Clarification for proof of $\mathbb{Q}$ being dense in $\mathbb{R}$ (Rudin's PMA)

Question

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: https://i.stack.imgur.com/U8hgO.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.

Answer 1

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 $m-1\leq nx< m$.  

Answer 2

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).

And using these results it's proven the following:

Hopefully with this approach now you can understand Rudins proof.

Answer 3

Let $ x < y $ be reals. We are to show there is a rational number $ \frac{m}{n} $ strictly between them, i.e. that $ x < \frac{m}{n} < y $ for some integer $ m $ and a positive integer $ n $.

Equivalently, we want a positive integer $ n $ such that $ (nx, ny) $ has at least one integer $ m $ in it.

So it suffices to pick an $ n $ such that the length of the interval $ ny - nx $ is $ > 1 $, and this is possible by the Archimedean property.