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.
Best Answer
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:
(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 $m-1\leq nx< m$.