. Caicedo Jul 13 '14 at 2:31
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    $\begingroup$ I suggest to read Robert Bartles proof of this result. I get the feeling that Rudin is proving several important results when proving (b). In Bartle and Sherberts "Introduction to Real Analysis" he first proves a couple of corollaries that will be used to prove the density of $\mathbb Q$ and it reads nicely. $\endgroup$ – taue2pi Jul 13 '14 at 4:54

Your confusion seems to arise because the Archimedes principle is stated in terms of $x,y$, and you have different $x,y$ in (b). Restate the Archimedean principle as:

(a) If $u,v$ are real numbers, with $u>0$ then there is a positive integer $k$ such that: $ku>v$.

(All I've done is change the variables, I hope.)

Now, $1$ is a real number, $y-x$ is a real number, and you've proven that $y-x>0$. So we know from (a) that if $u:=y-x$ and $v:=1$ that there is a positive integer, which we will call $n$, such that $(y-x)n>1$.

Similarly, since we know that $nx$ is a real number, and we know that $1$ is a real number and $1>0$, that from (a), setting $u:=1$ and $v:=nx$, that there is a positive integer we'll call $m_1$ such that $m_1\cdot 1 > nx$.

Finally, set $u:=1>0$ and $v:=-nx$ to show that there must be an $m_2$ so that $m_2\cdot 1>-nx$.

The last step is subtler, and doesn't use (a). Since $m_2>-nx$, $-m_2<nx$. So we know that $-m_2<m_1$.

Now, you need a property of the integers: If a non-empty set of integers has a lower bound, then it has a least element.

Take the set $S=\{m\in\mathbb Z: m> nx\}$. We know that $m_1\in S$, so $S$ is non-empty, and we know that $-m_2$ is a lower bound for $S$. So there is a least element $m\in S$. Then $m-1\notin S$, and therefore $m> nx$ and $m-1\leq nx$. So $m-1\leq nx< m$.  

  • $\begingroup$ Michael Albanese's explanation of the first part ($n(y-x) > 1$) was more intuitive. I don't really understand yours, so I'll just leave it at that. For the next lines, what does "$u = 1 > 0$" mean? Is that the same thing as $u = 1$ since $1 > 0$ is just a true equality? If so, how do you know to set $u = 1$? And how do you know to set $v = nx$ and $v = -nx$? Since we changed our variables to $u$, $k$, and $v$, what is $n$ then? Is it just from the first AP? And then we're using AP again on the first AP? $\endgroup$ – mr eyeglasses Jul 13 '14 at 2:05
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    $\begingroup$ But his "explanation" did not Archimedes property as you've stated it, so you can't use that as a step in your proof. The point is, AP applies to any pair of real numbers with the first positive, no matter what we call them. $\endgroup$ – Thomas Andrews Jul 13 '14 at 2:08
  • $\begingroup$ @nablablah Can you state the version of the Archimedian property as given in Rudin? Is it just (a)? $\endgroup$ – Thomas Andrews Jul 13 '14 at 2:25
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    $\begingroup$ Okay, so property (a) is what the book is calling the Archimedean property. Cool, that was my assumption, but I realized I might have been mistaken. $\endgroup$ – Thomas Andrews Jul 13 '14 at 2:34
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    $\begingroup$ I believe you require "we know $-m_2$ is a lower bound for $S$." $\endgroup$ – jII May 25 '16 at 23:34

In Bartle and Sherberts "Introduction to Real Analysis" this result is proved with preliminary corollaries (you'll note that are in some way results that Rudin also uses).

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And using these results it's proven the following:

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Hopefully with this approach now you can understand Rudins proof.

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    $\begingroup$ How come in corollary 2.4.6, the inequalities are both '$\leq$' ($n_y - 1 \leq y \leq n_y$), but in the proof, we have $m - 1 \leq nx < m$ (one '$\leq$' and one '$<$')? $\endgroup$ – mr eyeglasses Jul 13 '14 at 13:16
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    $\begingroup$ Your correct. In the proof of corollary 2.4.6, what is actually proven is that if $y>0$, then there exists $n_y\in\mathbb N$ such that $n_y-1\leq y\lt n_y$, hence $n_y-1\leq y\leq n_y$ $\endgroup$ – taue2pi Jul 13 '14 at 23:12
  • $\begingroup$ The proof is as follows. The Archimedean Property ensures that the subset $E_y:=\{m\in\mathbb N :y\lt m\}$ of $\mathbb N$ is not empty. By the Well-Ordering Porperty, $E_y$ has a least element wich will be denoted by $n_y$. Then $n_y-1$ does not belong to $E_y$, and hence we have $n_y-1\leq y\lt n_y$. this last inequality implies the inequality seen in the corollary. $\endgroup$ – taue2pi Jul 13 '14 at 23:18

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